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Series : SKS/1

Roll No.

Code No. 56/1/1Candidates must write the Code onthe title page of the answer-book.

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

Please check that this question paper contains 30 questions.

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

15 minutes time has been allotted to read this question paper. The question paper will be distributed at10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will notwrite any answer on the answer script during this period.

CHEMISTRY

[Time allowed : 3 hours] [Maximum marks : 70]

General Instructuions:

(i) All questions are compulsory.

(ii) Questions numbered 1 to 8 are very short-answer questions and carry 1 mark each.

(iii) Questions numbered 9 to 18 are short-answer questions and carry 2 marks each.

(iv) Questions numbered 19 to 27 are also short-answer questions and carry 3 marks each.

(v) Questions numbered 28 to 30 are long-answer questions and carry 5 marks each.

(vi) Use Log Tables, if necessary. Use of calculators is not allowed.

Studymate Solutions to CBSE Board Examination 2012-2013

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STUDYmate

1. How many atoms constitute one unit cell of a face-centered cubic crystal?

Ans. 8 × 1

8 + 6 ×

1

2 = 4

In one unit cell, 4 atoms are present.

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

Ans. Mond’s Process

3. What is the covalency of nitrogen in N2O5?

Ans. 4 (Four)

N – O – NO

O

O

O

4. Write the IUPAC name of CH – CH – CH – CH = CH3 22

Cl

Ans. 4-Chloropent-1-ene

5. What happens when CH3 – Br is treated with KCN?

Ans. Nucleophilic substitution occurs

CH3Br + K+C– N CH3CN + KBr

6. Write the structure of 3-methyl butanal.

Ans. H C – C – CH – C – H3 2

H

CH3 O

3-Methyl butanal

7. Arrange the folllwing in increasing order of their basic strength in aqueous solution:

CH3NH2, (CH3)3N, (CH3)2NH

Ans. (CH3)2NH > CH3NH2 > (CH3)3N

8. What are the three types of RNA molecles which perform different functions?

Ans. m – RNA

t – RNA

r – RNA

9. 18 g of glucose, C6H12O6 (Molar mass = 180 g mol–1) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil?

(Kb for water = 0.52 K kg mol–1, boiling point of pure water = 373.15 K)

Ans. Tb – Tb° = Kb × m

Tb – 373.15 = 0.52 × 18 1 kg

180 1 kg

Tb = 0.052 + 373.15

Tb = 373.202 K

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STUDYmate

10. The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm–1. Calculate its molarconductivity.

Ans. m = k 1000

M

; m is molar conductivity.

= 0.025 1000

0.20

= 125 S cm2 mol–1.

11. Write the dispersed phase and dispersion medium of the following colloidal systems:

(a) smoke (b) milk

OR

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

Ans.Dispersed phase Dispersion medium

Smoke Solid Gas or Air

Milk Liquid Liquid

OR

Ans. Lyophilic colloids are liquid loving. Certain substances mixed with a suitable liquid (dispersionmedium) directly forms the colloid. Such colloids are lyophilic colloids. Example, gum, gelatin,starch, etc.

Lyophobic colloids are liquid hating. When substances like metals or metal sulphides aremixed with the dispersion medium do not form the colloidal sol.

Such type of sols can be prepared by special methods.

Example: Metal sulphides.

On addition of small amount of electrolyte lyophobic collides can be precipitated.

12. Write the differences between physisorption and chemisorption with respect to the following:

(a) specificity (b) temperature dependence

(c) reversibility and (d) enthalpy change

Ans. Physisorption Chemisorption

1. Physisorption is non-specific in nature. As physisorption involves Vander Waal’s interaction between adsorbate and adsorbent.

2. On increasing temperature physisorption decreases. It is because desorption occurs rapidly. Physisorption occurs usually at low temperature.

