STUDY SUPPORT MATERIAL SESSION 2018-19 · study support material – session 2018-19 class xi subject – chemistry prepared during 02 day workshop (3rd august 2018 & 04th august

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STUDY SUPPORT MATERIAL – SESSION 2018-19

CLASS XI

SUBJECT – CHEMISTRY

PREPARED DURING 02 DAY WORKSHOP (3RD AUGUST 2018 & 04TH AUGUST 2018)

FOR PGT-CHEMISTRY

VENUE – KV CHITTARANJAN

COURSE DIRECTOR-

Mr. P. K. TAILOR, PRINCIPAL

KV CHITTARANJAN

RESOURCE PERSON(S):

Mr. KUNTAL BATABYAL, PGT-CHEMISTRY,

KV ASANSOL

Mrs. ANUDEEP SAINI, PGT-CHEMISTRY,

KV COMMAND HOSPITAL

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INDEX

SL NO NAME OF THE CHAPTERS PAGE NO FROM

1 Some basic concepts of chemistry

1

2 Structure of atom

13

3 Classification of elements and periodicity

19

4 Chemical Bonding

22

5 States of matter

27

6 Chemical Thermodynamics

35

7 Equilibrium

40

8 Redox reaction

44

9 Hydrogen

50

10 S block

59

11 P Block

73

12 Organic chemistry

86

13 Hydrocarbons

114

14 Environmental chemistry

147

15 Model Question Papers HY 155

16 Model Question Papers SEE 168

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Unit I CHAPTER-I-Some Basic Concepts of Chemistry (By Group-H)

Important points on “Some Basic Concepts of Chemistry”

Classification of Matter:-

Elements:

Smallest unit of an element is known as atom.

Total number of the known elements is 118 out of which 98 elements occur naturally and 20 are

formed by artificial transmutation.Examples: Na, K, Mg. Al, Si, P, C, F, Br etc.

Compound:

It is a non-elemental pure compound, Formed by chemical combination of two or more atoms of

different elements in a fixed ratio. .Examples: H2O, CO2, C6H12O6 etc.

Mixture:

Formed by physical combination of two or more pure substances in any ratio. Chemical identity of

the pure components remains maintained in mixtures.

Homogeneous mixtures are those whose composition for each part remains constant.

Example, Aqueous and gaseous solution.

Heterogeneous mixtures are those whose composition may vary for each and every part.

Example, Soil and concrete mixtures.

Physical Quantities and Their Measurement:

Fundamental Units:-

These units can neither be derived from one another nor can be further resolved into any other units.

Seven fundamental units of the S.I. system

Physical quantity Name of the unit Symbol of the unit

Time Second S

Mass Kilogram kg

Length Meter m

Temperature Kelvin K

Electric current Ampere A

Luminous intensity Candela Cd

Amount of Mole Mol

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substance

Derived Units:-

These units are the function of more than one fundamental unit

Quantity with Symbol Unit (S.I.) Symbol

Velocity (v) Metre per sec ms-1

Area (A) Square metre m2

Volume (V) Cubic metre m3

Density (r) Kilogram m-3

Kg m-3

Energy (E) Joule (J) Kg m2s

-2

Force (F) Newton (N) Kg ms-2

Frequency (n) Hertz Cycle per sec

Pressure (P) Pascal (Pa) Nm-2

Electrical charge Coulomb (C) A-s (ampere – second)

Measurement of Temperature

Three scales of temperature

Kelvin scale (K)

Degree Celsius scale (oC)

Degree Fahrenheit scale (oF)

Relations between the scales:

oF = 9/5(

oC) + 32

K = oC + 273

0 K temperatures is called absolute zero.

Dalton’s Atomic Theory:

Every matter consists of indivisible atoms.

Atoms can neither be created nor destroyed.

Atoms of a given element are identical in properties

Atoms of different elements differ in properties.

Atoms of different elements combine in a fixed ratio to form molecule of a compound.

Precision and Accuracy:

Precision: Closeness of outcomes of different measurements taken for the same quantity.

Accuracy: Agreement of experimental value to the true value

Significant figures:

Rules:

All non-zero digits are significant.

Zeroes preceding the first non-zero digit are not significant.

Zeroes between two non-zero digits are significant.

Zeroes at the end of a number are significant when they are on the right side of the decimal point.

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Counting numbers of objects have infinite significant figures.

Scientific Notation:

Numbers are represented in N × 10n form.

Where, N = Digit term & n = exponent having positive or negative value.Examples,

12540000 = 1.254 × 107

0.00456 = 4.56 ×10-3

Laws of Chemical Combination:

Law of conservation of mass:

―For any chemical change total mass of active reactants are always equal to the mass of the product

formed‖

Law of constant proportions:

―A chemical compound always contains same elements in definite proportion by mass and it does not

depend on the source of compound‖.

Law of multiple proportions:

―When two elements combine to form two or more than two different compounds then the different

masses of one element B which combine with fixed mass of the other element bear a simple ratio to

one another‖

Law of reciprocal proportion:

― If two elements B and C react with the same mass of a third element (A), the ratio in which they do so

will be the same or simple multiple if B and C reacts with each other‖.

Gay Lussac’s law of combining volumes:

―At given temperature and pressure the volumes of all gaseous reactants and products bear a simple

whole number ratio to each other‖.

Atomic and Molecular Masses:

Atomic Mass:Mass of an atom compared with one-twelfth mass of 1 atom of C 12 isotope taken as 12

u.

Molecular Mass:

Mass of a molecule of covalent compound compared with one-twelfth mass of 1 atom of C 12 isotope

taken as 12 u.

It is equal to the sum of atomic masses of all the elements present in the molecule.

Formula Unit Mass

Mass of a molecule of an ionic compound

It is also equal to the sum of atomic masses of all the elements present in the molecule

Mole Concept:

Mole:

One mole amount of substance that contains as many particles or entities as there are atoms in

exactly 12 g of the12C isotope.

Molar mass:

Mass of one mole of a substance in gram

Molar mass in gram is numerically equal to atomic/molecular/formula mass in amu or u.

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%Percentage composition:

Mass percentage of an element in a compound = (Mass of that element in the compound /

Molecular massof the compound)×100

Percentage yield:

It is the ratio of actual yield of the reaction to the theoretical yield multiplied by 100.

% yield = (Actual yield /Theoretical yield) ×100

Empirical formula and molecular formula:

Molecular Formula:-

Represents the actual number of each individual atom in any molecule is known as molecular

formula.

Empirical Formula:-

It expresses the smallest whole number ratio of the constituent atom within the molecule.

Molecular formula = (Empirical formula)x n

Molecular weight = n × Empirical formula weight

Also,Molecular weight = 2 × Vapour density

Limiting Reagent:

The reactant which is totally consumed during the course of reaction and when it is consumed

reaction stops.

For a balanced reaction reaction:

A +B → C + D

B would be a limiting reagent if nA / nB>nB/nA

Similarly, A is a limiting reagent if nA / nB<nB/nA

Concentration of the solutions:

Mass by Mass Percentage: or % by mass or w/w%

Amount of solute in gram present per 100 gm of the solution.

Mass percentage of solute = [(Mass of solute)/(Mass of solution)] x100

Mass by Volume Percentage or % by mass by volume or w/v %

Amount solute in gram present per 100 mL of the solution.

Volume by Volume Percentage or v/v %

Volume of solute per 100 mL of the solution

Volume by volume percentage of solute = [(Volume of solute)/(volume of solution)] x100

Parts per million ( ppm) :-

The amount of solute in gram per million (106) gram of the solution.

ppm = [(mass of solute/mass of solution)]x 106

Mole fraction:-

Ratio of the moles of one component of the solution to the total number of moles of solution

Total mole fraction of all the components of a solution is equal to 1.

For binary solutions having two components A and B

Mole fraction of A

XA = (nA)/(nA+nB)]

Mole fraction of B

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XB = (nB)/(nA+nB)]

or XB = 1- XA

Molarity(M):-

Number of moles of solute per 1000 mL of the solution.

M = (Number of moles of solute)/(Volume of solution in L)

It is temperature dependent.

Molality(m):-

number of moles of solute per 1000 gram of the solvent.

m = (Number of moles of solute)/(Weight of solvent in kg)

It is temperature independent.

(For Late Bloomers)

Some selected NCERT Exercise Questions with answers

Qu.1). Calculate the mass percent of different elements present in Sodium Sulphate (Na2SO4).

Answer

Molar mass of Na2SO4 = [(2 × 23.0) + (32.00) + 4 (16.00)] = 142 g

Mass percent of an element = (Mass of that element in compound/Molar mass of that compound) ×

100

∴ Mass percent of sodium (Na): (46/142) × 100 = 32.39%

Mass percent of sulphur(S): (32/142) × 100 = 22.54%

Mass percent of oxygen:(O): (64/142) × 100 = 45.07%

Qu.2). Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1%

dioxygen by mass.

Answer

% of iron by mass = 69.9 % [Given]

% of oxygen by mass = 30.1 % [Given]

Atomic mass of iron = 55.85 amu

Atomic mass of oxygen = 16.00 amu

Relative moles of iron in iron oxide = %mass of iron/Atomic mass of iron = 69.9/55.85 = 1.25

Relative moles of oxygen in iron oxide = %mass of oxygen /Atomic mass of oxygen =

30.01/16=1.88

Simplest molar ratio = 1.25/1.25 : 1.88/1.25

⇒ 1 : 1.5 = 2 : 3

∴ The empirical formula of the iron oxide is Fe2O3.

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Qu.3. Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer

The balanced reaction of combustion of carbon in dioxygens is:

C(s) + O2(g) → CO2 (g)

1mole 1mole(32g) 1mole(44g)

(i) In dioxygen, combustion is complete. Therefore 1 mole of carbon dioxide produced by burning 1

mole of carbon.

(ii) Here, oxgen acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it will react

with 0.5 mole of carbon to give 22 g of carbon dioxide.

(iii) Here again oxgen acts as a limiting reagent as only 16 g of dioxygen is available. It is a limiting

reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon

dioxide.

Qu.4). Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar

aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1.

Answer

0.375 M aqueous solution of sodium acetate means that 1000 mL of solution containing 0.375 moles

of sodium acetate.

∴ No. of moles of sodium acetate in 500 mL = (0.375/1000)×500 = 0.375/2 = 0.1875

Molar mass of sodium acetate = 82.0245g mol-1

∴ Mass of sodium acetate acquired = 0.1875×82.0245 g = 15.380g

Qu.5). Calculate the concentration of nitric acid in moles per litre in a sample which has a density,

1.41 g mL-1 and the mass per cent of nitric acid in it being 69%.

Answer

Mass percent of 69% means tat 100g of nitric acid solution contain 69 g of nitric acid by mass.

Molar mass of nitric acid (HNO3) = 1+14+48 = 63g mol-1

Number of moles in 69 g of HNO3 = 69/63 moles = 1.095 moles

Volume of 100g nitric acid solution = 100/1.41 mL = 70.92 mL = 0.07092 L

∴ Conc. of HNO3 in moles per litre = 1.095/0.07092 = 15.44 M

Qu.6) . How much copper can be obtained from 100 g of copper sulphate (CuSO4 )?

Answer

1 mole of CuSO4 contains 1 mole of copper.

Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)= 63.5 + 32.00 + 64.00 = 159.5 g

159.5 g of CuSO4 contains 63.5 g of copper.

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∴ copper can be obtained from 100 g of copper sulphate = (63.5/159.5)×100 = 39.81g

Qu.7). Calculate the atomic mass (average) of chlorine using the following data :

% Natural Abundance Molar Mass

35Cl 75.77 34.9689

37Cl 24.23 36.9659

Answer

Fractional Abundance of 35Cl = 0.7577 and Molar mass = 34.9689

Fractional Abundance of 37Cl = 0.2423 and Molar mass = 36.9659

∴ Average Atomic mass = (0.7577×34.9689)amu + (0.2423×36.9659)

= 26.4959 + 8.9568 = 35.4527

Qu.8) In three moles of ethane (C2H6), calculate the following :

(i) Number of moles of carbon atoms.

(ii) Number of moles of hydrogen atoms.

(iii) Number of molecules of ethane.

Answer

(i) 1 mole of C2H6 contains 2 moles of Carbon atoms

∴ 3 moles of of C2H6 will contain 6 moles of Carbon atoms

(ii) 1 mole of C2H6 contains 6 moles of Hydrogen atoms

∴ 3 moles of of C2H6 will contain 18 moles of Hydrogen atoms

(iii) 1 mole of C2H6 contains Avogadro's no. 6.02×1023

molecules

∴ 3 moles of of C2H6 will contain ethane molecule = 3×6.02 ×1023

= 18.06 ×1023

molecules.

Qu.9) What is the concentration of sugar (C12H22O11) in mol L-1

if its 20 g are dissolved in

enough water to make a final volume up to 2L?

Answer

Molar mass of sugar (C12H22O11) = (12×12)+(1×22)+(11×16) = 342 g mol-1

No. of moles in 20g of sugar = 20/342 = 0.0585 mole

Volume of Solution = 2L (given)

Molar concentration = Moles of solute/Volume of solution in L = 0.0585mol/2L = 0.0293 mol L-1

= 0.0293 M

Qu.10) If the density of methanol is 0.793 kg L-1

, what is its volume needed for making2.5 L of its

0.25 M solution?

Answer

Molar mass of methanol (CH3OH) = (1×12)+(4×1)+(1×16) = 32 g mol-1 = 0.032 kg mol-1

Molarity of the solution = 0.793/0.032 = 24.78 mol L-1

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Applying, M1V1 (Given Solution) = M2V2 (Solution to be prepared)

24.78×V1 = 0.25×2.5 L

V1= 0.02522 L = 25.22 mL

Qu.11). A sample of drinking water was found to be severely contaminated with chloroform,

CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

(i) Express this in percent by mass.

(ii) Determine the molality of chloroform in the water sample.

Answer:

(i) 15 ppm means 5 parts in million(106) parts.

∴ % by mass = 15/106 × 100 = 15×10

-4 = 1.5×10

-3 %

(ii) Molar mass of chloroform(CHCl3) = 12+1+(3×35.5) = 118.5 g mol-1

100g of the sample contain chloroform = 1.5×10-3

g

∴ 1000 g(1 kg) of the sample will contain chloroform = 1.5×10-2

g

= 1.5×10-2

/118.65 mole = 1.266×10-4

mole Hence: Molality = 1.266×10-4

m.

Qu.12). How many significant figures are present in the following?

(i) 0.0025(ii) 208(iii) 5005(iv) 126,000(v) 500.0(vi) 2.0034

Answer

(i) 2(ii) 3(iii) 4(iv) 3(v) 4(vi) 5

Qu.13). If the speed of light is 3.0 × 108ms

-1, calculate the distance covered by light in 2.00 ns.

Answer

Distance covered = Speed×Time = 3.0 × 108ms-1 × 2.00 ns

= 3.0×108ms-1×2.00 ns×10

-9s/1ns = 6.00×10

-1m = 0.600m

Qu.14). In a reaction: A + B2 → AB2

Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B(iv) 5 mol A + 2.5 mol B(v) 2.5 mol A + 5 mol B

Answer:

(i) According to the reaction, 1 atom of A reacts with 1 molecule of B.

∴ 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unreacted.

Hence, B is the limiting reagent.

(ii) According to the reaction, 1 mol of A reacts with 1 mol of B.

∴ 2 mol of A will react with only 2 mol of B leaving 1 mol of B. Hence, A is the limiting reagent.

(iii) 1 atom of A combines with 1 molecule of B.

∴ All 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is

stoichiometric and ther is no limiting reagent.

(iv) 1 mol of atom A combines with 1 mol of molecule B.

∴ 2.5 mol of B will combine with only 2.5 mol of A. and 2.5 mol of A will be left unreacted. Hence,

B is the limiting reagent.

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(v) 1 mol of atom A combines with 1 mol of molecule B.

∴ 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left.

Hence, A is the limiting reagent.

Qu.15). Dinitrogen and dihydrogen react with each other to produce ammonia according to the

following chemical equation:

N2(g) + 3H2(g) → 2NH3(g)

(i) Calculate the mass of ammonia produced if 2.00×103g dinitrogen reacts with 1.00×10

3g of

dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

Answer:

1 mole of dinitrogen (28g) reacts with 3 mole of dihydrogen (6g) to give 2 mole of ammonia (34g).

∴ 2000 g of N2 will react with H2 =( 6/28) ×200g = 428.6g.

Thus, here N2 is the limiting reagent while H2 is in excess.

28g of N2 produce 34g of NH3.

∴ 2000g of N2 will produce = (34/28)×2000g = 2428.57 g of NH3.

(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain unreacted.

(iii) Mass of dihydrogen left unreacted = 1000g - 428.6g = 571.4 g

Qu.16) How are 0.50 mol Na2CO3and 0.50 M Na2CO3 different?

Answer

Molar mass of Na2CO3 = (2×23)+12.00+(3×16) = 106 g mol-1

∴ 0.50 mol Na2CO3 means 0.50×106g = 53g

0.50 M Na2CO3 means 0.50 mol of Na2CO3 i.e. 53g of Na2CO3 are present in 1litre of the solution.

Qu.17). If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many

volumes of water vapour would be produced?

Answer:

Dihydrogen gas reacts with dioxygen gas as,

2H2(g) + O2(g) → 2H2O(g)

Thus, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of

water vapour. Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to

produce ten volumes of water vapour.

Qu.18). Which one of the following will have largest number of atoms?

(i) 1 g Au (s)(ii) 1 g Na (s)(iii) 1 g Li (s)(iv) 1 g of Cl2 (g)

Answer:(i) 1 g Au = 1/197 mol = (1/197)×6.022×1023

atoms

(ii) 1 g Na = 1/23 mol = (1/23)×6.022×1023

atoms

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(iii) 1 g Li = 1/7 mol = (1/7)×6.022×1023

atoms

(iv) 1 g Cl2 = (1/71) mol = (1/71)×6.022×1023

atoms, Thus, 1 g of Li has the largest number of

atoms.

Qu.19). Calculate the molarity of a solution of ethanol in water in which the mole fraction of

ethanol is 0.040 (assume the density of water to be one).

Answer:

Mole fraction of C2H5OH = No. of moles of C2H5OH/No. of moles of solution

n(C2H5OH) = n(C2H5OH)/n(C2H5OH)+n(H2O) = 0.040 (Given) ……..Eqn. 1

We have to find the number of moles of ethanol in 1L of the solution but the solution is dilute.

Therefore, water is approximately 1L.

No. of moles in 1L of water = 1000g/18g mol-1 = 55.55 mol

Substituting n(H2O) = 55.55 in eqn 1

n(C2H5OH)/n (C2H5OH) + 55.55 = 0.040

⇒ 0.96n(C2H5OH) = 55.55 × 0.040⇒ n(C2H5OH) = 2.31 molHence, molarity of the solution =

2.31M

Qu.20). What will be the mass of one 12C atom in g ?

Answer:

1 mol of 12C atoms = 6.022×1023

atoms = 12g

∴ Mass of 1 atom 12C = 12/6.022×1023

g = 1.9927×10-23

g

Qu.21) How many significant figures should be present in the answer of the following calculations?

(i) 0.02856×298.15×0.112/0.5785

(ii) 5 × 5.364

(iii) 0.0125 + 0.7864 + 0.0215

Answer:

(i) Least precise term i.e. 0.112 is having 3 significant digits.

∴ There will be 3 significant figures in the calculation.

(ii) 5.364 is having 4 significant figures.


(iii) Least number of decimal places in each term is 4.


Qu.22) Calculate the number of atoms in each of the following

(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.

Answer:

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(i) 1 mol of Ar = 6.022×1023

atoms

∴ 52 mol of Ar = 52×6.022×1023

atoms = 3.131×1025

atoms

(ii) 1 atom of He = 4 u of He

4 u of He = 1 Atom of He

∴ 52 u of He = (1/4) × 52 = 13 atoms

(iii) 1 mol of He = 4 g = 6.022×1023

atoms

∴ 52 g of He = (6.022×1023

/4) × 52 atoms = 7.8286×1024

atoms

Qu.23). A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in

oxygen gives 3.38 g carbon dioxide , 0.690 g of water and no other products. A volume of 10.0 L

(measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii)

molar mass of the gas, and (iii) molecular formula.

Answer

Amount of carbon in 3.38 g of CO2 = 12/44 × 3.38 g = 0.9218 gAmount of hydrogen in 0.690 g H2O

= 2/18 × 0.690 g = 0.0767 g

The compound contains only C and H, therefore total mass of the compound = 0.9218 + 0.0767 =

0.9985 g

% of C in the compound = (0.9218/0.9985)×100 = 92.32%

% of H in the compound = (0.0767/0.9985)×100 = 7.68%

(i) Calculation of empirical formula,

Moles of carbon in the compound = 92.32/12 = 7.69

Moles of hydrogen in the compound = 7.68/1 = 7.68

Simplest molar ratio = 7.69 : 7.68 = 1:1 (approx)

∴ Empirical formula CH

(ii) 10.0 L of the gas at STP weigh = 11.6 g

∴ 22.4 L of the gas at STP = 11.6/10.0 × 22.4 = 25.984 = 26 (approx)

∴ Molar mass of gass = 26 g mol-1

(iii) Mass of empirical formula CH = 12+1 = 13

∴ n = Molecular Mass/Empirical Formula = 26/13 = 2

∴ Molecular Formula = C2H2

Qu.24). Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the

reaction, CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)

What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Answer:

1000 mL of 0.75 M HCl have 0.75 mol of HCl = 0.75×36.5 g = 24.375 g

∴ Mass of HCl in 25mL of 0.75 M HCl = 24.375/1000 × 25 g = 0.6844 g

From the given chemical equation,

CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)

2 mol of HCl i.e. 73g HCl react completely with 1 mol of CaCO3 i.e. 100g

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∴ 0.6844 g HCl reacts completely with CaCO3 = 100/73 × 0.6844 g = 0.938 g

Qu.25). Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous

hydrochloric acid according to the reaction

4HCl (aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Answer

1 mol of MnO2 = 55+32 g = 87 g

87 g of MnO2 react with 4 moles of HCl i.e. 4×36.5 g = 146 g of HCl.

∴ 5.0 g of MnO2 will react with HCl = 146/87×5.0 g = 8.40 g.

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CHAPTER-II - STRUCTURE OF ATOM (By Group-H)

CLASS XI

Short Review of formulas (for one electron atom or ions):

1. Velocity of electron in nth

orbit = vn = 2.165 x 106 Z/n m/s

2. Radius of nth

orbit = rn = 0.53 x 10–10

n2/Z m

3. Energy of an electron in nth

state = En = –13.6 Z2/n

2 eV/atom

En = –2.18 × 10–18

Z2/n

2 J/atom = –13.6 Zn

2/n

2 eV/atom

4. Kinetic energy = KE = 1/2 mv2

n = kZe2 / rn

5. Potential energy = PE = –kZe2 / 2rn

6. Total energy of an electron = –En = –kZe2 / 2rn

PE = 2TE ; PE = –2KE ; TE = –KE

7. Binding energy of an electron in nth state

En = (–13.6 / n2) Z

2 eV

8. Ionisation Energy = – B.E.

I.E. = (+ 13.6 / n2 ) Z

2 eV

9. Ionisation Potential

Ionisation potential = I.E. / e = ( 13.6/n2 ) Z

2 V

10. Excitation Energy

The energy taken up by an electron to move from lower energy level to higher energy level. Generally

it defined from ground state.

• Ist excitation energy = transition from n1 = 1 to n2 = 2

• IInd excitation energy = transition from n1 = 1 to n2 = 3

• IIIrd excitation energy = transition from n1 = 1 to n2 = 4 and so on …

• The energy level n = 2 is also called as Ist excited state.

• The energy level n = 3 is also called as IInd excited state. & so on …

In general, excitation energy (ΔE) when an electron is excited from a lower state n1 to any higher state

n2 is given as:

ΔE = 13.6 Z2 (1/n1

2 – 1/n2

2) eV

11. Energy released when an electron jumps from a higher energy level (n2) to a lower energy level

(n1) is given as:

ΔE = 13.6 Z2 (1/n1

2 – 1/n2

2) eV

If v be the frequency of photon emitted and λ be the wavelength, then:

ΔE =hv = h c/λ

The wavelength (λ) of the light emitted an also be determined by using:

1/λ = v = R Z2

(1/n12 – 1/n2

2)

R = 1.096 x 107 /m

Important: Also remember the value of 1/R = 911.5 Å for calculation of λ to be used in objectives

only).

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12. The number of spectral lines when an electron falls from n2 to n1 = 1 (i.e. to the ground state) is

given by:

No. of lines = n2(n2–1) / 2

If the electron falls from n2 to n1, then the number of spectral lines is given by:

No. of lines = (n2 – n1 + 1) (n2–n1) / 2

FOR SLOW LEARNERS / LATE BLOOMERS

Q.1 Calculate the number of electrons which will together weigh one gram.

Ans. 9.1x10-28

g weighs 1 e-

1g weighs 1

9.1𝑥10−28 = 1.09x1027

Q.2 Calculate the mass and charge of one mole of electrons.

Ans Mass of 1 mol of e- = 9.1 x 10-31

x 6.022 x 1023

kg = 54.8 x 10-8

kg

Charge of 1 mol e- = 1.6x10-19

x6.022 x 1023

= 96487 C

Q.3 Write a relationship between angular momentum & Planck‘s constant

Ans. mvr = nh/2π

Q.4 State Aufbau principle , Hund‘s rule, Pauli exclusion principle and Heisenberg Uncertainty

principle.

Ans. Aufbau Principle: it states that atomic orbitals are filled by electrons in order of increasing

energies (according to n+l rule)

Hund‘s rule: Pairing of electrons in degenerate orbitals take place only after each orbital is filled in with

single electron with same spin.

Pauli‘s exclusion Principal: No two electrons in an atom have all the four quantum numbers same. Or

An orbital can accommodate a maximum of two electrons of opposite spin.

Heisenberg‘s Uncertainty Principle:

It is impossible to determine simultaneously the exact position & momentum of a microscopic particle

like electron.

FOR AVERAGE STUDENTS

Q.5 Name the spectral lines observed in visible region of line spectrum of hydrogen. Write its n1& n2

values.

Ans. Balmer series, n1 = 2 & n2 = 3,4,5,….

Q.6 Write electronic configuration of Cu(Z=29),Cr(Z =24)

Ans: Cu =1s2 2s

2 2p

6 3s

2 3p

6 3d

10 4s

1 Cr = 1s

2 2s

2 2p

6 3s

2 3p

6 3d

5 4s

1

Q.7) Find the maximum no. of electrons for n+l = 7?

Ans The possible orbitals are 7s,6p, 5d, 4f so maximum no. of electrons are 2+6+10+14 = 32

Q.8) What is significance of Heisenberg Uncertainty principle in daily life?

Ans. No. significance as in daily life we deal with macro particles (large mass) &

15

∆𝑥. ∆𝑣 ∝1

𝑚 so uncertainty in position of mass or velocity is least.

Q.9) Explain why quantum numbers are referred to as the address of electron in an atom.

Ans. It denotes the complete address of an electron as n= size, energy of orbit or shell.

l= shape, angular momentum,ml = orientation of orbitals & ms = electron spin around in its axis.

Q.10) More than two electrons can not be found in an orbital. Justify

Ans. It is in accordance to Pauli‘s exclusion principle.

Q.11) Why the energy of 4s is lower than 3d?

Ans: It is according to n+l rule: For 4s, n+l = 4+0=4 & for 3d, n+l = 3+2 = 5, so 4s has lower energy

than 3d orbital.

Q.12) Write electronic configuration of Cl-(Z=17),Cu

2+(Z=29),Cr

3+(Z=24)

Cl- =1s

2 2s

2 2p

6 3s

2 3p

6 , Cu

2+=1s

2 2s

2 2p

6 3s

2 3p

6 3d

9 Cr

3+= 1s

2 2s

2 2p

6 3s

2 3p

6 3d

3

Atomic Structure: For Bright students

Problem 1:Why Bohr‘s orbits are called stationary states?

Solution: This is because the energies of orbits in which the electrons revolve are fixed.

Problem 2: Explain why the electronic configuration of Cu is 3d10

4s1 and not 3d

94s

2.

Solution: In the 3d10

4s1 the d-sub shell is completely filled which is more stable.

Problem 3: Fe3+

ion is more stable than Fe2+

ion. Why?

Solution: In Fe3+

ion 3d sub shell is half filled hence more stable configuration.

Problem 4: Calculate the accelerating potential that must be applied to a proton beam to give it an

effective wavelength of 0.005 nm.

Solution: v = h/mλ & eV = 1/2mv2

Putting the values we get,

V = 32.85 volt

Problem 5: Give one example of isodiapheres.

Solution: Isodiapheres have same difference between the number of neutrons and protons.

For example 39

19K & 31

15P Here For 39

19K n-p=20-19=1

&31

15P, n-p=16-15=1

Problem 6: Which electronic transition in Balmer series of hydrogen atom has same frequency as that

of n = 6 to n = 4 transition in He+. [Neglect reduced mass effect].

Solution :

16

1/λ(He+) = RZ

2 [ 1/4

2 - 1/6

2]

= 4R [ 36 - 16/36 x 16 ] = 5R/36

1/λ(H) = R x 12 [ 1/2

2 - 1/n

2]

∴ 1/λ (He+) = 1/λ(H)

5R/36 = R/4 - R/n2

On solving above equation

n2

= 9

∴ n = 3

Or corresponding transition from 3 → 2 in Balmer series of hydrogen atom has same frequency as that

of 6 → 4 transition in He+.

Problem 7: Calculate ionization potential in volts of (a) He+ and (b) Li

2+

Solution:

I.E. = 13.6Z2/n

2

= 13.6 x Z2 /1

2 [Z =2 for He+]

= 13.6 x 4 = 54.4 eV

Similarly for Li2+

= 13.6 x 32/1

2

= 13.6 x 9 = 122.4 eV

Problem 8: Calculate the ratio of K.E and P.E of an electron in an orbit? (For competitive exam)

Solution:

K.E. = Ze2/2r

P.E. = -Ze2/r

∴ P.E. = –2K.E

∴ K.E/P.E = - 1/2

Problem 9: How many spectral lines are emitted by atomic hydrogen excited to nth energy level?

Solution:

Thus the number of lines emitted from nth energy level

17

= 1 + 2 + 3 +………… n – 1 = ∑(n – 1)

∑n = n(n+1)/2

∴ ∑ (n – 1) = ( n-1) (n-1+1)/2 = (n-1) (n)/2

Number of spectral lines that appear in hydrogen spectrum when an electron jumps from nth energy

level = n (n-1)/2

Problem 10: Calculate (a) the de Broglie wavelength of an electron moving with a velocity of

5.0x 105 ms

–1 and (b) relative de Broglie wavelength of an atom of hydrogen and atom of oxygen

moving with the same velocity (h = 6.63 x 10–34

kg m2 s

–1)

Solution:

(a) λ = h/mv = 6.63 x 10-34

kgm2s

-2/ (9.11 x 10

-31kg) ( 5.0 x 10

5ms

-1)

Wavelength λ = 1.46 x10–9

m

(b) An atom of oxygen has approximately 16 times the mass of an atom of hydrogen. In the formula λ =

h/mv, h is constant while the conditions of problem make v, also constant. This means that λ and m are

variables and λ varies inversely with m. Therefore, λ for the hydrogen atom would be 16 times greater

than λ for oxygen atom.

Problem 11: A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest

approach for head on collision. (For competitive exam)

Solution: K.E. = P.E.

1/2mv2 = 1/4πεo Ze

2/d Hence d = Ze

2/4πεo( 1/2mv

2)

= 9 x 109 x 79 x ( 1.6 x 10

-19)2 / ( 1.6 x 10

-13 ) [ as 1 MeV = 1.6 x 10

-13 J]

= 1.137 x 10-13

m

Problem 12: What is the wavelength associated with 150 eV electron

Solution:

λ = h/√2 x m x K.E.

= 6.626 x 10-34

Js / √ 2 x 9.1 x 10-31

kg x 150 x 1.6 x 10-19

J

=6.626 x 10-34

/ √4368 x 10-50

= 10–10

m = 1 Å

Problem 13: The energy of electron in the second and third Bohr orbit of the hydrogen atom is –

5.42 x 10–12

erg and –2.41 x 10–12

erg, respectively. Calculate the wavelengths of emitted radiation

when the electron drops from third to second orbit.

Solution: E3 – E2 = hv = hc/λ – 2.41 x 10–12

– (– 5.42 x 10–12

) = 6.626 x 10-27

x 3 x 1010

/ λ

∴ λ = 6.626 x 10-27

x 3 x 1010

/ 3.01 x 10-12

= 6.604 x 10–5

cm = 6.604 x 10–5

x 108 = 6604 Å

Problem 14: O2 undergoes photochemical dissociation into one normal oxygen and one excited oxygen

atom, 1.967 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen

atoms requires 498kJ mole–1. What is the maximum wavelength effective for photochemical

dissociation of O2? #(For competitive exam)

18

Solution:

O2 → ONormal+ Oexcited

O2 → ONormal + ONormal

E = 498 x 103 J / mole

= 498 x 103 / 6.023 x 10

23J per molecule = 8.268 x 10

–19 J

Total energy required for photochemical dissociation ofO2

= 8.268 x 10–19

+ 3.146 x 10–19

= 11.414 x 10–19

J hc/λ = 11.414 x 10–19

J

λ = 6.626 x 10-34

x 3 x 108 / 11.414 x 10

-19= 1.7415 x 10

–7 m = 1741.5 Å

Problem 15: Compare the wavelengths for the first three lines in the Balmer series with those which

arise from similar transition in Be3+

ion. (Neglect reduced mass effect).

Solution 1

𝜆H = R x 1

2 ( 1/2

2 - 1/n

2)

1

λBe

+3 = R x 4

2 ( 1/2

2 - 1/n

2)

∴1

λBe

+3/1

𝜆H = λH / 𝜆𝐵𝑒3+ = 16

So we can conclude that all transitions in Be3+

will occur at wavelengths 1/16 times the hydrogen

wavelengths.

19

CHAPTER-III

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES(by Group D)

Mendeleev’s periodic law: The properties of the elements are the periodic functions of their atomic

weights.

No of periods: 07

No of groups: Initially it was VIII. After the discovery of inert gases Zero group was added.

Modern periodic law: Physical and chemical properties of the elements are the periodic functions of their

atomic numbers.

No of periods: 07

No of groups: 18

Nomenclature of elements with atomic number more than 100:-

To write the name of the elements with atomic numbers more than 100 Latin words for various digits are

written and the name of the element ends at the suffix ‗ium‘.

Latin words for the numbers are:-

Nos. 1 2 3 4 5 6 7 8 9 0

Latin

word

Un bi tri quad pent hex sept oct enn nil

Ex:- Name of the element with atomic numbers:

111 Unununium Uuu

115 Ununpentium Uup

118 Ununoctium Uuo

Periodic trends in physical properties:

1. Atomic radius: Distance between the nucleus and the outermost shell of an atom.

Factors on which atomic radius depends:

(a) Nuclear charge: higher the value of nuclear charge smaller the size of atom.

(b) Screening effect/ Shielding effect: Electrons present in the inner shells protect outermost

shell from being pulled by the nucleus as a result size of the atom does not decrease as much

as expected. This is called shielding effect.

