CBSE Class 12 Chemistry Previous Year Question Paper 2012

Class XII Chemistry www.vedantu.com 1

CBSE Class 12

Chemistry

Previous Year Question Paper 2012

Series: SMA Code no. 56/1

Please check that this question paper contains 12 printed pages.

Code number given on the right-hand side of the question paper should be written on the title page of the answer-book by the candidate.

Please check that this question paper contains 30 questions.

Please write down the Serial Number of the question before attempting it.

15 minutes of time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question paper only and will not write any answer on the answer script during this period.

CHEMISTRY (Theory)

Time Allowed: 3 hours Maximum Marks: 70

General Instructions:

1. All questions are compulsory.

2. Marks for each question are indicated against it.

3. Questions number 1 to 8 are very short-answer questions and carry 1 mark each.

4. Questions number 9 to 18 are short-answer questions and carry 2 marks each.

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5. Questions number 19 to 27 are also short-answer questions and carry 3 marks each.

6. Questions number 28 to 30 are long-answer questions and carry 5 marks each.

7. Use Log Tables, if necessary. Use of calculators is not allowed.

1. How may the conductivity of an intrinsic semiconductor be increased? 1 Mark

Ans:

(i) Conductivity of intrinsic semiconductors can be increased by increasing the temperature as, when we increase the temperature, the electrons in the conduction band get enough energy to jump to the valence band. The electrons in this band can move freely in the material. This movement of electrons creates an electric current and thereby increases the conductivity of the semiconductor.

(ii) Conductivity can also be increased by adding a suitable impurity to the semiconductor. This process is known as Doping.

2. Define ‘Peptization’. 1 Mark

Ans: Peptization is the process in which the precipitate is converted into a colloidal sol by shaking it with a dispersion medium in the presence of a small amount of peptizing chemical. This procedure is used to turn a freshly generated precipitate into a colloidal sol in general. For example- using sodium carbonate as a peptizing agent the clay can form a colloid with the water.

3. How is copper extracted from a low-grade ore of it? 1 Mark

Ans: The method of leaching can be used to extract copper from low-grade ore. Leaching is the process of extracting chemicals from a solid by dissolving them in a liquid; the ore, but not the impurities, are dissolved in the liquid in this process.



4. Which is a stronger reducing agent, 3SbH or 3BiH , and why? 1 Mark

Ans: 3BiH is a stronger reducing agent than 3SbH because Bi lies below Sb in the periodic table, due to which the size of Bi is more than Sb . A good reducing agent is a substance that can donate hydrogen atoms easily. As the size of Bi is bigger the bond length between the Bi and hydrogen increases so the bond strength decreases, hence it can lose +H easily.

5. What happens when bromine attacks ≡2 2CH =CH-CH -C CH 1 Mark

Ans: Bromine is a strong reducing agent, if it is present in excess amount then the result will be bromination of double and triple bonds. The reaction will be:

2 2

2 2 2

2

CH =CH-CH -C CHCH Br-CHBr-CH -CBrCHBr

≡→→

The reddish color of bromine will disappear, else no change in color is observed, as both double and triple bonds get brominated.

If bromine is present in a lesser amount then it will be able to reduce only the double bonds and not the triple bond. The reaction will be:

2 2 2 2CH =CH-CH -C CH CH Br-CHBr-CH -C CH≡ → ≡

The reddish-brown color of bromine will not disappear as only double bonds are reduced, not the triple bonds.

6. Write the IUPAC name of the following: 1 Mark O||

3 2CH -CH -CH=CH-C-H

Ans: The IUPAC name of the compound is Pent-2-enal.



7. Write the structure of the product obtained when glucose is oxidized with nitric acid. 1 Mark

Ans: Nitric acid is a strong oxidizing agent, it converts the alcoholic and aldehydic group present in D- glucose into the carboxylic acid group and results in the formation of D-glucaric acid.