3. Physisorption is reversible in nature.

4. In case of physisorption enthalpy change is small.For example: 20-40 kJ mol–1 (approx).

1. Chemisorption is highly specific in nature as chemisorptions includes chemical bond formation between adsorbate and adsorbent.

2. Chemisorption increases with increase in temperature. After reaching a maximum value, extent of adsorption decreases with temperature.

3. Chemisorption is irreversible in nature.

4. In case of chemisorptions enthalpy change is high.For example: 40-400 kJ mol–1 (approx).

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STUDYmate

13. (a) Which solution is used for the leaching of silver metal in the presence of air in themetallurgy of silver?

(b) Out of C and CO, which is better reducing agent at the lower temperature range in theblast furnance to extract iron from the oxide ore?

Ans. (a) Aqueous NaCN

(b) CO is a better reducing agent at lower temperature.

rG° would be more negative for reduction in blast furnace to extract Fe from the oxideore by using CO.

14. What happens when

(a) PCl5 is heated? (b) H3PO3 is heated?

Ans. (a) 5 3 2PCl g PCl g Cl g

(b) 4H3PO3 3H3PO4 + PH3

15. (a) Which metal in the first transition series (3d series) exhibits +1 oxidation state mostfrequently and why?

(b) Which of the following cations are coloured in aqueous solutions and why?

Sc3+, V3+, Ti4+, Mn2+

(At. nos. Sc = 21, V = 23, Ti = 22, Mn = 25)

Ans. (a) Copper shows +1 oxidation state extensively. Cu+ has 3d10 configuration.

(b) V3+ = 3d2+ 4s0

Mn2+ = 3d5 4s0

V3+ as well as Mn2+ both have unpaired electrons in their d-orbitals which can show d-d transition.

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

Ans. Chlorobenzene is extremely less reative towards nucleophilic substitution reaction due toconjugation.

(a) Partial double bond character between carbon of benzene and halogen causes cleavageof bond with difficulty.

(b) Phenyl cation is quite unstable, So SN1 mechanism is not possible too.

17. Explain the mechanism of the following reaction:

H3 2 3 2 2 3 2413K

2CH CH OH CH CH O CH CH H O

Ans. Step 1: Formation of Protonated alcohol

CH CH3 2 – OH + H CH3 – CH – O – H2

H

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STUDYmate

Step 2: Formation of protonated ether

H C3 – CH – O2 CH3 – CH – O – C2

H

+ C – OH2

H

CH3 H H

HH

CH3

+ H2O

Step 3: Formation of ether

H C – CH – O – CH – CH3 2 2 3 CH – CH – O – CH – CH + H3 2 2 3

H

18. How will you convert:

(a) Propene to Propan-2-ol (b) Phenol to 2, 4, 6–trinitrophenol

Ans. Conversion

Propene Propan-2-ol

2H O/H3 2 3 3CH CH CH CH CH CH

|OH

Phenol 2,4,6 trinitrophenol (Picric acid)

OH

conc. HNO3

NO2

(Picric acid)

OH

NO2

NO2

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. (a) Silicon is an element of group 14, Boron belongs to group 13.

So, a p-type semiconductor is resulted.

(b) Ferromagnetism (as all magnetic moments lie in same direction)

(c) AgCl with CdCl2 causes impurity defect leading to cationic vacancies.

20. Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10–2 g of K2SO4 in2L of water at 25°C, assuming that it is completely dissociated.

(R = 0.0821 L atm K–1 mol–1, Molar mass of K2SO4 = 174 g mol–1)

Ans. K2SO4 2K+ + SO42–

i = 2 + 1 = 3

Volume of solution (v) = 2L

Mass of K2SO4 (WB) = 2.5 × 10–2

= 0.025 g

Molar mass of K2SO4 (MB) = 174 g/mol

T = 25°C = 298.15 K

= i CRT

= iB

B

W RT

M V

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STUDYmate

=

3 0.025 0.0821 298.15

174 2

= 5.27 × 10–3 atm.