(c) Effective nuclear charge: Zeff = Ze – S where S is the shielding constant. Higher the value of

Zeff smaller the size of atom.

Periodicity:

20

Across a period: Moving across a period from left to right atomic radius decreases.

Down the group: Moving down the group from top to bottom atomic radius increases.

2. Ionization potential:- Amount of energy required to remove an electron from the outermost shell

of an atom when it is in its isolated and gaseous state.

Factors on which ionization potential depends:

(a) Size of atom: Smaller the size of atom more the value of Ionization potential.

(b) Effective nuclear charge: Higher the effective nuclear charge more the IP

(c) Stability of orbitals: More stable the orbital more difficult to remove the electron and higher

the value of IP.

(d) Penetration effect: Greater the penetration effect more the IP. Penetration effect follows the

order: s>p>d>f.

Periodicity

Across the period:- Moving across a period from left to right IP increases.

Down the group:Moving down the group from top to bottom IP decreases.

Exceptions:

(a) IP of Be is more than that of B.

Explanation: Be (Z=4) 1s22s

2 B(Z=5) – 1s

22s

22p

1

As the penetration effect on s is more than that on p so it is difficult to remove the electron

from s orbital and therefore IP of Be is more than that of B.

(b) IP of N is more than that of O.

Explanation: N (Z=7) – 1s22s

22p

3 O(Z=8) – 1s

22s

22p

4

N being a half filled system is more stable than O so it is difficult to remove an electron

from N and hence IP of N is more than that of O

3. Electron gain enthalpy: Amount of energy released when an electron is added to the outermost

shell of an atom when it is in its isolated and gaseous state.

Periodicity

Across the period:- Moving across a period from left to right negative EGE increases.

Down the group:Moving down the group from top to bottom negative EGE decreases.

Exception: EGE of F is less negative than that of Cl.

Explanation: Due to very small size of F electron density around the nucleus is very high.

Incoming electron experiences an interelectronic repulsion. To add this electron some positive

energy is required and hence EGE of F becomes less negative than that of Cl.

4. Electronegativity: Tendency of an element to attract the shared pair of electrons towards it is

called electronegativity.

Periodicity

Across the period:- Moving across a period from left to right it increases.

Down the group:Moving down the group from top to bottom it decreases.

21

FREQUENTLY ASKED QUESTIONS WITH THEIR MODEL ANSWERS

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

1. What is the basic difference between Modern Periodic table and Mendeleev’s Periodic table?

Ans: Modern periodic table is based on atomic number of the elements whereas Mendeleev‘s

Periodic table on atomic masses.

2. Write the IUPAC name and symbol of the element with atomic number 114.

Ans: Ununquadium Uuq.

3. Why do elements in the same group have similar chemical properties?

Ans: Due to similar valence shell electronic configuration.

4. What do you mean by isoelctronic species? Name a species isoelctronic to F- and Ar.

Ans: Isoelectronic species: Species having same number of electrons.

F- ……..Ne Ar……..K

+

5. Consider the following species: N3-

, O2-

, F-, Na

+, Mg

2+, and Al

3+.

(a) What is common in them?

(b) Arrange them in the order of increasing ionic radii.

Ans: (a) They all have 10 electrons, i.e they are isoelectronic species.

(b) Al3+

< Mg2+

< Na+< F

-< O

2-< N

3-

6. Give reasons:

(a) Be has higher IP than B.

(b) F has less negative EGE than Cl.

Ans: (a) IP of Be is more than that of B.

Explanation: Be (Z=4) 1s22s

2 B(Z=5) – 1s

22s

22p

1

As the penetration effect on s is more than that on p so it is difficult to remove the electron

from s orbital and therefore IP of Be is more than that of B.

(b) Due to very small size of F electron density around the nucleus is very high.

Incoming electron experiences an interelectronic repulsion. To add this electron some positive

energy is required and hence EGE of F becomes less negative than that of Cl.

7. Define the following: (a) Ionization potential (b) Electronegativity.

Ans: (a) Ionization potential: Amount of energy required to remove an electron from the

outermost shell of an atom when it is in its isolated and gaseous state.

(b) Electronegativity: Tendency of an element to attract the shared pair of electrons towards it is

called electronegativity.

8. Assign position to the element with atomic number 40 in periodic table.

Ans: Electronic configuration: 1s22s

22p

63s

23p

64s

23d

104p

65s

24d

2

Period No : 5 Group No: 2+2= 4

Group members:- 1.Mrs.Mosarrat Jehan (KV Chittaranjan)

2. Mr. Raghib Roushan ( KV N0.2 Ishapore)

3. Mr. Amit Kumar ( KV Andal)

4. Mr.Navneet Singh ( KV AFS Salua)

5. Mrs.Gopi Priya ( KV Panagarh)

22

CHAPTER-4 CHEMICAL BONDING AND MOLECULAR STRUCTURE

(gp A)

CONTENT AREA OF CHEMICAL BONDING FOR LATE BLOOMERS

1) Octet Rule and its Limitation.

2) Lewis Structure of simple molecules.

3) Factors which favours Ionic Bonding.

4) Lattice Enthalpy and Factors on which it depends.

5) Formal charge and its calculation and advantages of formal charges.

6) Concept of VSEPR Theory and prediction of Geometry and Shape of simple molecules.

7) Overlapping of atomic orbitals and formation of simple molecules.

8) Types of overlapping and nature of covalent bond.

9) Comparison of sigma and pi bonds.

10) Dipole Moment and its application.

11) Concept of resonance.

12) Concept of hybridization and simple technique for predicting the hybrid state of central atom in a simple

molecule or in a polyatomic ion.

13) Bond order and calculation of bond order in a simple diatomic molecule.

14) Information conveyed from the knowledge of bond order.

15) Hydrogen bonding and its application.

Frequently asked question on chemical bonding with model answers

1. Why sodium chloride does not conduct electricity in solid state?

Ans. Solid sodium chloride has crystalline structure in which the ions are not free to move. The ions became

mobile when it is in molten state and thus, the electricity can be conducted.

2. Why H2O is a liquid while H2S is a gas at ordinary temperature?

Ans. Oxygen has high electronegativity than Sulphur. As a result, H2O forms hydrogen bonding. Consequently

molecules of water come near to each other through hydrogen bonding. This results in higher boiling point of

water hence it is a liquid.

3. The density of ice is less than water or ice floats over water. Explain.

Ans. In Ice, water molecule is associated with four other molecules through hydrogen bonding in a tetrahedral

manner i.e. it has open cage like structure in which lesser molecules are packed per unit volume. when the ice

melts, the molecules come closer to one another. Consequently the density of water in liquid state is more

than in solid state. Thus ice floats over water.

4. Arrange the following in order of increasing;

23

A)Dipole moment of H2O, H2S , BF3 B)Covalent character of LiCl, LiBr , LiI C)Covalent character of NaCl , MgCl2 , AlCl3

Ans. A) BF3 is a symmetrical molecule. It has zero dipole moment. Oxygen being more electronegative

than S, bond moment of O-H is more than S-H. So the dipole moment are in the order of BF3<H2S<H2O

B) The anion size in increasing order is Cl-<Br-<I-. Hence LiCl is least covalent than LiI . The order is

LiCl<LiBr<LiI.

C) Cation size in decreasing order is Na+ >Mg2+ >Al3+. Thus Al3+ ion has maximum polarization effect and

Na+ ion has least. Thus the covalent order is NaCl <MgCl2<AlCl3

5. Write the electronic configuration and calculate the bond order of H+2 , H2 and He2. Explain why

bondlength in H+2 is longer than in H2?

Ans. The numbers of electrons, their configuration, for the given species are:

Species No. of

electrons

Configuration Nb Na Bond order

H2+ 1 (Ϭ1S)1 1 0 0.5

H2 2 (Ϭ1S)2 2 0 1

He2 4 (Ϭ1S)2(Ϭ*1S)2 2 2 0

The bond length in H2+ is longer than in H2because only one electron is present to shield the two nuclei from

mutual repulsion. In H2,there are two electrons to hold the two nuclei ,thus nuclear repulsion is less than that

in H+2. Hence, nuclear separation in H+

2 is more than in H2 .

7. Write the MO electron distribution of O2. Specify its bond order and magnetic property.

Ans. O2 (No. of electrons=16) KKσ(2s)2σ*(2s)2 σ (2pz)2𝜋 (2px)2 π(2py)2 π*(2px)1π(2py)1

Bond order =0.5(8-4) =2

As two of the orbitals are singly occupied, the molecule is paramagnetic in nature.

8. Apply VSEPR theory to predict the shapes of SF4,ICl3,NH4+.

Ans. SF4:Bond pair: 4 , lone pair: 1 Geometry is- trigonalbipyramidal shape- see-saw

ICl3:Bond pairs-3 geometry- trigonal bipyramidal shape- T-shape Lone pairs-2

NH4+:Bond pair-4Shape- tetrahedral

9. Indicate the number of σ and π bonds in the molecule CH2=C=CH2

Ans. No. of sigma bonds = 6 No. of pi bonds = 2

10. Sodium chloride solution gives a precipitate of AgCl with AgNO3 whereas CCl4 does not. Why?

24

Ans. NaCl is an ionic compound and gives Cl- which results with Ag+ to form AgCl , whereas CCl4 is covalent

compound, therefore it does not give ions in aqueous solution.

11. Why does He2 not exist?

Ans. He2 does not exist because repulsive forces dominate attractive forces and bond order is 0 for He2.

12. out of NaCl and MgO which has higher value of lattice energy?

Ans. MgO has higher value of lattice energy due to stronger force of attraction between divalent Mg2+ and O2-

than monovalent Na= and Cl- ions.

13. Why is o-nitrophenol steam volatile whereas p-nitrophenol is not?

Ans. p-nitrophenol, is associated through intermolecular H-bonding and hence less volatile.

14. Which out of NH3 and NF3 has higher dipole moment and why?

Ans. In NH3 bond moments are towards lone pair whereas in NF3 bond moments are opposite to lone pair of

electrons.

15. Write resonating structure of O3

Ans.

16. Compare the H-N-H bond angles in the following molecules and ions: NH3, NH4+,NH2

-

Ans. NH4+>NH3>NH2

-

17. Out of oxygen and nitrogen which has greater bond disassociation enthalpy and why?

Ans. N2 has higher bond disassociation energy than O2 due to presence of triple bond in nitrogen whereas

oxygen has double bond.

18. What is the hybrid state of central atom in the following?

NO3-, BF4

-, PF5, IF5, CO2.

N in NO3- is sp

2hybridised, B in BF4

- is sp

3hybridised, P in PF5 is sp

3d hybridized, C in CO2 is sp

hybridized.

19. What is hybridized state of each carbon in

a. CH2=C=CH2 b.

Ans. a)Sp2Sp Sp

2 b) Sp

2

Q20) Which of the following Lewis structure of CO2 molecule is least significant resonating form?

A) :O:=C=:O: B) :O+= C-:O*

*: C) :O:=C

-= O

+

Ans.(C) is least contributing towards resonance hybrid because negative charge is at less electronegative

Carbon, positive charge is on Oxygen is more electronegative

Q21) The two O-O bond distance in ozone molecule are equal. Justify

Ans. It is due to resonance

Q22) How many types of bonds are present in NH4Cl ?

25

Ans. Ionic, Co-ordinate & Covalent bonds are present inNH4Cl

Q23) Give reasons for the following:

H2+ and H2

- ions have same bond order but H2

+ ions are more stable than H2

-.

Ans. It is because e- present in anti-bonding molecular orbital in H2

- destabilize the molecule .

Q24) What is the state of hybridisation of Carbon in CO32-

ion?

Ans. Sp2

Q25) Write a neutral molecule which is isoelectronic with ClO-

Ans. OF2

Q26) Give two difference between Ϭ &π bonds.

σ- Bond π- Bond

Ϭ- Bonds are formed by the overlapping of orbitals

at the internuclear axis.

π- Bonds are formed by the overlapping of

orbitals perpendicular to the internuclear axis.

Ϭ- Bonds are stronger than π-Bonds. Π-Bonds are weaker than Ϭ-Bonds.

Q27) Compare the relative stability of the following species on the basis of molecular

orbitaltheory & indicate their magnetic properties: O2+,O2

-.

Ans. Molecular orbital theory of O2 is

(Ϭ1s)2(Ϭ*1s)

2(Ϭ2s)

2(Ϭ*2s)

2(Ϭ2Pz)

2(π2Px

2=π2Py

2)(π*2Px

1=π*2Py

1)

Hence molecular orbital, Configuration of O2+ is

(Ϭ1s)2(Ϭ*1s)

2(Ϭ2s)

2(Ϭ*2s)

2(Ϭ2Pz)

2(π2Px

2=π2Py

2)(π*2Px

1)

Bond Order of O2+

= (10-5)/2=5/2=2.5

It is paramagnetic .

Molecular Orbital Configuration of O2-

is(Ϭ1s)2(Ϭ*1s)

2(Ϭ2s)

2(Ϭ*2s)

2(Ϭ2Pz)

2(π2Px

2=π2Py

2)(π*2Px

2=π*2Py

1)

Bond Order of O2- = (10-7)/2=3/2=1.5

It is paramagnetic.

Since, the bond order of O2+ is greater than O2

-therefore , O2

+ is more stable than O2

-.

Q28) Deduce the shape of SF4 molecule on the basis of VSEPR theory.

Ans. In SF4, there are four bonded pair of e-& one lone pair of e

- , It has see-saw shape so as to have

minimum repulsion. It has see-saw shape.

Q29) You are given the electronic Configuration of A,B,C,D,E , A1s22s

22p

63s

2 B- 1s

22s

22p

63s

1 C-

1s22s

22p

1 D-1s

22s

22p

5 E-1s

22s

22p

6 .Write the empirical formula for the substance containing:i)A

& D ii)B & D iii)Only D iv)Only E

Ans. i)AD2 ii)BD iii)D2 iv)E

26

Q30) Define octet rule. Write its significance & limitations.

Ans. Octet Rule- Every element tries to acquire 8 electrons in its outermost orbit by loosing or gaining

or sharing electron.

Significance – Most of the molecules are formed by following octet rule eg. N2,O2,Cl2,Br2etc it is

useful for understanding most of the organic compounds and it applies mainly to the second period

elements of the periodic table.

Limitations- Although octet rule is very useful but is not universal.In some compounds, the number of

electron surrounding the central atom is less than 8.In NO,NO2 octet rule is not satisfied they are odd

electron molecule In PF5,SF6,IF7 these expanded octets or super octets i.e. 10,12 and 14 electrons

respectively after sharing

Q31) Define hybridisation . Explain the structure of C2H2 .

Ans. Hybridisation is the process of mixing atomic orbitals of slightly different energies with give rise

to hybridised orbitals having exactly equal energy , identical shape and more stability

H-C=C-H each C is sp hybridised, therefore ,it has linear shape.

Q32) Why KHF2 exist but KHCl2 does not?

Ans. KF forms H-bond with HF while KCl cannot form H-bond with HCl.

Q33) Deduce the structure of BrF5

Ans. In BrF5 there are 5 bonded pairs of electrons & one lone pair of electrons, therefore, it is square

pyramidal.

27

CHAPTER-5, STATES OF MATTER( gp A)

Concept- Intermolecular forces and gas laws

KNOWLEDGE

One Mark Questions Q1. What will be the molar volume of nitrogen and argon at 273.15 K and 1atm pressure?( L-2)

Ans. At 273.15K and 1atm pressure every gas has molar volume = 22.4L.

Q 2.What do you mean by thermal energy? ( L-1)

Ans. Thermal energy is the energy of a body arising from motion of its atoms or molecules.

Q 3.Name two intermolecular forces that exist between HF molecules in liquid state.( L-2 )

Ans. Dipole-dipole interaction and hydrogen bonding.

Q 4.Define Boyle‘s law. How is it represented mathematically? (L-1)

Ans.The volume of a given mass of a gas is inversely proportional to its pressure at constant temperature

Mathematically V α 1/P. PV = constant.

Two Marks Questions Q 1.How is the strength of hydrogen bond determined? ( L-2)

Ans. Strength of H bonding determined by the coulombic interaction between the electronegative atom of

one molecule and H atom of other molecule.

Q 2 .Why liquids diffuse slowly as compared to gases? ( L-2)

Ans.In liquid the intermolecular forces (vander waal‘s forces) are greater than in gas.

Three Mark Questions Q 1 .Define (a) Dipole-Dipole Forces (b) Avogadro law (c) Intermolecular forces ( L-1)

Ans.(a) Dipole-dipole forces : The force of attraction which act between two molecules having

permanent dipoles.

(b) Avogadro Law : Equal volumes of all gases under the similar condition of temperature and pressure

contain equal number of molecules . V α n

(c) Intermolecular Forces: The forces of attractions and repulsion between interacting particles.

Q 2 .Describe Charle‘s Law. In terms of Charle‘s Law explain why -2730 C is the lowest temperature?

Ans. The volume of a fixed mass of a gas is directly proportional to its absolute temperature at constant

Pressure. V α T , constant P.

At -273 0 C, volume of the gas approaches to zero i.e the gas ceases to exist.

Q 3 .Define dispersion forces. Write characteristic features of dispersion forces.( L-2)

Ans. Dispersion forces: The forces of attraction between two temporary dipoles.

Characteristics of dispersion forces:

(i) These forces are always attractive.

(ii) These forces are effective only at short distance.

(iii) Their magnitude depends on the polarisability of the particles

UNDERSTANDING

One Mark Question Q1. Write various means to easily liquify the gases. (L-1)

Ans. By lowering the temperature and increasing the pressure.

Q2. According to you, what are the factors that determine the state of matter ( L-2)

Ans. The factors that determine the different states of the matter are pressure, temperature .

Q 3.Arrange solid, liquid and gas in order of energy. Give reason.(L-1)

Ans. Solid < liquid < gas. This is because solid absorb energy to change into a liquid which further

absorb energy to change into a gas.

Two Marks Questions

28

Q 1 Give one example for each of the following types of inter-particle forces (L-2)

(a) London dispersion forces

(b)Dipole-Dipole forces

(c) Hydrogen bond

(d) Ion-dipole forces

Ans. (a) noble gases.

(b) between HCl molecules.

(c) Between water molecules or HF molecules.

(d)Between NO3- and water molecules.

Q 2 .Why liquids do not have definite shape although they have definite volume? (L-3)

Ans.Because inter molecular forces are strong enough to hold the molecules but not so strong as to fix

them into definite position.

Q 3 .Which type of intermolecular forces are effected in each of the following process:( L-2)

(i)Sublimation of iodine.

(ii)Evaporation of water.

Ans. (i) Sublimation of iodine : Dispersion forces.

(ii)Evaporation of water : Hydrogen bond.

Three Mark Questions Q 1.Which type of intermolecular forces exists among the following molecules?( L-2)

(a) H2S molecules

(b) H2O

(c) Cl2 and CCl4

(d) SiH4

(e) Helium

(f) He atoms and HCl molecules

Ans. (a) Dipole-Dipole interactions.

(b)H-Bond.

(c)Dispersion forces.

(d)Dispersion forces.

(e)Dispersion forces.

(f)Induced dipole-dipole interactions.

Q 2 Elaborate the statement that physical state of existence at given conditions is a balance between the

thermal energy and intermolecular forces. (L-3)

Ans. Intermolecular forces tend to keep the molecules together but thermal energy of the molecules tends

to keep them apart. The three states of matter are the result of the balance between intermolecular forces

and thermal energy of the molecules.

Q 3 What are isochores? From the three isochores I, II, III given below, for a certain amount of ideal

Gas,What is the correct arrangement of volumes V1,V2 and V3. ( L-3)

Ans. Isochore is the plot beween P and T for a definite amount of a gas at a constant volume.

From different isochors at different volumes draw a line parallel to Temperature axis represent a constant

P and cutting the three isochors at T1,T2 and T3 respectively.Fr om the graph we find T1,>T2 >T3. Since

V α T , at constant P. Thus V1 > V2 > V3.

APPLICATION

One Mark Questions Q 1. What would happen if dipole develops momentarily in atoms and non polar molecules which are

electrically symmetrical? (L-2)

Ans. London forces or dispersion forces are developed.

29

Q 2. On the basis of Boyle‘s Law explain why mountaineers carry oxygen cylinder with them. (L-2)

Ans. At high altitude as the atmospheric pressure is low , the air is less dense. As a result less oxygen

is available for breathing .

Q 3. Predict what will happen if intermolecular forces between the molecules are very weak (L-2) Ans.

When intermolecular forces are very weak , molecules do not cling together t o make liquid and solid.

Two Marks Questions

Q 1. In a hospital an oxygen cylinder holds 10 L of oxygen at 200atm pressure. If a patient breathes in

0.50ml of oxygen at 1atm with each breath, for how many breaths the cylinder will be sufficient.

(Assume that all the data is at 300 C.) (L-3)

Ans. Apply Boyle‘s Law P1 V 1= P 2V 2

200 x 10 = 1 x V 2

V2 =2000 L

No. of breaths = Total volume/ vol. consumed per breath

=2000/0.5x10 -3L

=4x106

Q 2 . What will be the volume of hydrogen when 3 litres of it are cooled from 150 C to -730C at constant

pressure? (L-2)

Ans. T1 = 15+273 = 288K , V1 = 3L

T2 = -73 + 273 = 200 K , V2 = ?

V1/T1 = V2 / T2 , V2 = ( 3 x 200) / 288 = 2.08 L.

Q 3. Explain the absolute zero (in terms of volume) with the help of isobar. (L-3)

Ans: At absolute zero ,volume approach to zero and below this temperature, the volume will be negative,

which is meaningless. Hence absolute zero is the lowest possible temperature .

Three Marks Questions Q1 Use the information and data given below to answer the questions (L-3)

• Stronger intermolecular forces result in higher boiling point

• Strength of London forces increases with the number of electrons in the molecule

• Boiling point of HF ,HCl, HBr and HI are 293K,189K,206K and 238K respectively

(a) Which type of intermolecular forces are present in the molecules HCl, HBr and HI?

(b) Looking at the trend of boiling points of HCl, HBr and HI, explain, out of dipole interaction and

London interaction, which one predominant here?

(c) Why is the boiling point of HF highest while that of HCl lowest?

Ans :( a) In the molecules HCl, HBr and HI, dipole-dipole interaction and London forces are present.

(b) Dipole moment of HCl > HBr > HI but boiling point of HCl < HBr <HI.It means London forces are

predominant over dipole-dipole interaction which depends upon surface area.HCl has least surface area,

Whereas HI has maximum surface area.

(c) HF has highest dipole moment due to highest difference in electronegativity, therefore hydrogen

bonding is also present due to which force of attraction will increase, hence it has higher boiling point.

Q 2 . A gas is enclosed in a room .The temperature, pressure, density, and number of moles respectively

is T0C ,P atm, d g cm -3 n moles.

(a) What will be the pressure, temperature, density and number of moles in each compartment, if the

room is partitioned into four equal compartments?

(b)What will be the values of pressure, temperature, density and number of moles if an equal volume of

the gas at pressure ‗P‘ and temperature ‗T‘ is let in the same room? (L-2)

Ans: (a)Pressure =Patm ,

Temp. =ToC ,

density =d g/cm3 &

No. of moles=n/4 (b) Pressure =2Patm , Temp. = ToC ,

30

density =2d g/cm3 &

No. of moles=2n.

Q 3 . An open vessel contains 200mg of air at 170C.What weight percentage of air would be expelled if

the vessel is heated to 1170C? (L-3)

Ans: Suppose vol. of 200mg of air at 170C =V mL

Apply Charle‘s Law V1/ T1=V2/ T2

V1/ 290 = V2/390

V2 =1.34 V1 ,

volume of air expelled = 1.34V1 –V1 =o.34V1 mL

% of air expelled: 0.34V1x 100 /V1 = 34%

Q 4 . In a J tube partially filled with mercury the volume of an air column 4.2ml and the mercury level in

two limbs is same. Some mercury is now added to the tube so that the volume of the air is enclosed in

shorter limb is now 2.8ml.What is the difference in the levels of mercury in this situation (atm

pressure=1bar.) (L-3)

Ans: P1= 1bar= 750.12mmHg

P2 = ( P1 ₊ h)

V1=4.2mL ,V2 =2.8mL

P1V1= P2 V2

or P2 = P1V1/ V2

= 750.12 x4.2 /2.8 =1125.18mmHg

Now 1125.18 = 750.12 ₊ h

h = 1125.18 -750.12 = 375.06 mmHg or 37.506cm.

Q 5 An iron tank contains helium at a pressure of 2.5atm at 250C .The tank can withstand a maximum

pressure of 10atm.the building in which tank has been placed catches fire. Predict whether, the tank will

blow up first or melt (m.p of Fe=15350C.) (L-3)

Ans :. P1 = 2.5 atm

P2 = ?

T1= 25oC =298 K

T2 = 15350C =1808 K

According to Gay Laussac law,

P1/ T1 = P2 / T2

P2 = 2.5 x 1080/ 298 =15.16 atm.

Since pressure of gas in the tank is much more than 10 atm at the m.p. of Fe.Thus tank will blow up

before reaching m.p.

Concept – Ideal gas Equation, kinetic molecular theory of gases, ideal gas and real gases,

liquefaction of gases, liquid state

KNOWLEDGE

One mark questions Q1.What is Boyle‘s point?(L1)

Ans. The temperature at which the real gas obeys ideal gas law for an appreciable range of

pressure is called Boyle‘s point.

Q2.Why liquid drops are spherical in shape? (L2)

31

Ans. Because of surface tension, the molecules tend to minimize the surface area and sphere has

minimum surface area.

Q3.State with expression Dalton‘s law of partial pressure?(L1)

Ans. The total pressure exerted by the mixture of non reactive gases is equal to the sum of the partial

pressures of individual gases.

Ptotal = p1 + p2 + p3 +…………………

Two marks questions:- Q1. What would have the effect on the gas pressure if collisions between the gas molecules are not

elastic?(L2)

Ans. The gas pressure would gradually become zero as molecules will gradually slow down and

ultimately settle down due to constant loss of energy.

Q2.What are the two faulty assumptions of kinetic theory of gases?(L3)

Ans. i) There is no force of attraction and repulsion between the molecules of a gas.

ii) Volume of the molecules of a gas is negligibly small in comparison to the empty space

between them.

Q3. Give the units of Vander Waal‘s constants a and b.(L2)

Ans. Unit of a is bar L2 mol-2 and that of b is L mol-1.

Three mark questions Q1. Define i) surface tension ii) viscosity. Give their units.(L1)

Ans.i) The force acting parallel to the surface and perpendicular to a line of unit length anywhere in the

surface is called surface tension. Its unit is Nm-1.

ii) The force of friction which one part of the liquid offers to another part of the liquid is called

viscosity. Its unit is poise.

Q2.Give three differences between evaporation and boiling.(L1)

Ans.i) Evaporation occurs at all temperatures while boiling occurs at boiling point.

ii) Evaporation is surface phenomenon while boiling occurs throughout the bulk of liquid.

iii) Evaporation is slow while boiling is rapid.

Q3. Give Vander Waal‘s equation of state for real gases? What is the significance of constants a and b in

this equation?(L3)

Ans. ( P + an2/V2) (V – nb) = nRT

a gives the idea of magnitude of attractive forces between the molecules of the gas while b is a

measure of effective size of gas molecules.

UNDERSTANDING

One Mark questions Q1. Explain how can gases be liquefied?(L1)

Ans. Gases can be liquefied by decrease of temperature and increase of pressure.

Q2.What is the effect of temperature on surface tension and viscosity of a liquid?(L1)

Ans. Both decrease by increasing temperature.

32

Q3..Name the property of liquids which i)causes internal resistance to flow

ii)causes capillary action. Give their units.(L2)

Ans. i) viscosity. Its unit is poise.

ii) surface tension. Its unit is Nm-1.

Two marks questions Q1. In the plot of Z(compressibility factor) vs P, Z attains a value of unity at a certain pressure. What

does this signify?(L3)

Ans. It implies that at this value of pressure attractive and repulsive forces balance each other.

Below this pressure attraction dominates and Z < 1. Above this pressure repulsion dominates

and Z > 1.

Q2. Account for the following properties of gases on the basis of kinetic molecular theory of gases-

a) high compressibility b) gases occupy whole of the volume available to them.(L2)

Ans. a) High compressibility is due to large empty space between the gas molecules.

b) Due to absence of attractive forces between the molecules, they can easily separate from

one another.

Q3. Explain the pressure and volume corrections in ideal gas equation?(L2)

Ans. At low temperature and high pressure intermolecular attractions cannot be neglected, the

observed pressure P is smaller than the ideal pressure Pi.

Therefore, Pi = P + p = P + an2/V2

As volume of gas molecules is not negligible as compared to total volume of the gas

the ideal volume Vi is smaller than the observed volume.

Therefore, Vi = V + v = V - nb

Three Marks questions Q1.i) Can a gas with a = 0 be liquefied?(L2)

Ans. A gas with a = 0 means absence of intermolecular forces. Hence such a gas cannot be

Liquified.

ii) Out of NH3 and N2 which will have (L3)

a) larger value of ‗a‘

b)larger value of ‗ b‘ .Explain.

Ans. a) NH3 will have larger value of a because of H – bonding.

b) NH3 will have larger value of b due to larger molecular size.

Q2. What is meant by compressibility factor of gases? How does its value deviate from that of an ideal

gas in case of real gases and what does it indicate?(L2)

Ans. It is the ratio of the product PV to nRT.

Z = PV/nRT. For ideal gas its value is unity. Z < 1 indicates negative deviation i.e. gas is more

compressible due predominance of attractive forces. Z > 1 indicates positive deviation i.e.

gas is less compressible due predominance of repulsive forces.

33

Q3. Critical temperature for CO2 is 30.98 oC. What does this mean? Explain other two critical

constants.(L2)

Ans. It means that CO2 cannot be liquefied above 30.98oC which is its critical temperature(Tc).

Critical pressure(Pc) – it is the pressure required to liquefy the gas at critical temperature.

Critical volume( Vc) – it is the volume of one mole of the gas at Tc and Pc .

APPLICATION

One mark questions Q1. How is the partial pressure of a gas related to the total pressure in a mixture of gases?(L2)

Ans. Partial pressure = mole fraction x total pressure. For gas A

pA = xA x P

Q2. When do real gases deviate from ideality?(L2)

Ans. At high pressure and low temperature.

Q3. Why excluded volume ‗b‘ is four times the actual volume of the molecules?(L3)

Ans. The two molecules cannot come closer than distance 2r and the volume of sphere with radius

2r is four times the volume of one molecule.

Two marks questions Q1. Critical temperature of two gases A and B are 30o and -50o respectively. Which of them has strong

intermolecular forces and why?(L2)

Ans. A will have stronger inter particle forces as it can be liquefied at a higher temperature.

Q2.Explain: (L1)

i)Vapour pressure increases with increase in temperature.

ii)Giycerine is more viscous than water.

Ans. i) At higher temperature inter particle attraction weakens and more number of mole cules

escape to vapour.

ii) As interparticle forces are stronger in glycerine.

Q3. Derive relation between density and pressure of a gas.(L2)

Ans. Ideal gas equation PV= nRT

or n/V = P/RT

or m/MV = P/RT (as nm = m/M)

or d/M = P/RT

or P = dRT/M

Three marks questions:- Q1.At what temperature 128 g of SO2 confined in vessel of 5 dm3 capacity will exhibit a pressure of

10.0 bar? Given a = 6.7 bar L2 mol-2 and b= 0.0564 L mol-1.(L2)

Ans. w = 128 g

V = 5 dm3

P = 10.0 bar

MSO2 = 64 g n = w/M = 128/64 = 2

Putting values in equation : ( P + an2/V2) (V – nb) = nRT We get T = 329 K

34

Q2. Calculate the pressure exerted by 8.5 g of NH3 contained in a 0.5 L vessel at 300 K. For ammonia a

= 4.0 atm L2mol-2 and b = 0.036 L mol-1.(L2)

Ans. V =0.5 L , T = 300 K

n= 8.5/17 = 0.5 mol

( P + an2/V2) (V – nb) = nRT

Putting values , we have

P = 21.51 atm

Q3. 20 mol of Chlorine gas occupies a volume of 800 mL at 300 K and 5 x 106 Pa pressure. Calculate the

compressibility factor of the gas. Comment on the compressibility of the gas under these conditions(.L3)

Ans. P = 5 x 106 Pa = 5 x 106 / 105 bar

n = 20

T = 300 K

R = 0.083 L bar K-1 mol-1

Vreal = 800 mL

Videal = nRT/P

Putting values , we have

Videal = 1004 mL

Z = Vreal/ Videal= 800/1004 = 0.796

As Z is less than 1, it means gas is more compressible under these conditions.

Q4. Explain the effect of increasing the temperature of a liquid, on intermolecular forces operating

between its particles. What will happen to the viscosity of a liquid if its temperature is increased?

Arrange the following in increasing order of viscosity-water, hexane, glycerine.( L2)

Ans. On increasing the temperature , the kinetic energy of liquid molecules increases so that it can

overcome the attractive forces between the molecules and hence liquid can flow more easily

i.e. viscosity decreases. Order of increasing viscosity : hexane < water < glycerine.

Q5. i) Derive ideal gas equation.

ii) At 25o and 760 mm of Hg pressure a gas occupies 600 mL volume.

What will be its pressure at a height where temperature is 10o C and volume of the gas is 640

mL.(L2)

Ans. i)at constt. T and n ; V ∝ 1/P ( Boyle‘s law )

at constt. P and n ; V ∝ T (Charle‘s law )

at constt. P and T; V ∝ n ( Avogadro‘s law)

thus, V ∝ nT/P

or V = nRT/P

or PV = nRT

ii) Given p1 = 760 mm Hg V1 = 600 mL

T1 = 25 + 273 = 298 K

V2 = 640 mL and T2 = 10 + 273 = 283K

According to combined gas law : p1V1/T1 = p2V2/T2

or p2 = p1V1 T2/T1 V2

putting values and calculating, we have p2 = 676.6 mm Hg

35

CHAPTER-6, THERMODYNAMICS

FORMULAE :

1. ∆G = ∆H - T∆S

2. ∆S = dqrev/T

3. ∆Sv = ∆Hv/Tb,

4. ∆Sf= ∆Hf/Tf,

5. ∆G0

= - RTlnk

6. ∆G0

= - 2.303RTlogk

7. ∆H = ∆U + ∆nRT

8. ∆H = ∑B.ER - ∑B.Ep

9. ∆H = ∑HP - ∑HR

KEY POINTS:

1. System, surrounding and universe.

2. Extensive properties: depend of mass

3. Intensive properties`: independent of mass

4. State function: Depend on state variable.

5. Path functions: Depend on the mode of process.

6. Internal energy: Some of the energies possessed by the molecules.

7. Enthalpy: Heat content of the system.

8. Entropy: Randomness of the system.

9. Free energy: Useful work.

10. 1st law of thermodynamics.

11. 2nd law of thermodynamics.

12. Criteria for spontaneity.

13. Hess‘s law of constant heat summation.

14. Entropy: i) Increases , S= +ve ii) Decreases, S= -ve

ONE MARKS QUESTIONS:

Q1. State 1st law of thermodynamics.

Q2. State 2nd

law of thermodynamics.

Q3. State Hess‘s law of constant heat summation ?

Q4. Why is entropy of substance taken as zero at absolute zero of temperature ?

Q5. Distinguish between heat of reaction and heat of formation.

