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Series : SMA/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 titlepage of the 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 distributedat 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only andwill not write 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) Marks for each question are indicated against it.

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

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

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

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

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

Studymate Solutions to CBSE Board Examination 2011-2012

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STUDYmate

1. What is meant by ‘doping’ in a semiconductor?

Ans. Introducing impurities into elements of Group 13 {B} or Group 15 {P} into the elements

of Group 14 {Si or Ge}. This improves conductivity of Group 14 elements.

2. What is the role of graphite in the electrometallurgy of aluminium?

Ans. Graphite is used as anode and cathode during electrolysis of fused alumina in moltencryolite.

3. Which one of PCl +4 and PCl –

4 is not likely to exist and why?

Ans. –4PCl can not exist, because ‘P’ can not withdraw e– from more electronegative Cl.

4. Give the IUPAC name of the following compound.

CH = C – CH Br2 2

CH3

Ans. 3-Bromo-2-methyl propene.

5. Draw the structural formula of 2-methylpropanl-2-ol molecule.

Ans. CH – C – CH3 3

OH

CH3

6. Arrange the following compounds in an increasing order of their reactivity innucleophilic addition reactions: ethanal, propanal, propanone, butanone.

Ans. butanone < propanone < propanal < ethanal.

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

CH3NH2, (CH3)2NH, (CH3)3N and NH3

Ans. (CH3)2NH > CH3NH2 > (CH3)3N > NH3.

8. Define the term, ‘homopolymerisation’ giving an example.

Ans. Homopolymerization is defined as a polymerization reaction which involvespolymerization of single monomeric species. Example: polythene from ethene.

9. A 1.00 molal aqueous solution of trichloroacetic acid (CCl3COOH) is heated to itsboiling point. The solution has the boiling point of 100.18 °C. Determine the van’tHoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol–1)

OR

Define the following terms:

(i) Mole fraction (ii) Isotonic solutions

(iii) Van’t Hoff factor (iv) Ideal solution

Ans. m = 1 Tb = 100.18 °C

Tob = 100 °C Tb = 0.18 °C

i = ?

Kb = 0.512 K Kg/mol

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STUDYmate

Tb(theo) = Kb × m

Tb(theo) = 0.512 × 1 = 0.512

i =

obs

theo

b

b

T

T

i = 0.180.512

= 0.35156

OR

Ans. (i) Mole fraction: It is the ratio of number of moles of one component to the totalnumber of moles present in solution/mixture.

(ii) Isotonic solutions: Two solutions having same osmotic pressure at sametemperature are isotonic solutions.

(iii) Van’t Hoff factor: It is a ratio of observed colligative property and calculatedcolligative property.

So, Van’t Hoff factor ‘i’ is,

i = observed colligative property

calculated colligative property

i = Total number of moles of particle after association/dissociationNumber of moles of particles before association/dissociation

i = normal molar mass

abnormal molar mass

(iv) Ideal solution: The solutions which obey Raoult’s law over the entire range ofconcentration are known as ideal solutions.

10. What do you understand by the ‘order of a reaction’? Identify the reaction order fromeach of the following units of reaction rate constant:

(i) L–1 mol s–1

(ii) L mol–1 s–1

Ans. It is the sum of powers of concentrations of the reactants in the rate law expression.Order of a reaction can be 0, 1, 2, 3 or a fraction.

(i) Zero (ii) Second

11. Name the two groups into which phenomenon of catalysis can be divided. Give anexample of each group with the chemical equation involved.

Ans. The two groups in which catalysis can be divided are

(i) Homogeneous catalysis: Eg: SO2 (g) + O2 (g) NO(g) SO3 (g) or other examples.

(ii) Heterogeneous catalysis: Eg: Ni/Pt2 4 2 2 6

EthaneEtheneC H (g) H (g) C H (g) or other

examples.

12. What is meant by coagulation of a colloidal solution? Describe briefly any threemethods by which coagulation of lyophobic sols can be carried out.

Ans. The process of settling of colloidal particles is called coagulation or precipitation ofthe sol.

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STUDYmate

Methods by which coagulation of lyophobic sols can be carried out.