21. Calculate the emf of the following cell at 298 K:

Fe(s) | Fe2+ (0.001 M) || H+ (1 M) | H2(g) (1 bar), Pt(s)

(Given E°cell = +0.44 V)

Ans. Cell Reaction

22(n 2)

Fe 2H Fe H

According to Nernst equation

2o

cell cell 2

0.0591 [H ]E E log

2 [Fe ]

= 0.0591

0.44 (3)2

= 0.44 + 0.0887

= 0.528 V

22. How would you account for the following?

(a) Transition metals exhibit variable oxidation states.

(b) Zr (Z = 40) and Hf (Z = 72) have almost identical radii.

(c) Transition metals and their compounds act as catalyst.

OR

Complete the following chemical equations:

(a) 2 2+ +2 7Cr O + 6Fe + 14H

(b) 2 +42CrO + 2H

(c) 2 +4 2 42MnO + 5C O + 16H

Ans. (a) Transition metals exhibit variable oxidation states due to the participation of ns and (n– 1) d electrons in bond formation. Both ns and (n – 1) d electrons have almost equalenergies.

(b) Zr (160 pm) and Hf (159 pm) have almost identical radii due to Lanthanoid contraction.Lanthanoid contraction is steady decrease in the size of lanthanoid ions (M3+) with theincrease in atomic number. The reason for this is a gradual increase in the effectivenuclear charge experienced by the outer electron.

(c) Transition metal and their compounds act as catalysts,

(i) because of their variable valencies. Transition metals sometimes form unstableintermediate compounds and provide a new path with lower activation energy forthe reaction.

(ii) in some cases they provide suitable surface for the reaction to take place. Thereactants are adsorbed on the surface of the catalyst where reaction occurs.

OR

Ans. (a) Cr2O72– + 14H+ + 6Fe2+ 2Cr3+ + 6Fe3+ + 7H2O

(b) 2CrO42– + 2H+ Cr2 O7

2– + H2O

(c) 2MnO4– + 5C2O4

2– + 16H+ 2Mn2+ + 8H2O + 10CO2

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STUDYmate

23. Write the IUPAC names of the following coordination compounds :

(i) [Cr(NH3)3Cl3] (ii) K3[Fe(CN)6]

(iii) [CoBr2(en)2]+, (en = enthylenediamine)

Ans. (i) Triamminetrichloridochromium(III)

(ii) Potassium hexacyanoferrate(III)

(iii) Dibromidobis(ethane–1,2–diamine)cobalt(III)

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

(i) 2 3H O/H NHCuCN6 5 2C H N Cl A B C

(ii) 2 2NaNO HCl H O/HSn HCl6 5 2 273K

C H NO A B C

Ans. (i) CuCN

N Cl2 CN

(A)

H O/H2

COOH

(B)

NH3

C – NH2

(C)

O

(ii)Sn/HCl

NO2 NH2

Aniline

NaNO + HCl

273 K

2

Benzene diazoniumchloride

H O/H2

OH

Phenol

Nitrobenzene

N Cl2

25. Write the names and structures of the monomers of the following polymers:

(i) Buna – S (ii) Neoprene

(iii) Nylon – 6, 6

Ans. (i) Buna-S CH = CH – CH = CH2 2 +

CH = CH2

1,3–Butadiene

Styrene

(ii) Neoprene

Cl

CH = C – CH = CH2 2

2–Chloro–1,3–butadiene

(iii) Nylon 6,6

COOH

H N(CH ) NH + (CH )2 2 6 2 2 4

Hexamethylene diamine

COOH

Adipic acid

26. After watching a programme on TV about the adverse effects of junk food and soft drinks onthe health of school children, Sonali, a student of Class XII, discussed the issue with theschool principal. Principal immediately instructed the canteen contractor to replace the fastfood with the fibre and vitamins rich food like sprouts, salad, fruits etc. This decision waswelcomed 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.