Q6. Define adiabatic process.

36

MODEL ANSWERS

Ans1. Energy can neither be created nor destroyed although it can be converted from one form

to another.

Ans2. In any spontaneous process, the entropy of the universe always increases.

Ans3. The change in enthalpy of reaction remains same, whether the reaction takes place in one

step or several steps.

Ans4. At absolute zero of temperature , there is complete orderly molecular in the crystalline

substance. Therefore, there is no randomness at 0 K and entropy is taken to be zero.

Ans5. Heat of formation - The amount of enthalpy change per mole for the formation of a

compound from its constituent elements.

Heat of reaction – The enthalpy change per mole between the product and reactant of a

reaction.

Ans6. The process in which no heat change takes place in a reversible manner between system and

surrounding.

TWO MARKS QUESTIONS:

Q1. What is a thermodynamic state function ? Neither q nor w is a state function but q+w is a

state function .Explain.

Q2. What is entropy ?

Predict in which of the following entropy increases/decreases:

a)A liquid crystallizes into a solid.

b)Temperature of a crystalline solid is raised from 0 K to 100 K .

Q3. What do you understand by the term internal energy and enthalpy?

Q4. In a process , 300 J of heat is absorbed by a system and 200 J of work is done by the

system. What is the change in internal energy for the process ?

Q5. Define- Extensive properties and Intensive properties.

Q6. Heat capacity is an extensive property but specific heat capacity is intensive property.

Comment on the statement.

Q7. Calculate the number of kJ necessary to raise the temperature of 60 g of aluminium from

350 C to 55

0C. Molar heat capacity of Al is 24 J /mol/ K

MODEL ANSWERS

1. A thermodynamic state function is a property of the system whose value depends only on

thestate of the system and is independent of the path by which the state is reached.

As q+w=∆U, which is a state function.

2. The disorder or degree of randomness in a system.

a) Entropy decreases , as molecules attains an ordered state .

37

b) Entropy increases as temperature increases due to movement of particles.

3. Internal energy- The sum of all forms of energies stored within a substance (or a system) is

called its internal system. Enthalpy- It is the sum of the internal energy change of system

and the pressure volume work done.

4 . Change in internal energy = q + w

= +300 J + ( - 200) J

= 100 J

5. Extensive properties- Those properties whose value depends upon the quantity of matter

but are independent on the nature . Ex- mass, volume, internal energy, enthalpy , heat capacity.

Intensive properties- Those properties whose value depends upon the nature of matter but

are independent of its quantity in the system . Ex- Pressure, Temperature, concentration,

Density. Etc

6. Heat capacity depends upon the mass of the substance and therefore, it is an extensive

property.

But specific heat is the heat capacity per unit mass and therefore , becomes independent of

amount of substance. Hence specific heat is intensive property.

7 . Energy required q = C m x n x ∆ T

= 24(J/mol/K) X 2.22(mol) X 20 (K )

[∆ T = 55 0C-35

0 =20

0C=20K]

= 1066.8 J or 1.07 KJ

THREE MARKS QUESTIONS

Q1. The enthalpy of formation of CO(g) CO2(g), N2O(g), N2O4(g) are -110, -393, 81, and 9.7

KJ/mole respectively.Find the value of ∆H for the reaction,

N2O4 + 3CO → N2O + 3CO2

Q2. For a reaction at 298K – 2A + B → C

∆H = 400KJ/ mole and ∆S = 2KJ/K/mole

At what temperature the reaction becomes spontaneous considering ∆H &∆S to be constant over the

temperature range.

Q3. Calculate the standard enthalpy of formation of CH3OH from the following data.

i) CH3OH + 3/2 O2→ CO2 +2H2O ∆Hr = -726KJ/mole

ii) C + O2→ CO2 ∆Hc = -393KJ/mole

iii) H2 + ½ O2→ H2O ∆Hf = -286KJ/mole

38

Q4. The reaction of cyanamide, NH2CN with oxygen was affected in a bomb calorimeter and

∆U was found to be -742.7KJ/mole of cyanamide at 298K. Calculate the enthalpy change

For the reaction at 298K. NH2CN + 3/2 O2→ N2 + CO2 + H2O

Q5. Calculate the enthalpy change for the process – CCl4→ C(g) + 4Cl(g)

And calculate the bond enthalpy of C-C in CCl4(g)

Given ∆Hvap= 30.5kj/mole , ∆Hf(CCl4) = -135.5kj/mole

∆Hatomisation(Cl2) = 242kj/mole

MODEL ANSWERS:

Ans1. ∆Hr = ∑∆H0

f(product) - ∆H0

f(reactant)

= (3 x -393)+ 81 – (3 x -110) – 9.7

= - 1179 + 81 +330 – 9.7 = -777.7 kj/mole

Ans2. ∆G = 0 , T = ∆H/∆S = 400/2 = 200K

At T = 200K the reaction is at equilibrium. And will become spontaneous as the T>200K.

Ans3. Multiply equation(iii) by 2 and add to equation (ii).

∆H = -(393+522) = -965kj/mole

Subtract eq i from eq iv , ∆H = -965 – (-726) = -239kj/mole

Ans4. ∆H = ∆U + ∆n(g)RT

= -742.7+.5 X 8.314 X 10-3

X 298.15 = -741.5KJ

Ans5. i) CCl4(l) → CCl4(g)

ii) C + 2Cl2 → CCl4

iii) C(S) → C(g)

iv) Cl2 (g) → 2Cl(g)

Eq iii) + 2 x eq iv – eq i – eq ii

∆H = 715.0 + 2x 242 – 30.5 – (-135.5) = 1304kj/mole

= 1304/4 = 326kj/mole

MADE BY GROUP – C

MEMBERS: 1. A.K MANI 2. A.K LAHA 3. H. RAHMAN BISWAS 4. S. CHATTERJEE 5. M.

KARMAKAR

39

CHAPTER-7 EQUILIBRIUM

FORMULAE TO BE REMEMBERED:

1.A + B = C + D

KC = [C][D]/[A][B]

2. A(g) + B(g) = C(g) + D(g)

KP = pc(g) x pD(g)/ pA(g) x pB(g)

3. Kp = Kc (RT)∆n

4. Qc =[C][D] / [A][B]

5. Kc = [C]c[D]

d / [A]

a[B]

b

6. pH = -log[H+] For strong electrolyte.

pH= -log(√Ka x C) For weak electrolyte.

7. Buffer solution : pH= pKa + log([salt] /[ acid]) For acidic buffer.

pOH= pKb + log ([salt ] / [base])

pH= 14 – pOH For basic buffer.

8. Hydrolysis: a) Salt of strong acid and strong base – pH= 7

b) Salt of weak acid and strong base – Na2CO3 + H2O = H2CO3 + NaOH

pH = 7+1/2 pKa + 1/2logC

c) Salt of strong acid weak base – FeCl3 + H2O = Fe(OH)3 + HCl

pH = 7-1/2 pKa - 1/2logC

d) Salt of weak acid weak base – CH3COONH4 + H2O = CH3COOH + NH4OH

pH = 7+1/2( pKa ~ pKb)

9. Solubility & solubility product

AxBy = xA+y

+ Yb-x

KSP= Sx+y

x xx x y

y S = Solubility of salt in moles per litre

. KEY POINTS:

1. Equilibrium

40

2. Physical and chemical equilibrium

3. Law of mass action

4. Equilibrium constant and its magnitude to predict the extent of reaction

5. Factors affecting the equilibrium – Le Chatelier‘s Principles.

6. Ionic equilibrium

7. Ph , Henderson Equation

8. Hydrolysis of salt – Elementary idea

9. Buffer solution

10. Solubility nand solubility product

11. Common ion effect & its application in analysis

One marks questions with Answer

1. Define Equilibrium

2. What is Physical and Chemical Equilibrium?

3. What is equilibrium constant?

4. What do you mean Kc and Kp ?

5. Write the relationship between Kc and Kp for the reaction

H2 (g) + I2 (g) 2HI (g)

Ans:

1. It is the state of a system in which all the properties remain unchanged or same with

time.

2. Equilibrium between two or more different physical states is known as physical Equilibrium

Equilibrium between the reactant and product is known as Chemical Equilibrium

3. Ratio of the conc. of product and reactant at const. T and P.

4. Kc : Equilibrium constant involving the concentration of product ant reactant at Const. T and P.

Kp: constant involving the partial pressure of gaseous product ant reactant at Const.

T and P.

5. Kp = Kc ( As change in number of moles of reactant and product is Zero)

Two marks questions with Answer

1. State law of chemical equilibrium.

2. State Le- Chatelier principle.

3. Equilibrium is a dynamic process. Explain.

Ans:

1. It is the ratio of the concentration of products raised to the power of coefficient given in

the reaction and concentration of reactantraised to the power of coefficient given in the

reaction is constant at constatnt T and P.

41

2. In an equilibrium process, if any one of the factors (temp., pr., or concentration) changes

then the equilibrium will be shifted towards that direction in which the effect can be

nullified.

3. At equilibrium, the rate of forward process and backward process is equal. It means a

hidden dynamicity is maintained at equilibrium state.

Three marks questions with Answer

1. Establish the realationship between Kp and Kc in any gaseous chemical reaction.

2. Write three differences between homogeneous and heterogeneous equilibrium.

Ans:

1.

42

2.

MADE BY GROUP – C

MEMBERS: 1. A.K MANI 2. A.K LAHA 3. H. RAHMAN BISWAS 4. S. CHATTERJEE 5. M.

KARMAKAR

43

CHAPTER-8 Redox reaction (By Group – H)

Redox Reactions

Oxidation is defined as the loss of electrons by a chemical species (atom, ion or molecule).

Reduction is the gain of electrons by a chemical species (atom, ion or molecule).

An oxidising agent that chemical species which takes electrons thus it is an electron acceptor.

A reducing agent is the chemical species that gives electrons and thus acts as an electron donor.

When Fe2+

(aq) ions are being oxidised they are acting as reducing agents, and when Fe3+

(aq) ions are

being reduced they are acting as oxidising agents. In general ;

What are Redox Reactions?

Redox reactions are those chemical reactions in which both oxidation as well as reduction occur

simultaneously.

Oxidation and reductions go hand in hand.

The substance which undergo reduction is called oxidising agent while the substance which undergo

oxidation is called reducing agent. One can say that the substance that causes the oxidation of any

substance in reaction is called the oxidizing agent while the substance that causes the reduction is

called the reducing agent.

Neither reduction nor oxidation occurs alone. Both of them occur simultaneously. Since both these

reactions must occur at the same time they are often termed as "redox reactions". The oxidation or

https://files.askiitians.com/cdn1/images/2017126-182053655-7331-redox-reactions-.jpg

44

reduction portion of a redox reaction, including the electrons gained or lost can be determined by

means of a Half-Reaction

Types of Redox Reactions

Redox reactions are divided into two main types.

(i) Inter molecular Redox Reactions:

In such redox reactions, one molecule of reactant is oxidized whereas molecule of other reactant is

reduced.

(ii) Intra molecular Redox Reactions:

One atom of a molecule is oxidized and other atom of same molecule is reduced then it is intra

molecular redox reaction.

Molecular Equations

When the reactant and products involved in a chemical change are written in molecular form in a

chemical equation, it is termed as molecular equation.

https://files.askiitians.com/cdn1/images/2017126-182040156-9914-chemical-reactions.gif

45

Example: MnO2 + 4HCl → MnCl2 + 2H2O +Cl2

In above example the reactant and products have been written in molecular forms, thus it is a molecular

equation.

Ionic Equation

When the reactant and products involved in a chemical change are ionic compounds, these will be

present in the form of ions in the solution. The chemical change is written in ionic forms in the

chemical equation, it is termed as ionic equation.

Example: MnO2 + 4H++ 4Cl

- → Mn

2++ 2Cl

- + 2H2O +Cl2

In the above example, the reactant and products have been written in ionic forms, thus the equation is

termed as ionic equation.

Oxidising Agent

The substance (atom, ions or molecules) that gain electrons and is thereby reduced to a low valency

state is called an oxidising agent.

Reducing Agent

The substance that loses electrons and its valency thereby oxidised to a higher valency state is called a

reducing agent.

Very Short Answer type question (1 Mark)

1.Define oxidation in terms of electronic concept?

2.What is meant by reduction?

3.Calculate the oxidation no.of P in PO4-3

4.What is salt bridge?

5.Why do we need salt bridge?

6.What is meant by elcectrochemical series?

7.What is redox couple?

1 Marks answers

1. Loss of electron by an atom is called oxidation.

2. Gain of electron by an atom is called reduction.

Trick to remember

LEO the lion GER

Loss of Electrons Oxidation Gain of Electron Reduction

3. ON of P in PO43-

is x . so x +(-2)x4 = -3; or, x = +5

46

4. Salt bridge is inverted U tube like structure that contain strong electrolyte (KCl. NH4NO3) in

agar agar gel

5. It maintains the electrical neutrality of the half cells.

6. A series of elements arranged in decreasing order of their standard electrode potential.

7. The oxidised and reduced forms of a substance taking part in a half reaction is known as redox

couple.

Short Answer type question (2 Mark)

1.Balance the following ion electron method in acidic medium.

(i) NO3 -

+ Bi → Bi3+

+ NO2

(ii)Al + NO3 -

→ Al(OH)4 - + NH3

2.What is disproportionation reaction? Give an example.

3. Balance the following equation by oxidation no. method:

(a)MnO4- + Br

-→ MnO2 + BrO3

– (basic médium)

(b) Cr2O72-

+ SO32-

→ 2Cr3+

+ SO4 2-

(Acid Medium)

4. Find the oxidation state of underline element in the following species

H2SO5, CrO5

2 Marks answers

i) NO3- + Bi Bi

3+ + NO2

0 +3

Oxidation half Bi Bi3+

+ 3e

+5 +4

Reduction half (NO3- + e + 2H

+ NO2 +H2O )x3

Bi + 3NO3- + 6H

+ Bi

3+ + 3NO2 + 3H2O

ii) Al + NO3- Al(OH)4

- + NH3

0 +3

Oxidation half (Al + 4H2O Al(OH)4- + 4H

+ + 3e)x8

+5 +4

Reduction half (NO3- + 9H

+ + 8e NH3 + 3H2O)x3

8Al + 3NO3- + 23H2O 8Al(OH)4

- + 3NH3 + 5H

+

Those reactions in which single reactant undergoes oxidation as well as reduction are called

disproportionation reactions.

-1 -2 0

Ex. 2H2O2 2H2O + O2

a) Balanced equation will be: Br- + 2MnO4- + H2O → BrO3

- + 2MnO2 + 2OH

-

b) Balanced equation will be: 3SO32-

+ Cr2O72-

+ 8H+ → 3SO4

2- + 2Cr

3+ + 4H2O

2 × (+1) + x + 3 × (–2) + 2 × (–1)

(for H) (for S) (for O) (for O–O)

or 2 + x – 6 – 2 = 0 or x = + 6.

x + 4 × (–1) + 1 × (–2)

(for Cr) (for O--O) (for O)

Or x – 4 -2= 0 or x = + 6.

47

Short Answer II type question (3 Mark)

1.) Balance the following reaction byion electron method .(mention the oxidizing and reducing

agent)

Cr2O7-2

+ Fe+2

→ Cr+3

+ Fe+3

+ H2O ( Acidic medium )

OR

MnO4-

+ I-

→ MnO2 + I2 (Basic medium )

2.) For the following cell reaction,

Zn(s) + 2Ag+ → Zn+2

+ 2Ag

(i) Which of the electrode is negatively charged?

(ii) What is the carrier of the current in the cell?

(iii) Write the individual reaction at each electrode.

.

3 Marks answers

6Fe2+

+ Cr2O72-

+ 14H+ --------> 6Fe

3+ + 2Cr

3+ + 7H2O

OR,

2MnO4-(aq.) + 6I

-(aq.) + 4H2O 2 MnO2(s) +3I2(s) + 8OH

-

Zn I Zn2+

II Ag+ I Ag

Ag electrode

ZnSO4

Zn Zn2+

+2e

Ag + e Ag+

Question 1: Which of the following statements regarding redox reactions is incorrect?

a. Oxidation is the process of loss of electrons by a chemical species ( atom, ion or molecule).

b. Reduction is the process of gain of electrons by an atom, ion or molecule.

c. A reducing agent is one that undergo reduction

d. A reducing agent gives electrons;

Question 2: Which of the following rules to be followed for writing ionic equations is incorrect?

a. All soluble ionic compounds involved in a chemical changes are expressed in ionic symbols and

covalent substances are written in molecular form. H2O, NH3, NO2, NO, SO2, CO, CO2, etc., are

expressed in molecular form.

b. The ionic compound which is highly insoluble is expressed in molecular form.

c. The ions which are common and equal in number on both sides are written on product side.

d. Besides the atoms, the ionic charges must also be balanced on both the sides.

Question 3: Which of the following equations is not a redox reaction?

a. Fe2+

→ Fe3+

+ e-

b. MnO2 + 4HCl → MnCl2 + 2H2O +Cl2

48

c. Zn + 2H+ + 2Cl

- → Zn

2+ + 2Cl

- + H2

d. MnO2 + 4H++ 4Cl

- → Mn

2++ 2Cl

- + 2H2O +Cl2

Question 4: In the reaction MnO2 + 4HCl → MnCl2 + 2H2O +Cl2

Which of the following species acts as reducing agent?

a. MnO2

b. HCl

c. Cl-

d. None of above

Q.1 Q.2 Q.3 Q.4

c c a b

49

CHAPTER-9 HYDROGEN (By Group – H)

Hydrogen, the most abundant element in the universe and the third most abundant on the surface of the

globe, is being visualised as the major future source of energy

PREPARATION OF DIHYDROGEN, H2

Laboratory Preparation of Dihydrogen

Zn + 2H+ → Zn

2+ + H2 ; Zn + 2NaOH → Na2ZnO2 + H2

Commercial Production of Dihydrogen

The mixture of CO and H2 is called water gas.

As this mixture of CO and H2 is used forthe synthesis of methanol and a number ofhydrocarbons, it is

also called synthesis gas or 'syngas'.

Nowadays 'syngas' is producedfrom sewage, saw-dust, scrap wood,newspapers etc. The process of

producing'syngas' from coal is called 'coal gasification'.

water-gas shift reaction

Reactions of dihydrogen with organic compounds

Hydrogenation of vegetable oils using nickel as catalyst gives edible fats (margarine and vanaspati

ghee)

(ii) Hydroformylation of olefins yields aldehydes which further undergo reduction to give alcohols.

Uses of Dihydrogen

In the synthesis of ammonia / in the manufacture of vanaspati fat / in the manufacture of bulk organic

chemicals /used for the preparation of hydrogen chloride/ In metallurgical processes,/.Atomic hydrogen

and oxy-hydrogen torches find use for cutting and welding / used as a rocket fuel/ used in fuel cells for

generating electrical energy

HYDRIDES

Ex (e.g., MgH2) or EmHn (e.g., B2H6) E= any element except noble gas

The hydrides are classified into three categories :

(i) Ionic or saline or saltlike hydrides

(ii) Covalent or molecular hydrides

(iii) Metallic or non-stoichiometric hydrides

50

Ionic or saltlike hydrides Covalent or molecular hydrides Metallic or non-stoichiometric

hydrides

compounds of dihydrogen

formed with s-block elements

LiH, BeH2 and MgH2. The

ionic

hydrides are crystalline, non-

volatile and nonconducting

in solid state.

compounds with p-block

elements.

Three types

electron-deficient (B2H6)

electron-preciseCH4

electron-richNH3 , H2O

These are formed by many d-

block and f-block elements.

Unlike

saline hydrides, they are almost

always nonstoichiometric,

example, LaH2.87, YbH2.55,

WATER

The unusual properties of water in the condensed phase (liquid and solid states) are

due to the presence of extensive hydrogen bonding between water molecules. This leads to high

freezing point, high boiling point, high heat of vaporisation and high heat of fusion in comparison to

H2S and H2Se.

Structure of Water Structure of Ice

Hydrolysis Reaction

Q. How many hydrogen-bonded water molecule(s) are associated in CuSO4.5H2O?

Solution

Only one water molecule, which is outside the brackets (coordination sphere), is hydrogen-bonded. The

other four molecules of water are coordinated.

51

Hard and Soft Water

Presence of calcium and magnesium salts in the form of hydrogencarbonate, chloride and sulphate in

water makes water ‗hard‘.

Hard water does not give lather with soap.

Hard water forms scum/precipitate with soap.

It is harmful for boilers as well, because of deposition of salts in the form of scale. This reduces the

efficiency of the boiler

Temporary Hardness

Temporary hardness is due to the presence of magnesium and calcium hydrogencarbonates.

It can be removed by :

Boiling :

Clark‘s method

Permanent Hardness

It is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and

sulphates in water. It can be removed by the following methods:

Treatment with washing soda (sodium carbonate):

Ion-exchange method

Synthetic resins method:

Cation exchange resins contain large organic molecule with - SO3H group and are water insoluble. Ion

exchange resin (RSO3H) is changed to RNa by treating it with NaCl. The resin exchanges Na+ ions

with Ca2+

and Mg2+

ions present in hard water to make the water soft. Here R is resin anion.

Pure de-mineralised (de-ionized) water free from all soluble mineral salts is obtained by passing water

successively through a cation exchange (in the H+ form) and an ionexchange (in the OH

– form) resins:

HYDROGEN PEROXIDE (H2O2)

Preparation

NaZ is zeolite (NaAlSiO4)

52

H2O2 is miscible with water in all proportions and forms a hydrate H2O2.H2O

(mp 221K). A 30% solution of H2O2 is marketed as ‗100 volume‘ hydrogen peroxide. It means that one

millilitre of 30% H2O2 solution will give 100 V of oxygen at STP. Commercially, it is marketed as 10

V, which means it contains 3% H2O2.

10 volume solution of H2O2 means that 1L of this H2O2 will give 10 L of oxygen at STP (problem 9.4)

Structure

in gas phase in solid phase

Chemical Properties

It acts as an oxidising as well as reducing agent in both acidic and alkaline media

Oxidising action in acidic medium

Reducing action in acidic medium

Oxidising action in basic medium

Reducing action in basic medium

Storage:

H2O2 decomposes slowly on exposure to light. In the presence of metal surfaces or traces of alkali

(present in glass containers), the above reaction is catalysed. It is, therefore, stored in wax-lined glass or

plastic vessels in dark. Urea can be added as a stabiliser. It is kept awayfrom dust because dust can

induce explosive decomposition of the compound.

Uses

Hair bleach / mild disinfectant/ antiseptic/ to manufacture sodium perborate and per-

carbonate(detergents)/ bleaching agent for textiles, paper pulp, leather, oils. Nowadays it is also used in

Environmental (Green) Chemistry. For example, in pollution control treatment of domestic and

industrial effluents, oxidation of cyanides, restoration of aerobic conditions to sewage wastes, etc.

53

HEAVY WATER, D2O

used as a moderator in nuclear reactors and in the study of reaction mechanisms

It can be prepared by exhaustive electrolysis of water

DIHYDROGEN AS A FUEL

Dihydrogen releases large quantities of heat on combustion. dihydrogen can release more

energy than petrol (about three times) Moreover, pollutants in combustion of dihydrogen will be less

than petrol. In this view Hydrogen Economy is an alternative.

The basic principle of hydrogen economy is the transportation and storage of energy in the form of

liquid or gaseous dihydrogen. Advantage of hydrogen economy

is that energy is transmitted in the form of dihydrogen and not as electric power.

Unit IX Hydrogen

Very Short Answer type question (1 Mark)

1.Which isotope of hydrogen is radioactive?

2.What is the cause of hardness of water?

3.What is use of hydrogenation in manufacture of vanaspati ghee?

4.How is pure hydrogen obtained?

5.How does calcium carbide react with heavy water

6.What do you mean by hydrogen economy?

1 Marks answers

1. Tritium

2. Presence of calcium and magnesium salts in the form of hydrogencarbonate, Chloride and

Sulphate in water makes water hard.

3. To convert unsaturated fatty acid to saturatedfatty acid.

4. Electrolysis of water gives pure Hydrogen.

5. CaC2 + 2 D2O = Ca(OD)2 + C2D2

6. Use of Hydrogen as a fuel due to the Hydrogen releases large quantity of heat on combustions.

Short Answer type question (2 Mark)

1. Explain the term:

[a] Water gas shift reaction [b] Four resemblance of H2 with halogen

2.Dicuss the position of hydrogen in periodic table?

3.How is H2O2 manufactured?

4.Show how H2O2 functions both as a reducing agent and as a oxidizing agent?

5.In which respect hydrogen differs from alkali metals and halogens?

6.What causes the temporary and permanent hardness of water?

7.Calculate the volume strength of 3% solution H2O2 ?

8.Write the chemical reaction to show amphoteric nature of water?

54

9. What is meant by demineralised water? And how can it be obtained?

10.Write the two uses of interstitial hydrides?

2Marks answers

1. a) The water-gas shift reaction (WGSR) describes the reaction of carbon monoxide and water

vapour to form carbon dioxide and hydrogen (the mixture of carbon monoxide and hydrogen

(not water) is known as water gas):

CO + H2O CO2 + H2

b) i)Same valency , diatomic molecules

(ii) Forms hydrides like halides

iii) The ionization energy of Hydrogen and halogens are comparable.

iv) Forms covalent bonds like halogens

2. i) Hydrogen can be placed in gr. 1 as well as gr. 17

ii) Hydrogen shares many similarities with alkali metals, i.e. elements in group I-A. This is one of the

factors that dictates the position of hydrogen in the table.(some properties like electronic configuration,

forms halides, reducing agent).

iii) Hydrogen shares many similar properties with halogens like electronegativity, diatomic molecules,

covalent bonds.

3. BaO2.8H2O(s) + H2SO4(aq.) BaSO4(s) + H2O2(aq.) + 8H2O

4. It acts as an oxidising as well as reducing agent in both acidic and alkaline media

Oxidising action in acidic medium

Reducing action in acidic medium

Oxidising action in basic medium

Reducing action in basic medium

5. Differences with alkali Metals

Non-metal: Hydrogen is essentially not a metal like all alkali metals, but a non-metal

Loss of Electron: Although it has only one electron in its outer shell, hydrogen cannot easily lose this

electron to gain electropositivity. All other alkali metals can do this with ease.

State: At room temperatures where all alkali metals exist is the solid state, hydrogen is a gas.

size of atom: The H+ ion of hydrogen is much smaller than ions of alkali metals.

Ionization Potential: The ionization potential of hydrogen is over 300 kcal per mole, The maximum

ionization potential for metals is 147 kcal per mole

6. Temporary Hardness

Temporary hardness is due to the presence of magnesium and calcium hydrogencarbonates.

Permanent Hardness

https://en.wikipedia.org/wiki/Carbon_monoxide

https://en.wikipedia.org/wiki/Water_vapor



https://en.wikipedia.org/wiki/Carbon_dioxide

https://en.wikipedia.org/wiki/Hydrogen

https://en.wikipedia.org/wiki/Water_gas

55

It is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and

sulphates in water.

7. It is marketed as 10 V, which means it contains 3% H2O2. 10 volume solution of H2O2 means

that 1L of this H2O2 will give 10 L of oxygen at STP

8. Amphoteric substance is that substance which has ability to donate as well as accept a Proton.

According to Bronsted -Lowry concept, bases stronger than water, tend to accept proton from it. Thus

by donating proton, water acts as an acid.

H₂O + NH₃<==> OH- + NH₄+

While acid stronger than water, like HCl ,tend to donate proton to water. Thus by accepting proton

from such strong acids, water acts as base.

H₂O + HCl <===> H₃O+ + Cl

-

Thus water can donate as well as accept proton and show amphoteric nature.

9. Pure de-mineralised (de-ionized) water free from all soluble mineral salts is obtained bypassing

water successively through a cation exchange (in the H+ form) and an anion exchange (in the

OH– form) resins:

10. i. Interstitial hydrides are used in metallurgy, in manufacture of vacuum tubes.

ii. Metallic or interstitial hydrides are used as reducing agents.

Short Answer II type question (3 Mark)

1. What is the difference between terms hydrolysis and hydration?

2. What properties of water make useful as a solvent?

3. What do you understand by metallic hydrides?

4. What is syn gas ?

5. Give four uses of hydrogen peroxide?

6. What happens H2O2 is treated with

(a) acidified potassium permanganate (b) acidified ferrous sulphate

Write the chemical reaction involved in it .

7. Write notes on

(1) Ion exchange method for softening of hard water (2) Clark‘s method (3) Calgons‘s method

56

3Marks answers

1. Hydrolysis: Breaking of a compund using water.

Eg. NaCl + H2O NaOH + HCl

Hydration: Addition of a water molecule to a compound

Eg. Hydration of ethane C2H4 + H2O → CH3CH2OH

2. It is water's chemical composition and physical attributes that make it such an excellent solvent.

Water molecules have a polar arrangement of the oxygen and hydrogen atoms—one side

(hydrogen) has a positive electrical charge and the other side (oxygen) had a negative charge.

This allows the water molecule to become attracted to many other different types of molecules.

Water can become so heavily attracted to a different molecule, like salt (NaCl), that it can

disrupt the attractive forces that hold the sodium and chloride in the salt molecule together and,

thus, dissolve it.

3. These are formed by many d-block and f-block elements. Unlike saline hydrides, they are

almost always nonstoichiometric. example, LaH2.87, YbH2.55.

4. The mixture of CO and H2 is called water gas.

As this mixture of CO and H2 is used for the synthesis of methanol and a number of hydrocarbons, it is

also called synthesis gas or 'syngas'.

Nowadays 'syngas' is produced from sewage, saw-dust, scrap wood, newspapers etc. The process of

producing 'syngas' from coal is called 'coal gasification'.

5. Use of H2O2: Hair bleach / mild disinfectant/ antiseptic/ to manufacture sodium perborate and per-

carbonate (detergents)/ bleaching agent for textiles, paper pulp, leather, oils. Nowadays it is also used

in Environmental (Green) Chemistry.

6. a) H2O2 is used as a reducing agent and it reduces KMnO4 into MnO2

b) H2O2 is used as a oxidising agent and it oxidises Ferrous to Ferric.

7.

1) In this method sodium zeolite is used as ion-exchanger. Na-zeolite is passed through the pipes

containing hard water. Sodium zeolite is converted into calcium-zeolite or magnesium-zeolite. These

are insoluble in water and are separated from water by filtration.

Ca+2

+ Na2-zeolite Ca-zeolite + 2Na+1

Mg+2

+Na2-zeolite Mg-zeolite + 2Na+1

57

2) In Clark‘s method a calculated amount of calcium hydroxide Ca (OH)2 is added to hard water. Due

to reaction, insoluble carbonates are obtained which are separated by filtration.

Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2H2O

Mg(HCO3)2 + Ca(OH)2 CaCO3 + MgCO3 + 2H2O

3) Calgon is mainly used for removing the Hardness from the water , so basically what it does is that it

ionizes to give a complex anion:

Now as we know that Hard water contains

dissolved magnesium and calcium ions. These make it more difficult for the water to form a lather with

soap the addition of Calgon to hard water causes the calcium and magnesium ions to undergo a

displacement reaction with the anion produced by the Calgon !

The obtained product lowers the hardness of water by forming a complex with the hardness - providing

ions thus the relative hardness of water lowers down and the by product , the complex of calcium and

Magnesium ions don't form any precipitate with soap and hence readily produced lather with soap

.

58

58

CHAPTER-10, THE s – BLOCK ELEMENTS(by Gr D)

PROPERTIES GROUP 1 ELEMENTS GROUP 2 ELEMENTS

Name of

members Lithium, sodium, potassium,

rubidium, cesium and francium.

Beryllium, magnesium, calcium,

strontium, barium and radium.

Group 1 elements are collectively

known as the alkali metals. These

are so called because they form

hydroxides on reaction with water

which are strongly alkaline in

nature.

These elements with the exception of

beryllium are commonly known as the

alkaline earth metals. These are so

called because their oxides and

hydroxides are alkaline in nature and

these metal oxides are found in the

earth‘s crust.

General

outermost

electronic

configuration

ns1n from 2 to 7 ns

2n from 2 to 7

Ionization

Enthalpy The ionization enthalpies of the

alkali metals are considerably low

and decrease down the group from

Li to Cs. This is because the effect

of increasing size outweighs the

increasing nuclear charge, and the

outermost electron is very well

screened from the nuclear charge.

The alkaline earth metals have low

ionization enthalpies due to fairly large

size of the atoms.

Since the atomic size increases down

the group, their ionization enthalpy

decreases The first ionisation enthalpies

of the alkaline earth metals are higher

than those of the corresponding Group

1 metals. This is due to their small size

as compared to the corresponding alkali

metals. It is interesting to note that the

second ionisation enthalpies of the

alkaline earth metals are smaller than

those of the corresponding alkali

metals.

Hydration

Enthalpy Decrease with increase in ionic

sizes. Li+ > Na+ > K+ > Rb+ >

Cs+ Li+ has maximum degree of

hydration and for this reason

lithium salts are mostly hydrated,

e.g., LiCl· 2H2O

Decrease with increase in ionic size

down the group. Be2+

> Mg2+

> Ca2+

>

Sr2+

> Ba2+

Softness: All the alkali metals are silvery

white, soft and light metals due to

weak metallic bonding

Alkaline earth metals are are less softer

than alkali metals.

59

59

The melting and

boiling points: The melting and boiling points of

the alkali metals are low

indicating weak metallic bonding

due to the presence of only a

single valence electron in them.

The melting and boiling points of the

alkaline earth metals are more than

alkali metals.

Flame

colouration: The alkali metals and their salts

impart characteristic colour to an

oxidizing flame. This is because

the heat from the flame excites the

outermost orbital electron to a

higher energy level. When the

excited electron comes back to the

ground state, there is emission of

radiation in the visible region.

The electrons in beryllium and

magnesium are too strongly bound to

get excited by flame. Hence, these

elements do not impart any colour to

the flame. The flame test for Ca, Sr and

Ba is helpful in their detection in

qualitative analysis and estimation by

flame photometry.

Atomic and

Ionic Radii The alkali metal atoms have the

largest sizes in a particular period

of the periodic table and radii

increases moving down the group

due to increase in the number of

shell.

The atomic and ionic radii of the

alkaline earth metals are smaller than

those of corresponding alkali metals in

the same periods. This is due to the

increased nuclear charge in these

elements. Within the group, the atomic

and ionic radii increase with increase in

atomic number.

Reactivity

towards air The alkali metals tarnish in dry air

due to the formation of their

oxides which in turn react with

moisture to form hydroxides.

They burn vigorously in oxygen

forming oxides. Lithium forms

monoxide, sodium forms

peroxide, the other metals form

superoxides. The

superoxide O2 – ion is stable only

in the presence of large cations

such as K, Rb, Cs. Because of

their high reactivity towards air

and water, alkali metals are

normally kept in kerosene oil.

Beryllium and magnesium are

kinetically inert to oxygen and water

because of the formation of an oxide

film on their surface. However,

powdered beryllium burns brilliantly on

ignition in air to give BeO and Be3N2.

Magnesium is more electropositive and

burns with dazzling brilliance in air to

give MgO and Mg3N2. Calcium,

strontium and barium are readily

attacked by air to form the oxide and

nitride.

Reactivity

towards water The alkali metals react with water

to form hydroxide and

dihydrogen.