(D- glucose) (D-glucaric acid)

8. Differentiate between disinfectants and antiseptics. 1 Mark

Ans: The difference between disinfectants and antiseptics is:

Disinfectants Antiseptics

A disinfectant can be defined as an antimicrobial agent that can be applied on the surface of some objects in order to destroy the microorganisms present on them. Example- Phenol.

An antiseptic can be defined as an antimicrobial agent which can be applied to the body of living organisms to inhibit the action of microbes. Example-Dettol

9. Express the relation among cell constants, resistance of the solution in the cell, and conductivity of the solution. How is molar conductivity of a solution related to its conductivity? 2 Marks

Ans: Cell constant-G, Resistance-R, Conductivity=K, Molar conductivity= mΛ

Conductivity can be written as,



1 lK= ×R A

The relation between cell constant, conductivity, and resistance are:

1 lK= ×G G=R A

The relation between conductivity and molar conductivity is:

2 -1m

K×1000 Scm molM

=Λ

Or

The molar conductivity of a 1.5M solution of an electrolyte is found to be 138.9 2 -1Scm mol . Calculate the molar conductivity of this solution. 2 Marks

Ans: Molar conductivity, mConductivityConcentration

Λ =

mkc

∧ =

Conductivity, mk c= Λ ×

-1138.9 1.5 0.2085Scm1000

×= =

10. A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half? 2 Marks

Ans: Let the concentration of the reactant be [A] =x

Rate of reaction, 2 2R=k[A] =kx

(i) If the concentration of the reactant is doubled, i.e.



[A] =2x, then the rate of the reaction would be 2R=k[2x] =4kx=4R

Therefore, the rate of the reaction would increase by 4 times.

(ii) If the concentration of the reactant is reduced to half of the initial concentration, i.e.

1 x2

A = then the rate of the reaction would be

2 21 1 1R=k x = x = R2 4 4

Therefore, the rate of the reaction would be reduced to 14

times.

11. Which methods are usually employed for purifying the following metals: 2 Marks

(i) Nickel

Ans: Nickel is purified with the help of Mond’s process: Mond’s process also known as the carbonyl process, is used to extract and purify nickel. Carbon monoxide combines with nickel to give nickel carbonyl. The mixture of nickel carbonyl is heated at 220-250 , resulting in decomposition back to nickel and carbon monoxide.

( ) ( ) ( ) ( )4Ni s +4CO g Ni CO g→

( ) ( ) ( ) ( )4Ni CO g Ni s +4CO g→

(ii) Germanium

Ans: Germanium is purified by the Zone refining method, it is a group of similar methods of purifying crystals, in which the narrow region of a crystal is melted, the molten region melts impure solid at the forward edge, and leaves a pure material.



12. Explain the following facts giving an appropriate reason in each case: 2 Marks

(i) 3NF is an exothermic compound whereas 3NCl is not.

Ans: The size of the atom increases as we go down the group, the size of the chlorine atom is more than the size of the fluorine. The large difference in the size of N and Cl bond results in the weakness of the strength of N-Cl bond. whereas the size of N and F is small. therefore, the N-Fbond is stronger and cannot be broken easily, thus 3NF It is an exothermic compound.

(ii) All the bonds in 4SF are not equivalent.

Ans: The hybridisation of is 3sp d type. The five electrons around Sulphur in 4SFattain trigonal bi-pyramidal geometry in which one position is occupied by a lone pair. The lone pair finds a position that minimizes the number of repulsions it has with bonding electron pairs. It occupies an equatorial position with two repulsions. The axial S-F bonds are twisted away from the lone pair by a small amount. Therefore, all the bonds in 4SF are not equivalent.