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STUDYmate

Ans. (a) They are good educated citizens who believe in the saying “healthy mind in a healthybody.”

This shows they value education and are caring towards the school inmates.

(b) Vitamin B and Vitamin C

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. (a) Sodium Benzoate

(b) To impart antiseptic properties

(c) Narcotic Analgesics

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 concentrations of both A and B are doubled?

(b) A first order reaction takes 40 minutes for 30% decoposition. Calculate t1/2 for thisreaction.

(Given log 1.428 = 0.1548)

OR

(a) For a first order reaction, show that time required for 99% completion is twice the timerequired for the completion of 90% of reaction.

(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ accordig to the equation:

aE 1log k log A

2.303 R T

where Ea is the activation energy. When a graph is plotted for log k Vs. 1

T, a straight

line with a slope of –4250 K is obtained. Calculate ‘Ea’ for the reaction.

(R = 8.314 JK–1 mol–1)

Ans. (a) (i)2dx

k[A] [B]dt

(ii) The rate will become 9 times.

(iii) The rate will become 8 times.

(b) t = 40 min.

Finding rate constant by the given data.

0

t

A2.303k log

t A

2.303 1

k log40 0.7

2.303

log1.42840

2.303

0.154840

= 0.0089 min–1

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STUDYmate

Now, 1/20.693

tk

1/20.693

t0.0089

= 77.865 min.

OR

Ans. (a) 90%

2.303 at log

k a 0.9a

2.303

log10k

2.303

k

99.9%

2.303 at log

k a 0.999a

2.303 alog

k a 0.001a

99.9%2.303

t 3k

99.9% 90%t t 3

(b) The slope of the reaction = aE

2.303 R

Now,aE

4250 K2.303 R

Ea= 4250 × 2.303 × 8.314

= 81375.35 Joules

= 81.375 kJ

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.

(b) Draw the structures of the following molecules:

(i) BrF3

(ii) (HPO3)3(iii) XeF4

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 behaviour less than sulphur.

(b) Draw the structures of the following molecules:

(i) XeF2

(ii) H2S2O8

Ans. (a) (i) In F2, the bond between two F-atoms is formed by the head on overlapping of 2porbitals, while in Cl2 it is formed by the head on overlapping of 3p orbitals.

2p being smaller in size than 3p has more interelectronic repulsions, which makesit weaker, thus its bond enthalpy is lower than that of Cl2.

(ii) Due to presence of H-bonding in NH3 its boiling point becomes more than PH3.

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STUDYmate

(b) (i) BrF3F

Br

T-shapedF

F

(ii) (HPO3)3

O

POHHO

OO

P

P

O O

O OHCyclic trimer

(iii) XeF4

Xe

F

FF

F

Square Planar

OR

Ans. (a) (i) Deep sea divers depend upon compressed air for their oxygen supply. Thus, adeep sea diver has both N2 and O2 dissolve considerably in the blood and otherbody fluids.

Oxygen is used up for metabolism but due to high partial pressure and greatersolubility of N2, it will remain dissolved and will form bubbles when the divercomes to the atmospheric pressure. These bubbles affect nerve impulses andgive rise to a disease called bends or decompression sickness. To avoid bends andalso the toxic effects of high concentration of nitrogen in the blood, the cylindersused by the divers are filled with air diluted with helium.

(ii) Fluorine does not exhibit positive oxidation state because fluorine is the mostelectronegative element.

(iii) Because of stronger S–S bonds as compared to O–O bonds, sulphur has a strongertendency for catenation than oxygen.