2M+2H2O =2M++2OH

-+ H2

(M = an alkali metal)

They react with water with increasing

vigour even in cold to form hydroxides.

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Reactivity

towards

dihydrogen

The alkali metals react with

dihydrogen at about 673K

(lithium at 1073K) to form

hydrides. 2M+ H 2=2M H

All the elements except beryllium

combine with hydrogen upon heating to

form their hydrides.

M+ H 2=M H2

Reactivity

towards

halogens

The alkali metals readily react

vigorously with halogens to form

ionic halides, M+ X

–.

lithium halides are somewhat

covalent. It is because of the high

polarisation capability of lithium

ion among halides, lithium

iodide is the most covalent in

nature.

All the alkaline earth metals combine

with halogen at elevated temperatures

forming their halides.

Reducing nature The alkali metals are strong

reducing agents, lithium being the

most and sodium the least

powerful

the alkaline earth metals are strong

reducing agents. This is indicated by

large negative values of their reduction

potentials however their reducing

power is less than those of their

corresponding alkali metals.

Beryllium has less negative value

compared to other alkaline earth metals.

However, its reducing nature is due to

large hydration energy associated with

the small size of Be2+

ion and relatively

large value of the atomization enthalpy

of the metal.

Solutions in

liquid ammonia: The alkali metals dissolve in

liquid ammonia giving deep blue

solutions which are conducting in

nature.

M +(x +y)NH3= [M(NH3 )x ]+

+

[e (NH3 )y ]-

The blue colour of the solution is

due to the ammoniated electron

which absorbs energy in the

visible region of light and thus

imparts blue colour to the

solution.

The solutions are paramagnetic

and on standing slowly liberate

hydrogen resulting in the

formation of amide.

the alkaline earth metals dissolve in

liquid ammonia to give deep blue black

solutions forming ammoniated ions.

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iv)Reactivity

towards acids The alkali metals readily react

with acids liberating

dihydrogen.

2M + 2HCl →2 MCl + H2

The alkaline earth metals readily react

with acids liberating dihydrogen.

M + 2HCl → MCl2 + H2

Diagonal

Relationship between Lithium and

Magnesium:

Some of the similarities are: (i)

both of them forms nitrides and

monoxides(ii) both LiCl and

MgCl2 are deliquescent.

between Beryllium and Aluminium :

Some of the similarities are: (i) Like

aluminium, beryllium is not readily

attacked by acids because of the

presence of an oxide film on the surface

of the metal. (ii) Beryllium hydroxide

dissolves in excess of alkali to give a

beryllate ion, [Be(OH)4] 2–

just as

aluminium hydroxide gives aluminate

ion, [Al(OH)4] –

BIOLOGICAL

IMPORTANCE OF SODIUM AND

POTASSIUM:

Sodium ions are found primarily

on the outside of cells, being

located in blood plasma and in the

interstitial fluid which surrounds

the cells. These ions participate in

the transmission of nerve signals,

in regulating the flow of water

across cell membranes and in the

transport of sugars and amino

acids into cells.

OF MAGNESIUM AND CALCIUM:

All enzymes that utilise ATP in

phosphate transfer require magnesium

as the cofactor. The main pigment for

the absorption of light in plants is

chlorophyll which contains magnesium.

About 99 % of body calcium is present

in bones and teeth.It also plays

important roles in neuromuscular

function, interneuronal transmission,

cell membrane integrity and blood

coagulation.

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SOME IMPORTANT COMPOUNDS OF SODIUM

Sodium Carbonate (Washing Soda), Na2CO3·10H2O :

Sodium carbonate is generally prepared by Solvay Process. In this process, advantage is taken of the

low solubility of sodium hydrogencarbonate whereby it gets precipitated in the reaction of sodium

chloride with ammonium hydrogencarbonate. The latter is prepared by passing CO2 to a concentrated

solution of sodium chloride saturated with ammonia, where ammonium carbonate followed by

ammonium hydrogencarbonate are formed. The equations for the complete process may be written as :

2NH3+ H2O+CO2=(NH4)2CO3

(NH4)2CO3+H2O+CO2=2NH4HCO3

NH4HCO3+NaCl=NH4Cl+NaHCO3

Sodium hydrogencarbonate crystal separates. These are heated to give sodium carbonate.

2 NaHCO3 → Na2 CO3+CO2+H2O

It may be mentioned here that Solvay process cannot be extended to the manufacture of

potassium carbonate because potassium hydrogencarbonate is too soluble to be precipitated by

the addition of ammonium hydrogencarbonate to a saturated solution of potassium chloride.

Uses: (i) It is used in water softening, laundering and cleaning. (ii) It is used in the manufacture of

glass, soap, borax and caustic soda.

Sodium Chloride, NaCl:

Uses : (i) It is used as a common salt or table salt for domestic purpose. (ii) It is used for the

preparation of Na2O2, NaOH and Na2CO3.

(NaOH )Sodium hydroxide: Sodium hydroxide is generally prepared commercially by the

electrolysis of sodium chloride in Castner-Kellner cell.

Uses: It is used in (i) the manufacture of soap, paper, artificial silk and a number of chemicals, (ii) in

petroleum refining,

(Baking Soda), NaHCO3:Sodium hydrogencarbonate is known as baking soda because it decomposes

on heating to generate bubbles of carbon dioxide (leaving holes in cakes or pastries and making them

light and fluffy).

Sodium hydrogencarbonate is made by saturating a solution of sodium carbonate with carbon dioxide.

The white crystalline powder of sodium hydrogencarbonate, being less soluble, gets separated out.

Uses:(i)Sodium hydrogencarbonate is a mild antiseptic for skin infections.(ii) It is used in fire

extinguishers.

SOME IMPORTANT COMPOUNDS OF CALCIUM:

Important compounds of calcium are calcium oxide, calcium hydroxide, calcium sulphate, calcium

carbonate and cement.

Calcium Oxide or Quick Lime, CaO:

It is prepared on a commercial scale by heating limestone (CaCO3) in a rotary kiln at 1070-1270 K.

CaCO3 CaO+ CO2

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Uses: (i) It is an important primary material for manufacturing cement and is the cheapest form of

alkali. (ii) It is used in the manufacture of sodium carbonate from caustic soda.

Calcium Hydroxide (Slaked lime), Ca (OH) 2:

Calcium hydroxide is prepared by adding water to quick lime, CaO. The aqueous solution is known as

lime water and a suspension of slaked lime in water is known as milk of lime.

When carbon dioxide is passed through lime water it turns milky due to the formation of calcium

carbonate.

Ca (OH) 2 + CO2 CaCO3+ H2 O

On passing excess of carbon dioxide, the precipitate dissolves to form calcium hydrogencarbonate.

CaCO3+ CO2 +H2 O Ca(HCO3)2

Milk of lime reacts with chlorine to form hypochlorite, a constituent of bleaching powder.

2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl 2 +2H2O

Uses: (i) It is used in the preparation of mortar, a building material. (ii) It is used in white wash due to

its disinfectant nature.

Calcium Carbonate, CaCO3:

It can be prepared by passing carbon dioxide through slaked lime or by the addition of sodium

carbonate to calcium chloride.

Ca (OH) 2 + CO2 CaCO3+ H2 O

CaCl2 + Na2CO3 CaCO3+ 2NaCl

Excess of carbon dioxide should be avoided since this leads to the formation of water soluble calcium

hydrogencarbonate.

When heated to 1200 K, it decomposes to evolve carbon dioxide.

CaCO3 CaO+ CO2

Uses: (i)It is used as a building material in the form of marble and in the manufacture of quick lime.

(ii)Calcium carbonate along with magnesium carbonate is used as a flux in the extraction of metals

such as iron.

Calcium Sulphate (Plaster of Paris) CaSO4· 1/2H2O:

It is a hemihydrate of calcium sulphate. It is obtained when gypsum, CaSO4·2H2O, is heated to 393 K.

2(CaSO4.2H2O) 2(CaSO4).H2O + 3H2O

Above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, CaSO4 is formed.

This is known as ‗dead burnt plaster‘.

Uses: (i)The largest use of Plaster of Paris is in the building industry as well as plasters.(ii) It is used for

immoblising the affected part of organ where there is a bone fracture or sprain.

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Important question from s block elements:

1)Explain the following:

i)LiI is more soluble than KI in ethanol.

Ans: The lithium ion has a smaller size and as a result of that, it has a higher polarizing capability. This

enables it to polarize the electron cloud around an iodide ion thus resulting in a greater covalent

character in LiI than KI. Thus, LiI is easily soluble in ethanol.

ii)The solution of alkali metals in liquid ammonia is blue in colour.

Ans: The alkali metals dissolve in liquid ammonia giving deep blue solutions

M +(x +y)NH3= [M(NH3 )x ]+

+ [e (NH3 )y ]-

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible

region of light and thus imparts blue colour to the solution.

iii)Sodium carbonate solution is alkaline.

Ans: Sodium carbonate is a salt of strong base and weak acid so it on hydrolysis gives NaOH which is a

strong base so Sodium carbonate solution is alkaline.

iv)BeO is insoluble in water while BeSO4 is water soluble.

Ans: The sizes of Be2+

and O2-

are small and are highly compatible with each other. Due to this, a high

amount of lattice energy is released during its formation. The hydration energy, when it is made to

dissolve in water, is not enough to overcome the lattice energy. Thus, BeO is barely soluble in water.

Whereas the size of an SO42-

is large compared to Be2+

and there is lesser compatibility and lattice

energy which can be easily overcome by the hydration energy. Thus, BeSO4 is easily soluble in water.

2)The alkali and alkaline earth metals cannot be obtained by chemical reduction methods.

Ans: By using a stronger reducing agent, the oxides of metals gets reduced by the process called

chemical reduction. Alkaline earth metals and alkali metals are strong among the reducing agents. No

stronger reducing agent is available than them. Therefore alkaline earth metals and alkali cannot be

obtained by chemical reduction of their oxides.

3)Define diagonal relationship between the elements.List two similarities in chemical behavior of Li

with Mg.

Ans: some elements of certain groups in the second period resemble with the certain elements of the

next higher group in the third period. This is called diagonal relationship.

i)oxides of lithium and magnesium are less soluble in H2O. also the hydroxides of both decompose at

high temperature.

2LiOH→Li2O+H2O

Mg(OH)2→MgO+H2O

(ii) Nitrides is formed from both the lithium and magnesium when they react with N2.

6Li+N2→2Li3N

3Mg+N2→Mg3N2

iii)Neither Li nor Mg form superoxides or peroxides.

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4)Give chemical reaction when:

i)Quick lime is heated with silica.

Ans: Silica (SiO2) combines with Quick lime (CaO) resulting in formation of Slag.

CaO + SiO2 → CaSiO3

ii)Chlorine reacts with slaked lime.

Ans: Bleaching powder is formed when chlorine is made to react with slaked lime.

Ca(OH)2 + Cl2 → CaOCl2 + H2O

iii)Calcium nitrate is heated.

Ans: Calcium nitrate, when heated, undergoes decomposition to form calcium oxide.

2Ca(NO3)2(s) → 2CaO(s) + 4NO2(g) + O2(g)

5)Why K2CO3 cannot be prepared by Solvay process?

Ans: Solvay process is not applicable for the preparation of potassium carbonate because potassium

bicarbonate is soluble in water and it doesn‘t precipitates out like sodium bicarbonate.

6)Li is the strongest reducing agent in spite of its high ionization enthalpy .why?

Ans: Due to its high hydration energy Lithium results in strong reducing agent among all alkali metals.

7)Explain the following:

i)Be and Mg don‘t impart any colour to the flame.

Ans: The valence electrons get excited to a higher energy level when an alkaline earth metal is heated.It

radiates energy which belongs to the visible region when this excited electron comes back to its energy

level which is low. The colour is observed here. The electrons are strongly bound in the beryllium and

magnesium. The energy required to excite these electrons is very high. When the electron reverts back

to its original position, the energy released does not fall in the visible region. Hence, no colour is seen

in the flame.

ii)Mg is burnt in air.

Ans: Mg is burnt in air with dazzling sparkle For the formation of MgO and Mg3 N2.

8)Complete the reaction:

i)KO2+ H2O →

Ans: 2 KO2+2 H2O → 2KOH + H2O2+O2

9)Why Li2CO3 decomposed at lower temperature whereas Na2CO3 at higher temperature?

Ans: The electropositive character increases while moving down in the group of alkali metal which

results in increase in stability of alkali carbonates. Generally, lithium carbonate is not stable when it

reacts to heat because lithium carbonate is covalent. Due to the smaller size of lithium ion it polarizes

large carbonate ion which results in the formation of stable lithium oxide.

10)Why are lithium salts commonly hydrated and those of others alkali ions are usually anhydrous?

Ans: : The hydration enthalpies of alkali metal ions decrease with increase in ionic sizes. Li+ > Na+ >

K+ > Rb+ > Cs+ Li+ has maximum degree of hydration and for this reason lithium salts are mostly

hydrated.

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11)Complete the equation:

i)Mg3N2+ H2O →

Mg3N2+ 6H2O → 3Mg(OH)2+ 2NH3

ii)Ca(OH)2 + Cl2 →

2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl 2 +2H2O

iii)Ca(NO3)2 →

2Ca(NO3)2(s) → 2CaO(s) + 4NO2(g) + O2(g)

12) Arrange alkali metal ions in decreasing order of hydration enthalpies.

Ans: Li+ > Na+ > K+ > Rb+ > Cs+

13) Write the chemical formula of plaster of paris and baking soda.

Ans:(Plaster of Paris) Calcium Sulphate CaSO4· H2O

(Baking Soda), NaHCO3

14) What is milk of magnesia? Give its one important use.

Ans: The chemical formula for milk of magnesia is Mg (OH)2. Milk of magnesia is often used for

constipation, heartburn, and upset stomach.

15)Arrange LiH,NaH and CsH in order of increasing ionic character.

Ans: LiH<NaH< CsH

16)Write two uses of caustic soda.

Ans: It is used in (i) the manufacture of soap, paper, artificial silk and a number of chemicals, (ii) in

petroleum refining.

17) Why are alkali metals not found in nature?

Ans: Sodium, cesium, lithium, francium, potassium, rubidium all together comprises of alkali metals.

they consist of only one electron on its valence shell, which gets loosed easily due to their low

ionization enthalpy.

18)Find the oxidation state of sodium in Na2O2.

Ans: Let the oxidation state of Na be y.

In case of peroxides, the oxidation state of oxygen is -1.

Therefore,

2(y) + 2(-1) = 0

2y – 2 = 0

2y = 2

Y = +1

Therefore, the oxidation state of Na is +1.

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19)Explain why is sodium less reactive than potassium?

Ans: on moving down the group in the alkali metals, the size of the atom increases and the effect of the

nuclear charge gets decreased. Due to this factors, the electron of potassium which is located outer gets

lost easily as compared to Na. therefore, potassium reacts higher than Na.

20) Explain why alkali and alkaline earth metals cannot be obtained by chemical reduction methods?

Ans: By using a stronger reducing agent, the oxides of metals gets reduced by the process called

chemical reduction. Alkaline earth metals and alkali metals are strong among the reducing agents. No

stronger reducing agent is available than them. Therefore alkaline earth metals and alkali cannot be

obtained by chemical reduction of their oxides.

21)Why are potassium and cesium, rather than lithium used in photoelectric cells?

Ans: Lithium, potassium, and cesium, are all alkali metals. But still, potassium and cesium are used in

photoelectric cell and not Lithium because Li is smaller in size when compared to the other two.

On the other hand, cesium and potassium have low ionization energy. Therefore, they lose electrons

easily. This property is utilized in photoelectric cells.

22) When an alkali metal dissolves in liquid ammonia the solution can acquire different colours.

Explain the reasons for this type of colour change.

Ans: When the alkali metal is dissolved in liquid ammonia, a deep blue coloured solution is formed.

M +(x +y)NH3= [M(NH3 )x ]+

+ [e (NH3 )y ]-

The ammoniated electrons absorb energy corresponding to red region of visible light. Therefore, the

transmitted light is deep blue in colour.

Clusters of metal ions are formed at higher concentration (3M) which causes the solution to attain a

copper-bronze colour and a metallic lustre.

23)Discuss the various reactions that occur in the Solvay process.

Ans: complete process may be written as :

2NH3+ H2O+CO2=(NH4)2CO3

(NH4)2CO3+H2O+CO2=2NH4HCO3

NH4HCO3+NaCl=NH4Cl+NaHCO3

Sodium hydrogencarbonate crystal separates. These are heated to give sodium carbonate.

2NaHCO3=Na2 CO3+CO2+H2O

24)Starting with sodium chloride how would you proceed to prepare (i) sodium metal (ii) sodium

hydroxide (iii) sodium peroxide (iv) sodium carbonate?

Ans: ) i) sodium metal:Sodium chloride can be converted into sodium by Downs process.

It can be achieved by electrolysis of fused CaCl2 (60 %) and NaCl (40%) at 1123 K in a special

apparatus (Downs cell).

A graphite block is the anode while steel is made the cathode. Metallic Ca and Na are formed at the

cathode. Molten Na is supported by dipping in kerosene.

Nacl →Electrolysis Na+ + Cl

–

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(Molten)

At Cathode: Na+ +e

– → Na

At Anode: Cl– - e

– → Cl

ii)Sodium Hydroxide:

By electrolyzing a solution of sodium chloride, we can get Sodium hydroxide. This process is

commonly known as Castner- Kellner process.

The process is carried out using a mercury cathode and a carbon anode.

Sodium metal, deposited at cathode forms an Amalgam by combining with Mercury.

Cathode: Na+ +e

– →Hg Na-Amalgam

Anode; 2 Cl– → Cl2 + 2e

–

Cl + Cl → Cl2

(iii) Sodium peroxide

After Na metal is gotten from Downs process, it is heated on Aluminium trays in presence of

air(without CO2) to form Sodium peroxide.

2Na + O2(air) → Na2O2

iv) Sodium carbonate

Sodium hydrogen carbonate is obtained as a precipitate by reacting sodium chloride with ammonium

hydrogen carbonate. The resultant crystals can be heated to obtain Sodium Carbonate.

2NH2 + H2O → (NH4)2CO3

(NH4)2CO3 + H2O + CO2 → 2NH4HCO3

2NH4HCO3 + NaCl → NH4Cl + NaHCO3

The resultant crystals can be heated to obtain Sodium Carbonate.

2NaHCO3 → Na2CO3+ CO2 + H2O

25)Describe two important uses of each of the following: (i) caustic soda (ii) sodium carbonate (iii)

quicklime.

Ans:(i) Caustic soda

(a) Heavily used in soap industries.

(b) Common reagent in laboratories.

(ii) Sodium carbonate

(a) Finds uses in both soap and glass industries.

(b) Also finds use as a water softener.

(iii) Quick lime

(a)Finds use as a primary material for manufacturing slaked lime.

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(b) It helps in the manufacture of cement and glass.

26)Draw the structure of (i) BeCl2 (vapour) (ii) BeCl2 (solid).

Ans: In solid phase, BeCl2 is a polymer

BeCl2 has a linear structure and exists as a monomer in vapour state.

27) The hydroxides and carbonates of sodium and potassium are easily soluble in water while the

corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.

Ans: Since the atomic sizes of magnesium and calcium are smaller than that of sodium and potassium,

calcium and magnesium form carbonates and hydroxides with higher lattice energies. Thus, they are

only sparingly soluble whereas those of potassium and sodium are readily soluble due to low lattice

energies.

28)Describe the importance of the following (i) limestone (ii) cement

Ans:Uses of cement:

Bridge construction

Plastering

Most important ingredient in concrete

Uses of Plaster of Paris:

Used to make casts and moulds

Used to make surgical bandages

Uses of limestone:

Preparation of cement and lime

As a flux in iron ore smelting

29) Why is LiF almost insoluble in water whereas LiCl soluble not only in water but also in acetone.

Ans: LiF has a greater ionic character than LiCl which disturbs the balance between hydration energy

and lattice energy. This balance is crucial for the solvability of ions in solution. Due to greater covalent

character and lower lattice energy, dissolution of LiCl is more exothermic in nature than that of LiF.

30)Explain the significance of sodium, potassium, magnesium and calcium inbiological fluids.

Ans: (i) Sodium (Na):

They are found in our blood plasma and the interstitial fluids around the cells. They help in

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(a) Transmission of nerve signals.

(b) They regulate the flow of water across the membranes of the neighboring cells.

(c) Transport sugars and amino acids from and to cells.

(ii) Potassium (K):

They are found mostly in the cell fluids in greater quantities.

They help in

(a) Activating enzymes.

(b) Oxidising glucose to form ATP.

(c) Transmitting nerve signals.

(iii) Magnesium (Mg) and calcium (Ca):

They are also called as macro-minerals named so because of their abundance in our body. Mg helps in

(a) Relaxing nerves and muscles.

(b) Building and strengthening bones.

(c) Maintaining blood circulation in our body.

Ca helps in

(a) coagulation of blood

(b) Maintaining homeostasis.

31)What happens when

i)Sodium metal is dropped in water?

Ans:(i) Sodium reacts to form NaOH and H2 gas when it is dropped in water. The reaction occurs as

shown below:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

ii)Sodium metal is heated in free supply of air?

Ans: Sodium peroxide is formed when sodium reacts with oxygen while heating it in presence of air.

The reaction proceeds as shown below:

2Na(s) + O 2(s) → Na2O 2(s)

iii)Sodium peroxide dissolves in water?

Ans: NaOH and water are formed as a result of hydrolysis of Sodium peroxide when it is dissolved in

water.

Na2O 2(s) + 2H2O(l) → 2NaOH(aq) + H2O2(aq)

32)State as to why

i) alkali metals are prepared by electrolysis of their fused chlorides.

Ans: Chemical reduction cannot be used to prepare alkali metals since they themselves are reducing in

nature. Alkali metals are highly electropositive and thus cannot be prepared by displacement reactions.

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Since they also react with water, these alkali metals cannot be prepared by electrolysis of their aqueous

solutions. Thus, alkali metals are mostly prepared by electrolysis of their fused chlorides.

ii)sodium is found to be more useful than potassium?

Ans: Sodium ions are primarily found in the Blood plasma and the interstitial fluids around the cells

whereas Potassium ions are found within the cell fluids. Sodium

ions help in the transmission of nerve signals and also regulate the flow of water and transport sugars

and amino acids into the cells.

Thus, Sodium is more important for our survival than potassium.

33) Write balanced equations for reactions between

i)Na2O2 and water

Ans: Na2O2+2H2O 2NaOH + H2O2

ii) KO2 and water

Ans:2 KO2+2 H2O 2KOH + H2O2+O2

iii)Na2O and CO2

Ans: Na2O+ CO2

Na2CO3

Group members:- 1.Mrs.Mosarrat Jehan (KV Chittaranjan)

2. Mr. Raghib Roushan ( KV N0.2 Ishapore)

3. Mr. Amit Kumar ( KV Andal)

4. Mr.Navneet Singh ( KV AFS Salua)

5. Mrs.Gopi Priya ( KV Panagarh)

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CHAPTER – 11 p- Block Elements (Gr-E)

Boron Family (Group 13 Elements )

Members:B, Al, Ga, In & Tl

Melting Point: Decreases from B to Ga and then increases up to Tl.

Ionization Energies: 1st <<< 2nd< 3rd

Metallic Character: Increases from B to Tl. B is non-metal

Boron

Preparation of Boron:

From Boric Acid: B2O3(s) + 3Mg(s) → 2B(s) +3 MgO(s)

From Boron Trichloride

o (at 1270 k): 2BCl3+ 3H2 (g) → 2B(s) + 6HCl (g)

o (at 900 0C): 2BCl3(g)+ 3Zn (s) → 2B(s) + 3 ZnCl2 (s)

By electrolysis of fused mixture of boric anhydride (B2O3) and magnesium oxide (MgO) & Magnesium

fluoride at 1100 0C

o 2 MgO- → 2Mg + O2(g)

o B2O3 + 3Mg → 2B + 3MgO

By thermal decomposition of Boron hydrides & halides:

B2H6 (g) + Δ → 2B(s) + 3H2 (g)

Compounds of Boron:

Orthoboric acid (H3BO3)

Preparation of Orthoboric acid

From borax : Na2B4O7 + H2SO4 + 5H2O → Na2SO4 + 4H3BO3

From colemanite : Ca2B6O11 + 2SO2 + 11H2O → 2Ca(HSO3)2 + 6H3BO3

Properties of Orthoboric acid

Action of Heat:

Weak monobasic acidic behavior:

B(OH)3 ↔ H3BO3 ↔ H+ + H2O +

Thus on titration with NaOH, it gives sodium metaborate salt

H3BO3 + NaOH ↔ H++B(OH)4-

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Borax (sodium tetraborate) Na2B4O7. 10H2O

Preparation from Boric Acid

4H3BO3 + Na2CO3 --> Na2B4O7 + 6H2O + CO2

Properties of Borax

Basic Nature:-

Aqueous solution of borax is alkaline in nature due to its hydrolysis

Na2B4O7 + 3H2O → NaBO2 + 3H3BO3

NaBO2 + 2H2O → NaOH + H3BO3

Action of heat:

Diborabe( B2H6)

Preparation of Diborane:

Reduction of Boron Trifluoride:

BF3 + 3LiAlH4 → 2B2H6 + 3 LiAl F4

From NaBH4:

2NaBH4 + H2SO4 → B2H6 + 2H2 + Na2SO4

2NaBH4 + H3PO4 → B2H6 + 2H2 + NaH2PO4

Properties of Diborane:

Reaction with water: B2H6 + H2O -->2H3BO3 + 6H2

Combustion: B2H6 +2O2 --? B2O3 + 3H2O ΔH = -2615 kJ/mol

Compounds of Aluminium:

Aluminium Oxide or Alumina (Al2O3)

2Al(OH)3 +Heat → Al2O3 + 2H2O

2Al(SO4)3 +Heat → Al2O3 + 2SO3

(NH4)2Al2(SO4)3·24H2O --> 2NH3 +Al2O3 + 4SO3 + 25 H2O

Aluminum Chloride AlCl3:

Structure of Aluminium Chloride:

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Anomalous Behaviour of Boron

Boron shows anomalous behaviour with the other members of the group, due to the following

reasons:

(i)Smallest size in the group.

(ii)High ionisation energy.

(iii)Highest electronegativity in the group.

(iv) Absence of vacant d-orbital.

A few points of difference are:

1.It is a non-metal while other members of the group are metallic.

2.It shows allotropy while other members do not.

3. It has the highest melting point and boiling point in group 13.

4. It forms only covalent compounds while other members form both ionic and covalent

compounds.

5. The halides of boron exist as monomers while AlCI3 exists as a dimer.

6, The oxides and hydroxides of boron are weakly acidic while those of aluminium arc

amphoteric and those of other elements are basic.

7. It can be oxidised by concentrated HNO3 while aluminium becomes passive due to the

formation of oxide layer on the surface.

Diagonal Relationship between Boron and Silicon:

Boron exhibit resemblance with its diagonal element silicon of group 14.

1.Both Boron and Silicon are non-metals.

2.Both are semi-conductors.

3. Both B and Si form covalent hydrides, i.e.. boranes and silanes respectively.

4. Both form covalent, and volatile halides which fume in moist air due to release of HCI gas.

BCI3 + 3H2O → H3 BO3 + 3HCl i

SiCl4 + 4H2O → Si(OH)4 + 4HCl

5. Both form solid oxides which get dissolve in alkalies forming borates and silicates

respectively. ..

6. Both react with electropositive metals and give binary compounds, which yield mixture of

boranes and silanes on hydrolysis.

1. Borax or Sodium Tetraborate Decahydrate [Na2B4O7 * 1OH2O]

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Carbon Family (Group 14 Elements):

Members:C, Si, Ge, Sn, & Pb

Ionization Energies:Decreases from C to Sn and then increases up to Pb.

Metallic Character: C and Si are non metals, Ge is metalloid and Sn and Pb are metals

Catenation: C and Si show a tendency to combine with its own atoms to form long chain polymers

Compounds of Carbon:

Carbon Monoxide

Preparation of Carbon Monoxide

By heating carbon in limited supply of oxygen: C + 1/2O2 --> CO.

By heating oxides of heavy metals e.g. iron, zinc etc with carbon.

o Fe2O3 + 3C → 2Fe + 3CO

o ZnO + C → Zn + CO

By passing steam over hot coke: C + H2O → CO + H2 (water gas)

By passing air over hot coke: 2C + O2 + 4N2 → 2CO + 4N2 (Producer gas)

Properties of Carbon Monoxide:

A powerful reducing agent : Fe2O3 + 3CO → 2Fe + 3CO2 CuO + CO → Cu + CO2

Burns in air to give heat and carbon dioxide: CO + 1/2O2 → CO2 + heat.

Tests For Carbon Monoxide:

Burns with blue flame

Turns the filter paper soaked in platinum or palladium chloride to pink or green.

Carbon di-oxide

Preparation of Carbon di-oxide

By action of acids on carbonates: CaCO3 + 2HCl → CaCl2 + H2O + CO2

By combustion of carbon: C + O2 → CO2

Properties of Carbon di-oxide

It turns lime water milky Ca(OH)2 + CO2 → CaCO3 ¯ + H2O,

Milkiness disappears when CO2 is passed in excess

CaCO3 + H2O + CO2 → Ca(HCO3)2

Solid carbon dioxide or dry ice is obtained by cooling CO2 under pressure. It passes from the soild state

straight to gaseous state without liquefying (hence dry ice).

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Compounds of Silicon:

Sodium Silicate (Na2SiO3):

?Prepared by fusing soda ash with pure sand at high temperature:

Na2CO3+ SiO3 → Na2SiO3 +CO2

Silicones:

Silicon polymers containing Si – O – Si linkages formed by the hydrolysis of alkyl or aryl substituted

chlorosilanes and their subsequent polymerisation.

Silicates:

Salts of silicic acid, H4SiO4 comprised of SiO44- units having tetrahedral structure formed as result of sp3

hybridization

Q.1Explain the variation pattern in oxidation states of

(i) C to Pb

(ii) B to Tl

Ans:

(i) C to Pb -- The electronic configuration of group 14 elements is ns2 np

2. Hence, the most common

oxidation state exhibited by them should be +4. On moving down the group, the +2 oxidation state

becomes more and more common and the higher oxidation state becomes less stable because of the

inert pair effect.Si and c mostly show the +4 state. Although Sn,Ge and Pb shows both the +4 and +2

states, the stability of higher oxidation states decreases than lower oxidation state on moving down the

group.

(ii) B to Tl -- Group 13 elements have their electronic configuration of ns2 np

1 and the oxidation state

exhibited by these elements should be 3. Apart from these two electrons boron and aluminum, other

elements of this group exhibit both +1 and +3 oxidation states. Boron and aluminum show oxidation

state of +3. This is because of the inert pair effect. The two electrons, which are present in the S-shell

do not participate in bonding as they are strongly attracted by the nucleus. As we move down the group,

the inert pair effect become more prominent. Therefore Ga (+1) is unstable and TI (+1) is very stable

On moving down the group, the stability of the +3 oxidation state gets decreased.

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Q .2Compare--higher stability of BCl3 to TlCl3 and explain.

Ans: Thallium and boron belongs to group 13 of the periodic table and +1 oxidation state becomes

more stable as we move down the group. Boron is more stable than thallium because +3 state of

thallium is highly oxidizing and it reverts back to more stable +1 state.

Q .3Why boron trifluoride behave as lewis acid?

Ans: The electronic configuration of boron is ns2 np

1. It contains 3 electrons in its valence shell. Thus,

it can form only 3 covalent bonds which means that there are only 6 electrons around boron and its

octet remains incomplete. When 1of the boron‘s atom combines with 3 fluorine atoms, its octet (8)

remains incomplete. Therefore, boron trifluoride remains electron-deficient and acts as Lewis acid.

Q .4Consider the compounds CCl4, BCl3 and explain their behaviour with water.

Ans: Being a Lewis acid, BCl3 readily undergoes hydrolysis. Boric acid is formed as a result.

BCl3+3H2O→3HCl+B(OH)3

CCl4 completely resists hydrolysis. Carbon does not have any vacant orbital. Therefore, it cannot

accept electrons from water to form an intermediate. Separate layers are formed when CCl4 and water

are mixed.

CCl4+H2O→NoReaction

Q .5Is boric acid a protic acid? Explain.

Ans: Boric acid is a weak monobasic acid which behaves as a Lewis acid. So, it is not a protic acid.

B(OH)3+2HOH→*B(OH)4+−+H3O+

It behaves as an acid by accepting a pair of electrons from –OH ion.

Q .6Explain the reaction while heating boric acid.

Ans: On heating orthoboric acid at 370 K or above, it changes to metaboric acid and On further

heating, this yields boric oxide B2O3.

H3BO3→HBO2→B2O3

Q .7Explain the shapes of BH4–

and BF3 and assign the hybridisation of boron in these species.

Ans:(i) BH4 – Boron-hydride ion (BH4

–) is formed by the sp

3 hybridisation of boron orbitals.

Therefore, it is a tetrahedral structure.

(ii) BF3 -- As a result of its small size and high electronegativity, boron tends to form monomeric

covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by

the overlap of three sp2 hybridized orbitals of boron with the sp orbitals of 3 halogen atoms. Boron is

sp2 hybridized in BF3.

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Q .8Justify amphoteric nature of aluminum with reactions.

Ans: A substance which displays both characteristics of acids and bases is known as amphoteric.

Aluminium gets dissolved in both acids and bases, showing amphoteric behaviour.

(i) 2Al(s)+6HCl(aq)→2Al3+(aq)+6Cl−(aq)+3H2(g)

(ii) 2Al(s)+2NaOH(aq)+6H2O(l)→2Na+[Al(OH)4]−(aq)+3H2(g)

Q .9What are electron deficient compounds? Are SiCl4and BCl3electron deficient species?

Ans: In an electron-deficient compound, the octet of electrons is not complete, i.e., the central metal

atom has an incomplete octet. Hence, it needs electrons to complete its octet.

(i) SiCl4 -- The electronic configuration of silicon is ns2 np

2 . This indicates that it has 4 valence

electrons. After it forms 4 covalent bonds with 4 chlorine atoms, its electron count increases to 8. Thus,

SiCl4 is not an electron-deficient compound.

(ii) BCl3 -- It is an appropriate example of an electron-deficient compound. B has three valence

electrons. After forming 3 covalent bonds with chlorine, the number of electrons around it increases to

six. However, it is still short of 2 electrons to complete its octet.

Q .10 Mention the states of hybridization of carbon in

(a) Graphite (b) CO2−

3(c) Diamond

Ans: The state of hybridization of carbon in:

(a) Graphite -- Each carbon atom in graphite is sp2 hybridized and is bound to 3 other carbon atoms.

(b) CO2−

3 - C is sp2 hybridized and is bonded to 3 oxygen atoms.

(c) Diamond -- Each carbon in diamond is sp3 hybridized and is bound to 4 other carbon atoms.

Q .11 Rationalize the given statements and give chemical reactions:

a) Lead (II) chloride reacts with Cl2 to give PbCl4. b) Lead (IV) chloride is highly unstable towards heat. c) Lead is known not to form an iodide, PbI4.

Ans:(a) Lead belongs to group fourteen of the periodic table. The two oxidation states displayed by this

group is +2 and +4. On moving down the group, the +2 oxidation state becomes more stable and the +4

oxidation state becomes less stable. This is because of the inert pair effect. Hence, PbCl4 is much less

stable than PbCl2. However, the formation of PbCl4 takes place when chlorine gas is bubbled through a

saturated solution of PlCl2.

