CBSE/12 Class/2013/ CHEMISTRY/SET-1 · CBSE/12th Class/2013/ CHEMISTRY/SET-1 Poornima University S.No Questions Answers Q.1 How many atoms constitute one unit cell of a face-centered

CBSE/12th Class/2013/ CHEMISTRY/SET-1

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S.No Questions Answers Q.1 How many atoms constitute one unit cell of a face-centered cubic crystal?

Ans-1 8 corner atoms. 6 face atoms. Each corner atom exists in 8 adjacent cells.. so 1/8 of each corner atom is in any particular unit cell. Each face atom exists in 2 adjacent cells. so 1/2 of each face atom is in any given unit cell. total atoms / cell = 8 x 1/8 + 6 x 1/2 = 4. so atoms / cell = 4

Q.2 Name the method used for the refining of Nickel metal.

Ans-2 Monds process-The impure nickel reacts with carbon monoxide at

50–60 °C to form the gas nickel carbonyl, leaving the impurities as solids.

Ni(s) + 4 CO(g) → Ni(CO)4(g) The mixture containing nickel carbonyl (and synthesis gas) is heated to 220–250 °C, resulting in decomposition back to nickel and carbon monoxide: Ni(CO)4(g) → Ni(s) + 4 CO(g)

Q.3 What is the covalency of nitrogen in N2O5?

Ans-3 The covalence of nitrogen in nitrogen pentoxide is 4.

Q.4 . Write the IUPAC name of

Ans-4-Chloro1-Pentene 5 4 3 2 1

CH3-CH-CH2-CH=CH2 Cl

Q.5 What happens when CH3-Br is treated with KCN? Ans-5 CH3Br + KCN ------------> CH3CN + KBr

Nucleophilic Substitution reaction Q.6 Write the structure of 3-methyl butanal. Ans-6

Q.7 Arrange the following in increasing order of their basic strength in aqueous

solution: CH3.NH2,(CH3)3N,(CH3)2NH

Ans-7 The increase in Kb from methylamine to dimethylamine may be attributed to +I effect; however, there is a decrease from dimethylamine to trimethyl amine due to the predominance of steric hindrance offered by the three methyl groups to the approaching Brönsted acid. (CH3)3N<CH3.NH2<( CH3)2NH

Q.8 What are three types of RNA molecules which perform different functions?

Ans-8 Three different types of RNA molecules: Messenger RNA (mRNA), Transfer RNA (tRNA) and Ribosomal RNA (rRNA).

Q.9 18g of glucose, C6H12O6 (Molar Mass= 180g mol-1) is dissolved in 1Kg of water in Ans-9 weight of solvent= 1000gm, weight of solute18 gm, molar mass.


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a sauce pan. At what temperature will this solution boil?

of solute= 180 gm/mole, b= Kbx1000xwt of solute/molar mass of solutetxwt of solvent =0.52x1000x18/180x1000 =0.052K Tb = Tb-Tb0 = Tb-373.15 0.052=Tb-373.15 Tb =373.202K

Q.10 The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm-1. Calculate its molar conductivity.

Ans-10 Molar conductivity= K/1000xmolarity =0.025/1000x0.02=1.25x10-4Scm-2mol-1

Q.11 Write the dispersed phase and dispersion medium of the following colloidal system: (i) Smoke (ii) Milk

OR What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes?

Ans-11. (i) Smoke Dispersed phase: Solid and dispersion medium: Gas (ii) Milk Dispersed phase: Liquid and dispersion medium: Liquid

OR Lyophilic : Lyophilic colloids are liquid loving colloids (Lyo means solvent and philic means loving). Sols of organic substances, eg. gelatin, gum, starch and proteins. Lyophobic : Lyophobic colloids are liquid hating colloids (Lyo means solvent and phobic means hating). Sols of inorganic substances, eg Arsenic (As2S3), Iron (Fe(OH)3) and Platinum.

Q.12 Write the differences between physisorption and chemisorption with respect to the following: (i) Specificity (ii) Temperature dependence (iii) Reversibility and (iv) Enthalpy change

Ans -12 physisorption chemisorption specificity Not specific in nature Specific in nature Temperature dependence

Occurs at temperature below the B.p

Occurs at all temperature

Reversibility reversible irreversible Enthalpy change Adsorption enthalpy

is low It is high.

Q.13 a) Which solution is used for the leaching of silver metal in the presence of air in the metallurgy of silver? (b) Out of C and CO, which is a better reducing agent at the lower temperature range in the blast furnace to extract iron from the oxide ore?

Ans-13.(a) NaCN(dil) or KCN(dil) is used for leaching of silver metal in the presence of air in the metallurgy of silver. (b)from C and CO , CO is better reducing agent at low temperature range because gibbs free energy(CO,CO2)<gibbs free energy of (Fe,FeO) therefore, CO will reduce FeO and will itself will oxidize to CO2.