13. Complete the following chemical reaction equations: 2 Marks

(i) →2- - +2 7Cr O +I +H

Ans: It is a redox reaction, in which the iodine oxidation number increases from -1 to 0 and chromium oxidation number decreases from +6 to +3. The balanced equation will be:

2- - + 3+2 7 2 2Cr O +6I +14H 2Cr +3I +7H O→



(ii) →- - +4 2MnO +NO +H

Ans: The oxidation number of Mn decreases from +7 to +4. And the oxidation number of Nitrogen increases from +3 to +5.

- - + - 2+4 2 3 22MnO +5NO +6H 5NO +2Mn +3H O→

14. Explain the mechanism of acid catalyzed hydration of alkene to form corresponding alcohol. 2 Marks

Ans: The mechanism involves electrophilic addition of the proton (or acid) to the double bond to produce a carbocation intermediate during hydration. In the second step, water is added, resulting in the creation of an oxonium ion, which is deprotonated to give the alcohol.

Step 1:

The mechanism of hydration involves the electrophilic addition of the proton to the double bond to form a carbocation intermediate.

Step 2:

Now, the addition of water in the second step results in the formation of an oxonium ion.

Step 3:

The oxonium ion gets deprotonated by adding the water molecule to it, forming alcohol.



15. Explain the following behaviors: 2 Marks

(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.

Ans: Alcohols form hydrogen bonds with water molecules due to the presence of oxygen atoms. Water is a polar solvent, while other hydrocarbons of comparable molecular masses do not form hydrogen bonds due to their non-polar nature.

(ii) Ortho-nitrophenol is more acidic than ortho-methoxy phenol.

Ans: Ortho-nitrophenol is more acidic because the nitro group is an electron-withdrawing group, the presence of a nitro group at the ortho position decreases the electron density in the O-H bond, thus releasing the proton easily.

While the methoxy group is an electron-donating group, which enhances the electron density of the O-H bond. As a result, protons are difficult to release.

Hence ortho nitrophenol is more acidic than ortho methoxy phenol.

16. Describe the following giving the relevant chemical equation in each case: 2 Marks

(i) Carbylamine reaction



Ans: Carbylamine, also known as isocyanides, are generated when aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, This test has no effect on secondary or tertiary amines

Δ2 3 2R-NH +CHCl +3KOH R-NC+3KCl+3H O→

(Primary amine) (Chloroform) (isocyanide)

(ii) Hofmann’s bromamide reaction

Ans: Hoffmann’s bromamide reaction: When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. The Hoffmann bromamide reaction is the name for this type of breakdown.

2 2 2 2 3 2R-CO-NH +Br +4NaOH R-NH +Na CO +2NaBr+H O→

(Amide) (Primary Amine)

17. Complete the following reaction equations: 2 Marks

(i) →6 5 2 3 2 2C H N Cl+H PO +H O

Ans: Benzenediazonium chloride is a very reactive compound which oxidizes hypophosphorous acid to hydrophosphoric acid and the reactant is reduced to benzene.

6 5 2 3 2 2 6 6 2 3C H N Cl+H PO +H O C H +N PO +HCl→

(ii) ( )→6 5 2 2C H NH +Br aq.

Ans: Aniline undergoes bromination to form 2, 4, 6-tribromoaniline and hydrogen bromide.

( )6 5 2 2 6 5 3C H NH +Br aq. C H NBr +3HBr→

18. What are food preservatives? Name two such substances. 2 Marks



Ans: A preservative is a substance that is added to food substances or beverages to prevent the decomposition of microbes and to avoid chemical changes taking place undesirably. Furthermore, a variety of environmental conditions, including heat and the presence of bacteria, can cause meals to change in ways that are potentially harmful.

Sodium benzoate and sodium sorbate are two examples of chemical food preservatives.