(b) On moving down group IV, the higher oxidation state becomes unstable because of the inert pair

effect. Pb(IV) is highly unstable and when heated, it reduces to Pb(II).

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(c) Lead is known not to form PbI4. Pb (+4) is oxidizing in nature and I is reducing in nature. A

combination of Pb(IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV)

oxidizes I– to I

2 and itself gets reduced to Pb(II).

PbI4→PbI2+I2

Q .12 Suggest reasons why the B–F bond lengths in BF3 (130 pm) and 𝐵𝐹4– − (143 pm) differ.

Ans: The B–F bond length in BF3 is shorter than the B–F bond length in 𝐵𝐹4– .

BF3is an electron deficient species. With a vacant p-orbital on boron, the fluorine and boron atoms

undergo pn–pn back-bonding to remove this deficiency. This imparts a double bond character to the B–

F bond.

This double-bond character causes the bond length to shorten in BF3 (130 pm). However, when

BF3 coordinates with the fluoride ion, a change in hybridization from sp2 (in BF3) to sp

3 (in 𝐵𝐹4

–)

occurs. Boron now forms 4ζ bonds and the double-bond character is lost. This accounts for a B–F bond

length of 143 pm in 𝐵𝐹4– ion.

Q .13 If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.

Ans: As a result of the difference in the electronegativities of Cl and B, the B–Cl bond is naturally

polar. However, the BCl3 molecule is non-polar. This is because BCl3 is trigonal planar in shape. It is a

symmetrical molecule. Hence, the respective dipole moments of the B–Cl bond cancel each other,

thereby causing a zero dipole moment.

Q. 14Aluminum trifluoride is insoluble in anhydrous HF but dissolves when NaF is added. It

precipitates out of the resulting solution when gaseous BF3(boron trifluoride) is bubbled through.

Give reasons.

Ans:

Hydrogen fluoride is a covalent compound and has a very strong intermolecular hydrogen-bonding.

Thus, it does not provide ions and aluminum fluoride does not dissolve in it. Sodium fluoride is an

ionic compound and when it is added to the mixture, Alf dissolves. This is because of the availability of

free F– . The reaction involved in the process is:

AlF3+3NaF→Na3[AlF6]

Aluminum fluoride gets precipitated out of the solution when boron trifluoride is added to the solution.

This happens because the tendency of boron to form complexes is much more than that of aluminum.

Therefore, when boron trifluoride is added to the solution, B replaces Al from the complexes according

to the following reaction:

Na3[AlF6]+3BF3→3Na[BF4]+AlF3

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Q 15.Explain the reason why Carbon Monoxide is poisonous.

Ans: Carbon monoxide is highly poisonous due to its ability to form a complex with hemoglobin. The

former prevents Hb from binding with oxygen. Thus, a person dies because of suffocation on not

receiving oxygen The CO–Hb complexly is more stable than the O2–Hb complex. It is found that the

CO–Hb complex is about 300 times more stable than the O2–Hb complex.

Q .16How is an excessive content of CO2 responsible for global warming?

Ans: Carbon dioxide is an essential gas for our survival. However, an increased content of CO2 in the

atmosphere poses a serious threat. An increment in the combustion of fossil fuels, decomposition of

limestone, and a decrease in the number of trees has led to greater levels of carbon dioxide. Carbon

dioxide has the property of trapping the heat provided by sun rays. Higher the level of carbon dioxide,

higher is the amount of heat trapped. This results in an increase in the atmospheric temperature, thereby

causing global warming.

Q .17What happens when

(i) Aluminum is treated with dilute NaOH,

(ii) BF3 is reacted with ammonia,

(iii) Boric acid is added to water,

(iv)Borax is heated strongly.

Ans:(i) Aluminium is treated with dilute NaOH

Aluminium reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is

liberated in the process.

2Al(s)+2NaOH(aq)+6H2O(l)→2Na+[Al(OH)4+−(aq)+3H2(g)

(ii) BF3 is reacted with ammonia

BF3 (a Lewis acid) reacts with NH3 (a Lewis base) to form a product. This results in a complete octet

around B in BF3.

F3B+:NH3→F3B←:NH3

(iii) Boric acid is added to water

When boric acid is added to water, it accepts electrons from –OH ion.

B(OH)3+2HOH→*B(OH)4+−+H3O+

(iv) Borax is heated strongly

When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it

turns into a transparent liquid, solidifying to form a glass-like material called borax bead.

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Q .18 Explain the following reactions

(i) CO is heated with ZnO;

(ii) Silicon is heated with methyl chloride at high temperature in the presence of copper;

(iii) Hydrated alumina is treated with aqueous NaOH solution;

(iv) Silicon dioxide is treated with hydrogen fluoride.

Ans:(i) CO is heated with ZnO

When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.

ZnO(s)+CO(g)→Zn(s)+CO2(g)

(ii) Silicon is heated with methyl chloride at high temperature in the presence of copper

When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of

about 537 K, a class of organosilicon polymers called methyl substituted chlorosilane MeSiCl3,

Me2SiCl2, Me3SiCl, and Me4Si) are formed.

(iii) Hydrated alumina is treated with aqueous NaOH solution

When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the

formation of sodium meta-aluminate.

Al2O3.2H2O+2NaOH→2NaAlO2+3H2O

(iv)Silicon dioxide is treated with hydrogen fluoride.

When silicon dioxide (SiO2) is heated with hydrogen fluoride (HF), it forms silicon tetrafluoride (SiF4).

Usually, the Si–O bond is a strong bond and it resists any attack by halogens and most acids, even at a

high temperature. However, it is attacked by HF.

SiO2+4HF→SiF4+2H2O

The SiF4 formed in this reaction can further react with HF to form hydro-fluorosilicic acid.

SiF4+2HF→H2SiF6

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Q .19Provide reasons:

(i) Diamond is used as an abrasive

(ii)A mixture of dilute NaOH and aluminum pieces is used to open drain.

(iii) Aluminum alloys are used to make aircraft body.

(iv) Conc.HNO3 can be transported in an aluminum container.

(v) Aluminum wire is used to make transmission cables.

(vi) Graphite is used as the lubricant.

(vii) Aluminum utensils should not be kept in water overnight.

Ans:(i) Diamond is used as an abrasive

In diamond, carbon is sp3 hybridized. Each carbon atom is bonded to 4 other carbon atoms with the

help of strong covalent bonds. These covalent bonds are present throughout the surface, giving it a very

rigid 3-D structure. It is very difficult to break this extended covalent bonding and for this reason,

diamond is the hardest substance known. Thus, it is used as an abrasive and for cutting tools.

(ii)A mixture of dilute NaOH and aluminum pieces is used to open drain.

Sodium hydroxide and aluminum react to form sodium tetra hydroxy aluminate(III) and hydrogen gas.

The pressure of the produced hydrogen gas is used to open blocked drains.

2Al+2NaOH+6H2O→2Na+[Al(OH)4+−+3H2

(iii) Aluminum alloys are used to make aircraft body.

Aluminum has high tensile strength and it is light weight. It can also be alloyed with various metals

such as Si, Mg, Cu, MnandZn. It is very malleable and ductile. Therefore, it is used in making of

aircraft bodies.

(iv) Conc.HNO3 can be transported in an aluminum container.

Concentrated HNO3 can be stored and transported in aluminum containers as it reacts with aluminum to

form a thin protective oxide layer on the aluminum surface. This oxide layer renders aluminum passive.

(v) Aluminum wire is used to make transmission cables.

Silver, copper, and aluminum are among the best conductors of electricity. Silver is an expensive metal

and silver wires are very expensive. Copper is quite expensive and is also very heavy. Aluminum is a

very ductile metal. Thus, aluminum is used in making wires for electrical conduction.

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(vi) Graphite is used as the lubricant.

Graphite has a layered structure and different layers of graphite are bonded to each other by weak van

der Waals‘ forces. These layers can slide over each other. Graphite is soft and slippery. Therefore,

graphite can be used as a lubricant.

(vii) Aluminum utensils should not be kept in water overnight.

The oxygen present in water reacts with aluminum to form a thin layer of aluminum oxide. This layer

prevents aluminum from further reaction. However, when water is kept in an aluminum vessel for long

periods of time, some amount of aluminum oxide may dissolve in water. As aluminum ions are

harmful, water should not be stored in aluminum vessels overnight.

Q .20How would you explain the lower atomic radius of Ga as compared to Al?

Ans: Although Ga has one shell more than Al, its size is lesser than Al. This is because of the poor

shielding effect of the 3d-electrons. The shielding effect of d-electrons is very poor and the effective

nuclear charge experienced by the valence electrons in gallium is much more than it is in the case of Al.

(A) Classify the following oxides asAcidic, Neutral, Amphoteric or basic.→ SiO2, Tl2O3 , CO ,

CO2 , B2O3 , PbO2 , Al2O3

(B) Write suitable chemical equations to show their nature.

Ans: SiO2= Acidic. Being acidic, it reacts with bases to form salts. It reacts with NaOH to form

sodium silicate.

SiO2+2NaOH→2Na2SiO3+H2O

→ Tl2O3 = Basic. Being basic, it reacts with acids to form salts. It reacts with HCl to form thallium

chloride.

Tl2O3+6HCL→2TlCl3+3H2O

→ CO = Neutral

→ CO2 = Acidic. Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium

carbonate.

Co2+2NaOH→Na2CO3+H2O

→ B2O3 = Acidic. Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium

metaborate.

B2O3+2NaOH→2NaBO2+H2O

→ PbO2 = Amphoteric. Amphoteric substances react with both acids and bases. PbO2 reacts with both

NaOH and H2SO4.

PbO2+2NaOH→Na2PbO3+H2O 2PbO2+2H2SO4→2PbSO4+2H2O+O2

→ Al2O3= Amphoteric. Amphoteric substances react with both acids and bases. Al2O3 reacts with both

NaOH and H2SO4.

Al2O3+2NaOH→NaAlO2 Al2O3+3H2SO4→Al2(SO4)3+3H2O

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Q .21When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which

is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl

to form compound (C). The compound (A) when heated strongly gives (D), which is used to

extract the metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their

identities.

Ans: The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves

in excess of sodium hydroxide. Hence, X must be aluminum. The white precipitate (compound A)

obtained is aluminum hydroxide. The compound B formed when an excess of the base is added is

sodium tetrahydroxy aluminate(III)

Al + NaOH → Al(OH)3 + 3Na +, Al(OH)3 + NaOH → Na [Al(OH)4 ]

Now, when dilute hydrochloric acid is added to aluminum hydroxide, aluminum chloride (compound

C) is obtained. 2Al(OH)3 + 6HCL → 2AlCl3 + 3H2 O, 2Al(OH)3 → Al2O3 + 3H2 O,

Also, when compound A is heated strongly, it gives compound D. This compound is used to extract

metal X. Aluminium metal is extracted from alumina. Hence, compound D must be alumina(Al2O3)

Q. 22What do you understand by

(a) Inert pair effect

(b) Allotropy and

(c) Catenation?

Ans:(a) Inert pair effect

As one moves down the group, the tendency of s-block electrons to participate in chemical bonding

decreases. This effect is known as inert pair effect. In the case of group 13 elements, the electronic

configuration is ns2 np1 and their group valence is +3. However, on moving down the group, the +1

oxidation state becomes more stable. This happens because of the poor shielding of the ns2 electrons by

the d- and f- electrons. As a result of the poor shielding, the ns2 electrons are held tightly by the

nucleus and so, they cannot participate in chemical bonding.

(b) Allotropy

Allotropy is the existence of an element in more than one form, having the same chemical properties

but different physical properties. The various forms of an element are called allotropes. For example,

carbon exists in three allotropic forms: diamond, graphite, and fullerenes.

(c) Catenation

The atoms of some elements (such as carbon) can link with one another through strong covalent bonds

to form long chains or branches. This property is known as catenation. It is most common in carbon and

quite significant in Si and S.

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CHAPTER-12

Organic Chemistry:Some Basic Principlesand Techniques

The hydrides of carbon (hydrocarbons) and their derivatives are called organic compounds.The branch of chemistry

which deals with these compounds is called organic chemistry.Berzelius (1808) defined organic chemistry as the

chemistry of substances found in living matter and gave the vital force theory. Synthesis of urea. the first organic

compound synthesised in laboratory, by Wohler. gave death blow to the vital force theory.

Reasons for Large Number of Organic Compounds (a)Catenation

It is the tendency of self combination and is maximum in carbon. A carbonatom can combine with other carbon atoms by

single, double or triple bonds. Thus, it formsmore compounds than the others.

(b)Tetravalency and small size

Carbon being tetravalent, is capable of bonding with fourother C atoms or some other monovalent atoms. Carbon can

form compound with oxygen.hydrogen. chlorine, sulphur, nitrogen and phosphorus. These compounds have

specific properties depending upon the nature of the element or group attached with the

carbon.Furthermore, these compounds are exceptionally stable because of the small size of carbon.

General Characteristics of Organic Compounds 1. These are the compounds of carbon with H, 0, N, S, P, F, CI, Br and 1.

2. These are generally found in living organisms. e.g., carbohydrates, proteins etc.

3. These may be gases, liquids or solids

.4. Being covalent in nature, these have low boiling point and melting point and soluble inorganic solvents.

5. These are generally volatile and inflammable.

6. They do not conduct electricity because of the absence of free ions.

7. They possess distinct colour and odour.

Classification of Organic Compounds

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Classification of Carbon Atoms

1.On the Basis of Number of C Attached

(i) Primary carbon atom When carbon atom is attached with one other carbon atom only, it is called

primary or 1° carbon atom.

(ii) Secondary carbon atom When carbon atom is attached with two other carbon atoms, it is called

secondary or 2°carbon atom.

(iii) Tertiary carbon atom When carbon atom is attached with three other carbon atoms, it iscalled

tertiary or 3° carbon atom.

iv) Quaternary carbon atom When carbon atom is attached with four other carbon atoms, it is

called quaternary or 40 carbon atom.Reactivity order of carbon atoms is as follows 3° > 2° > 1°

On the Basis of Position of Functional Group (i) α – carbon Carbon which is directly attached to the functional group.

(ii) β- carbonCarbon which is directly attached to the n-carbon

Organic compounds can also be classified on the basis of functional groups into families or

homologous series.

Functional group

The functional group can be defined as an atom or a group of atoms that are joined together in a

specific manner which is responsible for the characteristic chemical properties of organic compounds.

Examples in this case are hydroxyl group (−OH), aldehyde group (−CHO) and carboxylic acid group

(−COOH).

Homologous series

A group or a series of organic compounds in which each member contains the same characteristic

functional group and differs from each other by a fixed unit form a homologous series and therefore its

members are known as homologous. The members of the homologous series can be represented by a

general formula and the successive members differ from each other in the molecular formula by a –

CH2 unit. There are a number of homologous series in organic chemistry such as alkanes, alkenes,

alkynes, haloalkanes, alkanols, amines etc.

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Organic Chemistry Nomenclature

Chemical Nomenclature – IUPAC Rules:

According to the IUPAC system, the nomenclature of organic compounds consists of the following

parts:

Steps involved:

1. Longest Chain Rule: Identify the parent hydrocarbon and name it. The parent chain of the

compound is considered as the longest chain of carbon atoms. This chain can either be straight or of

any other shape.

2. Lowest Set of Locants: The numbering of the carbon atoms in the longest chain starts from the end

which gives the lowest number to the carbon atoms carrying the substituents.

3. Presence of same substituent more than once: Prefixes such as di, tri, etc. are given to the

substituents which are present twice, thrice respectively in the parent chain.

4. Naming different substituents: If more than one substituent is present then the substituents are

arranged in alphabetical order of their names.

5. Naming different substituents at equivalent positions: If two different substituents are present on

the same position from the two ends then the substituents are named such that the substituent which

comes first in the alphabetical order gets the lowest number.

6. Naming The Complex Substituents: Naming of a complex substituent is done when the

substituent on the parent chain has a branched structure (i.e complex structure). These substituents

are named as a substituted alkyl group and the carbon atom of this substituent attached to the parent

chain is numbered as 1. Name of these type of substituents is written in brackets.

The final name will be in format : Locant + Prefix + Root + Locant + Suffix

Word root: It indicates the number of carbon atoms in the longest carbon chain that is selected. For

example, C1 is Meth and C5 is Pent.

The suffix – primary (1°) and secondary (2°): A suffix is generally a functional group in the

molecule which follows the word root. It is further divided into two types.

Primary suffix: It is written immediately after the word root. For example, in alkanes the suffix is

ane.

Secondary suffix: It is written after the primary suffix. For instance, if a compound has an alkane

and alcohol group attached to it, the naming will be alkanol, -ol being the suffix for alcohol.

Prefix – primary (1°) and secondary (2°): It is added before the word root while naming the

compound. It indicates the presence of substituent groups or side chains in the organic molecule. It

reveals the cyclic and acyclic nature of the compound.

Primary prefix: Indicates whether the molecule is cyclic or not. For example, for cyclic compounds

the prefix used is cyclo.

Secondary prefix: Indicates the presence of substituent groups or any side chain. For example -

CH3is known as Methyl and -Br is Bromo.

Chemical Nomenclature- Types

1. Compositional Nomenclature

The term is used to denote the named constructions based on the composition of species or

substances being named, against the systems that involve structural composition or information.

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One among them is the generalized stoichiometric name. Substances or the elements are named

with multiple prefixes in order to give the overall stoichiometry of an element or a compound.

When there are more components, then they are divided into 2 classes namely electropositive

and electronegative components.

These names will sound like salt names and this does not imply the chemical nature and behavior of

species that are named.

The rules ordered are required to the using up of multiple prefixes, ordering of components and

proper endings in names for electronegative components.

Examples– Sodium Chloride – NaCl, Trioxygen – O3, Phosphorous trichloride – PCl3

2. Substitutive Nomenclature

It is based on the approach where parent hydride is changed by replacement of hydrogen atoms

with atoms or a group of atoms.

It is a system where organic compounds are named using functional groups as the suffix or prefix to

the name of the parent compound.

This system is also used in naming compounds derived from hydrides of specific group elements in

the periodic table.

Similar to that of carbon, these elements may form rings and chains that will have many

derivatives.

Rules come in handy in naming the parent or main compounds and their substituents.

Hydrides belong to group 13-17 of the periodic table are named suffix – ane. For example –

Borane, Phosphane, and oxidane etc.

Examples – 1, 1-difluoro trisilane (SiH3 SiH2 SiHF2), Trichlorophosphine, PCl3

3. Additive Nomenclature

This method is mainly formulated for the coordination compounds even though it has wide

applications. An example for its application is pentaamminechlorocobalt(III) chloride –

[CoCl(NH3)5]Cl2.

Chloride will have the prefix chloro while ligand will have chlorido.

Examples– PCl3 – trichloridophosphorus, [CoCl3 (NH3)3] – tri-ammine-trichloridocobalt.

Names of some important aliphatic compounds

1. Alkanes: General formula = CnH2n+2

Suffix – ane

Examples: CH4 – methane and C4H10 – Butane

2. Alkenes: General formula = CnH2n

Suffix – ene

Examples: C2H4 – ethene and C3H6 – Propene

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3. Alkynes: General formula = CnH2n-2

Suffix – yne

Examples: C2H2 – Ethyne and C4H6 – But-1-yne

Example:

Let us understand it with the help of an example:

In this case, we have 9 carbon atoms in the straight chain. 5th Carbon atom from both the ends of the

straight chain consists of a substituent having 3 carbon chains. On the first two carbon atoms of the

substituent group, there is one additional carbon atom attached.

Now if we consider this as a new parent chain, it has a substituent which has one additional carbon

each. For naming them we will first number the parent chain. In this case, we have 9 carbon atoms in

the straight chain which is also the parent chain. Then we find that the substituent is on the fifth carbon

atom.

Now we have 3 substituent carbons and out of these three, two substituents have additional carbons

atoms attached to them. We find that the longest chain, in this case, can be the first four carbon atom

chain but this is wrong as the last carbon is not attached to the parent chain.

So we will consider only three carbon atom chain as the main chain. Thus it can be named as propane

and on the first and second position, we have methyl group. We can write the name as 1-2 Dimethyl

propane, but it will be written as 1-2 Dimethyl propyl as it is a substituent group.

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Now taking the substituent with the parent chain we will get 5-(1-2-Dimethyl Propyl) and as the parent

chain has 9 carbon atoms so, it will be named as nonane. Thus the final name of the compound will be

5-(1-2-Dimethyl Propyl)nonane.

Nomenclature of Organic Compounds having Functional Groups

In an organic compound, firstly, the functional group is identified which gives us the appropriate suffix.

Then the longest carbon chain having the functional group is chosen in such a manner that the

functional group gets the lowest number in the chain.

In case of multiple functional groups in a compound, one of the functional groups is chosen on the

basis of the priority list and is referred to as the principal functional group. The compound is named on

the basis of that functional group. Remaining functional groups are known as the substituents and are

named using the appropriate prefixes.

The priority list of the functional group can be given as:

-COOH > -SO3H > -COOR (R= alkyl group) > -COCl > -CONH2 > -CN > -HC=O >>C=O > -OH > -

NH2 > >C=C<> -CC-

Functional groups like -R, C6H5– , halogens (F, Cl, Br, I), -NO2, alkoxy (-OR) etc., are always used as

prefix substituents.

If more than one same functional group is present in the compound, then they are indicated as di, tri, …

etc. before the suffix and full name of parent alkane is written.

1. Name the compound given below.

In this compound, the functional compound present is –OH.

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The longest chain which contains the functional group has 8 carbon atoms. Hence, the saturated

hydrocarbon is octane.

The alcohol functional group is present on the 3rd position and a methyl group is in the 6th

position.

Hence, the IUPAC name will be 6- Methyloctan-3-ol.

2. Name the functional group given below:

This compound has the functional group as ketone (>C=O). Hence the suffix will be one. And there are

two ketone groups. So we will use di- before suffix as dione. Continuing in the same manner as above

we get the name as Hexane-2, 4-dione.

Substituted Benzene Compounds: Nomenclature

Mono-substituted Benzene Compounds

For the benzene compounds consisting of a single substituent, we simply prefix the name of the

substituent to benzene. Some examples along with their common names are listed below.

Methylbenzene or Toluene

Toluene

Hydroxybenzene or Phenol

Phenol

Aminobenzene or Aniline

Aniline

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Di-substituted Benzene Compounds

When there are two substituents present in the compound, we number each of the carbon atoms in such

a manner that the substituents are attached to the lowest possible numbered carbon atom.

1,3-dinitrobenzene is the name of the following compound. Naming it as 1,5-dinitrobenzene is

incorrect because the carbon atom would not be numbered lowest.

1,3-dichlorobenzene

Moreover, when there is more than one substituent, we also name their positions as ortho- (o), meta-

(m) and para- (p). They refer to the positions 1,2-;1,3- and 1,4- respectively. Thus, we can name the

compound 1,3-dichlorobenzene as m-dichlorobenzene.

Poly-substituted Benzene Compounds

In case of poly-substituted compounds, if there is a base compound present, we assign it as the position

1. We then choose the next compound for numbering such that it gets the lowest number. However, if

there are no special or base group present, we list them in the alphabetical order, giving them the lowest

numbers.

1-Bromo-2,4-dinitrobenzene

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For the following compound, aniline is the base compound. So, we name it as 3-Chloro-2-

nitroaniline.

To learn more about benzene, its compounds and their nomenclature, download Byju‘s The Learning

App.

Isomerism

Isomerism is a phenomenon in which two or more compounds have the same chemical formula but

possess different properties because of different structural or spatial arrangements. Compounds which

exhibit isomerism are called isomers. There are two types of isomerism:

Structural isomerism

Stereoisomerism

Structural isomerism

The compounds which exhibit this type of isomerism have the same molecular formula but different

structures, as in how they are linked to each other. Structural isomerism is further classified into

Chain isomerism

Position isomerism

Functional group isomerism

Metamerism

Chain isomerism: The phenomenon is termed as chain isomerism when two or more compounds have

the same molecular formula but differ in the branching of carbon atoms. These compounds are called

chain isomers. Example C5H12 can be represented as three compounds.

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CH3CH2CH2CH2CH3-pentane

Position isomerism: The isomerism in which two or more compounds differ in the position of

functional group or substituent atoms is called position isomerism. Example: C3H7OH can be

represented in two arrangements.

CH3CH2CH2OH -Propan-1-ol

Functional isomerism: The isomerism in which two or more compounds have identical molecular

formula but differ in the functional group present is called functional isomerism and these isomers are

called functional isomers. Example C3H6O can be represented as a ketone and as an aldehyde.

Metamerism: This isomerism is exhibited by compounds due to the presence of different alkyl chains

on either sides of the functional group.Example:C4H10O can be represented as ethoxyethane

(C2H5OC2H5) and methoxypropane (CH3OC3H7)

Stereoisomerism

Stereoisomerism is a phenomenon in which compounds have the same molecular formula but differ in

the relative positioning or orientation of atoms in space. The compounds which exhibit this type of

isomerism are called as stereoisomers. These are further classified into:

Geometrical isomerism

Optical isomerism

Geometric isomerism: This is being exhibited by molecules in which their special positions are locked

to each other due to the presence of a ring structure or a double bond.

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Optical isomerism: The isomerism exhibited by two or more compounds which have the same

molecular arrangement but differ in the optical activity.

Fission of a covalent bond:

A covalent bond is formed when electrons are shared between two atoms in the classical sense. A

single bond (sigma bond) is thus made up of two electrons. Now a chemical reaction takes place when

old bonds are broken and new ones are created. So how can one break a single bond—there are plainly

two ways to go about breaking a bond as shown below.

Heterolytic cleavage: In this cleavage the bond breaks in such a way that the sharedpair of electron remains with one of the fragments.

H3C– Br +CH3+ Br

-

Homolytic Cleavage: In this cleavage the shared pair of electron goes with each ofthe bonded atom.

R– X R. +X

.

Alkyl free radical

Nucleophiles : A reagent that brings an electron pair is called nucleophile ie nucleusseeking e g -OH , -CN

Electrophiles: A reagent that takes away electron pair is called electrophile I eelectron seeking e g > C= O , R3C – X

Inductive Effect: The displacement of the electron along the chain of the carbonatoms due to presence of an electro negative atom or group at the end of the chain.

ddd+ dd++ d+

CH3 C H2 CH2Cl

1. Electron withdrawing groups:

Electron withdrawing groups decreases the electron density on the adjacent carbon atom. These

stabilize the carbanion formed through the inductive effect. For example nitro (-NO2), carboxy (-

COOH), cyano (- CN), ester (-COOR) etc.

2. Electron donating groups:

Electron donating groups increases the electron density on the adjacent carbon atom. These

stabilize the carbocation formed through the inductive effect. For example, methyl (–CH3) and

ethyl (–CH2–CH3) etc.

Resonance Effect or Mesomeric effect:

The resonance effect is defined as the polarity produced in the molecule by the interaction of two π-

bonds or between a π- bond and lone pair of electrons present on an adjacent atom.The effect is

transmitted through the chain.


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Types of Resonance Effects

There are two types of Resonance effects namely positive resonance effect and negative resonance

effect.

1. Positive Resonance Effect- Positive resonance effect occurs when the groups release electrons to

the other molecules by the process of delocalization. The groups are usually denoted by +R or +M.

In this process, the molecular electron density increases. For example- -OH, -SH, -OR,-SR.

2. Negative Resonance Effect- Negative resonance effect occurs when the groups withdraw the

electrons from other molecules by the process of delocalization. The groups are usually denoted by

-R or -M. In this process, the molecular electron density is said to decrease. For example- -NO2,

C=O, -COOH, -C≡N.

Electromeric effect:

The electromeric effect is a temporary effect, mainly experienced in the presence of an attacking

reagent in the vicinity of an organic compound having multiple bonds(a double or triple bond). In this

effect, the complete transfer of a shared pair of π-electrons to one of the atoms joined by multiple bonds

on the demand of an attacking reagent takes place. The effect ceases as soon as the attacking reagent is

removed from the domain of the reaction. The electromeric effect is mainly categorized into two

categories.

1. Positive Electromeric Effect (+E effect):

The positive electromeric effect is defined as the transfer of π−electrons of the multiple bonds to the

atom with which the reagent gets attached.

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Positive Electromeric Effect

2. Negative Electromeric Effect (–E effect):

The positive electromeric effect is defined as the transfer of π−electrons of the multiple bonds to the

atom with which the reagent does not get attached.

Negative Electromeric Effect

Hyperconjugation:

Hyperconjugation effect is a permanent effect in which localization of ζ electrons of C-H bond of an

alkyl group directly attached to an atom of the unsaturated system or to an atom with an unshared p

orbital takes place.

From the above figure, we observe that one of the three C-H bonds of the methyl group can align in the

plane of the empty p orbital and the electrons constituting the C-H bond in a plane with this p orbital

can then be delocalized into the empty p orbital.

We also observe that the hyperconjugation stabilizes the carbocation as it helps in the dispersal of

positive charge. Thus, we can say that greater the number of alkyl groups attached to a positively

charged carbon atom, the greater is the hyperconjugation interaction and stabilization of the

carbonation.The relative stability on the basis of hyperconjugation is given as,

Hyperconjugation stability


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Methods of Purification of Organic Compounds

Sublimation:

This method can be applied to only those compounds that change their state directly from solid to vapor

on heating. Hence this method can be used for separating the sublimable compounds from the non-

sublimable impurities.This process is known as sublimation.

Crystallization:

The difference in the solubilities of an organic compound and that of the impurities present in it, in a

suitable solvent, is the basis for this method of separation. A concentrated solution is prepared by

dissolving the compound in a suitable solvent; the pure compound crystallizes out on cooling. The

dissolved impurities are removed through filtration. The filtrate contains some amount of pure

compound too. A compound containing impurities of comparable solubilities are purified by

repeated crystallization.

Distillation:

This method is mainly used for separating liquids with sufficient difference in their boiling points. It

can also be used for separating volatile liquids from non-volatile impurities. There are two kinds

of distillation methods widely used in the industry for purifying organic compounds. These are:

Fractional distillation:

It is the process in which the components of a mixture are separated by heating the chemical mixture.

Boiling point plays a vital role in fractional distillation. The component with lower boiling point

vaporizes first and then the components with higher boiling points. In this way the components of

Steam distillation:

In this method substances which are steam volatile and immiscible with water can be separated. It

basically lowers the distillation temperature of the organic compounds which are immiscible with water

having a high boiling point.

Differential Extraction:

It is used for the separation of organic compound present in an aqueous solution. The compound is

removed from the aqueous solution by dissolving the compound in a highly soluble solvent forming a

layer with the aqueous solution.The aqueous solution and the organic solvent should be immiscible and

are separated using a separatory funnel. Later on, the organic solvent and the compound is separated by

evaporation or distillation.

Chromatography:

Chromatography is an important separation technique used to separate constituent particles of a mixture

of substances, to purify the compounds and check the purity of compounds. In this technique on a

stationary phase (solid or a liquid) a mixture of substance is applied. The mixture of gas or the pure

solvent is allowed to move slowly on the stationary phase. Due to which the components of the mixture

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start separating from one another. Chromatography is of two types namely Adsorption chromatography

and Partition chromatography.

Qualitative Analysis Of Organic Compounds

Detection of carbon and hydrogen:

On heating, the compounds in the presence of copper (II) oxide, the presence of carbon and hydrogen

can be detected. Carbon is oxidized to carbon dioxide (tested by lime water, it turns the solution turbid)

and hydrogen to water (tested by anhydrous copper sulphate, turns it blue).

C + 2CuO → 2Cu + CO2

2H +CuO → Cu + H2O

CO2 + Ca(OH)2 → CaCO3 + H2O

5H2O +CuSO4 → CuSO4.5H2O

Detection of other elements:

Lassaigne‘s test is used to detect the presence of other elements in an organic compound. Conversion of

covalent bonds into sodium metals takes place in this method by fusing the compound with sodium

metal. The extract obtained after the fusion of elements with sodium metal is known as sodium fusion

extract.

Test for nitrogen:

o Sodium fusion extract is reacted with iron (II) sulphate and it is then mixed with concentrated

sulphuric acid. Formation of Prussian blue colour confirms the presence of nitrogen.

6CN– + Fe

2+ → [Fe(CN)6]

4-

xH2O

3[Fe(CN)6]4-

+ 4Fe3+

Fe4[Fe(CN)6]3.xH2O

Prussian blue

Test For sulphur:

a)The reaction of sodium fusion extract with acetic acid followed by the addition of lead acetate

yields a black precipitate of lead sulphide which confirms the presence of sulphur.

S2-

+ Pb2+

→ PbS

b)The reaction of sodium fusion extract with sodium nitroprusside yield a violet colour which confirms

the presence of sulphur.

S2-

+ [Fe(CN)5NO]2-

[Fe(CN )5NOS]4-

Violet colour

Test for halogens:

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o Formation of a white precipitate which is soluble in ammonium hydroxide on reacting sodium

fusion extract with nitric acid and then with silver nitrate confirms the presence of halogens in

the compound.

X– + Ag

+→ AgX (s)

X represents a halogen I, Br or Cl

For Cl,Br and I the colourof the ppt. are white, pale yellow and yellow respectively.

Test for phosphorus:

o Formation of a yellow coloured precipitate on treating the organic compound with nitric acid

and ammonium molybdate shows the presence of phosphorus.

Na3PO4 + 3HNO3à H3PO4 + 3NaNO3

H3PO4 + 12(NH4)2MoO4 + 21HNO3 → (NH4)3PO4.MoO3 + 21NH4NO3 + 12H2O

Ammonium molybdate Ammonium phosphomolybdate

Quantitative Analysis Of Organic Compounds

QUANTITIVE ANALYSIS FOR Carbon and Hydrogen

Let the mass of organic compound be m g. Mass of water and carbon

dioxide produced be m1 and m2 g respectively;

% of carbon = 12 x m2 x 100

44 x m

% of hydrogen = 2 x m1 x 100

18 x m

QUANTITIVE ANALYSIS FOR NITROGEN

DUMAS METHOD: A known mass of organic compound is heated with excess ofCuO in an

atmosphere of CO2, when nitrogen of the organic compound is

converted into N2 gas. The volume of N2 thus obtained is converted into STP and the percentage of nitrogen determined by applying the equation:

Volume of Nitrogen at STP = P1V1 x 273

760 x T1

%N= 28 x vol of N2 at STP x100

22400 x mass of the substance taken

KJELDAHL’S METHOD: A known mass of organic compound is heated

withconc. H2SO4 in presence of K2SO4 and little CuSO4 or Hg in a long necked flask called

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Kjeldahl‗s flask when nitrogen present in the organic compoundis

quantitatively converted into (NH4)2SO4. (NH4)2SO4 thus obtained is boiled with excess of

NaOH solution to liberate NH3 gas which is absorbed in a known excess of a standard acid

such as H2SO4 or HCl.

The vol of acid unused is found by titration against a standard alkali solution. From the vol of the acid used, the percentage of nitrogen is determined by applying the equation,

%N= 1.4 x Molarity of the acid x Basicity of the acid x Vol of the acidused

Mass of the substance taken

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Quantitative analysis of halogens:

1. Carius method: A mixture of organic compound and fuming nitric acid is heated in the

presence of silver nitrate contained in a tube (hard glass) known as Carius tube in a furnace.