(i) By electrophoresis: The colloidal particles move towards oppositely chargedelectrodes, get discharged and precipitated under an applied electric field.

(ii) By mixing two oppositely charged sols: Oppositely charged sols when mixedin almost equal proportions, neutralise their charges and get partially orcompletely precipitated. Mixing of hydrated ferric oxide (+ve sol) and arsenioussulphide (–ve sol) bring them in the precipitated forms. This type of coagulationis called mutual coagulation.

(iii) By boiling: When a sol is boiled, the adsorbed layer is disturbed due to increasedcollisions with the molecules of dispersion medium. This reduces the charge onthe particles and ultimately lead to settling down in the form of precipitate.

13. Describe the principle involved in each of the following processes.

(i) Mond process for refining of Nickel.

(ii) Column chromatography for purification of rare elements.

Ans. (i) Mond process for refining of Nickel: In this process, nickel is heated in astream of carbon monoxide forming a volatile complex, nickel carbonyl whichdecomposes to form pure nickel.

Ni + 4CO330–350 K

Impure

Ni(CO)4Tetracarboxyl

nickel

450–470 KNi + 4CO

Pure

(ii) Column chromatography: Different components of mixture are adsorbed todifferent extents, depending upon their polarity.

14. Explain the following giving an appropriate reason in each case.

(i) O2 and F2 both stabilize higher oxidation state of metals but O2 exceeds F2 indoing so.

(ii) Structures of Xenon fluorides cannot be explained by Valance Bond approach.

Ans. (i) The ability of O2 to stabilise higher oxidation states exceeds that of fluorinebecause oxygen can form multiple bonds with metals.

(ii) According to VBT, if electrons in an orbital are paired (especially of noble gases)then they do not participate in bond formations (or undergo hybridisation).

15. Complete the following chemical equations:

(i) 2 –2 7Cr O H I

(ii) 4 2MnO NO H

Ans. (i) 2 – 32 7 2 2Cr O 14H 6I 2Cr 3I 7H O

(ii) 2 –4 2 3 22MnO 5NO 6H 2Mn 5NO 3H O

16. What is meant by (i) peptide linkage (ii) biocatalysts ?

Ans. (i) Peptide linkage: Peptide is an amide formed between –COOH group and –NH2group of -amino acids in proteins.

(ii) Biocatalysts: Biocatalysts are enzymes, which are complex nitrogenouscompounds produced by living species and catalyse certain biological reactions.

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STUDYmate

17. Write any two reactions of glucose which cannot be explained by the open chainstructure of glucose molecule.

Ans. Any two

(i) Despite having the aldehyde group, glucose does not give 2,4–DNP test, Schiff’stest and it does not form the hydrogensulphite addition product with NaHSO3.

(ii) The pentaacetate of glucose does not react with hydroxylamine indicating theabsence of free –CHO group.

(iii) Glucose is found to exist in two different crystalline forms which are named as and . The -form of glucose (m.p. 419 K) is obtained by crystallisation fromconcentrated solution of glucose at 303 K while the -form (m.p. 423 K) isobtained by crystallisation from hot and saturated aqueous solution at 371 K.

18. Draw the structure of the monomer for each of the following polymers:

(i) Nylon 6 (ii) Polypropene

Ans. (i)

H

N

C = OH C2

H C2

H C2

CH2

CH2

-caprolactam

(ii) H3C – CH = CH2 Propene

19. Tungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is316.5 pm, what is the radius of tungsten atom?

ORIron has a body centred cubic unit cell with a cell dimension of 286.65 pm. Thedensity of iron is 7.874 g cm–3. Use this information to calculate Avogadro’s number.(At. mass of Fe = 55.845u)

Ans. For a body centred cubic, unit cell arrangement,

W( 3) 4a r

W( 3) 316.5

4

r

rW = 137.04 pmOR

Ans. d = 3

A

Z MN a

NA = 3

Z Md a

NA =

10 3

2 55.8457.874 (286.65 10 cm)

NA = 5.931 × 1023 atoms per mol

20. Calculate the amount of KCl which must be added to 1 kg of water so that the freezingpoint is depressed by 2K. (Kf for water = 1.86 K kg mol–1)