19. Copper crystalline with face centered cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal.(Atomic mass of Cu=63.55 u and Avogadro’s number 23 -1

AN =6.02×10 mol ) 3 Marks

Ans: Number of atoms in the unit cell, Z=4

Radius=127.8pm

Face centered cubic (r) 2 2

a

127.8 2 2 a× =

-10a=361.41pm=361×10 cm

The atomic mass of cell=a

Density= 3A

Z Ma N××

( ) ( )-1

-1 23 -1

4×63.55gmol=361.4110 × 6.022×10 mol

3-14= . g8 9× 0 /cm1

Or



Iron has a body centered cubic unit cell with the cell dimension of 286.65pm. Density of iron is -37.87gcm . Use this information to calculate Avogadro’s number (Atomic mass of Fe=56.0u) 3 Marks

Ans: a=286.65pm 10=286.65×10 cm

Density= -37.87cm

Atomic mass of Fe=56.0u

Z=2 (For body centered cubic unit cell)

Density3

A

Z×M=a ×N

( )( ) ( )

-1

A -10 3 -3

2× 56.0 gmolN =

286.65×10 cm × 7.874gcm

23 1AN 6.022 10 mol−= ×

20. The electrical resistance of a column of 0.05M NaOH solution of diameter 1 cm and length 50 cm is 35.5×10 ohm . Calculate its resistivity, conductivity and molar conductivity. 3 Marks

Ans: Resistance= 35.5×10 ohm

Length=50 cm

Concentration= 0.05M

2

2

2

A = πr

1A =3.14×2

A =0.785cm

Now, we will calculate the resistivity as follow: RA

= l



3 505.55 100.785

× =

3 0.7855.55 1050

= × ×

30.087 10 cm= × Ω

Now the conductivity is equals to,

3

2

1k0.087 10 cm0.01147Scm

=× Ω

=k

Now the molar conductivity is,

m1000×k

C=λ

m1000 0.001147

0.05×

=λ

2 -1m 229.4Scm mol=λ

21. The reaction ( ) ( ) ( )2 2N g +O g 2NO g , contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is -51.0×10 . Suppose in a case and [ ] -1

2O =0.20molL before any reaction occurs. Calculate the equilibrium concentration of the reactants and the product after the mixture has been heated to 1500 K. 3 Marks

Ans: ( ) ( ) ( )2 2N g +O g 2NO g→

Initial conc. 0.8M 0.20M 0

At equilibrium 0.80-x 0.20-x 2x



[ ][ ][ ]

( )( )( )

2 2

c2 2

NO 2xK = =

N O 0.80-x 0.20-x

-5cK =1.0×10 (Given)

( )2 -5 24x =1.0×10 0.16-0.80x-0.20x+x

5 2 24×10 x -x +x-0.16=0

Or 5 24×10 x +x-0.16=0

Since, the above equation is a quadratic equation, hence

2b b 4acx2a

− ± −=

5A 4 10= × , b=1 and c= -0.16

( ) ( )2 5

5

1 1 4 4 10 016x

2 4 10− ± − × × × −

=× ×

5

5 5

1 2.56 10 1 506x8 10 8 10

− ± × − ±= =

× ×

4 4x 6.3 10 ,6.3 10−= + × ×

As concentration cannot be negative we will be taking the positive value of x, 4x 6.3 10−= + ×

[ ] 4 3 1NO 2x 2 6.3 10 1.23 10 molL− − −= = × × = ×

[ ] 4 12N 0.80 6.3 10 0.799molL− −= − × =

[ ] 4 12O 0.20 6.3 10 0.199molL− −= − × =

22. Explain the following terms giving a suitable example for each: 3 Marks



(i) Aerosol:

Ans: An aerosol is known as a suspension of small solid particles or liquid droplets in air or another gas. It is a colloidal solution having gas as a dispersion medium and solid or liquid as dispersion phase. Example- Fog, mist, etc.

(ii) Emulsion:

Ans: An emulsion can be defined as a colloid consisting of two or more non-homogeneous types of liquids where one of the liquids contains the dispersion of the different forms of liquids. Example- oil and water mixture.