Q.14 14. What happens when (i) PCl5 is heated? (ii) H3PO3 is heated? Write the reaction involved

Ans.14 (i)PCl5 PCl3 + Cl2. (ii)4 H3PO3 3H3PO4 + PH3 Orthophosphoric acid Phosphine

Q.15 (a)Which metal in the first transition series (3d series) exhibits +1 oxidation state Ans-15 (a) Cu shows +1 oxidation state because Cu have configuration 3d10,


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most frequently and why? (b) Which of the following cations are colored in aqueous solutions and why? Sc3+,V3+,Ti4+,Mn2+ (At. nos. Sc = 21, V= 23, Ti = 22, Mn = 25)

4s1. (b)colour of solution is due to presence of unpaired e- in d-orbitals. The electronic configuration of the following cations is as follows: Sc (Z = 21) = 3d1 4s2 and Sc3+ =3d0 4s0. d-orbital is empty, it is colourless. V (Z = 23) =3d3 4s2 and V3+ =3d2 4s0. d-orbital is having 2 unpaired electrons, it undergoes d-d transition and shows green colour. Ti (Z = 22) = 3d2 4s2 and Ti4+ = 3d0 4s0. d-orbital is empty, it is colourless. Mn (Z =25) = 3d2 4s2 and Mn2+ =3d5 4s0. d-orbital is having 5 unpaired electrons, it show pink colour.

Q.16 Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same.

Ans-16

The contributing structure iii,iV,V indicate that C-Cl bond has partial double character as a result C-Cl bond in chlorobenzene is shorter and stronger thus cleavage of C-Cl bond becomes difficult which makes it less reactive towards nucleophilic substitution reaction. Moreover, the sp2 hybridized carbon atom involved in C-Cl bond in chlorobenzene is more electronegative than the sp3 hybrid carbon atom in alkyl halide. Therefore,this sp2 hybridised carbon atom has less tendency to release electrons to the Cl atom. Thus, lower the polarity of C-Cl, lesser is the reactivity.

Q.17 Explain the mechanism of the following reaction:

Ans-17 MECHANISM FOR ALCOHOL CONDENSATION TO GIVE AN ETHER Step 1: An acid/base reaction. Protonation of the alcoholic oxygen to make a better leaving group. This step is very fast and reversible. The lone pairs on the oxygen make it a Lewis base Step 2: The O of the second alcohol molecule functions as the nucleophile and attacks to displace the good leaving group, a neutral water molecule, by cleaving the C-O bond. This creates an oxonium ion intermediate. Step 3: Another acid / base reaction. The proton is removed by a suitable base (here a water molecule, ROH is another alternative) to give the ether product.


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Q.18 How will you convert: (i) Propene to Propan-2-ol? (ii) Phenol to 2, 4, 6 – trinitrophenol?

Ans-18(i)

(ii)

Q.19 (a) What type of semiconductor is obtained when silicon is doped with boron? (b) What type of magnetism is shown in the following alignment of magnetic moments?

(c) What type of point defect is produced when AgCl is doped with CdCL2?

Ans-19 (a) p-type semiconductor. (b) magnetic moments is ferromagnetism (c) Impurity defect.


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Q.20 Determine the osmotic pressure of a solution prepared by dissolving 2.5 102 g of K2SO4 in 2L of water at 25 C ○ , assuming that it is completely dissociated. (R=0.0821 L atm K-1mol-1, Molar mass of K2SO4=174g mol-1).

Ans-20 w2 = 2.5 × 10–2 g (Mass of K2SO4) and M2 = 174 g mol–1 (Molar mass of K2SO4) V = 2L, R = 0.0821 L atm K–1 mol–1 and T = 25°C = 298 K Osmotic pressure(π)=w2×RT/M2V =2.5 × 10–2 ×0.0821×298/174×2 =1.76×10-3atm.

Q.21 Calculate the emf of the following cell at 298 K: Fe(s) Fe+2 (0.001M) H(g) H2 (g)(1bar), Pt(s) ( 0.44 ) cell GivenV

Ans-21 Anodic reaction-Fe Fe+2 +2e- Cathodic reaction2H+ +2e- H2 According to Nernst equation-

Ecell=0.44-0.0591/2log0.001/1 E=0.44(-0.02955x-3) =0.44+0.08865 =0.534V

Q.22 How would you account for the following? (i) Transition metals exhibit variable oxidation states. (ii) Zr (Z=40) and Hf (Z=72) have almost identical radii. (iii) Transition metals and their compounds act as catalyst. OR Complete the following chemical equations:

Ans-22 Ans- (i) Highest Oxidation State for a Transition metal = Number of Unpaired d-electrons + Two s-orbital electrons . So they have variable oxidation states . (ii) It is due to lathanides contraction . Lanthanide contraction is a term used to describe the greater than expected decrease in ionic radii of the elements in the lanthanide series from atomic number 57, lanthanum, to 71, lutetium, which results in smaller than otherwise expected ionic radii for the subsequent elements starting with 72, hafnium (i)Transition metals are good metal catalysts because they easily lend and take electrons due to vacant d-orbitals.