Ans. For KCl, presuming complete dissociation, (i = 2)Tf = i × kf × m

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STUDYmate

= 2 × 1.86 × mass of KCl

174.5

2 = 2 1.86 mass of KCl

74.5

Mass of KCl = 74.5

1.86 = 40.05 gm

21. For the Reaction

2NO(g) + Cl2(g) 2 NOCl(g)

the following data were collected. All the measurements were taken at 263 K :

Experiment Initial [NO] Initial [Cl2] (M) Initial rate of disaappearance

No of Cl2 (M/min)

1 0.15 0.15 0.60

2 0.15 0.30 1.20

3 0.30 0.15 2.40

4 0.25 0.25 ?

(a) Write the expression for rate law.

(b) Calculate the value of rate constant and specify its units.

(c) What is the initial rate of disappearance of Cl2 in exp. 4 ?

Ans. (a)2 1

2d[R]

R k[NO] [Cl ]dt

(b) R = k [NO]2 [Cl2]1

k = 2 1 2 1 32

R 0.6 0.6

[NO] [Cl ] [0.15] [0.15] (0.15) = 177.75 [Using experiment (1)]

(c) R = k [NO]2 [Cl2]

R = 177.78 × (0.25)2 × (0.25)

R = 2.78 M / min.

22. How would you account for the following?

(i) Many of the transition elements are known to form interstitial compounds.

(ii) The metallic radii of the third (5d) series of transition metals are virtually thesame as those of the corresponding group members of the second (4d) series.

(iii) Lanthanoids form primarily +3 ions, while the actinoids usually have higheroxidation states in their compounds, +4 or even + 6 being typical.

Ans. (i) They have interstitial spaces on their surface, in which small atoms such as H,C or N can be trapped For e.g. steel.

(ii) This is due to lanthanoid contraction.

(iii) This is attributed to the fact that 5f, 6d and 7s levels are of comparable energiesin case of actinoids.

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STUDYmate

23. Give the formula of each of the following coordination entities.

(i) Co3+ ion is bound to one Cl–, one NH3 molecule and two bidentate ethylenediamine (en) molecules.

(ii) Ni2+ ion is bound to two water molecules and two oxalate ions.

Write the name and magnetic behaviour of each of the above coordination entities.

(At Nos.Co = 27, Ni = 28)

Ans. (i) [CO (NH3) (en)2Cl]+2 – IUPAC name Amminechloridobis(ethane-1,2-diamine)cobalt (III) ion

(ii) [Ni (H2O) (C2O4)2]2– Diaquadioxalatonickelate (II) ion

Magnetic behaviour

(i) Diamagnetic (d2sp3) hybridisation

(ii) Paramagnetic (sp3d2) hybridisation

24. Although chlorine is an electron withdrawing group, yet it is ortho-, para-directing inelectrophilic aromatic substitution reactions. Explain why it is so?

Ans. Chlorine withdraws electrons through inductive effect and releases electrons throughresonance. Through inductive effect, chlorine destabilises the intermediate carbocationformed during the electrophilic substitution.

Through resonance, halogen tends to stabilise the carbocation and the effect is morepronounced at ortho– and para– positions. The inductive effect is stronger thanresonance and causes net electron withdrawal and thus causes net deactivation. Theresonanace effect tends to oppose the inductive effect for the attack at ortho– andpara– positions and hence makes the deactivation less for ortho– and para– attack.Reactivity is thus controlled by the stronger inductive effect and orientation iscontrolled by resonance effect.

25. Draw the structure and name the product formed if the following alcohols are oxidized.Assume that an excess of oxidizing agent is used.

(i) CH3CH2CH2CH2OH

(ii) 2-Butenol

(iii) 2-Methyl-l-propanol

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STUDYmate

Ans. (i) CH3CH2CH2CH2OH 2 2 7K Cr O /H CH3CH2CH2COOH

Butanoic acid

(ii) CH – CH = CH – CH3 2

OH

CH – CH = CH – CHO3

But-2-en-1-al

PCC

(iii) CH – CH – CH OH3 2

CH3

CH – CH – COOH3

CH3

2-Methylpropanoic acid

K Cr O /H2 2 7

+

26. Write chemical equations for the following conversions:

(i) Nitrobenzene to benzoic acid.