(iii) Micelle:

Ans: In an aqueous solution, a micelle is generated by the accumulation of amphipathic molecules. when a variety of solutions are added to water. A micelle in water forms an aggregate with the hydrophilic head region in contact with the solvent.

23. How would you account for the following: 3 Marks

(i) Among lanthanoids, Ln (III) compounds are predominant. However, occasionally in solution or in solid compounds, +2 and +4 ions are also obtained.

Ans: Because of the increased stability of the empty, half-filled, and fully filled 4f-subshells, lanthanides occasionally produce +2, +4 ions. As a result, even Ln (III) compounds predominate.

(ii) The 2+

oM

ME copper is positive (0.34 V). Copper is the only metal in the first

series of transition elements showing this behavior.

Ans: The 2

oM

ME + The metal depends on the energy change involved in sublimation,

ionization and hydration as copper has a high enthalpy of atomization and low enthalpy of hydration. Hence, the 2

oM

ME + The value of copper is positive.

(iii) The metallic radii of the third (5d) series of transition metal are nearly the same as those of the corresponding members of the second series.



Ans: This is due to lanthanide contraction. In moving down a group, there is an increase in the number of shells leading to increase in size. So, the size of elements of the 4d series is larger than those of the 3d series. It occurs due to increasing nuclear charge and electrons entering the inner (n-2) f orbitals. Hence, the metallic radii of the third (5d) series of transition metal are nearly the same as those of the corresponding members of the second series.

24. Name the following coordination entities and draw the structures of their stereoisomers: 3 Marks

(i) ( ) +

22Co en Cl (en=ethane-1, 2 -diamine)

Ans: The name of the compound is Dichloridobis – (ethane-1, 2-diamine) cobalt (III) chloride.

As there are 2 chlorine atoms, the name would start as ‘dichlorido’ and there are 2-en groups attached, so it would be bis (ethane-1, 2-diamine) and the charge on the cobalt is +3.

(ii) ( ) 3-

2 4 3Cr C O

Ans: Trioxalatochromate (III) ion



There are 3 oxalate ions so the name would be trioxalato and the central atom is chromium with the charge +3.

(iii) ( )3 3Co NH Cl

Ans: Triamminetrichloridocobalt (III)

There are 3 ammine atoms and 3 chlorine atoms, so the name starts as ‘triamminetrichlorido’. And the central atom cobalt has +3 charge.

25. Answer the following questions: 3 Marks

(i) What is meant by the chirality of compounds? Give an example.

Ans: A compound is said to be chiral if it exists in two stereoisomeric forms which are non-superimposable mirror images of each other. This property of non-superimposability is called chirality. And the compounds which are non-



superimposable mirror images of each other are known as enantiomers. Example- Butan-2-ol is a chiral compound.

(ii) Which one of the following compounds is more easily hydrolyzed by KOH and why?

( )3 2 3CH CH Cl CH CH Or 3 2 2CH CH CH Cl

Ans: Hydrolysis of 2-chloro butane by KOH will result in the formation of secondary carbocation whereas the hydrolysis of chloropropane results in primary carbocation. A secondary carbocation is more stable than primary carbocation due to the inductive effect. Hence, hydrolysis of 2-chloro butane by KOH will be much easier.

(iii) Which one undergo S 2N substitution reaction faster and why?

Ans: It is a primary halide thus it will undergo S 2N reaction faster, iodine is a better leaving group than chlorine because of its large size it will be released at a much faster rate in the presence of incoming nucleophiles.

26. What is essentially the difference between α–glucose and β –glucose? What is meant by the pyranose structure of glucose? 3 Marks

Ans: Theα –glucose and β –glucose can be distinguished by the position of the hydroxyl group on the first carbon atom. In open chainα –glucose, the hydroxyl group on the first carbon atom is towards the right whereas, in the closed ringα –glucose, the hydroxyl group on the first carbon atom is below the plane of the ring.