OR (i)Cr2O7

2- + 6Fe+2 +14H+ 2Cr+3 +6Fe+3 +7H2O (ii)2CrO4

2- +2H+ Cr2O72- +H2O

(iii)2MnO7- +5C2O4

2- +16H+ 2Mn2+ +10CO2 +8H2O Q.23 Write the IUPAC names of the following coordination compounds:

(i) [Cr(NH3)3Cl3] (ii) K3[Fe(CN)6] (iii) [CoBr2(en)2]+, (en = ethylenediamine)

Ans-23 (i) Triamminetrichlorochromium (III) (ii) Potassium hexacyanoferrate (III) (iii)Dibromidobis (ethane-1, 2-diammine) cobalt (III) ion

Q.24 Give the structures of A, B and C in the following reactions: Ans-24


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CuCN H2O/H+ C6H5N2

+Cl- C6H5CN (A) C6H5COOH(B) heat NH3

C6H5CONH2(C) Sn+HCl NaNO2+HCl H2O/H+ (ii) C6H5NO2 C6H5NH2 (A) C6H5N2

+Cl-(B) C6H5OH(C)

Q.25 Write the names and structures of the monomers of the following polymers: (i) Buna-S (ii) Neoprene (iii)Nylone-6, 6

Ans-25 Ans-(i)

(ii)

(iii)


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Q.26 After watching a programme on TV about the adverse effects of junk food and soft

drinks on the health of school children, Sonali, a student of Class XII, discussed the issue with the school principal. Principal immediately instructed the canteen contractor to replace the fast food with the fibre and vitamins rich food like sprouts, salad, fruits etc. This decision was welcomed by the parents and the students. After reading the above passage, answer the following questions: (a) What values are expressed by Sonali and the Principal of the school? (b) Give two examples of water-soluble vitamins.

Ans - 26-(a) Values are adverse effect of the junk food and concern towards health of the students Principal showed agood response towards the Sonali’s opinion and concern towards the students. (b)Vitamin-B and Vitamin-C complex.

Q.27 a) Which one of the following is a food preservative? Equanil, Morphine, Sodium benzoate (b) Why is bithional added to soap? (c) Which class of drugs is used in sleeping pills?

Ans-27(i)Sodium benzoate - food preservative Equanil - tranquillizer Morphine - analgesic. (ii) Bithional is an antiseptic added to soaps to reduce odours produced by bacterial decomposition of organic matter on the skin. (iii)In sleeping pills Tranquillizers are used because it relieve stress, fatigue by inducing sense of well being.

Q.28 (a) A reaction is second order in A and first order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of A three times? (iii) How is the rate affected when the concentration of both A and B are doubled? (b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 for this reaction. (Given log 1.428=0.1548)

OR (a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. (b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the

Ans-28(a) (i)Differential rate equation: Rate=-d[R]/dt =K[A]2[B] (ii)Because the rate=K[A]2[B] So if on increasing the concentration of A three times Rate=K[3A]2[B] =9K[A]2[B]=9Rate Or 9 times the intial rate. (iii)Rate=K[A]2[B] If the concentration of both A and B are doubled then Rate =K [2A]2 [2B] =8K[A]2[B] =8Rate Or 8 times the initial rate. (b)K=2.303/t log a/a-x Given values- T (time)=40 minutes, a=concentration of reactant at time 0 min.(a-x)= concentration of reactant at time t min. X given 30% so(a-x)=70/100 a K=2.303/40 log a/(70/100)a K=2.303/40 log 100/70


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K=0.00891 min-1 T1/2=0.693/K T1/2=0.693/0.00891=77.78 min-

OR (a)given condition for first order reaction (i) when 99% reaction is completed time=? (ii)When 90% reaction is completed time=? For condition (i) a=concentration of reactant at time 0 min. (a-x)= concentration of reactant at time t min. X=99% so (a-x) =a/100 K=2.303/t log a/(a/100) =2.303/t log 100 =2.303/t log 102

Or T1/2=2(0.693/K) For condition (i) a=concentration of reactant at time 0 min. (a-x)= concentration of reactant at time t min. X=90% so (a-x)=a/10 K=2.303/t log a/(a/10) =2.303/t log10 Or T1/2=0.693/K So for first order reaction time required is just double when 99% reaction is completed than for 90 % reaction is completed. (b) log K=log A Ea/2.303R (1/T) Ea=Activation energy For above reaction Y=mx+c Where plot Y Vs X give a straight line with slope ‘P’ & intercept ‘Q’ Slope P=-4250 Log K Slope=-4250 K Slope=-Ea/2.303 R =-Ea/2.303R=-4250 R Ea=4250×2.303×8.314=81,375.3535 mol-1

Ea=81.3753 mol-1 1/T Q.29 a) Give reasons for the following:

(i) Bond enthalpy of F2 is lower than that of Cl2. (ii) PH3 has lower boiling point than NH3.