(ii) Benzyl chloride to 2-phenylethanamine.

(iii) Aniline to benzyl alcohol.

Ans. (i)

NO2

Sn/HCl

NH2

NaNO + HCl2

O°C

N Cl2

CuCN

CN

H O3+

COOH

(ii)

CH Cl2

KCN (aq)

CH CN2

LiAlH4

CH CH NH2 2 2

(iii)

NH2

NaNO , HCl2

N Cl2

CuCN

CN CH NH2 2

0°C

HNO2

CH OH2

LiAlH4

27. What are the following substances? Give one example of each one of them.

(i) Tranquilizers

(ii) Food preservatives

(iii) Synthetic detergents

Ans. (i) Tranquilizers are a class of compounds used for the treatment of stress, mild oreven severe mental diseases. (e.g.), veronal / luminal / seconal.

(ii) Food preservatives prevent spoilage of food due to microbial growth (e.g.) sodiumbenzoate.

(iii) Synthetic detergents are cleansing agents which have all the properties of soapsbut which actually do not contain any soap. (e.g.)Sodium laurylsulphate.

28. (a) What type of a battery is the lead storage battery? Write the anode and thecathode reactions and the overall reaction occurring in a lead storage batterywhen current is drawn from it.

(b) In the button cell, widely used in watches, the following reaction takes place

Zn(s) + Ag2O(s) + H2O(l) 2(aq)Zn + 2 Ag(s) + (aq)2OH

Determine E° and G° for the reaction.

(given: 2o o

Ag /Ag Zn /ZnE 0.80V, E 0.76V )

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STUDYmate

OR

(a) Define molar conductivity of a solution and explain how molar conductivitychanges with change in concentration of solution for a weak and a strongelectrolyte.

(b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is1500 . What is the cell constant if the conductivity of 0.001 M KC1 solution at298 K is 0.146 × 10–3S cm–1?

Ans. (a) It is a secondary cell.

Anode: Pb + SO2–4 PbSO4 + 2e–

Cathode: PbO2 + 4H+ + SO2–4 + 2e– PbSO4 + 2H2O

Overall reaction: Pb + PbO2 + 4H+ + 2SO2–4 2PbSO4 + 2H2O

(b) Zn is oxidized and Ag2O is reduced.

cellE = Ag O,Ag (r eduction) 2Zn / Zn (oxidation)2E E

= 0.8 – (–0.76) = 1.56 V

G° = – cellnFE = – 2×96500 × 1.56 = 301080 J.

ORAns. (a) Molar conductivity of a solution at a given concentration is the conductance of

the volume V of solution containing one mole of electrolyte kept between twoelectrodes with area of cross-section A and distance of unit length.

m = 2 1K 1000S cm mol

C

Molar conductivity increases marginally with decrease in concentration/increasein dilution for a strong electrolyte.

KCl

CH COOH3

(C)½

m

For a weak electrolyte, m increases steeply with decrease in concentration/increase in dilution. This is in accordance with Le chatelier principle.

(b) K = 1

R A

or K = 1

G *R

G*

A = Cell constant

G* = R × K

= 1500 ohms × 0.146 × 10–3 Ohm–1 cm–1

= 1500 × 0.146 × 10–3 cm–1

= 219 × 10–3 cm–1.

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STUDYmate

29. (a) Complete the following chemical equations:

(i) P4 + SO2Cl2 (ii) XeF6 + H2O

(b) Predict the shape and the asked angle (90° or more or less) in each of the followingcases

(i) 23SO and the angle O – S – O

(ii) ClF3 and the angle F – Cl – F

(iii) XeF2 and the angle F – Xe – FOR

(a) (i) 2(hot and conc.)

NaOH Cl

(ii) XeF4 + O2F2

(b) Draw the structures of the following molecules:

(i) H3PO2

(ii) H2S2O7

(iii) XeOF4

Ans. (a) (i) P4 + 8SO2Cl2 4PCl3 + 4SO2 + 2S2Cl2

(ii) XeF6 + H2O 2HF + XeOF4

(b) (i) Shape of SO2–3 – pyramidal

Bond angle – More than 90°.