On the other hand, in open chain β –glucose, the hydroxyl group on the first carbon atom is towards the left whereas, in the closed ring, the hydroxyl group on the first atom is above the plane of the ring.



∝-D-(+)-Glucose β-D-(+)-Glucose

(Open ring structure)

(Closed ring structure)

The pyranose structure represents a six-membered cyclic structure of glucose. In the pyranose structure of glucose, the oxygen atom is present in the six-membered ring. Due to this, the pyranose form resembles pyran, hence it is named as pyranose form.

27. Differentiate between thermoplastic and thermosetting polymers. Give one example of each. 3 Marks

Ans: The differentiation between thermoplastic and thermosetting is:

Thermoplastic Thermosetting



The process of addition polymerization can be used to make thermoplastics.

Thermosetting plastics are synthesized by condensation polymerization.

Thermoplastic has low melting points and low tensile strength.

Thermosetting plastics have high and tensile strength and a high melting point.

When opposed to thermosetting polymers, thermoplastics have a lower molecular weight.

Thermosetting plastics are high in molecular weight.

Examples of thermoplastics are- polystyrene, Teflon, and Nylon.

Examples of thermosetting plastics are- Vulcanized rubber, Bakelite, vinyl ester resin.

28. (a) Define the following terms: 2 Marks

(i) Mole fraction

Ans: The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components. Example- The mole fraction of A and B are given as:

Mole fraction of A

A B

nAn n

=+

where An is the number of moles of A.

(ii) Ideal solution

Ans: An ideal solution is a mixture in which the molecules of different species are distinguishable, however, unlike the ideal gas, the molecules in the ideal solution exert forces on each other.

(b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at o-0.34 C. What is the molar mass of the material? (

fK For water -1=1.86K Kgmol ). 3 Marks

Ans: Weight of solute= 15 g



Weight of solvent (Water)=450 g

The formula for freezing point depression is:

f fT K m∆ = × (Where m is molality)

The freezing point of pure water is o0 C

( )fT 0 0.34∆ = − −

ofT 0.34 C∆ =

We can calculate molality by: f

f

TmK∆

=

1

0.34Km1.86K g mol−

=

1m 0.183molKg−=

Convert the amount of water in kg=0.450 kg

The formula of molality is: mole of solutemKg of solvent

=

Mol of solute0.1830.450

=

Mol of solute= 0.0823mol=

Using the mole formula to determine the molar mass of unknown material.

15.00.0823Molar mass

=

gMolar mass 182.26 mol= .

Or

(a) Explain the following: 2 Marks



(i) Henry’s law about dissolution of a gas in a liquid.

Ans: Henry’s law about the dissolution of a gas in liquid: Henry’s gas law states that the amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid when the temperature is kept constant.

HP=K ×x

Where ‘P’ denotes the partial pressure of the gas

‘x’ denotes the concentration of the dissolved gas.

‘ HK ’ is Henry’s law constant of the gas.

(ii) Boiling point elevation constant for a solvent.

Ans: Boiling point elevation constant for a solvent: The boiling point elevation refers to the increase in the boiling point of a solvent upon the addition of a solute. When a nonvolatile solute is added to a solvent, the resulting solution has a higher boiling point than that of the pure solvent.

(b) A solution of glycerol ( )3 8 3C H O in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of o100.42 C. What mass of glycerol was dissolved to make the solution? ( bK for water

-1=0.512K Kg mol ) 3 Marks

Ans: Weight of solvent (water)=500g=0.5 Kg

Molecular mass of glycerol= 12 3 1 8 16 3 92u= × + × + × =

The boiling point of water is

bT 100.42 100∆ = −

bT 0.42K∆ =

The formula for boiling point depression is:

b bT K m∆ = × (Where m is molality)