(b) (i) Xe

F

F

(ii) H – O – S – O – O – S – O – H

O

O

O

O

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STUDYmate

30. (a) Although phenoxide ion has more number of resonating structures than carboxylateion, 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

(a) Complete the following reactions:

(i) Conc. KOH2H C H||O

(ii) 2Br /P3CH COOH

(iii)

CHOHNO /H SO

273-283 K

3 2 4

(b) Give simple chemical tests to distinguish between the following pairs of compounds:

(i) Ethanal and Propanal (ii) Benzoic acid and Phenol

Ans. (a) The two reasons are :

(i) In carboxylate anion, the resonating structures are exactly equal in energy,equivalent resonating forms. In phenol the energies of the resonating forms differ.

R – C – O

O

R – C = O

OO O etc.

(ii) In the resonating structures of phenol, the negative charge is placed on O-atomin one of the resonating form and on C-atom in another resonating form. Incarboxylic acid, the negative charge is only on O-atom.

We know that the stability of resonating form is more if negative charge is onmore electronegative atom. So the resonating forms of carboxylate anion are morestable than phenol.

(b) (i) Zn Hg/conc. HCl3 3 3 2 3

O||

CH C CH CH CH CH

This is known as Clemmenson’s Reduction.

(ii)

C—Cl

Pd-BaSO , S4

boiling xylene

O

CHO+ HCl

This reaction is called Rosenmund reduction

(iii) dil NaOH/100 C3 3 3

H|

CH CH O CH C O CH CH CH CHO

The above reaction is aldol condensation

OR

Ans. (a) (i) 2H – C – H

O

conc.

KOHHCOOK + CH OH+

3[Canizzaro reaction]

(ii) CH – C – O3 – H

OP/Br2 CH – C – O2 H

O

Br-Bromoacetic acid

[HVZ reaction]

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STUDYmate

× · × · × · × · ×

(iii)CHO

HNO /H SO

273-283 K

3 2 4

CHO

NO2

+ H O2

m-Nitrobenzaldehyde

(b) (i) We can use iodoform test to distinguish between ethanal and propanal.

CH – C = O + 3I + NaOH CHI + HCOONa + 3HI2 3 3

H

yellow ppt

This test is not given by propanal.

(ii) We can use sodium bicarbonate to distinguish between benzoic acid and phenol.

Benzoic acid is a stronger acid than phenol, so it reacts with NaHCO3 (weak base)liberating CO2.

C6H5–COOH + NaHCO3 C6H5COONa + H2O + CO2 (brisk effervescence)

Phenol shows no reaction with NaHCO3.

We can also used neutral FeCl3. It reacts with phenol to give violet colouredsolution.

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STUDYmate

1. What type of stoichiometric defect is shown by AgCl?

Ans. Frenkel Defect

2. Write the IUPAC name of CH CH = CH 3 – C – CH3

CH3

BrAns. 4-Bromo-4-methylpent-2-ene.

4. What type of bonding helps in stabilising the -helix structure of proteins?

Ans. Hydrogen bonding between – C –

O

and –NH– groups of the peptide bond.

6. What inspired N. Bartlett for carrying out reaction between Xe and PtF6?

Ans. Neil Bartlett prepared a red compound having the formula O+2PtF

6–. He realised that the first

ionization enthalpy of molecular oxygen (1175 kJ mol–1) is similar to Xe (1170 kJ mol–1). So, hewas inspired to synthesise a similar compound using Xenon.

7. What happens when ethyl chloride is treated with aqueous KOH?

Ans. CH3 – CH

2 – Cl + aq. KOH CH

3 – CH

2 – OH + KCl

Nucleophillic substitution reaction takes place and ethanol is formed.

8. Write the structure of 4-chloropentan-2-one.

Ans. H – C – C – C – C – CH3

H

H O H H

H Cl

11. What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions?Give an example of each type.

Ans. In oil/water type emulsion, water is the dispersion medium. Example: milk and vanishingcream.

In water/oil type emulsion, oil is the dispersion medium. Example: butter and cream.

17. (a) Which of the following ores can be concentrated by froth floatation method and why?

Fe2O

3, ZnS, Al

2O

3

(b) What is the role of silica in the metallurgy of copper?