(ii) ClF3 Bent T-shape

Bond angle – Less than 90°

(iii) XeF2 Linear shape

Bond angle – More than 90°OR

Ans. (a) (i) 6NaOH + 3Cl2 5NaCl + NaClO3 + 3H2O(hot and conc.)

(ii) XeF4 + O2F2 XeF6 + O2

(b) (i) H3PO2P

H

O

H OH

(ii) H2S2O7S

OOOH

O

S

OOH

O

(ii) XeOF4Xe

O

F F

F F

30. (a) Illustrate the following name reactions giving suitable example in each case:

(i) Clemmensen reduction (ii) Hell-Volhard-Zelinsky reaction

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STUDYmate

(b) How are the following conversions carried out?

(i) Ethylcyanide to ethanoic acid

(ii) Butan-1-ol to butanoic acid

(iii) Benzoic acid to m-bromobenzoic acid

OR

(a) Illustrate the following reactions giving a suitable example for each.

(i) Cross aldol condensation (ii) Decarboxylation

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

(i) Pentan-2-one and Pentan-3-one

(ii) Benzaldehyde and Acetophenone

(iii) Pehnol and Benzoic acid

Ans. (a) (i) Clemmensen Reduction

CH – C – H3

O

H C – CH3 3

Zn(Hg) inconc. HCl

Ethanal Ethane

H C – C – CH3 3

O

H C – CH – CH3 2 3

Zn(Hg) inconc. HCl

Propanone Propane

(ii) Hell-Volhard-Zelinsky reaction

H C – CH – COOH3 2 H C – CH – COOH3

(i) Cl /Red P.

(ii) H O2

2

Cl-Chloropropanoic acidPropanoic acid

(b) (i) CH CH CN3 2 CH CH CONH3 2 2

H O3

+

CH CH COOH3 2

NH , 3

NaOH, Br2

CH CH NH3 2 2

HONOCH CH OH3 2

[O]

CH COOH3

(ii) CH – CH – CH – CH – OH3 2 2 2 H C – CH – CH – C – OH3 2 2

K Cr O /dil. H SO

oxidation2 2 7 2 4

O

Butanoic acidButan-1-ol

(iii)

C – OH

O

+ Br2

AlBr or FeBr3 3

C – OH

O

BrBenzoic acid m-Bromobenzoic acid

OR

Ans. (a) (i) Cross Aldol condensation: When aldol condensation is carried out betweentwo different aldehydes and / or ketones, it is called cross aldolcondensation. If both of them contain -hydrogen atoms, it gives a mixtureof four products. This is illustrated below by aldol reaction of a mixture ofethanal and propanal.

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STUDYmate

CH CHO +CH CH CHO

3

3 2

CH – CH = CH – CHO3

(i) NaOH

(ii) But-2-enal

CH CH – CH = C – CHO3 2

2-Methylpent-2-enal from twomolecules of propanal

+

CH3

from two molecules of ethanal

simple or self aldol products

+

CH – CH = C – CHO3

CH3

CH CH – CH = CHCHO3 2+

2-Methylbut-2-enal Pent-2-enal

cross aldol products

from one molecule of ethanal and one molecule of propanal

(ii) Decarboxylation: Carboxylic acids lose carbon dioxide to formhydrocarbons when their sodium salts are heated with sodalime (NaOHand CaO in the ratio of 3 : 1). The reaction is known as decarboxylation.

R–COONa NaOH & CaOHeat R–H + Na2CO3

Alkali metal salts of carboxylic acids also undergo decarboxylation onelectrolysis of their aqueous solutions and form hydrocarbons having twicethe number of carbon atoms present in the alkyl group of the acid. Thereaction is known as Kolbe electrolysis.

2CH3COONa (aq) + 2H2O electrolysis CH3CH3 + 2CO2 + 2NaOH + H2

(b) (i) CH – C – CH CH – CH3 2 2 3

O

Pentan-2-one

gives yellow precipitate of Iodoform with NaOH

and iodine.

H C – CH – C – CH – CH3 2 2 3

O

Pentan-3-one

does not give yellow precipitate.