So, we can calculate molality by: b

b

TmK∆

=

1

0.42Km0.512K Kg mol−

=

1m 0.820molkg−=

The formula of molality is: mole of solutemKg of solvent

=

mol of solute0.8200.5

=

Mol of solute= 0.41 mol

Using the mole formula to determine the molar mass of unknown material.

mol of solute0.4192

=

Mass of solute = 37.72 g

29. (a) Draw the molecular structure of the following compounds: 2 Marks

(i) 2 5N O

Ans: The name of the compound is Dinitrogen pentoxide. Its molecular structure is:



(ii) 4XeOF

Ans: The name of the compound is Xenon oxytetrafluoride, its geometry is square pyramidal. The molecular structure is:

(b) Explain the following observations: 3 Marks

(i) Sulphur has a greater tendency for catenation than oxygen.

Ans: Catenation is defined as the property of self-linking. Sulphur has a greater tendency to catenate than oxygen because oxygen belongs to the period second, therefore it does not have any d-orbital vacant where filling of electrons can take place. But Sulphur belongs to the period third and has a vacant d-orbital where the extra incoming electrons can be accommodated easily.

(ii) ICl is more reactive than 2I

Ans: I-Cl have interhalogen bond whereas 2I have covalent bond. The interhalogen bond is slightly ionic in nature which makes the bond weaker. A weaker bond can be broken easily and is considered to be more reactive than a complete covalent bond.

(iii) Despite lower value of its electron gain enthalpy with negative sign, fluorine ( )2F is a stronger oxidizing agent than 2Cl .



Ans: The electron gain enthalpy of fluorine is less than that of chlorine. However, the bond dissociation energy of fluorine is lesser than chlorine because of its small size and higher hydration enthalpy. So, fluorine is a much stronger oxidizing agent than chlorine.

Or

(a)Complete the following chemical equations: 2 Marks

(i) ( )3Cu+HNO dilute →

Ans: ( ) ( )3 3 223Cu 8HNO dilute 3Cu NO 2NO 4H O+ → + +

When copper reacts with dilute nitric acid, the copper get oxidized to 2Cu + ions form copper nitrate, whereas nitric acid is reduced to nitrogen dioxide.

(ii) 4 2 2XeF +O F →

Ans: 4 2 2 6 2XeF O F XeF O+ → +

The chemical name of 4XeF is xenon tetrafluoride. When 4XeF reacts with 2 2O F the two O F− bonds present in 2 2O F break. The compound formed will be xenon hexafluoride.

(b) Explain the following observations: 3 Marks

(i) Phosphorus has a greater tendency for catenation than nitrogen.

Ans: Because of the limited bond length, the single N-N bond is weaker than the P-P bond due to the significant interelectronic repulsion of the non-bonding electrons. As a result, in nitrogen, the catenation tendency is weaker.

(ii) Oxygen is a gas but Sulphur a solid.

Ans: Oxygen has a small size and high electron density, P Pπ = π the oxygen atom can be linked to each other by double bond because of bonding. Due to this the intermolecular forces present in oxygen are weak van der Waals forces. As result, oxygen exists as a gas. Whereas Sulphur does not form a strong S=S double bond because of its large size. Thus, Sulphur exists as a solid held together by strong covalent bonds.



(iii) The halogens are colored. Why?

Ans: The amount of energy required by the outer electrons to higher energy levels for every halogen is different. The color of halogen is due to the absorption of different quanta of radiations in the visible region which results in the excitation of electrons to higher energy levels. Thus, colors are observed.