Ans. (a) ZnS because sulphide ores have wettabilty with pine oil which adsorb ore particles onthem.

(b) It acts as flux so that another impurity slag may be formed.

Studymate Solutions to CBSE Board Examination 2012-2013

UNCOMMON QUESTIONS ONLY

Series : SKS/1 Code No. 56/1/2

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STUDYmate

18. (a) Why does p-dichlorobenzene have a higher m.p. than its o– and m–isomers?

(b) Why is () – Butan-2-ol is optically inactive?

Ans. (a) p-dichlorobenzene have a higher m.p. because its packing is more efficient due tosymmetry.

(b) Racemic mixture.

23. Write the names and structures of the monomers of the following polymers:

(a) Polystyrene

(b) Dacron

(c) Teflon

Ans. (a) Styrene

CH = CH2

(b) Ethylene glycol + Terephthalic acid/Benene-1, 4-dicarboxylic acid

HOH2C – CH

2OH + HOOC COOH

(c) Tetra fluoroethene

CF2 = CF

2

× · × · × · × · ×

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STUDYmate

1. What type of substances would make better permanent magnets, ferromagnetic or

ferrimagnetic?

Ans. Ferromagnetic substances

3. What is the composition of ‘Copper matte’?

Ans. Copper matte chiefly consists of Cu2S and some uncharged FeS.

5. What is a glycosidic linkage?

Ans. Glycosidic linkage formed when two mono-saccharides are held together with the loss of

small molecule’s like H2O. It is an oxide linkage which joins two monosaccharides.

7. Which compound in the following pair undergoes faster SN1 reaction?

Cl

and

Cl

Ans.Cl

8. Write the structure of p-Methylbenzaldehyde molecule.

Ans.

CHO

CH3

9. What is the difference between multi-molecular and macromolecular colloids? Give one

example of each.

Ans. When large number of atoms/molecules of a substance on dissolution aggregate together to

form species having size in the colloidal range, then species thus formed is called

multimolecular colloids. E.g. sulphur sol consisting of S8 particles.

When macromolecules in a suitable solvents form solutions in which the size of the

macromolecules is in colloidal range, then such systems are macromolecular colloids e.g.

starch, cellulose.

17. Account for the following:

(i) The C–Cl bond length in chlorobenzene is shorter than that in CH3 – Cl.

(ii) Chloroform is stored in closed dark brown bottles.

Studymate Solutions to CBSE Board Examination 2012-2013

UNCOMMON QUESTIONS ONLY

Series : SKS/1 Code No. 56/1/3

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STUDYmate

Ans. As resonance imparts partial double bond character

(i) to C–Cl bond in chlorobenzene.

Cl Cl

(ii) As it get oxidised in presence of sunlight

2 3 2sunlightPoisonous gas

O 2CHCl 2COCl 2HCl

23. Give the structures of products A, B and C in the followng reactions:

(i) 24 HNOLiAlHKCN

3 2 0 CCH CH Br A B C

(ii)

23 3NaOH BrNH CHCl Alc. KOH

3CH COOH A B C

Ans. (i)

24 HNO , 0 CLiAlHKCN

3 2 3 2 3 2 2 2 3 2 2Prop 1 ol

CH CH Br CH CH CN CH CH CH NH CH CH CH OH(A) (B)

(ii)

23 3NaOH/BrNH CHCl3 3 2 3 2 3Alc. KOH

CH COOH CH CONH CH NH CH NC(A) (B) (C)

27. Write the names and structures of the monomers of the following polymers:

(i) Bakelite (ii) Nylon-6

(iii) Polythene

Ans. (i)

OH

Phenol

HCHOFormaldehyde

&

(ii)

N

CH2

C

O

CH2

CH2

CH2

CH2

H

Caprolactam

(iii) CH2 = CH2 Ethene

× · × · × · × · ×