Ans-29 (a) (i) The bond dissociation energy of F2 is lower than that of Cl2 because F atom is small in size and due to this the e- & e- repulsions between the lone


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(b) Draw the structures of the following molecules: (i) BrF3 (ii) (HPO3)3

OR (a) Account for the following: (i) Helium is used in diving apparatus. (ii) Fluorine does not exhibit positive oxidation state. (iii) Oxygen shows catenation behavior less than sulphur. (b) Draw the structures of the following molecules. (i) XeF2 (ii) H2S2O8

pairs of F-F are very large. (ii)NH3 molecule possess intermolecular hydrogen bonding which binds its molecules strongly whereas PH3 has weaker Vander Waal's forces . Thus, PH3 has lower boiling point than NH3. (b)(i) BrF3

Bent ‘T’ structure

(ii) (HPO3)3

Cyclic structure

OR (i)Air contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as He is sparingly less soluble in blood. (ii) Fluorine being the most electronegative atom does not exhibit


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positive oxidation state because the electrons in fluorine are strongly attracted by the nuclear charge because of small size of fluorine atom and therefore, removal of an electron is not possible. (i)Oxygen is a rather boring element. It has only two allotropes, Oxygen and Ozone. Oxygen has a double bond, and Ozone has a delocalised cloud, giving rise to two "1.5 bonds". On the other hand, Sulphur has many stable allotropes, and a bunch of unstable ones as well. The variety of allotropes, is mainly due to the ability of Sulphur to catenate.But, Sulphur does not have a stable diatomic allotrope at room temperature. I, personally would expect diatomic sulphur to be more stable than diatomic Oxygen, due to the possibility of pπ−dπ back-bonding. So Sulphur and Oxygen have such opposite properties with respect to their ability to catenate.

(b)(i) XeF2

(ii) Linear structure

Q.30 (a) Although phenoxide ion has more number of resonating structures than Carboxylate ion, Carboxylic acid is a stronger acid than phenol. Give two reasons. (b) How will you bring about the following conversions? (i) Propanone to propane (ii) Benzoyl chloride to benzaldehyde (iii) Ethanal to but-2-enal

OR

Ans-30(a)Phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. This can be nexplained by considering the resonating structures of carboxylate ion and phenoxide ion.


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(a) Complete the following reactions:

(b) Give simple chemical tests to distinguish between the following pairs of compounds: (i)Ethanal and Propanal (ii) Benzoic acid and Phenol

In case of phenoxide ion structures III to V carry a negative charge on the less electronegative carbon atom.Therefore, their contribution towards the resonance stabilization of phenoxide ion is very small .Now in resonance structures of carboxylate ion the negative charge is on the more electronegative oxygen atom.Also in phenoxide , in structures I and II the negative charge is on the more negative oxygen atom.However in carboxylate ion the negative charge on the carboxylate ion is delocalized over the two oxygen atom while in the first two structures of phenoxide the negative charge is localized.Therefore the release of proton from carboxylic acid is much easier than phenols.Therefore carboxylic acid is stronger than phenols. (b)(i)Conversion of Propanone to Propane:

(ii)


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(iii)

Conc. KOH

2HCHO CH3OH + HCOOK (ii) Br2 / P CH3COOH Br-CH2-COOH (i) C6H5CHO HNO3/H2SO4 m- Nitrobenzaldehyde 273-283K

(b)(i)Distinguish test between ethanal and propanal: Iodoform Test: Ethanal gives iodoform test. CH3CHO + 4NaOH + 3I2 CHI3 (Yellow ppt.) + HCOONa + 3NaI + 3H2O Propanal does not give this test. CH3CH2CHO + 4NaOH + 3I2 No Reaction. (ii) Distinguish test between Benzoic acid and Phenol: NaHCO3 Test: When


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Benzoic acid reacts with NaHCO3, brisk effervescence of CO2 gas evolved. C6H5COOH + NaHCO3 C6H5COONa + H2O + CO2 Phenols do not give any effervescence with NaHCO3.

CBSE/12 Class/2013/ CHEMISTRY/SET-1 · CBSE/12th Class/2013/ CHEMISTRY/SET-1 Poornima University S.No Questions Answers Q.1 How many atoms constitute one unit cell of a face-centered

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