(ii)

C – OH

O

Benzaldehyde

C – CH3

O

Acetophenone

No reaction Gives yellow precipitate of Iodoform withNaOH and iodine.

(iii)

OH COOH

(I) Gives violet colouration No reactionwith neutral FeCl3 solution.

(II) No reaction Gives brisk effervescence with NaHCO3.

× · × · × · × · ×

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STUDYmate

1. Write a point of distinction between a metallic solid and an ionic other than metalliclustre.

Ans. Metallic solids are good conductors but an ionic solid is a a bad conductor of electricity.

11. Describe a conspicuous change observed when

(i) a solution of NaCl is added to a sol of hydrated ferric oxide.

(ii) a beam of light is passed through a solution of NaCl and then through a sol.

Ans. (a) Addition of NaCl to a sol of hydrated Fe2O

3 causes coagulation.

(b) When a beam of light is passed through NaCl solution, no tyndall effect isproduced. But when it is passed through a sol tyndal effect is produced.

13. Describe the following :

(i) The role of cryolite in electro metallurgy of aluminium.

(ii) The role of carbon monoxide in the refining of crude nickel.

Ans. (i) In the metallurgy of Aluminium, purified Al2O

3 is mixed with cryolite.

Cryolite lowers melting point of the mix and brings conductivity.

(ii) CO forms a volatile complex. (Nickle tetralarbonyl)

300 350 K4Ni 4CO Ni [CO]

The carbonyl is subjected to higher temperture so that it is decomposed giving apure metal.

450 470 K4Ni [CO] Ni 4CO

18. Write the main structural difference between DNA and RNA. Of the two bases, thymineand uracil, which one is present in DNA ?

Ans. The main structural difference between DNA and RNA is

(i) DNA has double helical structural

RNA has single strand structure

(ii) DNA contains cytosine and thymine as pyrimidine bases

RNA contains cytosine and uracil as pyrimidine bases.

20. A solution of glycerol (C3H

8O

3) in water prepared by dissolving some glycerol in 500 g

of water. This solution has a boiling point of 100.42 °C while pure water boils at 100°C. What mass of glycerol was dissolved to make the solution ?

(Kb for water = 0.512 K kg mol–1).

Studymate Solutions to CBSE Board Examination 2011-2012

UNCOMMON QUESTIONS ONLY

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

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STUDYmate

Ans.2H OW = 500 g = 0.5 kg T

b = 100.42°C

0bT = 100°C T

b 0.42 ºC

Wglycerol = ? Tb = K

b . m

Tb = K

b. m

Tb =

2H O

Wglycerol

W Mglycerol

0.42 = 0.512 × Wglycerol

0.5 62

Wglycerol = 0.42 0.5 62

0.512

Wglycerol = 25.429 gm

22. State a reaon for each of the following situations :

(i) Co2+ is easily oxidized to Co3+ in presence of a strong ligand.

(ii) CO is a stronger complexing reagent than NH3.

(iii) The molecular shape of Ni(CO)4 is not the same as that of [Ni(CN)

4]2–

Ans. (i) Co (III) has greater tendency to from coordination complexes than CO (II). Hence,in the presence of ligands, Co (II) changes to Co (III), i.e., is easily oxidized.

(ii) As co creates a synergic effect which strengthens the bond between Co and themetal. NH

3 doesn’t exhibit any synergic effect

(iii) In Ni(CO)4, Ni si in zero oxidation state

NiG.S.

(28)4p4s

Ni0(ES )

sp3 hydbridisation tetrahedral shape×× ×× ×× ××

2–4In (Ni(CN) ]

Ni(28)4p4s3d

2(G.S.)Ni

2ESNi

dsp2

( )Square planar shape

Thus the shapes are different.

23. How would you account for the following ?

(i) With the same d-orbital configuration (d4) Cr2+ is a reducing agent while Mn3+ isan oxidizing agent.

(ii) The actinoids exhibit a larger number of oxidation states than the correspondingmembers in the lanthanoid series.

(iii) Most of the transition metal ions exhibit characteristic in colours in aqueoussolutions.

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STUDYmate

Ans. (i) Cr2+ is reducing as its configuration changes from d4 to d3, the latter having ahalf-filled t

2g level. On the other hand, the change from Mn2+ to Mn3+ results in

the half-filled (d5) configuration which has extra stability.

