Iit Jee 2012 Paper 2 Solutions

8/2/2019 Iit Jee 2012 Paper 2 Solutions

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MAHESH JANMANCHI IIT JEE 2012 PAPER 2

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PAPER-2

Maximum Marks: 66

Question paper format and Marking scheme:

1. In Section I ( Total Marks: 24), for each question you will be awarded 3 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases,

minus one (1) mark will be awarded.

2. In Section II (Total Marks: 18), for each question you will be awarded 3 marks if you darken ALL the

bubble(s) corresponding to the correct answer(s) ONLY and zero marks other wise. In all other cases,

minus one (1) mark will be awarded.

3. In Section III (Total Marks: 24), for each question you will be awarded 4 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks if no bubble is darkened. There are no negative

marks in this section.


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SECTION - I

(Single Correct Answer Type)

This section contains 8multiple choice questions, Each question has four choices, (A), (B), (C) and (D) out

of which ONLY ONE is correct.

21. The Shape of 2 2XeO F molecule is

(A) trigonal bypyramidal (B) square planar

(C) tetrahedral (D) see-saw

Sol. (D)

XeO2F2 has trigonal bipyramidal geometry due to sp3d hybridisation of the central atom Xe but due to

the presence of lone pair at equatorial position, its shape is distorted and becomes see-saw.

22. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the

elevation in boiling point at 1 atm pressure is 20

C. Assuming concentration of solute is muchlower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take

10.76K kgmolbK

= )

(A) 724 (B) 740 (C) 736 (D) 718

Sol. (A)

b bT K m =

2 0.76 m=

m 2.63 =

o

0

P P 1000molality

P ''xM

= ( where M is the mol wt of solvent)

760 P 10002.63

760 18x

=

P 724 torr =


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23. The compound that undergoes decarboxylation most readily under mild condition is

(A)

COOHCH COOH

2

(B)

COOHO

(C)

COOH

COOH

(D)

CH COOH2

O

Sol. (B)

In decarboxylation, -carbon acquires partial ve charge. Whenever partial ve charge is stabilized,

decarboxylation becomes easy.

In option (B), it is stabilized by - M & - I of C = O group

24. Using the data provided, calculate the multiple bond energy ( )1kJmol of a C = C bond in C2H2.

That energy is (take the bond energy of a C-H bond as 350 1kJmol )

( ) ( ) ( )2 2 22C +H g C H gs 1H 225 kJ mol =

( ) ( )2C 2C gs 1H 1410kJmol =

( ) ( )2H g 2H g 1

H 330 kJ mol

= (A)1165 (B) 837 (C) 865 (D) 815

Sol. (D)

( ) ( ) ( )2 2 22C +H g C H gs 1H 225 kJ mol =

( ) C C225 [1410 330] [ 350 2 ] + = + +

C C 815 KJ/mol = +


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25. The major product H of the given reaction sequence is

95%H SOCN 2 4CH CH CO CH G H3 2 3 Heat

(A)

CH CH C COOH3

=

CH3 (B)

CH CH C CN3

=

CH3

(C)

CH CH C COOH3 2

CH3

OH

(D)

CH CH C CO NH3 2

=

CH3

Sol. (A)

26. ( ){ }6 5 22

NiCl P C H ( )2 2 5 2

C H exhibits temperature dependent magnetic behavior

(paramagnetic/diamagnetic). The coordination geometries of 2Ni + in the paramagnetic and

diamagnetic states are respectively

(A) tetrahedral and tetrahedral (B) square planar and square planar

(C) tetrahedral and square planar (D) square planar and tetrahedral

Sol. (C)

( ){ }6 5 22

NiCl P C H ( )2 2 5 2

C H contains 2+Ni

In high spin state, it is paramagnetic, sp3

hybridized, tetrahedral.

In low spin state, it is diamagnetic, dsp2, square planar.


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27. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents

used are(A) O2 and CO respectively (B) O2 and Zn dust respectively

(C) HNO3 and Zn dust respectively (D) HNO3 and CO respectively

Sol. (B)

In extraction of silver, Ag2S is leached with KCN in presence of air:

( )2 2 2 2 32Ag S+NaCN+O Na Ag CN +Na S O

Thus, O2 is oxidizing agent.