Carbon and hydrogen present in the organic compound are oxidized to carbon dioxide and

water respectively. The halogens present in the organic compound reacts with the silver

nitrate to from the corresponding silver halide (AgX). After this it is filtered, washed, dried

and weighed.

Calculations:

Let the mass of the given organic compound be m g.

Suppose mass of AgX formed = m1 g.

We know that 1 mol of AgX consists of 1 mol of X.

So, in m1 g of AgX , mass of halogen = (atomic mass of X × m1 g)(molecular mass of AgX)

Percentage of halogen = (atomic mass of X × m1 × 100)(molecular mass of AgX × m)

Quantitative analysis of sulphur:

A fixed mass of an organic compound containing sulphur is heated in a Carius tube containing

sodium peroxide or fuming nitric acid. Sulphur present in the organic compound is oxidized to

sulphuric acid. It is then precipitated as barium sulphate by the addition of barium chloride

solution in water. The precipitate is then washed, filtered, dried and weighed. The mass

of barium sulphate is used in calculating the percentage of sulphur in the given organic

compound.

Calculations:

Suppose the mass of organic compound = m g.

Let the mass of barium sulphate formed = m1 g.

We know that 32 g sulphur is present in 1 mol of BaSO4

Therefore, 233 g BaSO4 contains 32 g sulphur

⇒ M1 g of BaSO4 contains 32 × m1233g of sulphur

Percentage of sulphur = 32 × m1 × 100233 × m

Quantitative analysis of OXYGEN:

The % of O = 100 – (sum of the percentages of all other elements in the organic compound)

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LEVELWISE QUESTIONS WITH ANSWERS:

ONE MARK QUESTION:

1. State the principle involved in the process of sublimation.

Ans:. It is based on the principle that the solids are directly converted into

gaseous state without passing through the liquid state.

2. Suggest a method to purify a liquid which decomposes at its boiling point.

Ans: The process Distillation Under reduced pressure is used to purify a liquid which decomposes at its boiling point.

3. How will you separate a mixture of O-nitrophenol and p- nitrophenol ?

Ans: O-nitrophenol is steam volatile therefore it can be separated by Steam distillation.

4. Lassaigne‗s test is not shown by diazonium Salt. Why?

Ans:On heating diazonium Salts loses Nitrogen and could not fuse with the Sodium metal

therefore diazonium Salt do not show Positive Lassaigne‗s test for nitrogen.

5. Alcohols are weaker acids than Water, Why ?

Ans:The alkyl group in alcohols has + I effect due to which electron density is increases on Oxygen atom which makes the release of hydrogen ion more difficult from alcohol. R → O →H

6. Why is nitric acid is added to Sodium extract before adding Silver nitrate for testing

halogens ?

Ans: Nitric acid is added to decompose NaCN and Na2S

NaCN + HNO3 → NaNO3 +HCN

Na2S+2HNO3 → 2NaNO3 +H2S

7. which of the two O2NCH2CH2-or CH3CH2O

– is expected to be morestable and why ?

Ans: NO2 group has –I effect and disperse the negative charge on Oxygen atom

O2N←CH2← CH2O-

8. Arrange the following in increasing Order of Stability ;

(CH3 )3C +

, CH3CH2CH2C+

H2 , CH3CH2C+

HCH3 ,CH3C+

H2 , CH3CH2C+

H2

Ans:: CH3C+

H2< CH3CH2C+

H2< CH3CH2CH2C+

H2<CH3CH2C+

HCH3< (CH3 )3C +

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9. Write the IUPAC name of thefollowing

CHCHCHCH2CH3

CH3CH3

Ans. 2,3Dimethylpentane

10. Write the hybridized state of C atoms in the following

CH2 = CH - C Ξ N

Ans: sp2

sp2

sp

CH2 = CH - C Ξ N

11. Give the IUPAC name of the following compound.

Ans: 2,5Dimethylheptane

Two Marks Questions

1. Write the Principle involved in the Lassaigne’s Test . Also write reaction if Sulpure is present in

organic compound.

Ans:It is based on the principle that the elements present in the organic compound are

converted from covalent bond into ionic bond on fusing the compound with sodium metal.

2Na + S Na2S

Q 2 Draw the Structures of the following compounds.

A) Hex-3-enoic acid b) 2-chloro-2-methylbutan-1-ol

A

a)

CH3-CH2-CH=CH-CH2-COOH

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b) Cl

CH3 - CH2 – C - CH2 OH

CH3

Q 3. Explain Inductive effect with example.

Ans: Inductive Effect: The displacement of the electron along the chain of the carbon atoms due to presence of an atom or group at the end of the chain.

δ+++ δ++ δ+

CH3 → CH2→CH2→Cl

Q 4. Explain why (CH3 )3C+

is more stable than CH3C+

H2.

Ans:(CH3 )3C+

has nine alpha hydrogens and has nine hyperconjugation structures while

CH3C+

H2has three alpha hydrogens and has three hyperconjugation structures, therefore

(CH3 )3C+

is more stable than CH3C+

H2

Q

4

Q 5. Give the number of Sigma and pi bonds in the following molecules

a) CH3NO2 b)HCONHCH3

A

4

a) 6 Sigma and 1 pi bond

b) 8 Sigma and 1 pi bond

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Q 5

Q 6. Write the condensed and bond line formula of 2,2,4-Trimethylpentane.

Ans:

CH3

CH3―C― CH2―CH―CH3

CH3 CH3

Q 7. How Sodium fusion extract is prepared ?

Ans: small piece of dry Sodium metal is heated with a organic compound in a fusion tube for 2 -3 minutes and the red hot tube is plunged in to distilled water contained in a china dish. The contained of the china dish is boiled ,cooled and filtered. The filtrate is known as Sodium fusionextract.

Q 8. Explain the principle of paper chromatography.

Ans:Paper chromatography is based on the difference in the rates at which the components of a mixture are adsorbed. The material on which different components are adsorbed is called Stationary phase which is generally made up of alumina, silica jel or activated charcoal. The mixture to be separated is dissolved in a suitable medium and it is called moving phase. The moving phase is run on the Stationary phase , the different compounds are adsorbed on stationary phase at different rates.

Q 9. Why is an organic compound fused with Sodium fortesting nitrogen,halogens and sulphur?

Ans: On fusing with sodium metal the elements presents in an organic compounds are converted in to sodium salts which are water soluble which can be filtered and detected by the respectivetests.

Q 10. It is not advisable to use sulphuric acid in place of acetic acidfor acidification while testing sulphur by lead acetate test. Give reason

Ans: Lead acetate on reacting with sulphuric acid will give a white ppt of lead sulphatewhih interfere in the detection of sulphur.

(CH3COO)2Pb +H2SO4 PbSO4 + 2CH3COOH

Q 11. Under what conditions can the process of steam distillation is used ?

Ans:Steam distillation is used to purify the liquids which are steam volatile and water and the liquid are not miscible with each other.

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Three Marks Questions 1. (a) How would you separate glycerol from spent- lye in soap industry?

(b). What is retardation factor (Rf)? How do you separate red and blue ink?

Ans: (a) By distillation under reduced pressure

(b) It is the ratio of the distance travelled by the substance from base line to the distance

travelled by the solvent from the base line.

2. In an estimation of sulphur by carius method 0.468 g of an organic compound gave 0.668 g of barium sulphate. Find the percentage of sulphur in the compound.

Ans: Mass of the compound = 0.468 g

Mass of the barium sulphate = 0.668 g

% of sulphur = 32 X Mass of barium sulphate X 100

233 X Mass of the compound

= 32 x 0.668x100 233 x0.46

= 19.60 %

3. Which bond is more polar in the following pairs of molecules.

a) H3C-H,H3C-Br b)H3C-NH2,H2C-OH c) H3C-OH,H3C-SH

Ans:a) C-Br because Br is more electronegative than H

b) C-O because O is more electronegative than N

c) C-O because O is more electronegative thanS

4. Define Isomerism.Explain position Isomerism and Functional Isomerism with examples.

Ans: Two or more compounds having the same molecular formula but different physical and chemical properties are called isomers and this phenomenon is called isomerism. Position Isomerism : Compounds which have the same structure of carbon chain but differ in position of double or triple bonds or functional group are called position isomers and this phenomenon is called Position Isomerism. e g

CH3-CH2-CH=CH2 CH3-CH = CH –CH3

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Functional Isomerism :Compounds which have the same molecular formula but different functional group are called functional isomers and this phenomenon is called functional Isomerism. e g

CH3 – CH2– OH CH3 – O –CH3

5. write the IUPAC names of the following compounds.

O O

a) CH3 – CH2 – C – CH2 – C – CH3

b) HC Ξ C – CH = CH – CH –CH2

c) Cl CH2CH2CH2CH2Br

Ans: a) hexane2,4dione

b) hexa-1,3-dien-5-yne

c) 1-bromo-4-chlorobutane

6. Define Homologous series. Write the general formula of alkane, alkene and alkynes.

Ans: Homologous Series : It is defined as group of similar organic compounds which contains the similar functional groups and the two adjacent members of the series is differ by a –CH2 group.

Alkanes CnH2n+2

Alkenes CnH2n

Alkynes CnH2n-2

7. How many Sigma and pi bonds are present in the following molecules .

a) HC Ξ CCH =CHCH3

b) CH2 = C =CHCH3

Ans:a)Sigma bonds = 10, pi bonds = 3

b) Sigma bonds= 9, pi bonds =2

8. Define functional groups. Write the general formula of Carboxylic acids acid chlorides. Ans:Functional Groups :It is an atom or group of atoms bonded together ina unique manner which is usually the site of chemical reactivity in an organic molecule. e gCH3OH

General formula of Carboxylic acids : CnH2n+1COOH General formula of acid chlorides

:RCOCl

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9. Write a shirt note on differential extraction.

Ans:When an organic compound is present in an aqueous medium it is separated by shaking it with organic solvent in which it is more soluble than in water. The aqueous solution is mixed with organic solvent in a separating funnel and shaken for sometimes and then allowed to stand for some time .when organic solvent and water form two separate layers the lower layer is run out by opening the tap of funnel and organic layer is separated. the process is repeated several times and pure organic compound isseparated.

10. How carbon and Hydrogen is detected in a organic compounds.

Ans: The Carbon and Hydrogen present in the Organic compound is detected by heating the compound with Copper II oxide in a hard glass tube when carbon present in the compound is oxidized to CO2 which can be tested with lime Water and Hydrogenis converted to water which can be tested with anhydrous copper sulphate which turnsblue.

C +CuO →2Cu + CO2

2H+CuO → Cu+H2O

CO2 +Ca (OH)2→CaCO3 + H2O

5H2O+ CuSO4→CuSO4.5H2O

11. Write a short note on Resonance effect .

Ans:Resonance Effect : The polarity produced in the molecule by the interaction of two pi bonds or between a pi bond and lone pair of electron present on an adjacent atom.

There are two types of resonance effect: a) Positive resonance effect: In this effect the transfer of electrons is away from an atom or substituent group attached to the conjugatedsystem.The atoms or groups which shows +R effect are halogens,-OH , -OR,- NH2

b) Negative resonance effect: In this effect the transfer of electrons is towards the atom or substituent group attached to the conjugated system.The atoms or groups which shows -R effect are –COOH , -CHO , -CN

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FiveMarks Questions

1. Explain hyperconjugation effect. How does hyperconjugationeffect explain the stability of alkenes?

Ans The relative stability of various classes of carbonium ions may be explained by the number of no-bond resonance structures that can be written for them. Such structures are obtained by shifting the bonding electrons from an adjacent C-H bond to the electron deficient carbon so the positive charge originally on carbon is dispersed to the hydrogen. This manner of electron release by assuming no bond character in the adjacent C-H bond is called Hyperconjugation. Greater the hyperconjugation greater will be the stability of alkenes.

CH3 – CH = CH – CH3<CH3 – C = CH – CH3 <CH3 – C = C – CH3

CH3 CH3 CH3

2. In DNA and RNA nitrogen is present in the ring system. Can kjeldahl method be used for the estimation of nitrogen present in these ?give reasons Q 1 Explain hyperconjugation effect. How does hyperconjugationeffect explain the stability of alkenes?

Ans The relative stability of various classes of carbonium ions may be explained by the number of no-bond resonance structures that can be written for them. Such structures are obtained by shifting the bonding electrons from an adjacent C-H bond to the electron deficient carbon so the positive charge originally on carbon is dispersed to the hydrogen. This manner of electron release by assuming no bond character in the adjacent C-H bond is called Hyperconjugation. Greater the hyperconjugation greater will be the stability of alkenes.

CH3 – CH = CH – CH3<CH3 – C = CH – CH3 <CH3 – C = C – CH3

CH3 CH3 CH3

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3. In DNA and RNA nitrogen is present in the ring system. Can kjeldahl method be used for the estimation of nitrogen present in these ?give reasons

Ans:In DNA and RNA nitrogen is present in hetrocyclicrings.Kjeldahl method can not be used to estimate nitrogen present in the ring because cannot be completely

convertedinto(NH4)2SO4duringdigestion.ThereforeKjeldahlmethodcannotbe

used to estimate nitrogen present in DNA and RNA.

In DNA and RNA nitrogen is present in hetrocyclicrings.Kjeldahl method can not be used to estimate nitrogen present in the ring because cannot be completely

convertedinto(NH4)2SO4duringdigestion.ThereforeKjeldahlmethodcannotbe

used to estimate nitrogen present in DNA and RNA.

4. Differentiate between the principle of estimation of nitrogen in an organic compound by i) Dumas method ii) Kjeldahl‗s method.

Ans: DUMAS METHOD: A known mass of organic compound is heated with excess of

CuO in an atmosphere of CO2, when nitrogen of the organic compound is

converted into N2 gas. The volume of N2 thus obtained is converted into STP and the percentage of nitrogen determined by applying the equation:

Volume of Nitrogen at STP = P1V1 x 273

760 x T1

%N= 28 x vol of N2 at STP x100

22400 x mass of the substance taken

KJELDAHL‗S METHOD: A known mass of organic compound is heated with conc. H2SO4 in presence of K2SO4 and little CuSO4 or Hg in a long necked flask called

Kjeldahl‗s flask when nitrogen present in the organic compound is quantitatively converted into (NH4)2SO4. (NH4)2SO4 thus obtained is boiled with excess of

NaOH solution to liberate NH3 gas which is absorbed in a known excess of a standard acid

such as H2SO4 or HCl.

The vol of acid unused is found by titration against a standard alkali solution. From the vol of the acid used, the percentage of nitrogen is determined by applying the equation,

%N= 1.4 x Molarity of the acid x Basicity of the acid x Vol of the acid used

Mass of the substance taken

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5. A sample of 0.50g of organic compound was treated according to Kjeldahl‗s method. The

ammonia evolved was absorbed in 50mL of 0.5M H2SO4. The residual acid required 60mL

of 0.5M solution of NaOH for neutralization. Find the percentage composition of nitrogen in

the compound.

Ans:

Vol of acid taken=50mL of 0.5M H2SO4= 25mL of 1M H2SO4

Vol of alkali used for neutralization of excess acid= 60 mL of 0.5m NaOH=30mL of 1M NaOH

Now 1 mole of H2SO4 neutralizes 2 moles of NaOH (i.e. H2SO4 +2 NaOH Na2SO4

+2H2O)

… 30mL of 1M NaOH = 15mL of 1MH2SO4

% of nitrogen.

1 mole of H2SO4 neutralizes 2 moles of NH3 … 10mL of 1M H2SO4 = 20mL of 1M NH3

But 1000mL of 1M NH3 contain N=14g.

20 ml of 1M NH3 will contain nitrogen =14 x20

1000

But this amount of nitrogen is present in 0.5 g of organic compound

... % of N = 14 x 20 x 100 =56.0

1000x 0.5

6. You have a mixture of three liquids A, B , C. there is a large difference in the boiling point of A and the rest two liquids. Boiling points of liquids B and C are quite close. Liquid A boils at higher temperature than B and C and boiling point of B is lower than C. How will you separate the components of the mixture.

Ans:.Since the boiling point of liquid A is much higher than those of liquids B and C , therefore separate liquid A by simple distillation. Since boiling points of liquids B and C are quite close but much lower than liquid A therefore mixture of B and C will distil together leaving behind A. on further heating A will distil over.

Now place the mixture of liquids B and C in a flask fitted with fractionating column. Since the b.p. of liquid B is lower than that of C , on fractional distillation first liquid B will distil over and than liquid C.

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CHAPTER 13(GrB)

HYDROCARBON

KEY AREAS OF THE CHAPTER

Nomenclature and Structure

Preparation of alkanesphysical and chemical properties of Alkanes

Conformation of ethene

Preparation of Alkenes, physical and Chemical properties of alkenes

Geometrical Isomers of alkenes

Preparation of alkynes, physical and chemical properties of Alkynes

Aromaticity and Aromatic Compounds

Electrophilic substitution reaction of benzene and benzenoid compound

Directive influences of functional group

KEY POINTS Hydrocarbons are composed of Carbon and hydrogen. The important fuels like Petrol, kerosene, coal gas, CNG, LPG etc. are all hydrocarbons or their mixture.

Sources:Petroleum and natural gas are the major sources of aliphatic hydrocarbon while coal is an important source of aromatic hydrocarbons. The oil trapped inside the rocks is known as petroleum. PETRA –ROCK, OLEUM –OIL. The oil in the petroleum field is covered with a gaseous mixture known as natural gas. The main constituents of the natural gas are methane, ethane, propane andbutane.

lkanes:- Paraffins

General formula CnH2n+2

SP3hybridisation

C–C bond length 1.15 4 Ao

Chemically unreactive

Show chain, position and optical isomerism.

Heptane has 9 isomer, Octane 18 and Decane 75.

Nomenclature:

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Preparation and Chemical Properties of Alkanes

Physical Properties:-

1. Nature:- Non-Polar due to covalent nature of C—C bond and C—H bond. C— C bond enrgy = 83

kj/mole and C—H bond energy = 99kj/mole.

C1—C4 = gases, C5—C17 = colourless odourless liquid and > C17 = Solid.

2. Solubility:- Like dissolvelike

Viz, Polar compounds dissolve in polar solvent and Non-Polar compound dissolve in non polar solvent. Boiling point:- Low boiling point due to non polar in nature.

The molecules are held together only by since we known that the magnitude of proportional to the

molecular size. Therefore, the boiling point increases with increase the molecular size i.e. with increase

in number of carbon atoms.

Noted:-the boiling points of the branched chain Alkanes are less than the straight chain isomers.

This is due to the fact that branching of the chain makes the molecule more compact and thereby decreases the surface aria and consequently, the magnitudes of Van der Waall‘sforces also decrease.

Important chemical properties in detail

Chemical properties Combustion :

CH4 + 2O2 CO2 + 2H2O

Oxidation: CH4 + O2Cu/573K 2CH3OH

CH4 + O2Mo2O3/MethanalHCHO + H2O

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Substitution:

Halogenation : CH4 + Cl2 hv CH3Cl + HCl CH3Cl hv CH2Cl2 hv CH3Cl hv CCl4

Noted:- Iodination is a reversible reaction. So it is carried out by heating alkane in the presence of some oxidizing agent like iodic acid (HIO3) or nitric acid (HNO3) or mercuric oxide (HgO) which oxidizes HI formed during the reaction.

CH4 +I2 CH3I +HI

5HI +HIO3 3H2O +3I2

2HI +2HNO3 2H2O+I2 +2NO2

Noted:- Fluorination of alkane takes place explosively resulting even in the rupture of C—C bond in

higher alkanes.

Features of Halogenations:-

(i) The reactivity of Halogens:- F2> Cl2> Br2> I2. (ii) The rate of replacement of Hydrogens of

alkanes is: 3° > 2° > 1°

2. Nitration:- The reaction takes places by free radicals mechanism at high temp (450

0C).

At high temp C—C bond is also broken so that mixture of nitroalkanes is obtained.

CH3CH2CH3 450°C

CH3CH2CH2NO2 + CH3CHCH3 + CH3CH2NO2+ CH3NO2

Conc. HNO3

NO2

25% 40% 10% 25%

The reaction occurs as: 4500C

HO-NO2 Homolyticfission HOo + oNO2

RH +0OH

Ro + HOH

Ro + oNO2 RNO2

3.Aromatization:-

H3C(CH2)4CH3 Cr2O3

773 K

Hexane 10-20 atm Benzene

This method is also called dehydrogenation or hydroforming

Similarly, heptane gives toluene, n-Octane give o-xylene and 2, methyl heptane give m-xylene. 5. Thermal decomposition or Pyrolysis or cracking or Fragmentation: - when higher alkanes are

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heated at high temp (about 700-800k) in the presence of alumina or silica catalysts, the alkanes break down to lower alkanes andalkenes. CH3-CH2-CH2CH2-CH-CH2 heat CH3-CH3 + C2H4 + CH4

Action of steamcatalyst: nickel, alumina Al2O3 at 1000 oC

CH4 +H2O(Steam) CO +3H2

This reaction is used for the industrial preparation of hydrogen from natural gas.

Isomerisation reaction

CONFORMATIONAL ISOMERISM:

The different molecular arrangements arising as a result of rotation around carbon carbon single bonds are called conformational isomers or rotational isomers and the phenomenon is called conformational isomerism.

Numerous possible arrangements of ethane are possible. Two extremeconformations are known. These are eclipsed conformation and staggeredconformation.

Alkenes

Unsaturated hydrocarbon which have doublebond.

General molecular formulaCnH2n

C–C bond hybridization 1.34A0

sp2

hybridization When we treated Alkene with chlorine, oily products are obtained. So Alkenes are also known as

Olefins. (Greek olefiant meaning oil forming).

Show chain, positional and geometrical isomerism

Stucture of double bond:-

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Preparation:-

a. From Alkynes:-Alkynes on partial reduction with Partiallydeactivatedpalladised charcoal known as Lindlar’s catalyst givealkynes.

b. From Haloalkanes: -dehydrohalogenation

CH3CH2-Br Alc. KOH

CH2=CH2

Predominantly formation of a more substituted alkene is formed according to Saytzeff‗s rule

Preparation of Alkene from Dihaloalkane

Preparation of alkene from alcohols

Mechanism

Geometrical Isomers in Alkenes

If two atoms attached to carbons are different then the alkene shows Cis- Trans isomerism

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While ALKENES does not show conformation due to presence of double bond which restricts the rotation.

Chemical Properties of Alkenes

Addition reactions i) Addition of Hydrogen (H2) to alkenes

ii) Addition of Halogens to Alkenes

iii) Addition of HX to Alkenes

iv) Addition of H2O

Mechanism of Acid catalyzed reaction

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Markownikov rule:-negative part of the addendum (adding molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms.

Peroxideeffect or Kharasch-In1933(Anti Markovnikov’s addition or Kharasch effect)

You may observed that when HBr is added to an unsymmetrical double bond in the presence of organic

peroxide, the reaction take places opposite to the Markovnikov rule.

Anti Markovnikov’s rule:Anti Markovnikov addition or peroxide effect or Kharasheffect :

In the presence of peroxide, addition of HBr to unsymmetrical alkenes like propene takes place

contrary to theMarkovnikov rule. This happens only with HBr but not with HCl and Hl. This addition

reaction was observed by M.S. Kharash and F.R. Mayo . This reaction is known as peroxide or

Kharash effect or addition reaction anti toMarkovnikov rule.

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The secondary free radical obtained in the above mechanism (step iii) is more stable than the

primary. This explainsthe formation of 1-bromopropane as the major product.The peroxide effect is

not observed in addition of HCl and HI. This may be due to the fact that the H–Cl bond being

stronger (430.5 kJ mol–1) than H–Br bond(363.7 kJ mol–1), is not cleaved by the free radical, whereas

the H–I bond is weaker(296.8 kJ mol–1) and iodine free radicals combine to form iodine molecules

instead of adding to the double bond

v) Addition of sulphuric acid : Cold concentrated sulphuric acid adds to alkenes in accordance with Markovnikov rule to form alkyl hydrogen sulphate bythe electrophilic addition reaction.

Ozonolysis :Ozonolysis of alkenes involves the addition of ozone molecule to alkene to form ozonide,

and then cleavage of the ozonide by Zn-H2O to smaller molecules.This reaction is highly useful in detecting the position of the double bond in alkenes or other unsaturated compounds.

Polymerisation: Polythene is obtained by the combination of large number of ethene molecules at

hightemperature, high pressure and in the presence of a catalyst. The large molecules thus obtained

are called polymers.This reaction is known as polymerisation. The simple compounds from which

polymers are made are called monomers. Other alkenes also undergo polymerisation.

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Uses of Polymers :Polymers are used for the manufacture of plastic bags, squeeze bottles, refrigerator

dishes,toys, pipes, radio and T.V. cabinets etc. Polypropene is used for the manufacture of milk crates,

plastic buckets andother moulded articles.

ALKYNES : General f ormula-CnH2n-2

Nomenclature and isomerism

1.H C≡C-CH2-CH2-CH3 Pent-1-yne

2.H3C-C≡C-CH2-CH3 Pent-2-yne

3.H3C-CH-C≡CH 3-Methylbut-1-yne

CH3

Structure of triple bond:

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Orbital picture of ethyne showing Sigma and pie overlaps

The C-C triple bond length=120 pm, C-C double bond length=133 pm, C-C single bond length=154 pm

Preparation of alkynes

PHYSICAL PROPERTIES :

1.First three members are gases, the next eight are liquids and the higher ones are solids.

2. Alkynes are weakly polar.

3.They are lighter than water and immiscible with water and soluble in organic solvents like ethers,

carbon Tetrachloride and benzene.M.p and b.p and density increases with increase in molecular mass.

Chemical properties of alkynes

Alkynes contain a triple bond. A triple bond has one sigma bond and two pie

bonds.Some characteristic reactions of alkynes are,

1. Combustion :Alkynes burns in air or oxygen with smoky flame.

1. Electrophilic addition reactions

Carbon-carbon triple bond, C=C, is a combination of one and two bonds. Alkynes give

123

electrophilic addition reactions as they show reactivity due to the presence of bonds.

This property is similar to alkenebut alkynes are less reactive than alkenes towards

electrophilic addition reactions due to the compactCC electron cloud. Some typical

electrophilic addition reactions given by alkynes are:

Addition of hydrogen An alkyne reacts with hydrogen in the presence of catalyst (Pt or Ni) at 250°C, first

forming alkenes and finally alkane.

For example, ethyne gives ethane in two steps.

Ethyne ethene ethane

Ethane is obtained in good yields if hydrogenation is done with a calculated amount of hydrogen in the

presence of palladium or barium sulphate. Propyne gives,

Addition of halogens Alkynes react with halogens (Cl2 or Br2) in the dark, forming dihaloalkenes first and finally

tetrahaloalkanes. The reaction gets accelerated in the presence of light or halogen carriers.

Ethyne dichloroethene tetrachloroethane

Dilute bromine water with ethyne gives dibromo, while liquid bromine gives tetrabromo derivative.

The Order of reactivity is Cl2>Br2>I2

Addition of halogen acids Alkynes reacts with halogen acids according to the Markownikov's rule i.e. the carbon atom carrying the

least number of hydrogen atoms will have the negative part of the

addendum attached to it

For example, ethyne (acetylene) with HBr gives,

With diluted HCl at 65°C and in the presence of Hg2+

(mercuric ion) ethyne gives vinyl chloride.

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propyne 2-bromopropene 2,2-dibromopropane

The rate of addition of halogen acids follows the order, HI >HBr>HCl

Alkynes add up two molecules of sulphuric acid. For example, ethyne gives

Nucleophilic addition reactions :

Alkynes also give the following nucleophile addition reactions. Addition of water

In the presence of sulphuric acid (42%) and 1 % mercuric sulphate at 60°C, alkynes

add on one watermolecule to give aldehydes or ketones. For example,

alkyne ketone

Ethyne gives ethanal and propyne gives acetone.

Addition of ozone Ozone adds up across the triple bond to give ozonised. After hydrolysis, ozonised give di-ketones &carboxylic acids.

Ethyne gives glyoxalin and formic acid,

glyoxalin formic acid

Substitution reactions

Due to their acidic nature, alkynes form metallic salts called alkynides e.g., sodium, silver and

copper(ous)

salts. Examples are,

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Acidic Character of alkynes Hydrogen atoms in ethyne and 1-alkynes, linked to the carbon atom having a triple bond on it, are acidic

in nature. For example, ethyne (acetylene) is a weak acid: weaker than water but stronger than ammonia.

This may be explained as follows:

The pie -electrons are more weakly bound than sigma electrons. Thus, in those compounds containing

carbon-carbon double or triple bonds, the electron density around such carbon atoms will be lesser than

the carbon atoms linked only through bonds. Thus, electronegativity of differently hybridized carbon

atoms will follow the orders sp1> sp

2> sp

3

i.e., the electronegativity will increase with the character in the hybrid orbitals. This increase in the

electronegativity of an alkyne carbon, (relative to the carbon atoms in

alkenes and alkanes) will polarize the C-H electron bond towards carbon and facilitate the release of

proton(s). Accordingly the acid strength of hydrogen's will follow the order. Alkynes > Alkenes >

Alkanes.

The stabilities of the anion left after the removal of proton, i.e. carbanions follow the order,

RC≡C-> RCH = CH

-> R-CH2-CH2

-

Thus, the acid strength follows the order, HC ≡CH > H2C = CH2> H3C-CH3

Compared to the organic acids e.g..ethanoic acid (CH3OOH), ethyne is about 1020

time less acidic,

while ethane is 1040

times less acidic.

HC≡CH +Na HC ≡C-Na+ +1/2H2

HC ≡C-Na+ +Na Na+C≡C Na+ 1/2H2

CH3-C≡C-H +Na-NH2 CH3-C≡C-Na +NH3

Polymerization 1. Linear Polymerisation :Ethyne gives polyethyne which is a high molecular weight polyene containing

2. repeating unit of (CH=CH-CH=CH) and can be represented as -(CH=CH-CH=CH)-n .

2 . Cyclic polymerisation: On heating alkynes undergo polymerization in the presence of

catalyst.

The nature of products depends upon the conditions.

For example,

When ethyne (acetylene) is passed through a hot copper tube, it polymerizes to

benzene.

With red hot iron tube:

3 CH≡CH Red hot iron tube,873k C6H6

Ethyne benzene

Aromatic hydrocarbon Aroma meaning pleasant smelling. The compounds having pleasant smell are called aromatic

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compounds. The aromatic compounds having benzene ring are called benzenoids and those not

containing a benzene ring arecalled non benzenoids.

STRUCTURE OF BENZENE:

1.The molecular formula of benzene is C6H6 which indicates high degree of unsaturation.

2. Benzene is found to be a stable molecule and found to form a triozonide which indicates the presence

of three double bonds.

3.Benzene is found to produce one and only one monosubstituted derivative which indicates that all the

six carbon

and six hydrogen atoms of benzene are identical.

Kekule structures:

Kekule structure indicates the possibility of two isomeric 1, 2-dibromobenzenes Or two isomeric

substituted benzene. But benzene is found to form only one ortho substituted product. This problem

was overcome by Kekule by suggesting the concept of oscillating nature of double bonds in benzene as

given below :

Even with this modification, Kekule structure of benzene fails to explain the unusual stability and

preference to substitution reactions than addition reactions, which could later on be explained by

resonance.

RESONANCE AND STABILITY OF BENZENE

According to VBT, the concept of oscillating double bonds in benzene is now explained by resonance.

Benzene is a hybrid of various resonating structures. The two structures given by Kekule are the main

contributing structures. The hybrid structure is represented by inserting a dotted circle in the hexagon

.The circle represents the six e- which are delocalised between the the six C-atoms of the benzene

(C6H6) ring as shown below:

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Orbital picture of benzene: The orbital overlapping gives us better picture about the structure of

benzene All the six carbon atoms in benzene are sp2 hybridised. Two sp2 hybridised orbitals of each

carbon atoms overlap with sp2 hybrid orbitals of adjacent carbon atoms to form six C-C sigma bonds

which are in the hexagonal plane. The remaining sp2 H.O. of each carbon atom overlaps with s orbital of

a hydrogen atom to form six C-H sigma bonds .Each carbon atom is now left with one unhybridised p -

orbital perpendicular to the plane of the ring. The hybridised p orbital of carbon atoms are close enough

to form a π bond by lateral overlap. There are two equal possibilities of forming three π bonds by overlap

of p orbitals. Structures shown below correspond to two Kekule‘s structure with localisedπ bonds. There

is equal probability for the p orbital of each carbon atom to overlap with the p orbital of adjacent carbon

atoms. This can be represented in the form of two rings of electron clouds ,one above and one below the

plane of the hexagonal ring as shown below.

: The six pie electrons are thus delocalised and can move freely about the six carbon nuclei , instead of

any two.Thedelocalised π electron cloud is attracted more strongly by the nuclei of the carbon atoms

than the electron cloud localised between two carbon atoms .Therefore presence of delocalised π

electrons in benzene makes it more stable than the hypothetical cyclohexatriene.

X-ray diffraction reveals that benzene is a planar molecule .Two types of C-C bond length are expected

.X-ray data indicates that all the six c-c bond lengths are of the same order (139 pm) which is

intermediate between c-c single bond (154 pm) and c-c double bond(133 pm).Thus the absence of pure

double bond in benzene accounts for the reluctance of benzene to show addition reactions under normal

conditions, thus explaining the unusual behaviour of benzene.

ORBITAL PICTURE OF BENZENE

DELOCALICED PI (π) ELECTRON CLOUD ABOVE AND BELOW THE PLANE

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AROMATICITY :Benzene is considered as parent aromatic compound.

Characteristics of aromatic (aryl) compounds

An aromatic (or aryl) compound contains a set of covalently bound atoms with specific characteristics:

1. A delocalized conjugated π system, most commonly an arrangement of alternating single and

double bonds

2. Coplanar structure, with all the contributing atoms in the same plane

3. Contributing atoms arranged in one or more rings

A number of π delocalized electrons that is even, but not a multiple of 4. That is, 4n + 2 number of π

electrons, where n=0, 1, 2, 3, and so on. This is known as Hückel's Rule.

Whereas benzene is aromatic (6 electrons, from 3 double bonds), cyclobutadiene is not, since the number

of π delocalized electrons is 4, which of course is a multiple of 4. The cyclobutadienide (2−) ion,

however, is aromatic (6 electrons). An atom in an aromatic system can have other electrons that are not

part of the system, and are therefore ignored for the 4n + 2 rule. In furan, the oxygen atom is sp²

hybridized. One lone pair is in the π system and the other in the plane of the ring (analogous to C-H bond

on the other positions). There are 6 π electrons, so furan is aromatic.

Some examples of aromatic compounds

Benzene and benzenoid

Preparation of benzene :1. Cyclic polymerisation of ethyne

1. Decarboxylation of aromatic acids

Reduction of phenol: Phenol is reduced to benzene by passing its vapour over heated zinc dust.

RESONANCE HYBRID OF TWO CANONICAL FORMS

http://en.wikipedia.org/wiki/Aryl

http://en.wikipedia.org/wiki/Covalent_bond

http://en.wikipedia.org/wiki/Delocalized_electron

http://en.wikipedia.org/wiki/Pi_electron

http://en.wikipedia.org/wiki/Covalent_bond

http://en.wikipedia.org/wiki/Coplanar

http://en.wikipedia.org/wiki/Singly_even

http://en.wikipedia.org/wiki/H%C3%BCckel's_rule

http://en.wikipedia.org/wiki/Cyclobutadiene

http://en.wikipedia.org/wiki/Furan

129

Properties :

Physical properties :

1.Aromatic hydrocarbons are nonpolar and are usually colour less liquids or solids.