(ii) There is a greater range if oxidation states, which is attributed to the fact thatthe 5f, 6d and 7s levels are of comparable energies.

(iii) Due to presence of unpaired electrons in d-orbitals.

30. (a) Give a plausible explanation for each one of the following :

(i) There are two – NH2 groups in semicarbazide. However, only one such

group is involved in the formation of semicarbazones.

(ii) Cyclohexanone forms cyanohydrin in good yield but 2, 4,6–trimethylcyclohexanone does not.

(b) An organic compound with molecular formula C9H

10O forms 2, 4, – DNP derivative,

reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorousoxidation it gives 1, 2-benzene-di-carboxylic acid. Identify the compound.

Ans. (a) (i) In semicarbazides, out of two – NH2 groups, one of the –NH

2 group bonded

to – C –

O is involved in conjugation through its lone pair.

H N – C – NH – NH2 2

O

: :

H N = C – NH – NH2 2

: :

O

: ::

(ii)H C3 CH3

CH3

O

highly sterically hindered ketone.

2, 4, 6 – Trimethylcyclohexanone

Due to steric hindrance posed by three methyl groups around carbonylgroup in 2, 4, 6-Trimethyl cyclohexanone, it will not give good yields ofcyanohydrin.

(b) (i) As C9H

10O forms 2, 4-DNP recation it is an aldehyde or ketone.

(ii) As C9H

10O reduces Tollen’s reagent, it is an aldehyde

(iii) As C9H

10O undergoes cannizaro’s reaction it is substitued benzaldehyde.

(iv) As C9H

10O gives 1, 2-Benzene-di-carboxylic acid, the compound is

O

C – HCH – CH2 3

2-Ethylbenzaldehyde

O

C – OH

C – OH

O1, 2-Benzene di-carboxylicalid

OR

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STUDYmate

× · × · × · × · ×

(a) Give chemical tests to distinguish between

(i) Phenol and Benzoic acid

(ii) Benzophenone and Acetophenone

(b) Write the structures of the main products of following reactions :

(i) 3

2

Anhydrous AlCl6 5 CS

C H COCl

(ii)

2

2 4Hg ,H SO3H C – C C – H

(iii)

CH3

NO2

2 2

3

1. CrO Cl

2. H O

Ans. (a) (i) Same as Set 1 Q 30 (b) (iii)

(ii) Acetophenone gives yellow crystals of iodoform on reaction with I2 and

NaOH

Benzophenone doesn’t give yellow crystals of iodoform on reaction with I2

and NaOH

(b) (i)

O

C

(ii) H C – C – CH3 3

O

(iii)

CHO

NO2

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STUDYmate

12. Explain the following terms giving one example for each:

(i) Micelles (ii) Aerosol

Ans. (i) Micelles : These are associated colloids.

For e.g. soap. cleansing action of soap is explained through formation of micellesaround the oil droplet on a fabric, in such a way that hydrophobic part of thesoap molecule is attached to oil droplet and hydrophilic part projects out of theoil droplet.

C

O

ONa+

hydrophobichydrophillic

Since the polar groups can interact with water, the oil droplet is pulled in waterand removed from the dirty surface. Thus soap helps in emulsification andwasing away of oils from the fabric.

(ii) Aerosol : It is a colloidal state. It may be of two types :

Aerosol of solid (e.g., smoke) and Aerosol of liquid (e.g. fog).

20. 15.0 g of an unknown molecular material was dissolved in 450 g of water. The resultingsolution was found to freeze at – 0.34 °C. what is the molar mass of this material (Kffor water = 1.86 K kg mol–1).

Ans. Tf= Kf × molality

0 – (– 0.34) = 1.86 K kg mol–1 ×

15g 1000

M 450

M = 11.86 K kg mol 15g 1000

0.34 K 450 kg

= 182.35 gm/mol

22. Explain the following observations giving an appropriate reason for each.

(i) The enthalpies of atomization of transition elements are quite high.

(ii) There occurs much more frequent metal-metal bonding in compounds of heavytransition metals (i.e., 3rd series).