30. (a) Write the suitable chemical equation to complete each of the following transformations: 2 Marks

(i) Butan-1-ol to butanoic acid

Ans: Butan-1-ol to butanoic acid

The formation of butanoic acid from Butan-1-ol is an oxidation reaction. The commonly used oxidation reagents are potassium dichromate and dilute sulphuric acid. The reaction take place as:

( )( )

2 2 7 2 4

2 4

i K Cr O /H SOii Dil.H SO3 2 2 2 3 2 2CH CH CH CH OH CH CH CH COOH→

(ii) 4-methylacetophenone to benzene-1, 4-dicarboxylic acid

Ans: The 4-methylacetophenone to benzene-1, 4-dicarboxylic acid is an oxidation reaction, in which the 4-methyl group gets oxidized with 4KMnO / KOH to carboxylic acid. The reaction is:

(b) An organic compound with the molecular formula 9 10C H O forms 2,4-DNP derivatives, reduces tollen’s reagent and undergoes Cannizzaro’s reaction on vigorous oxidation it gives 1,2-benzenedicarboxylic acid. Identify the compound. 3 Marks

Ans: The given compound reduces Tollen’s reagent and forms 2,4-DNP derivatives. Therefore, the compound must have an aldehyde group. Furthermore, the chemical



undergoes Cannizzaro's reaction, which results in the formation of 1, 2-benzenedicarboxylic acid upon oxidation.

Only the aldehyde group (-CHO) directly oxidized to the carboxylic acid. As a result, the aldehyde group is linked to a benzene ring, resulting in the formation of 2-ethylbenzaldehyde.

The above reactions can be shown as:

Or

(a)Give chemical tests to distinguish between 2 Marks

(i) Propanol and propanone

Ans: Propanol and propanone: To distinguish between alcohol and ketone group iodoform tests can be used. Propanone gives an iodoform test while propanol does not give this test.

3 2 2 3 3 2 2CH COCH 3I 2Na CO CH COONa 2CO H O+ + + + +

(Propanone) (Yellow ppt)



3 2 2 2CH CH CH OH 3I reaction+ →

Thus, no yellow precipitate is formed when reacted with propanol.

(ii) Benzaldehyde and acetophenone

Ans: Benzaldehyde and acetophenone: Benzaldehyde contains an aldehyde and acetophenone is having a ketone group, it can be identified by the iodoform test. Acetophenone in reaction with sodium hypoiodite forms a yellow-colored precipitation of iodoform.

6 5 3 6 5C H COCH 3NaOI C H COONa 2NaOH+ → +

(Acetophenone) (Sodium Benzoate)

Whereas, benzaldehyde on reaction with sodium hypoiodite does not form a yellow colored precipitate.

6 5C H CHO+2NaOI No reaction→

(Benzaldehyde)

(b) Arrange the following compounds in increasing order of their property as indicated: 3 Marks

(i) Acetaldehyde, acetone, methyl tert-butyl ketone (reactivity towards HCN)

Ans: The steric hindrance determines a compound reactivity. With the increase in +I effect of the alkyl group or increase in the steric hindrance, the reactivity towards HCN addition decreases. Hence, the increasing order of the reactivity towards HCN is

Di-tert-butyl ketone Methyl tert-butyl ketone Acetone Acetaldehyde

(ii) Benzoic acid, 3, 4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

Ans: When an electron-withdrawing group attached to a ring increases its acidic strength, while an electron releasing group decreases the acidic strength.

Nitro group is electron-withdrawing while the methoxy group is an electron releasing group. Therefore increasing order of acid strength is:



4 methoxybenzoic acid< benzoic acid< 3,4-Dinitro benzoic acid.

(iii) ( ) ( ) ( )3 2 3 2 3 2CH CH CH Br COOH.CH Br CH COH. CH CHCOOH (Acid

strength)

Ans: The group having +I effect will decrease the strength of the acids and groups having –I effect will increase the strength of the acids. After losing a proton, carboxylic acids gain a negative charge. Now any group that will help stabilize the negative charge will increase the stability of the carbonyl ion and as a result, will increase the strength of the acid. Therefore increasing order of acid strength:

( ) ( ) ( )3 3 2 2 3 2 3 22CH CHCOOH CH CH CH COOH CH CH Br CH COOH CH CH Br COOH< < <


CBSE Class 12 Chemistry Previous Year Question Paper 2012

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