2. They have charactericaroma.

3.Aromatic hydrocarbons are immiscible with water but are readily miscible with organic solvents.

4. They burn with a sooty flame.

Chemical properties :

The common reaction of arenes are (i) electrophilic aromatic substitution. (ii)Addition reactions

(iii) Oxidation reaction

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131

132

133

134

135

136

137

ONE MARK QUESTIONS

1. Give the IUPAC name of lowest molecular weight alkane that contains

a quaternarycarbon.

Ans.2,2dimethylpropane.

2. Arrange the following in the increasing order of C-C bond

length- C2H6 C2H4C2H2

Ans. C2H2<C2H4<C2H6

3. Out of ethylene and acetylene which is more acidic andwhy?

Ans. Acetylene, due to greater electonegativity of the sp hybrid carbon. 4. Name two reagents which can be used to distinguish between ethene

and ethyne. Ans.Tollen‗s reagent and ammoniacal C

5. Arrange the following in order of decreasing reactivity towards alkanes. HCl, HBr, HI,HF

Ans.HI>HBr>HCl>HF

6. How will you detect the presence of unsaturation in an

organiccompound? Ans. Unsaturated organic compound decolorize the

Brominewater.

7. Why carbon does have a larger tendency of catenation than silicon although they have same number of valence electrons?Ans .It is due to the smaller size C-C bond which is stronger (335 KJ mol-1) than in Si bond (225.7 KJ mol-1).

8. Write IVPAC names of the following CH3 (CH2)4 CH (CH2)3 CH3 CH2 – CH (CH3)2. Ans 09.5-(2 – Methyl propyl) – decane.

9. What is hydrogenation? Ans .Dihydrogen gas gets added to alkenes and alkenes in the presence of finely

divided catalysts like Pt, Pd or Ni to form alkanes. This process is called

hydrogenation.

10. How would you convert ethene to ethanemolecule?

11. Methane does not react with chlorine in dark. Why? Ans .Chlorination of methane is a free radical substitution reaction. In dark, chlorine is unable to be converted into free radicals, hence the reaction does not occur.

12. Which conformation of ethane is more stable? Ans .Staggered conformation.

13. What is β-elimination reaction? Ans. When hydrogen atom is eliminated from the β-carbon atom (carbon atom next

to the carbon to which halogen is attached).

ANS:

138

14. What is the number of σ and π bond in N ≡ C – CH = CH – C ≡ N? Ans .There are 7σ bonds and 5 π-bonds.

15. Name the type of hybridization in C (2) and C (3) in the following molecule

Ans..C(2) is sp-hybridized and C(3) is sp2 hybridized.

16. Why do alkynes not show geometrical isomerism? Ans.Alkyneshave linear structure. So they cannot show geometrical isomerism.

17. How will you convert ethyne to benzene? Ans.

Or

18. Although benzene is highly unsaturated; it does not undergo addition reactions. Give reason. Ans .Unlike olefins, π-electrons of benzene are delocalized (resonance) and hence these are uncreative towards addition reactions.

19. The boiling point of alkanes shows a steady increase with increase in molecular mass. Why? Ans .This is due to the fact that the intermolecular van der walls forces increase with increase of the molecular size or the surface area of the molecule.

20. All the four C-H bonds in methane are identical. Give reasons. Ans .The four C-H bonds of methane are identical because all of these are formed by the overlapping of the same type of orbital’s i.e; hybrid orbital’s of carbon and s-orbital of hydrogen.

21. How would you convert cyclohexane to benzene? Ans . Cyclohexane when treated with iron or quartz in a red hot tube undergoes oxidation to form benzene.

22. How is alkene produced by Kolbe’s electrolytic method?

Ans.

139

23. How will you demonstrate that double bonds of benzene are somewhat different from that of olefins? Ans.The double bonds of olefins decolourize bromine water and discharge the pink colour of Bayer‘s reagent while

24. How will you separate propene from propyne? Ans. By passing the mixture through ammonical silver nitrate solution when propyne reacts while propene passes over.

25. Write is the structure of the alkene which on reductive ozonolysis gives butanone and ethanal. Ans.-CH3CH2C(CH3)=C- CH3

26. How is alkene prepared from alcohol by acidic dehydration? Ans .Alcohols on heating with concentrated sulphuric acid form alkenes with the

elimination of one water molecule.

27. How are trans alkenes formed from alkynes?

Ans. Alkynes on reduction with sodium in liquid ammonia form trans alkenes.

140

28. How are cis – alkenes formed by alkynes? Ans. Alknes on partial reduction with calculated amount of dihydrogen in the presence of palladised charcoal partially deactivated with poisons like sulphur compounds or quinoline give cis-alkene.

TWO MARKS QUESTIONS

1. Write the IUPAC names of the following-

a. b. Ans.a.Pent-en-3-yne b.2-methylphenol

141

2. Write chemical equations for combustion reaction of (i) Butane (ii) TolueneAns.

.

(ii)

Toluene

3. What are the necessary conditions for any system to bearomatic?

Ans. A compound is said to be aromatic if it satisfies the following three conditions: (i) It should have a planar structure.

(i) The–electronsπof the compound are completely delocalized in the ring.(iii)The total–electronsnumberpresentinoftheringπshould be equalto

(4n + 2), wheren = 0, 1, 2 … etc. This is kn

(iv) What effect does branching of an alkane chain has on its boilingpoint?

Ans.As branching increases, the surface area of the molecule decreases whichresults in a small area of contact. As a result, the Van der Waals force also decreases which can be overcome at a relatively lower temperature. Hence, the boiling point of an alkane chain decreases with an increase in branching.

(v) How would you convert the following compounds intobenzene?

a. Ethyne b. Ethene

Ans. (i)Benzene fromEthyne:

(ii) Benzene from Ethene:

142

4. Suggest the name of Lewis acids other than anhydrous aluminiumchloride

which can be used during ethylation of benzene. Ans.

anhydrous FeCl3, SnCl4, BF3etc.

5. Write the name of all the possible isomers of C2H2Cl2 and indicate whichof them is non-polar.

Ans.(i) cis-1,2-dichloroethene (ii) trans-1,2-dichloroethene (iii) 1,1- dichloroethene. trans-1,2-dichloroethene is non-polar.

(ii) Although benzene is highly unsaturated, it does not undergo addition reactions, why?

Ans.Because of extra stability due to delocalization of π-electrons. (iii)What are alkanes? Why are they called paraffins?

Ans. Those hydrocarbons which contain single bond between carboncarbon are called alkanes. They are called paraffin because they are very less reactive (Latin- Parum= little, affins = affinity)

6. How will you convert the following compounds into benzene?

(i) ethene (ii) hexane.

Ans:

7. The boiling point of hydrocarbons decreases with increase in branching. Give reason.

Ans. Branching result into a more compact (nearly spherical) structure. This reduces the effective surface area

and hence the strength of the Vander wall’s forces, thereby leading to a decrease in the boiling point.

8. Unsaturated compounds undergo addition reactions. Why?

Ans. Unsaturated hydrocarbon compounds contain carbon – carbon double or triple bonds. The π-bond is

multiple bond is unstable and therefore addition takes place across the multiple bonds.

9. Sodium salt of which acid will be needed for the preparation of propane? Write chemical equation for the

reaction.

Ans. Butanoic acid,

CH3CH2CH2COO-Na+NaOH CaO CH3CH2CH3+Na2CO3.

143

10. How can ethene be prepared from (i) ethanol (ii) ethylbromide?

Ans. (i) Ethene from ethanol- by acidic dehydration of alcohols

ii)Ethene from ethyl bromide- by dehydrohalogenation of ethyl bromide CH3CH2Br +

KOH(alc) → 2HC = CH2 + KBr +H2O

THREE MARKS QUESTIONS

1. What is Wurtz reaction? How can it be used to preparebutane?

Ans- When alkyl halides is treated with metallic Na in presence of dry ether, alkanes are formed. This reaction is called Wurtzreaction.

Butane is prepared by the reaction of bromoethane with metallic Na in presence of dry ether

2. An alkene ‗A‘–C,containseightC–Hσ bondsthree–CandCπ one bond. ‗A‘ onozonolysis gives a compound of molarmass 44 u.

Deduce IUPACnameof ‗A‗.

Ans..The formation of two moles of an aldehyde indicates the presence ofidentical structural units on both sides of the double bond containingcarbon atoms. Hence, the structure of A (alkene)

XC = CX

There are eight C–H ζ bondsHence, there are 8 hydro there are three C–C bonds. Hence, there are four

carbon atoms present in the structure of ‗A

Combining the inferences, the structure

the IUPAC name-2-ene .of ‗A‗ is But Ozonolysis of ‗A‗ takes place as:

The final product is ethanal with molecular mass

144

3. In the alkane H3C –CH2 –C(CH3)2 –CH2 –CH(CH3)2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these. Ans.

The given structure has five 1° carbon atoms and fifteen hydrogen atoms attached toit.

The given structure has two 2° carbon atoms and fourhydrogen atoms attached toit.

The given structure has one 3° carbon atom and only one hydrogen atom is attached to it FIVE MARKS QUESTIONS

1. Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism

Ans. Addition of HBr to propene is an example of an electrophilic substitution reaction.

Hydrogen bromide provides an electrophile, H+

. This electrophile attacks the double bond to form 1° and 2° carbocations as shown:

Secondary carbocations are more stable than primary carbocations. Hence, the former

predominates since it will form at a faster rate. Thus, in the next step, Br–

attacks the carbocation to form 2 –bromopropane as the major product.

Thisreaction follows Markovnikov‗s

In the presence of benzoyl peroxide, an addition reaction takes place anti to Markovnikov‗srule. The reaction fol

145

Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. Thus, 1 – bromopropane is obtained as the major product.

146

CHAPTER-14

ENVIRONMENTAL CHEMISTRY-

(Gr-G)

BASIC CONCEPTS

Environmental chemistry deals with the study of the origin,transport,reactions,effects,fates of chemical

species in theenvironment.

ENVIRONMENTAL POLLUTION:-Environmental pollution is the effect of undesirable changes in

our surroundings that have harmful effects on plants, animals and human beings.A substance

which causes pollution is called a pollutant. They can be solid,liquid or in the gaseousstate.

ATMOSPHERIC POLLUTION:-The atmosphere that surrounds the earth is not of the same thickness

at different heights.Atmospheric pollution is generally studied as tropospheric and stratospheric

pollution.The ozone layer prevents about 99.5%of the sun‘s UVrays.

TROPOSPHERIC POLLUTION:-Tropospheric pollution occurs due to the presence of undesirable

solid or gaseous particles in the air. The following are the major gaseous and particulate pollutants

present in thetroposphere;

Gaseous air pollutants:These are oxides of sulphur, nitrogen and carbon, hydrogen

sulphide, hydrocarbons, ozone and other oxidants.

Particulate pollutants; these are dust, mist, fumes, smoke, smog etc

Global warming and green house effect

solar energy reaching the earth is absorbed by the earth‘s surface,which increases it‘s temperature.The

rest of the heat radiates back to the atmosphere.Some of the heat is trapped by the gases such as carbon

dioxide,methane,ozone,CFCS and Water vapour. They add to the heating of the atmosphere causing

Global warming

In a greenhouse,visible light passes through the transparent glass and heats up the soil and the

plants.The warm soil and plants emit infrared rays,it partly reflects and partly absorbs these

radiations,this mechanism keeps the energy of the sun trapped in the greenhouse

ACID RAIN: When the pH of the rain water drops below5.6, it is called acid rain.Acid rain is harmful for

agriculture, trees and plants as it dissolves and

washesawaynutrientsneededfortheirgrowth.Itcausesrespiratoryailments in human beings and animals.

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When acid rain falls and flows as ground water to reach rivers, lakes etc. it affects plants and animal life

in aquatic ecosystem

SMOG: The word smog is derived from smoke and fog.There are two types of smog:classical and

photochemical smog. Classical smog occurs in cool humid climate. It is a mixture of smoke, fog and

sulphur dioxide. It is also called reducing smog. Whereas photochemical smog occurs in warm and dry

sunny climate. It has high concentration of oxidizing agents and therefore ,it is also called as

oxidizingsmog

OZONE HOLE:Depletion of ozone layer is known as ozonehole.

EFFECTS OF DEPLETION OF THE OZONE LAYER: With the depletion of ozone layer, more UV

radiation filters into troposphere. UV radiations lead to ageing of skin, cataract, sunburn, skin

cancer, killing of many phytoplanktons, damage to fish productivityetc

WATER POLLUTION:-contamination of water by foreign substances which make it harmful for

health of animals or plants or aquatic life and make it unfit for domestic, industrial and agriculture

use.

SOURCES/ CAUSES OF WATER POLLUTION-

Sewage and domestic wastes Industrial

effluents Agriculture effluents

Siltation-mixing of soil or rock into water Thermal pollutants

Radioactive discharge

EUTROPHICATION:The process in which nutrient enriched water bodies support a dense plant

population, which kills animal life by depriving it of oxygen and results in subsequent loss of

biodiversity is known as Eutrophication

BOD: The amount of oxygen required by bacteria to break down the organic matter present in a certain

volume of a sample of water, is called Biochemical Oxygen Demand(BOD)

SOIL POLLUTION: Insecticides, pesticides and herbicides cause soil pollution.

SOME IMPORTANT QUESTION

One mark question

1. Name one insecticide?

A. DDT

2. Which acid is not present in acid rain?

HNO3, H2SO4, CH3COOH, H2CO3

148

A. CH3COOH

3. Define the term pollutants?

A. It is a substance present in the environment in great proportion than its natural abundance and

resulting in harmful damage effect.

4. Name two gases which are responsible for green house effect?

A. CO2 and CH4 gases.

5. Which part of the atmosphere contains ozone layers?

A. Stratosphere contains ozone layers.

6. What is full form BOD and DDT?

A. BOD‐Biochemical oxygen Demand and DDT‐Dichchloro Diphenyl Trichloro ethane.

7. What are PCBs?

A. Poly chlorinated biphenyls (PCBs) are used as cleansing solvent, detergents and fertilizers cause water

pollution and it is carcinogenic compound.

8. What is PAN?

A. Peroxyacetyl nitrate (PAN) is one of the components of photo chemical smog and it is powerful eye

irritant.

9. What is desirable concentration of F ‐ ions and pH of drinking water?

A. Desirable concentration of F‐ ions is 1ppm or 1mgdm

‐3 and PH is 5.5 to 9.5

10. Name the oxides of nitrogen?

A. Nitric oxide (NO) and Nitrogen dioxide (NO2).

11. Which gas caused Bhopal gas tragedy? Give its formula.

A. Methyl isocyanate (MIC) and its molecular formula CH3N=C =0

12. Write any two common chemicals of photochemical smog?

A. Acrolein and formaldehyde

13. Which can damage the great historical monument Tajmahal?

A. Acid rain (CaCO3 (marble) +H2SO4 CaSO4+H2O+CO2)

149

14. What is effect of excess of SO42‐

ion in drinking water?

A.Excess of SO42‐

ion in drinking water causes laxative effect (>500ppm)

15. What is troposphere?

A. The lower regions of atmosphere in which the human beings along with other organisms live are

called troposphere. It extends up to the high of ~10KM from sea level

16. What is stratosphere?

A. Above the troposphere, between 10 and 50km above sea level lies is called stratosphere.

17. Name the harmful radiation emitted from sun?

A.UV radiation

18. Which type of harmful radiations absorbed in ozone layers?

A. UV radiation

19. Name the types of pollutants cause troposphere pollution?

A. 1. Gaseous air pollutants 2.Particulate pollutants

20. What are the sources of dissolved oxygen inwater?

A. In water, the source of oxygen is either atmospheric oxygen or photosynthesis carried in plants during

day light.

21. Which of the following gases is not a green house gas? CO, CO3, CH4, H2O vapours

A. CO is not a green house gas.

22. What type of radiation are absorbed by CO2 in the atmosphere

A.IR radiations

23. Name the oxide of carbon?

A. CO & CO2

24. What is green house effect?

A. The increase in temperature of atmosphere due to presence of gases like CH4 , CO2 and water vapours,

which absorb infrared radiation is called green house effect.

150

II) Two marks questions and answers:

1. Name two herbicides?

A.NaClO3 (sodium chlorate) andNa3AsO3 (Sodium Arsenite)

2. List any two harmful effects of smog?

A. 1. Act as powerful eye irritants

2. Irritate the nose and throat and their high concentration causes headache, chest pain, and dryness of the

throat, cough and difficulty in breathing.

3. Write any two achievement of green chemistry?

A. 1.Development of polystyrene foam sheet packaging material ThisTechnology allows eliminations

CFCS which contribute to ozone depletion, global warming and ground level smog.

2. Catalytic hydrogenation of diethanolamine in which environmental friendly herbicide is produced in

less dangerous ways.

4. Define environmental chemistry?

A. It is the study of chemical and biochemical process occurring in nature

(OR)

It deals with the study of origin, transport relation, effects and fates of various chemical species in the

environment

5. What do you mean by ozone hole? What are itsConsequences

A. Ozone hole implies distribution of the ozone layer by the Harmful UV radiations the depletion will

virtually result. In creating some sort of holes in the blanket of ozone which surround us. As a result, the

harmful radiations cause skin cancer, loss of sight and also affect our immune system

6. What do you mean by Biochemical oxygen demand?

A. Biochemical oxygen demand is the amount of oxygen required by bacteria to decompose organic

matter in a certain volume of sample of water. Clean water would have BOD value of less than 5ppm,

where as highly polluted water has a BOD of 17ppm or more

7. Write the methods for management of waste material?

1. Recycling:‐materials are recycled which saves the cast of raw material and waste disposal.

2. Sewage treatment

151

3. Burning and Incineration

4. Digesting

5. Dumping

8. A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic

dumping by you find an abundance of phytoplankton, suggest a reason for the fish kill.

A. The amount of dissolved oxygen present in water is limited. The abundance of phytoplankton causes

depletion of this dissolved oxygen. This is because phytoplanktons are degraded by bacteria present in

water. For their decomposition, they require a large amount of oxygen. hence, they consume the oxygen

dissolved in water. As a result, the BOD level of water drops below 6ppm, inhibiting the growth of fish

and causing excessive fish kill.

9. What are harmful effects as depletion of ozone layer?

A. 1) The ozone layer protects the earth from the harmful UV radiation of the sun, with the depletion of

the layer, more radiation will enter the earth‘s atmosphere. UV radiations are harmful because they lead

to the skin cancer and sun burns.

2) They cause death of many phytoplanktons which lead to a decrease of fish productivity.

3) Increase in UV radiation, decreases the moisture content of the soil and damages both plants and

fibres.

10. What are pathogens? Mention its harmful effect?

A. Pathogens are water pollutants include bacteria and other organism. They enter water from animal

excreta and domestic sewage.

11. What are harmful effects of acid rain?

A. 1. It is harmful for crops

2. It damages buildings made up of marble.

12. Write by two harmful effects of oxides of nitrogen?

1. Damage the leaves of plants and retard the rate of photosynthesis.

2. Nitrogen dioxide is a lung irritant that can lead to an acute respiratory disease in children.

13. Write any two harmful effects of oxide of sulphur?

1. It causes respiratory diseases e.g. Asthama bronchitis in human beings

152

2. It causes irritation to the eyes, resulting in tears and redness.

14. Write the harmful effects of hydrocarbons pollutants? Mention its sources? Harmful effects:

1. Hydrocarbons are carcinogenic i.e. they cause cancer

2. They harm plants and shedding of leaves flowers & twinges Sources: Incomplete combustion of fuel

used in automobiles.

15.What are the harmful effects of CO? Mention its sources? Harmful effects:

1. It is high poisonous to living beings

2. It causes, headache, weak eyesight, nervousness and cardiovascular disorder

Sources: 1. incomplete combustion of coal, firewood, petrol etc…

2. by automobile exhaust

16 Write the harmful effect of CO2 ?mention its sources.

A.Harmful effect:-It causes global warming

Sources: 1. Respiration

2. Burning of fossil fuels for energy

3. By volcanic eruptions

4. Deforestation

5. By decomposition of limestone during the manufacture of cements.

17. During ward war II DDT, was found to be of great use for which purpose was is used? Why has its

use been banned in India now?

A. It is used to control of malaria and other insect borne diseases. After the war, DDT is used in

Agriculture to control the damages caused by insects, weeds and various crop diseases However, due to

adverse effects, its use has been banned India and it is nonbiodegradable and carcinogenic.

18. Acid rain is known to contain some acids. Name these acids and where from they come in rain?

A. The acids commonly present in acid rain are nitric acid, sulphuric acid and carbonic acid.Nitric acid is

formed by oxidation of nitric oxide present in air to NO2& NO3, which dissolve in water also present in

air. Sulphuric acid is formed by the oxidation of SO2 result in air to SO3 followed by dissolution in water.

CO2 present in the acid dissolves in water to give carbonic acid.

153

III) Three marks questions and answers

1. Explain tropospheric pollution?

A. Troposphere pollution occurs due to the presence of undesirable gases and the solid particles in the air

the major gaseous and the Particulate pollutants presents in the troposphere as follows.

1. Gaseous air pollutants: These include mainly oxides of sulphur (SO2&SO3), oxide of nitrogen

(NO&NO2) and oxides of carbon (CO&CO2) in addition to hydrogen sulphide (H2S), hydrocarbons and

other oxidants.

2. Particulate pollutions: These include dust, mist, fumes, smoke, smog etc

2. What is smog? How is classical smog different from photo chemical smog?

A. Smog is a mixture of smoke, dust partials and small drops of fog.

Classical smog photochemical smog

It occurs in cool humid and Climate It occurs in warm dry sunny climate

It is called reducing smog It is called oxidizingsmog

It is a mixture of smoke, Fog and sulphurdioxide It is a mixture of unsaturated hydrocarbons and

Oxides of nitrogen

1. Carbon monoxide gas is more dangerous than carbon dioxide gas. Why? Explain

A. It is highly poisonous to living beings because of its ability to block the delivery of oxygen to

the organs and tissues. It binds to haemoglobin complex to form carboxyl haemoglobin

(COHb) which is about 300 times more stable than oxy‐hemoglobin complex. In the blood

when the concentration of carbon haemoglobin reaches about 3‐4 percent, the oxygen carrying

capacity blood is greatly reduced. This oxygen deficiency, results in to headache, weak

eyesight, nervousness and cardiovascular disorder CO2 does not combine with haemoglobin

and hence is less harmful as pollutant but it is the main contributor forwards green house

effect & global warming.

2. What are pesticides and herbicides? Give examples

A. Pesticides: These are a mixture of two or more substances. They are used for killing pests.

Pests include insects, plants pathogens, weeds, mollusks etc…., that destroys the plant crop

and spread diseases

ex:-Aldrin and Dieldrin

Herbicides: These are chemicals specially meant for killing weeds. Ex:‐Sodium chlorate

(NaClO3) Sodium Arsenite (Na3 AsO3)

3. Define green chemistry? Explain with one example.

Chemistry and chemical process involving the minimum use and generation of harmful

substances is called green chemistry Ex:‐ Earlier tetrachloroethene themes was used as solvents

for dry cleaning. This compound is carcinogenic; therefore it has been replaced by liquefied CO2

along with a suitable detergent which is less harmful.

154

KENDRIYA VIDYALAYA SANGATHAN (GrE)

KOLKATA REGION

HALF YEARLY QUESTION PAPER 2018-19

CLASS – XI

SUBJECT: CHEMISTRY

BLUE PRINT

Sl Name of the topic VSA (1) SA (2) SA (3) LA (4) LA (5) Total

1. Some Basic Concept of Chemistry 1 (1) 1 (2) 2 (3) *** *** 4 (9)

2. Structure of Atom **** 1 (2) 1 (3) *** 1 (5) 3 (10)

3. Classification of elements and periodicity

in properties

**** 1 (2) 2 (3) *** *** 3 (8)

4. Chemical Bonding and Molecular

structure

**** 1 (2) 2 (3) *** *** 3 (8)

5. States of matter: Gases and liquid & solid 1 (1) 1 (2) 2 (3) *** *** 4 (9)

6. Thermodynamics *** 1(2) 1 (3) *** 1 (5) 3 (10)

7. Equilibrium 2 (1) *** 1 (3) *** 1 (5) 4 (10)

8. Redox Reactions 1 (1) 1 (2) 1 (3) *** *** 3 (6)

Total 5 (1) 7 (2) 12 (3) *** 3 (5) 27

(70)

155

KENDRIYA VIDYALAYA SANGATHAN, KOLKATA REGION

SUBJECT: CHEMISTRY

CLASS – XI

FULL MARKS: 70 TIME ALLOWED: 3 Hrs

General Instructions:

i) All questions are compulsory

ii) Q. 1 – Q. 5- 1 mark each

Q. 6 – Q. 12 – 2 marks each

Q. 13 – Q. 24 – 3 marks each

Q. 25 – Q. 27 – 5 marks each

iii) Calculators are not permitted

Q. 1. Write the conjugate acid of 𝐻𝐶𝑂3− ion.

Q. 2. Express the solubility of Ag2SO4 in terms of KSP.

Q. 3. Calculate the oxidation state of Sin H2S2O8

Q. 4. How many significant figures are there in the following:

(i) 0.00453 (ii) 2.0005

Q. 5. Write Van der Waal gas equation for n moles of a gas.

Q. 6. Define Disproportionation reaction with an example.

Q. 7. Write the Lewis dot structure of 𝐶𝑂32− ion and NH3 molecule.

Q. 8. Using s, p, d and f notations, describe the orbitals with following quantum numbers:

(a) n = 1, l=0 (b) n = 2, l= 1 (c) n = 4, l=2 (d) n = 3, l=1

Q. 9. Prove the relations:

(i) ∆H = ∆U + ∆ngRT OR (i) ∆G = ∆H - T∆S

Q. 10. Consider the following species N3-

, O2-

, Na+, Mg

2+, Al

3+

(i) What is common in them?

(ii) Arrange them in the increasing order of ionic radius

Q. 11. Calculate molarity of NaOH solution prepared by dissolving 4 g of it in enough water to form 250 ml of

the solution.

Q. 12. i) A cubic solid is made up of two elements P and Q. Atoms of Q are present at the corners of the cube

and atoms of P at the body centre. What is the formula of the compound?

ii) What type of defect is responsible for the pink colour of LiCl?

Q. 13. Calculate the number of atoms present in:

i) 52 moles of He ii) 52 u of He iii) 52 g of He

Q. 14. In three moles of ethane, calculate the following:

i) No. of moles of carbon atoms.

ii) No. of moles of hydrogen atoms

iii) No. of molecules of ethane

Q. 15. The mass of an electron is 9.1 x 10-31

kg. If its K. E. is 3.0 x 10-25

J, calculate its wavelength.

Q. 16. Explain:

(i) Why do ionization enthalpies decrease down a group of the periodic table?

(ii) Ionisation enthalpy of nitrogen is more than Oxygen

(iii) Electron gain enthalpy of chlorine is more negative than that of fluorine

156

Q. 17. i) What is the difference between ζ and π bonds.

ii) Explain the formation of ζ and π bonds in ethyne molecule (C2H2) in terms of hybridization

Q. 18. Answer the following

(i) Under what conditions do real gases tend to show ideal gas behaviour?

(ii) Why does a drop of liquid assume spherical shape?

(iii) If the no of moles of a gas are doubled by keeping T & P constant, what will happen to the

volume?

Or

What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm

3 at 30ᵒC?

Q. 19. (i) Distinguish between ideal gas and real gas.

(ii) Write the two faulty assumptions of kinetic theory of gases.

(iii) State Dalton‘s law of partial pressure

Q. 20. Calculate the enthalpy of formation of C6H6 (i) given that the enthalpy of combustion of benzene is -

3267.7 KJ and enthalpies of formation of CO2 (g) and H2O (I) are -393.3 KJ and -285.8KJ mol-1

respectively.

Q. 21. State Le-Chatelier‘s principle. What is the effect of (i) increasing pressure and (ii) increasing

temperature on the following state of equilibrium:

N2 (g) + 3H2 (g) 2NH3 (g)

Q. 22. Balance the following equations in ion electron method. (Any One)

i) 𝑀𝑛𝑂4− + Fe

2+ + H

+ Mn

2+ + Fe

3++ H2O

ii) Cr2𝑂72− + 𝐶2𝑂4

2− + 𝐻 + CO2 + Cr3+

+ H2O

Q. 23. (1) Which out of NH3 and NF3 has a higher dipole moment and why?

(2) Using VSEPR theory draw the shapes of PCI5 and SF6

Q.24. (i) What is the general electronic configuration of –d- block elements.

(ii) Write IUPAC name and symbol of element having atomic number 120

(iii) What is the difference between electron gain enthalpy and electronegativity.

Q. 25. Answer the following:

(1) Give the electronic configuration of Cr (Z=24) and find no of unpaired electrons

(2) Draw the shape of dz2 orbital

(3) State Heisenberg‘s Uncertainty principle.

(4) How many orbitals are associated in n = 4?

Or

(1) The energy associated with first orbit in hydrogen atom is -2.18x10-18

J atom-1

. What is the energy

associated with the fifth orbit?

(2) Calculate the radius of Bohr‘s fifth orbit for hydrogen atom.

(3) What is the maximum no. of emission lines when an excited electron of a H-atom in n=6 drops to the

ground state.

157

Q. 26. (i) In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is

the change in internal energy for the process?

(ii) How is standard free energy change related with standard cell potential?

(iii) Under what conditions of ∆H and ∆S, the reaction will be spontaneous i.e. ∆G will be negative?

(iv) State second law of thermodynamics.

(v) What is the enthalpy of element in its standard state?

Or

(i) State Hess‘s Law or Law of constant heat summation.

(ii) Define molar enthalpy of formation.

(iii) Predict in which of the following, entropy increases/decreases.

a) A liquid crystallizes into a solid

b) H2 (g) - 2H (g)

(iv) Write two examples of extensive property.

Q. 27. Answer the following:

(1) Which of the following are Lewis acids and Lewis bases?

H2O, BF3, H+, 𝑁𝐻4

+

(2) Define buffer solution

(3) What do you mean by common ion effect? Give example.

(4) Why is there a fizz when a soda water bottle is opened?

Or

(1) What is ionic product of water?

(2) For reaction N2 (g) + 3H2 (g) 2NH3 (g) at 400 K, Kp = 41. What is the value of Kp for

1

2𝑁2 𝑔 +

3

2𝐻2 𝑔 NH3 (g)

(3) At certain temperature and total pressure of 105 Pa, iodine vapours contain 40% by volume of iodine

atoms in the equilibrium l2 (g) - 2l (g),

158

KENDRIYA VIDYALAYA SANGATHAN

KOLKATA REGION

HALF YEARLY QUESTION PAPER 2018-19

CLASS – XI

SUBJECT: CHEMISTRY

MARKING SCHEME

Q. 1. Conjugate acid of 𝐻𝐶𝑂3− is H2CO3 1

Q. 2. Correct expression 1

Q. 3. +6 1

Q. 4. (i) 3 (ii) 5 ½ + ½

Q. 5. Correct equation 1

Q. 6. Correct definition + correct example 1 + 1

Q. 7. Correct Lewis dot structure 1 + 1

Q. 8. (i) IS (ii) 2p (iii) 4d (iv) 3P ½ + ½ + ½ + ½

Q. 9. Correct proof of relation 2

Q. 10. (i) Isoelectronic species 1

(ii) correct order 1

Q. 11. Moles of NaOH = 4/40 = 0.1 mole

Volume of solution = 250 ml = 0.250 L

Molarity = moles of solute / volume of solution = 0.1/0.250 = .4

1

1

Q. 12. (i) PQ 1

(ii) Metal excess defect due to anionic vacancies 1

Q. 13. (i) 3.13 x 1025

(ii) 13 (iii) 78.286 x 1023

1+1+1

Q. 14. (i) 6 moles (ii) 18 moles (iii) 18.066 x 1023

1+1+1

Q. 15. Correct formula + correct substitution of values

V = 812 ms-1

= 8967 x 10-10

m

1

1

1

Q. 16. Correct explanation 1+1+1

Q. 17. (i) Two correct difference 1

(ii) Correct explanation + diagram 1+1

Q. 18. Correct answer

Or

Correct formula + correct substitution of values + correct answer – 2.5 bar

1+1+1

Q. 19. (i) Correct point of distinction 1+1+1

(ii) Correct answer

(iii) Correct statement

Q. 20. (i) C6H6 + 15/2 O2 (g) 6CO2 (g) + 3H2O, H = -3267.7 KJ mol-1

1

(ii) C (S) + O2 (g) CO2 (g), H = -393.3 KJ mol-1

1

(iii) H2 (g) + ½ O2 (g) H2O (g), H = -285.8 KJ mol-1

Correct thermochemical equation

Correct answer J H = - 48.51 KJ mol-1

1

1

1

Q. 21. Correct Principle

(i) Towards right (ii) Towards left

1

1

159

Q. 22. Correct balanced equation 1+1+1

Q. 23. (i) NH3 + Explanation

(ii) Correct structure

1+1

2

Q. 24. (i) Correct configuration 1

(ii) Unbinilium, Ubn ½ + ½

(iii) correct difference 1

Q. 25. (i) correct configuration + no of unpaired electron 6

(ii) correct drawing

(iii) correct statement

(iv) n2 = 42 = 16 orbitals

Or

(i) E5 = −2.18 𝑥 10−8

52 = 8.72 x 10-20

J

(ii) r5 = 0.529 x 52 = 13.225 Aᵒ

(iii) no of emission lines = 𝑛 (𝑛+1)

2 =15

1+1

1

1

1

1½ + 1½

2

Q. 26. (i) U = 9 + w = 701 – 394 = 307J

(ii), (iii) correct answer

(iv) Correct statement

(v) Correct answer

Or

(i) Correct statement

(ii) Correct statement

(iii) a) decreases b) increases

(iv) Any two correct examples

1

1+1

1

1

1

1

2

1

Q. 27. (i) Lewis acid BF3, H+, 𝑁𝐻4+

Lewis base H2O

(ii) Correct answer

(iii) Correct answer

(iv) Correct answer

Or

(i) Correct answer

(ii) Kp = 41

(iii) PI = 0.4 x 105 Pa

PI2 = 0.6 X 105 Pa

Correct expression of Kp

Correct answer = 2.67 x 104 Pa

½ + ½ + ½

½

1+1+1

1

1

½

½

1

1

160

SAMPLE QUESTION PAPER HALF YEARLY EXAMINATION(Gr-H)

CLASS – XI MAX. MARKS - 70 SUBJECT – CHEMISTRY TIME – 3 HOURS

General Instructions:

(i) All questions are compulsory.

(ii) Questions number 1 to 5 are very short answer questions and carry 1 mark each.

(iii) Questions 6 to 12are short answer questions and carry 2 marks each.

(iv) Question number 13 to 24are also short-answer questions and carry 3 marks each.

(v) Question number 25 to 27are long-answer questions and carry 5 marks each.