(iii) Mn2+ is much more resistant than Fe2+ towards oxidation.

Ans. (i) This is due to strong metallic bonding.

(ii) Greater the number of unpaired electrons, stronger and more frequent is theresultant bonding.

Studymate Solutions to CBSE Board Examination 2011-2012

UNCOMMON QUESTIONS ONLY

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

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STUDYmate

(iii) Mn2+ has stable half filled (3d5) configuration. Hence does not get oxidised. ButFe2+ has 3d6 configuration hence it can lose 1 electron to become the morestable Fe3+ of 3d5 configuration (or) Fe2+ easily gets oxidised.

23. Write the name, the structure and the magnetic behaviour of each one of the followingcomplexes:

(i) [Pt(NH3)Cl(NO2)] (ii) [Co(NH3)4Cl2]Cl

(iii) Ni(CO)4

(Atomic numbers : Co = 27, Ni = 28, Pt = 78)

Ans. (i) [Pt(NH3)Cl(NO2)]

IUPAC : Amminechloridonitrito-N-platinum (II)

Structure : Trigonal planar

Magnetic behaviour : diamagnetic

(ii) [Co(NH3)4Cl2]Cl

IUPAC : Tetraamminedichloridocobalt (III) chloride

Structure : Octahedral

Magnetic behaviour : diamagnetic

(iii) Ni(CO)4

IUPAC : Tetracarbonylnickel (0)

Structure : Tetrahedral

Magnetic behaviour : diamagnetic

27. Explain the following terms giving one example of each type:

(i) Antacids (ii) Disinfectants

(iii) Enzymes

Ans. (i) Antacids : Antacids are mild bases which neutralise the excess production ofacid in the stomach. E.g., NaHCO3

(ii) Disinfectants : Disinfectants are chemical substances which either kill or presentthe growth of microorganisms. These are applied to inanimate objects such asfloor, drainage system. E.g., chlorine

(iii) Enzymes : Enzymes are proteins/macromolecules of biological origin whichperform various functions in the metabolic reactions. They are biologicalcatalysts. E.g., maltase, amylase

30. (a) Draw the molecular structures of following compounds:

(i) XeF6 (ii) H2S2O8

(b) Explain the following observations:

(i) The molecules NH3 and NF3 have dipole moments which are of oppositedirection.

(ii) All the bonds in PCl5 molecule are not equivalent.

(iii) Sulphur in vapour state exhibits paramagnetism.

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STUDYmate

OR

(a) Complete the following chemical equations:

(i) XeF4 + SbF5

(ii) Cl2 + F2 (excess)

(b) Explain each of the following:

(i) Nitrogen is much less reactive than phosphorus.

(ii) The stability of + 5 oxidation state decreases down group 15.

(iii) The bond angles (O – N – O) are not of the same value in NO2– and NO2

+.

Ans. (a) (i) XeF6 :

F

XeF

F

F

F

F

(ii) H2S2O8 : S

OOH HO

Peroxodisulphuric acid(H S O )2 2 8

O

O O

O

OS

(b) (i) N

HHH

N

FFF

Since ‘N’ is more electronegative than H hence the dipole is pointing towards‘N’ in NH3.

In NF3, F is more electronegative than ‘N’ hence the dipole is pointingtowards ‘F’.

(ii) The three equatorial P–Cl bonds are equivalent, while the other two axialbonds are longer than equatorial bonds. This is due to the fact that theaxial bond pairs suffer more repulsion as compared to equatorial bondpairs.

P

Cl Cl

Cl

Cl

240 pm

Cl

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STUDYmate

(iii) In vapour state sulphur partly exists as S2 molecule which has two unpairedelectrons in the antibonding * orbitals like O2 and, hence, exhibitsparamagnetism.

OR

(a) (i) XeF4 + SbF5 [XeF3]+ [SbF6]–

(ii) Cl2 + 3F2 (excess) 2ClF3

(b) (i) Nitrogen is associated with high bond dissociation energy and henceunreactive.

(ii) This is due to inert pair effect.

(iii) In NO2+, due to linear geometry the bond angle is 180°, but in NO–

2, due tobent shape, it is greater than 90° and thus different.

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