1. What is Stark effect? 2. Write the type of hybridization involved in C atoms CH4, C2H4, and C2H2. 3. In terms of Charle’s law explain why -2730C is the lowest possible temperature. 4. A + B AB K = 1 x 102 E + F EF K = 1 x 10-3 Out of AB and EF, which one is more stable? 5. In a binary compound of two non-metals, the positive oxidation state is assigned to which one? 6. How many molecules of water of hydration are present in 39.2 mg of Mohr’s salt? *Molecular Formula:FeSO4. (NH4)2SO4.6H2O ] 7. Write the atomic number of the element present in the third period and seventh group of the periodic table. 8. Among the elements of the third period Na to Ar pick out the element: (a) with the highest first ionization enthalpy. (b) with largest atomic radius. (c) that is most reactive non-metal. (d) that is most reactive metal. 9. Prove that the change in enthalpy is equal to the heat exchanged between the system and the surrounding at constant T and P.

OR, Calculate the number of KJ of heat necessary to raise the temperature of 60.0g of Aluminium from 350C to 550C. Molar heat capacity of Al is 24 J.mol-1.K-1

10. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

11.What is disproportionation reaction? Give an example. 12. Balance the following redox reaction by ion- electron method: MnO4-(aq.) + I-(aq.) MnO2(s) +I2(s) [basic medium] 13. 500cc of 0.250 M Na2SO4 solution added to an aqueous solution of 15.00g of BaCl2. How many moles and how many grams of BaSO4 are formed? (Mol. Wt. of BaSO4 = 233, and BaCl2 = 208)

OR,

161

4 g Copper chloride on analysis was found to contain 1.890g of Cu and 2.110g of Cl. What is the empirical formula of copper chloride? (At. Mass of Cu = 63.5u, Cl = 35.5u) 14.i) An atomic orbital has n = 3. What are the possible values of l and m? ii) List the quantum numbers of an electron in valence shell of K(At. No.= 19 u) iii) Which of the following orbitals are not possible? 1s, 2p, 1p, 3f 15. Explain the following: i) Electronegativity of elements increase on moving from left to right in the periodic table. ii) Ionisation enthalpy decrease in a group from top to bottom. 16.i) Use molecular orbital theory to explain why the Be2 molecule does not exist? ii) What type of hybridization are involved with central atom of a molecule having following shapes:

a) Trigonal planar, b) a regular tetrahedral, c) liner OR,

i) Give hybridization and shape of a) XeF4 b) BrF3 c) SF6 17. What names are given to the following ideal gas relationships? a) Volume and moles at constant T and p. b) Pressure of non-reacting gases in mixture of constant T and V. c) Volume and temperature in Kelvin at constant p and n. 18. Calculate the total pressure in 10L cylinder which contains 0.4g of Helium, 1.6g of oxygen and 1.4g of nitrogen at 270C. Also calculate the partial pressure of He gas in the cylinder. Assume ideal behavior of gases. (R = 0.082 L.atm. K-1.mol-1)

19. Prove G = -TStotal 20. Give two practical implications of Boyle’s law. 21. The ionization constant of HF, HCOOH and HCN at 298K are 6.8 X 10-4, 1.8 X 10-4, 4.8 X 10-9 respectively. Calculate the ionization constants of the corresponding conjugate base. 22. Describe the effect of : a) Addition of H2 b) Addition of CH3OH c) Removal of CO on the equilibrium of the reaction:

2H2 + CO CH3OH 23. Calculate the oxidation number of underlined atom in the following species: i) [Fe(CN)6]3- ii) C6H12O6 iii) KBrO4 24. A solution has the capacity to change its pH when a small amount of strong base is added to it. These solutions maybe acidic or basic or neutral in nature. i) Name the solution and give a general example. ii) Select the components to prepare this type of solution having pH between 4 and 5. 25. i) Write the outer electronic configuration of Cr atom. Why are half-filled orbitals more stable? ii) State Heisenberg’s uncertainty principle. An electron has a velocity of 50 ms-1 accurate upto 99.99%. Calculate the uncertainty in locating its position.(Mass of electron = 9.1 X 10-31 kg; h = 6.6 X 10-

34Js) OR,

162

Calculate (a) the de Broglie wavelength of an electron moving with a velocity of 5.0x 105 ms–1 and (b) relative de Broglie wavelength of an atom of hydrogen and atom of oxygen moving with the same velocity (h = 6.63 x 10–34 kg m2 s–1) 26. i) Among NH3 and NF3 which has greater value dipole moment and why? ii) Define lattice energy. How is Lattice energy influenced by (a) charge on the ions (b) size of the ions?

OR, a) Distinguish between a sigma and a pi bond. b) Give two compounds showing bond pairs and lone pairs.

27. Calculate the enthalpy of formation of anhydrous aluminium chloride, Al2Cl6 from the following data.

i) 2Al(s) + 6HCl(aq) Al2Cl6(aq.) + 3H2 (g) H = -1004.0 kJ

ii) H2(g) + Cl2(g) 2HCl(g) H = - 183.9 kJ

iii) HCl(g) + (aq.) HCl(aq.) H = -73.2 kJ

iv) Al2Cl6(s)+ (aq.) Al2Cl6(aq.) H = -643 kJ OR,

i) Comment on following statements: a) The entropy of a substance increases in going from liquid to the vapour state at any temperature.

b) Reaction with G0< 0 always have an equilibrium constant >1. ii) Under what conditions will the reaction occur, if

a) bothH and S are positive

b) bothH and S are negative

163

BLUEPRINT FOR SAMPLE QUESTIONPAPER

CLASS: XI MAX. MARKS: 70 SUBJECT: CHEMISTRY(THEORY) TIME: 3 Hours

UNIT NO.

TITLE VSA (1 mark)

SA I (2 marks)

SA II (3 marks)

LA (5

marks)

TOTAL

I Some Basic Concepts Of Chemistry

----- 2(1) 6(2) ----- 08

II Structure Of Atom

1(1) ----- 3 (1) 5 (1) 09

III Classification Of Elements and Periodicity in Properties

------ 4(2) 3(1) ----- 07

IV Chemical Bonding and Molecular Structure

1(1) ----- 3(1) 5(1) 09

V States of Matter: Gases and Liquids

1(1) ----- 9(3) ----- 10

VI Thermodynamics

----- 2(1) 3(1) 5(1) 10

VII Equilibrium 1(1) 2(1) 6(2) ----- 09

VIII Redox Reactions 1(1) 4(2) 3(1) ----- 08

TOTAL

5(5) 14(7) 36(12) 15(3) 70(27)

NOTE: 1) Number of questions is indicated inside the brackets and marks are indicated outside the brackets. 2) Internal Choice: There is no overall choice in the paper. However, there is internal choice in one question of 2 marks weightage, one question of 3 marks weightage and all the questions of 5 marks weightage.

164

Marking SchemeOfThe Sample Paper

Class- XI Subject- Chemistry

Sl. No. Answers Marks

1 Splitting of spectral lines under electric field in H spectra. 1

2 sp3, sp2, sp 1

3 Volume of the gas becomes zero. It ceases to exist. 1

4 AB 1

5 To the one with lower electronegativity. 1

6 392 g of FeSO4. (NH4)2SO4. 6H2O contain 6 X 6.023 X1023 molecules of cryst. water Hence 39.2 X 10-3 contains (6 X 6.023 X1023 X 39.2 X 10-3 /392) = 3.61 X 1020 molecules of cryst. water

2

7 Atomic no. = 17(explanation) 2

8 i)Ar,ii)Na,iii)Cl,iv) Na 2

9 H = U + PV (at const. P),

Applying 1st law of thermodynamics and w = - PV; H = qp

OR, q = n X C X DT = (60 mol/ 27) X (24 J.mol-1.K-1) X (328 - 308)K = 1.07 kJ

2

2

10 [Pure iiquid or Pure solid]=(density/mol. mass) This quantity is constant, thus not included in equilibrium constant.

2

11 Those reactions in which single reactant undergoes oxidation as well as reduction are called disproportionation reactions. -1 -2 0 Ex. 2H2O2 2H2O + O2

2

12 Balanced equation will be,2MnO4-(aq.) + 6I-(aq.) + 4H2O 2 MnO2(s) +3I2(s) + 8OH-

13 Na2SO4 + BaCl2BaSO4+ 2NaCl Given, no. of moles of Na2SO4= 0.125 (calculated) no. of moles ofBaCl2 = 0.072 (calculated) From the reaction, BaSO4formed = 0.072 moles and (0.072 X 233)g = 16.776g

OR, % of Cu = (1.890 X 100/4) = 47.25 % of Cl = (2.11 X 100/4) = 52.75 By calculation of number of moles and simplest ratio, Empirical formula = CuCl2

3

3

14 i) For n=1, l = 0, m = 0, s = +1/2 or – 1/2 ii)n = 3, l = 2, m = -2,-1,0,+1,+2 iii) 1p, 3f

1 1 1

15 i) Decrease in atomic size increase nuclear charge. ii) Increase in atomic size.

11/2 11/2

16 i)Bond order = 0(after calculation) Hence, Be2 does not exist.

2

165

ii) a) sp2 b) sp3c) sp

OR, a) sp3d2, Octahedral or Sq.planar. b) sp3d, T shaped. c) sp3d2, Octahedral

1

1 1 1

17 a) Avogadro’s Law b) Dalton’s Law c) Charle’s Law

1 1 1

18 First calculation of no. Of moles of each gas. P(total) = n(total) X RT/V = 0.492 atm Partial pressure of He = mole fraction [n(He)/n(total)] X P(total) = 0.246 atm

3

19 Stotal = Ssystem + Ssurrounding

Ssurrounding = - H/T (at constant P)

G = -TStotal (Detailed derivation needed)

3

20 a) Altitude sickness. b) Compression of gas.

11/2 11/2

21 For F-, Kb = Kw/Ka= 1.5 X 10-11

HCOO-, Kb = 5.6 x 10-11 CN-, Kb = 2.08 X 10-6

3

22 a) The equilibrium will shift to the forward direction. b) The equilibrium will shift to the backward direction. c) The equilibrium will shift to the backward direction.

3

23 i) +3, ii) 0, iii) +7 1

24 i) Buffer solution, ex.- Acetic acid-Sodium acetate or any other right example. ii) Acetic acid and Sodium acetate.(pH= 4.74)

2 2

25 i) 3d104s1, Explanation with symmetry and exchange energy.

ii) v = 50 X [(100-99.99%)] = 5 X10-3 m.s-1

x = h/4mv= 1.154 X 10-2 m OR,

(b) λ = h/mv = 6.63 x 10-34 kgm2s-2/ (9.11 x 10-31kg) ( 5.0 x 105ms-1) Wavelength λ = 1.46 x10–9m (b) An atom of oxygen has approximately 16 times the mass of an atom of hydrogen. In the formula λ = h/mv, h is constant while the conditions of problem make v, also constant. This means that λ and m are variables and λ varies inversely with m. Therefore, λ for the hydrogen atom would be 16 times greater than λ for oxygen atom.

2

3

2

3

26 i) NH3, greater electronegativity of F than H ii) Definition of lattice energy. a) Greater lattice energy. b) Greater lattice energy.

OR, a)Three suitable differences of Sigma and Pi bonds. b) Any two suitable examples

2 1 1 1

3 2

27 By adding, subtracting, multiplying the given equations in a suitable way, following equation is obtained:

2Al(s) + 3Cl2(g) Al2Cl6(s) and the 1351.9kJ

5

166

OR, i) a)At vapour state have greater freedom of motion, thus greater entropy.

b)- G0= RTlnK; hence, if G0<0, K>1

ii) Explaination with G = H -TS a) Temperature should be high b) Temperature should be low

2

3

Trayee Ray PGT Chemistry(Cont.) KV, Ranaghat

167


SESSION ENDING EXAMINATION (CLASS XI) SESSION 2018-19

SUBJECT- CHEMISTRY

BLUE PRINT OF QUESTION PAPER

SL NO NAME OF THE CHAPTER MARKS ALLOTED

MARKS DISTRIBUTED

1 Some basic concepts of chemistry

08 1+3

2 Structure of atom

1+3

3 Classification of elements and periodicity

04 1+3

4 Chemical Bonding

20 2+3

5 States of matter

2+3

6 Chemical Thermodynamics

2+3

7 Equilibrium

2+3

8 Redox reaction

20 1+2

9 Hydrogen

3

10 S block

3+5

11 P Block

3+3

12 Organic chemistry

18 2+5

13 Hydrocarbons

3+5

14 Environmental chemistry

1+2

TOTAL MARKS 70

168


SAMPLE SESSION ENDING EXAMINATION QUESTION PAPER (2018-19)

CLASS- XI SUBJECT- CHEMISTRY TOTAL MARKS 70 TIME -3 HRS

General Instructions

1. All questions are compulsory.

2. Question nos. 1 to 5 are very short answer questions and carry 1 mark each.

3. Question nos. 6 to 12are short answer questions and carry 2 marks each.

4. Question nos. 13 to 24are also short answer questions and carry 3 marks each

5. Question nos. 25 to 27are long answer questions and carry 5 marks each

6. Use log tables if necessary, use of calculators is not allowed.

Q1. Write the electronic configuration of Cu+2 . (z=29) (1)

Q2.write the name and symbol of the element with z=107. (1)

Q3.What is the oxidation number of Mn in K2MnO4. (1)

Q4.what is meant by Eutrophication. (1)

Q5. State Law of Multiple Proportion with eg. (1)

Q6. Draw the shape and predict the geometry of ClF3 and SF4 according to VSEPR theory (2)

OR

What is the hybridization of Pcl5 and why are axial bond longer than equatorial bond. Explain the shape with diagram.

Q7.Give reason for the following-a)Water boils at lower temperature at high altitude. (2)

b)Alkali halides give color to the flame.

Q8.Balance the following ionic equation by ion electron method in acidic medium- (2)

C2O42- + MnO4

- -------Mn2+ + CO2

Q9.Calculate the ΔH rxn for C3H8(g) +5O2(g)------3CO2(g) + 4H2O (g).Given the average (2) bond enthalpies of C-C, C-H, C=O, O=O, O-H are 347,414, 741, 498, 464 Kj/mol respectively.

Q10.a)Write the conjugate base of H2PO4 – and H3O +. (1)

169

b)StateLechatliers principle and hence predict the direction of reaction when temperature is increased for N2(g) + H2(g) => 2NH3 (g) + Energy (1)

Q11.Write the IUPAC name of a) CH2=CH-C CH (2)

b)

Q12.Define Green house effect and mention the green house gases. (2)

Q13.a)out of 0.5M and 0.5m, which is more concentrated and why? (1+2)

b)Calculate the concentration of HNO3 in moles/lt in a sample which has a density of 1.41g/ml and mass%of nitric acid is 69%.(molar mass HNO3 is 63g/mol)

Q14 a).Write 2 limitations of Bohrs Model of an atom. (2+1)

b)The 3d subshell of an atom has 8 electron.What is the maximum number of electron

having spin in the same direction.

Q15.Give reason- i) a) Electron gain enthalpy of S is more negative than that of oxygen. (3)

b) Anions are bigger in size than parent atom.

ii) Assign proper group and period number to the element with Z=23.

Q16.a) Distinguish between sigma bond and pi bond. (2+1)

b)All the bond length of C-O in CO32- are equal in length.Explain.

Q17.a) Why do gases deviate from ideal behavior.And write real gas equation for

n moles of gas. (3)

b) What is the type of semiconductor obtained when Si is doped with P.

Q18.For the reaction at 298K, 2A+BC, ΔH= 400kj/mol and ΔS =0.2J/k/mole. At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over temperature? (3)

Q19.a) at 473k,

theKcfordecompositionofPCl5is8.3x103.AndH=124kj/mol.WhatisthevalueofKcforreversereactionatthesametemperature. (3)

b)What would be the effect of Kci) if more PCl5 is added. Ii) if temperature is increased. Justify your answer.

OR

CN

170

a)Define solubility product and mention one application. (1)

b)The concentration of H+ in a sample of soft drink is 3.8x10-3M.What is the pH. (2)

Q20.i)Justify H2O2 is a good oxidising agent. (3)

ii)Distinguish between hydrolysis and hydration.

iii)By means of chemical equation show how temporary hardness of water can be removed .

Q21 i).Arrange according to increase in ionic character- NaH, LiH, CsH (3)

ii)Explain to why a solution of Na2CO3 is alkaline in nature.

iii)Li is strongest reducing agent though ionisation energy is high.Give reason.

Q22.Compare the properties of alkali and alkaline earth metal withrespect to (3)

i)Basicity of oxides ii)Thermal stability of carbonbates iii)Solubility of hydroxides.

Q23.Complete and balance the equation- (3)

i)BF3 + LiH--

II)SiO2 + NaOH----

iii)NH4NO3 (on heat)----

Q24.a)Distinguish between Ethyne and Ethene by chemical test. (3)

b)Propanal and pentan-3-one are the ozonolysis products of an alkene.What is the

structural formula and name of the alkene.

Q25.a)DefineHyperconjugation.Why(CH3)3C+ is more stable than CH3+. (2+3)

b)In Carius estimation of sulphur 0.468g of S gave 0.668g of BaSO4 .Find out the percentage of S in the compound.(at mass of Ba=137)

OR

a)Write a chemical reaction involved in Lassaignestestof organic compound having nitrogen.(2+3)

b)Define isomerism. Write the structure and IUPAC name ofall the isomers of C3H8O.

Q26.a)Write the chemical equation for following- i)Friedal craft alkylation ii)-elimination.(2+3)

171

b)Convert - i)Benzene to para nitro toluene

ii)Ethyne to benzene

iii)EthylChloride to ethane.

OR

a)Explain - i)AntiMarkownikoff Rule ii)Kolbes Electrolysis

b)Convert- i)Methyl bromide to ethane

ii)propene to 1-bromo propane

iii)Benzene to Toluene.

Q27.Explain the following-a)BeO is almost insoluble but BeSO4 is soluble in water (5)

b)Alkali metal ions acquire blue color in aqueous solutions.

c)Be and Mg do not respond to flame test.

d)Table salt gets wet in the rainy season.

e)Alkali and alkaline earth metals cannot be obtained by reduction process.

Or

I)Draw the structure of BeCl2 in the solid state.

2)K2CO3 cannot be prepared by Solvaysprocess.Explain.

3)K,Rb,Cs forms superoxides in preference to oxides and peroxides. Justify.

4)Arrange the mobilities of alkali metal ions in aqueous solutions-

Cs+, Na+, K+, Li+,Rb+

5)Mention one reason why Li show diagonal relationship with Mg.

172

MARKING SCHEME FOR CLASS XI SESSSION ENDING EXAMINATION (2018-19)

SUBJECT- CHEMISTRY

A1. [Ar]3d9 (1)

A2.Uns,Unnilseptium (1/2 +1/2)

A3. Mn= +6 (1)

A4.correct statement. (1)

A5.correct definition. (1)

A6. ClF3- T shape,&geometry-distortedtrigonalbipyramidal(with lone pair)

SF4- see saw shape & geometry- distorted trigonalbipyramidal (1/2 x4=2)

Or

Pcl5- sp3d,because of the repulsion between axial and equatorial bonds,diagram (1+1)

A7.a)high altitude low pressure, b)unpaired electron absorbs energy to a higher excited level and returns back with emission of light in visible range. (1+1)

A8.oxidation half rxn, reduction half rxn, balance electron number, overall rxn (1/2 x4=2)

A9.correct formula, substitution,correct answer & unit [-1662kj/mol] (1/2 x4=2)

A10.a) HPO42- ,H2O

b)correct statement, backward (1+1)

A11.but-1-en-3-yne , 3-methyl pentanitrile (1+1)

A12. Correct definition ,two gases (1+1)

A13.a)0.5 M as no. of moles in per lt of solution (1+2)

b)no. of moles=69/63= 1.095mol, volume of 100g nitric acid=100/1.41=70.92g/ml

concentration of nitric acid= 1.095/70.92x10 -3 = 15.44M

A14.a) any two, b) five (2+1)

A15.a) S has big size 3p orbital so less repulsion between incoming electron and existing electron than small size 2p orbital of oxygen. (1+1+1)

b)effective nuclear charge on the valence shell electron decreases.

173

c)4thpd and 5th group

A16.a) correct difference (2 points) (2+1)

b)resonance structures and stabilization

A17.a)attraction and volume correction factor, correct equation for n moles (1+1+1)

b)n type

A18.ΔG=ΔH-TΔS, correct calculation, change of units and substitution, correct answer (3)

A19.Kc for reverse reaction=1/8.3x10 -3 =120.5 (1+1+1)

b)Kc remains unchangedas it is independent on initial conco f reactant

ii)increase with increase in temperature.

Or

a) correct definition, one application (1+2)

b)pH= - log[H+], =-log3.8+3 =0.5798 +3=2.42

A20.a) correct eg of any equation (1+1+1)

b)correct reaction

c)correct definition with eg

A21)a)LiH<NaH<CsH (1+1+1)

b)Na2CO3 +H2O--NaOH

c)high hydration enthalpy and high negative electrode potential

A22. Correct trend of basicity of oxides, thermal stability & solubility between gpI and gpII (3)

A23. A) B2H6 + LiF B)Na2SiO3 +H2O C)N2O +H2O (all balanced) (1+1+1)

A24.a) one distinguish answer (1+2)

b) correct structure, and name (3-ethyl hex-3-ene)

A25. A) correctdefn, more hyperconjugation structures than CH3 + with correct expln. (1+1)

b)correct formula, substitution ,calculation

% of S = (32x0.668x 100)/(233 x0.468 x233) = correct answer. (1+1+1)

Or

174

a) correctequn. With sodium cyanide and iron II sulphate and formation of complex. (1+1)

b) correct definition, two isomers with name and structure (1+1+1)

A26.a) correct equation (1+1)

b)i) first nitration then friedal craft alkylation (1+1+1)

ii)cyclic polymerization with red hot iron

iii) elimination with alc KOH, and further reduction with H2

OR

a) correct equations (1+1)

b)i)wrutz reaction ii)anti markownikoffaddn iii) friedal craft alkylation (1+1+1)

A27.a) Hydration enthalpy more for BaSO4 (1x5=5)

b)ammoniated electron

c)paired electron in valence shell

d)CaCl2,MgCl2 as they absorb moisture

e)they are very reactive

or

a)correct structure. (1x5=5)

b)KHCO3 is too soluble to be precipitated.

c)O2– is stable with large cations.

d)Cs>Rb>K>Na>Li ions

e)similar sizes (charge to size ratio similar)

BY. S CHATTERJEE (KV1 SL)

175

BLUEPRINT OF QUESTION PAPER

Class 11

Sl. No.

Unit VSA (1 MARKS)

SA I (2 MARKS)

SA II (3 MARKS)

LA (5 MARKS)

TOTAL NUMBER

1 SOME BASIC CONCEPTS OF CHEMISTRY

1(A) 03

2 STRUCTURE OF ATOM 1(U) 1(E) 05

3 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

1(K) 1(U) 04

4 CHEMICAL BONDING AND MOLECULAR STRUCTURE

1(E) 1(A) 06

5 STATES OF M,ATTER 1(H) 1(A) 04

6 CHEMICAL THERMODYNAMICS 1(A) 1(U) 05

7 EQUILIBRIUM 1(U) 1(H) 05

8 REDOX REACTIONS 2(A+E) 04

9 HYDROGEN 1(K) 03

10 s BLOCK ELEMENTS 1(E) 1(A) 05

11 p BLOCK ELEMENTS 1(H) 1(K) 1(U) 08

12 ORGANIC CHEMISTRY: SOME BASIC PRINCIPLES AND TECHNIQUES

2(A+E) 06

13 HYDROCARBON 1(U) 1(H) 08

14 ENVIRONMENTAL CHEMISTRY 1(K) 1(U) 04

176

Sample Question Paper

SUB: CHEMISTRY

CLASS - XI

TIME: 3 HOURS M.M: 70

GENERAL INSTRUCTIONS:-

All questions are compulsory, however internal choice is given in some questions.

Questions NOs 1 to 5 carry ONE MARK each.

Question NOs 6 to 12 carry TWO MARKS each.

Question NOs 13 to 24 carry THREE MARKS each.

Question NOs 25 to 27 carry FIVE MARKS each.

Q1 Choose the correct answer. A thermodynamic state function is a quantity

(i) used to determine heat changes (ii)whose value is independent of path (iii) used to determine pressure

volume work (iv) whose value depends on temperature only.

Q2. What are isoelectronic species? write isoelectronic of O2-2

Q3. Write a balanced chemical equation for the reaction whose KC expression is as:

KC = [ CH3OH ]

[CO] [H2]2

Q4. Calculate the oxidation number of:

a) Cr in CrO4-2

b) C in C12H22O11

Q5. Why hydrogen peroxide is stored in wax lined dark colour bottle.

Write the equation involved?

Q6. What is Kc for the following equilibrium when the equilibrium concentration of each

substance is: [SO2]= 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M ?

SO2 + O2 2 SO3 Q7.

What is green chemistry? write down two advantage of it.

Q8. Why is the first ionization energy of Mg greater than that of both Na and Al taken

separately?

177

OR

Why is the first electron gain enthalpy of oxygen negative, while second one is

positive?

Q9. Explain how Charles‘ law helps us in finding out the absolute zero value.

Q10. Which gases are responsible for ozone layer depletion? write reactions involve in it.

Q11. At 5000C the equilibrium constant KC for the reaction:

N2 + 3H2 ↔ 2NH3 is 6.02x 10-2

litre2mol

-2 .What is Kp at this temperature?

Q12. State and Explain Law of Multiple Proportions with the help of an example.

Q13. Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and density is 10.5 g

cm−3, calculate the atomic mass of silver.

Q14 3 grams of H2 react with 29 grams of O2 to yield water

a). Which is the limiting reagent?

b). Calculate the maximum amount of water that can be formed?

c). Calculate the amount of one of the reactants which remains un reacted?

Q15. Calculate the wave length of radiation emitted when an electron in a hydrogen

atom makes a transition from an energy level with n=3 to n=2.

Q16. which of the following has larger value of the property indicated:

a) F or Cl (electron affinity )

b) F or F- (electro negativity)

c) Na or K (electro positivity)

Q17 a) Explain the principle of paper chromatography.

b) What is chemistry involve in Lassaigne'stest.

Q18. Based on VSEPR theory predict the geometry and shape of:

a) ClF3

b) NH4+

Draw the figure also.

OR

Based on molecular orbital theory compare the stability and magnetic property of

O2+& O2

- .

Q19. What will be the pressure exerted by a mixture of3.2 g of CH4 and 4.4 g of CO2

178

contained in a 9 dm3 flask at 270C.

Q20. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide

contained in a 9 dm3 flask at 27 °C ?.

Q21. For the reaction 2 Cl(g) → Cl2 (g) , what are the signs of ΔH and ΔS. Give

suitable reasons for your answer.

Q22. Balance the following ionic equation by ion electron method in acidic medium:

Cr2O72-

+ Fe2+

→ Fe3+

+Cr3+

Q23. Differentiate clearly giving examples:

a) Hydrolysis and hydration

b) Ionic hydrides and covalent hydrides

Q24 a) write structure of organic compound by IUPAC name :- 2, 3–dimethyl butanal b)

draw geometric isomers of cis and trans but-2-ene.

c) why t-butyl carbocation is more stable than isopropyl carbocation?

Q25. (i) a) Draw the shape of px orbital.

b) Write the orbitals represented by the following quantum numbers

( n = 2; l = 3 ) ( n = 5; l = 2 )

c) State the limitations of Bohr‘s model of an atom.

OR

(ii) a) Draw the shape of a dxy orbital.

b) Write the values of all the quantum numbers of the valence electron of sodium atom.

c) State Pauli‘s Exclusion Principle.

d) Distinguish between orbit and orbital. (Any 4 points)

Q26. (i) give reason:-

a) boric acid considered as a weak acid.

b) diborane is electron deficient compound.

c) [SiF6]-2

is known where as [SiCl6]-2

not.

(ii) write a reaction for preparation of silicon and also write their uses.

or

(i) arrange following molecules and give reason :-

179

a) N2O, NO, N2O3, N2O4, N2O5 (decreasing order of acidic nature)

b) NH3, PH3, AsH3, SbH3, BiH3 (increasing order of basic nature)

(ii) complete following reactions :-

a) 4 CH3COOH + P4O10

1.453-473 K, 220 atm, 2. HEAT

b) NH3 + CO2

Q 27) (i) explain following reactions :-

a) Friedel craft acylation

b) Wurtz Reaction

(ii) Draw resonating structures of phenol.

OR

(i) (a) Addition of HBr to propene yields 2-bromopropane while in the presence of peroxide, the

same reaction gives 1-bromopropane. explain and give mechanism.

(b) Draw the resonating structures of nitrobenzene.

180

MARKING SCHEME

Class - XI Science

Subject - Chemistry

Q.NO. ANSWER MARKS

1 A thermodynamic state function is a quantity whose value is independent

of a path.

Functions like p, V, T etc. depend only on the state of a system and not on

the path.

Hence, alternative (ii) is correct.

1 M

2 Molecules, atoms or ions which contain same no. of electrons are called as

isoelectronic species.

F2, Ca +2

½ M

½ M

3 CO(g)+2H2(g)↔ CH3OH 1M

4 6,0 ½+ ½ M

5 It decomposes to give H2O & O2 in presence of light, May result in

explosion if it contact with organic compounds

H2O2→ H2 O + 1/2O2

½+ ½ M

6 The equilibrium constant (Kc) for the give reaction is:

Hence, K for the equilibrium is 12.239 M–1.

1M

1M

7 Green chemistry, also called sustainable chemistry, is an area of

chemistry and chemical .... In addition to the human health benefits of

eliminating trans-fats, the process has reduced the use of toxic chemicals.

Benefits of Green Chemistry. Human health: Cleaner air: Less release of

hazardous chemicals to air leading to less damage to lungs. ... Increased

safety for workers in the chemical industry; less use of toxic materials;

less personal protective equipment required; less potential for accidents

(e.g., fires or explosions)

1M

1M

181

8 Na→3s1

Mg→3s2

Al→3s 23p

1

Mg has its valence electrons paired, requires more energy than Al 1M

Na nuclear charge is lesser & atomic size is greater 1M

OR

Repulsion due to –ve charge on O- requires energy to overcome the

repulsion. So electron gain enthalpy is +ve Relevant explanation

1M

1M

2M

9 by extrapolating the CHARLE‘S law graph

Graph:, EXPLANATION:

1M

1M

10 Once released CFCs mix with atmospheric gases and reach the

stratosphere, where they

are decomposed by UV radiations.

The chlorine free radical produced in reaction (iii) reacts with ozone as:

The radicals further react with atomic oxygen to produce more chlorine

radicals as:

(v) The regeneration of causes a continuous breakdown of ozone present

in the

stratosphere, damaging the ozone layer.

1M

½ M

½M

11 KP = KC (RT) Δn

Δn = -2

KP=6.02X10-2

(0.082X773)-2

=1.5X10-5atm-2

½M

½ M

1 M

12 When two elements combine to form more than one compound, the mass

of one element, which combine with a fixed mass of the other element will

always be ratio of whole number. eg. CO and CO2

1 M

1M

13 It is given that the edge length, a = 4.077 × 10−8 cm

Density, d = 10.5 g cm−3

As the lattice is fcc type, the number of atoms per unit cell, z = 4

We also know that, NA = 6.022 × 1023 mol−1 Using

the relation:

½ M

½ M

½ M

182

= 107.13 gmol−1

Therefore, atomic mass of silver = 107.13 u

½ M

½ M

½ M

14 2H2+O2→ 2H2O

Moles of hydrogen=1.49

E per atom=180.2/6.023x1023=3.03x10-19J/atom

Moles of oxygen=0.906

1.49moles of hydrogen reacts with1.49/2=.745moles of oxygen

H2 is the limiting reagent

Amount of water formed =1.49x18=26.82g

Amount of O2 un reacted=0.906-0.745=0.161g

1 M

2 M

15 ΔE=-1312(1/9-1/4)=180.2KJ/mol

λ=hc/E

6.6x10-34x3x108/3.03x10

-19=6.56x10

-7 m=656nm

1 M

1 M

1 M

16 a)Cl b)F C)K 1 X 3 = 3M

17 A) Chromatography Principle: It is based on the difference in

movement of individual components of a mixture

through the stationary phase under the influence of mobile phase.

For example, a mixture of red and blue ink can be separated by

chromatography. A drop of the mixture is placed on the chromatogram.

The component of the ink, which is less

adsorbed on the chromatogram, moves with the mobile phase while the

less adsorbed component remains almost stationary. B)

CHEMSITRY involves in Lassaigne's test.

1 x 3 =

3 M

183

18 a) AB3L2 , T shape diagram

b) AB4 , Tetrahedral diagram

½ + 1 M

½ + 1 M

19 No of moles of methane= 3.2/16=0.2

No of moles of CO2 = 4.4/44=0.11 M

P= (nCH4+nCO2) RT/V= (0.2+0.1)X0.0821X300/9=0.82 atm

1 M

1 M

1M

20

1M

1M

1M

21 Both –ve;

bond formation releases energy (ΔH –ve),bond formation decreases

randomness (ΔS-ve)

½ + ½

1 M + 1 M

22 Assigning correct oxidation state and splitting reaction into 2 correct half

part

Each correct step and final balanced reaction

1 M

1 +1 M

23 a) a bond is broken in hydrolysis in the presence of water, where water is 1 M

184

added in hydration.

any example

b) ionic hydrides called salt like hydride but covalent hydride called

molecular hydride.

s- block elements form ionic hydrides but p- block elements form covalent

hydrides.

Ionic hydrides are powerful reducing agent but covalent hydrides are weak

reducing agent. or any other

Any other

½ M

1 M

½ M

24

a)

b)

c) due to more no. of alpha hydrogen (hyperconjugation)

1M

1M

1 M

25 a) correct shape

b) Not possible, 5d

c) correct limitations(any three)

or

a) Correct shape

b) 3s1 : n=3, l=0,m=0, s= +½ or - ½. 1M

c) correct statement

1 M

½ + ½ M

3 M

1M

1M

1M

2M

185

d)correct difference(4 points)

26 a) because it is not able to release H+ ion on its own. it receives OH-

from

water to complete its octet and in turn release H+

b) B2H6 is an electron-deficient compound. B2H6 has only 12 electrons –

6 e– from 6 H atomsand 3 e– each from 2 B atoms. Thus, after combining

with 3 H atoms, none of the boronatoms has any electrons left.

C) steric hindrance of chloride ion, only small electronegative atom

fluorine can occupy 6 place around Si.

(ii) When silicon reacts with methyl chloride in the presence of copper

(catalyst) and at atemperature of about 537 K, a class of organosilicon

polymers called methylsubstitutedchlorosilanes (MeSiCl3, Me2SiCl2,

Me3SiCl, and Me4Si) are formed.

Any 2 uses

Or

i) a) N2O5, N2O4, N2O3, NO , N2O (due to increase in o.s. of nitrogen)

b) BiH3, SbH3, AsH3, PH3, NH3 ( due to diffusion of lp on large atom)

(ii) a) 2H4P2O7 + 2C3O2

b) urea

1M

1M

1M

1M

1M

OR

1 + ½ M

1 + ½ M

1M

1M

27

(i) a)

b)

1 + ½ M

1 + ½ M

186

c)

or

a) markovnikov and antimarkovnikov's mechanism

2 M

3M

2M

STUDY SUPPORT MATERIAL SESSION 2018-19 · study support material – session 2018-19 class xi subject – chemistry prepared during 02 day workshop (3rd august 2018 & 04th august

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