( ) ( )2

2 42Ag CN +Zn Zn CN +2Ag

So, Zn is reducing agent

28. The reaction of white phosphorus with aqueous NaOH gives phosphine along with another

phosphorus containing compound. The reaction type; the oxidation states of phosphorus in

phosphine and the other product are respectively

(A) redox reaction -3 and -5 (B) redox reaction; +3 and +5

(C) disproportion reaction; -3 and +5 (D) disproportion reaction; -3 and +3

Sol. (None of the above Zero marks to all )

( ) ( )P S +NaOH PH NaH PO aq4 3 2 2

+

(-3) (+1)

Na PO +PH3 4 3(+5)

Oxidation states of P in Na PO &PH3 4 3

are +5 & -3 respectively.

It is a disproportionation reaction.


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SECTION . II

(Paragraph Type)

This section contains 6 multiple choice questions relating to three paragraphs with two questions on each

paragraph. Each question has four choices (A), (B) (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 29 and 30

The electrochemical cell shown below is a concentration cell.

2+M|M (saturated solution of a sparingly soluble salt, MX2) ( )32+||M 0.01mol dm |M

The emf of the cell depends on the difference in concentration of 2M + ions at the two electrodes. The

emf of the cell is 298 K is 0.05 V.

29. The solubility product 3 9K mol dm,sp

of MX2 at 298 K based on the information available for

the given concentration cell is (take 2.303 R 298 / F = 0.059 V )

(A) 151 10 (B) 154 10 (C) 121 10 (D) 124 10

Sol. (B)

( ) ( )2+ 2+M|M aq ||M aq |M

0.001 M

Anode : ( )2+M M aq 2e +

Cathode: ( )2+M aq 2e M+

( ) ( )2+ 2c a

M aq M aq+

( )2+a

cell 3

M aq0.059E 0 log

2 10

=

( )2+a

3

M aq0.059

0.059 log2 10

=

( )2+a

3

M aq2 log

10

=

( )2 3 2+a

10 10 M aq Solubility = S = =

For MX3, ( )3

3 5 15

spK 4S 4 10 4 10 = = =


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30. The value of ( )1G kJ mol for the given cell is (take 1 F = 96500 1Cmol )

(A) -5.7 (B) 5.7 (C) 11.4 (D) -11.4

Sol. (D)3

cellG nFE 2 96500 0.059 10 kJ/mole

= =

= -11.4 kJ/mole


In the following reaction sequence, the compound J is an intermediate.

( ) ( ) CH CO O i H , Pd / C3 2 2I J KCH COONa

3 ( )ii SOCl2( )iii anhyd.AlCl

3

( )J C H O9 8 2 gives effervescence on treatment with NaHCO3 and a positive Baeyers test.31. The compound I is

(A) (B) (C) (D)

Sol. (A)

32. The compound K is

(A) (B)

(C) (D)


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Sol. (C)

Sol. (31 and 32)

Step 1 is perkin reaction.


Bleaching powder and bleach solution are produced on a large scale and used in several household products.

The effectiveness of bleach solution is often measured by iodometry.

33. 25 mL of household bleach solution was mixed with 30 mL of 0.50 MKIand 10 mL of 4 N acetic

acid. In the titration of the liberated iodine, 48 mL of 0.25 N2 3

Na SO was used to reach the end

point. The molarity of the household bleach solution is

(A) 0.48 M (B) 0.96 M (C) 0.24 M (D) 0.024 M

Sol. (C)

Milli mole Hypo 0.25 48= 22 milli mole of Cl=

Milli mole of Cl20.25 48

6millimole2

= =

= Milli mole of Cl2 = milli mole of CaOCl2

So, molarity =6

M=0.24M25


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34. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of

that oxoacid is

(A) 2Cl O (B) 2 7Cl O (C) 2ClO (D) 2 6Cl O

Sol. (A)

( )2CaOCl = Ca OCl Cl

OCl - Hydrochlorite ion, anion of HOCl

Anhydride of HOCl = Cl2O

SECTION - III

(Multiple Correct Answer(s) Type)

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONE orMORE are correct.

35. With respect to graphite and diamond, which of the statement(s) given below is (are) correct?

(A) Graphite is harder than diamond.

(B) Graphite has higher electrical conductivity than diamond.

(C) Graphite has higher thermal conductivity than diamond.

(D) Graphite has higher C Cbond order than diamond.

Sol. (B, D)

(A) Diamond is harder than graphite.

(B) Graphite is a better conductor of electricity than diamond.(C) Diamond is a better conductor of heat than graphite.

(D) Bond order of graphite ( )1.5 > Bond order of diamond ( = 1)

36. With reference to the scheme given, which of the given statement(s) about T, U, V and Wis

(are) correct ?

(A) Tis soluble in hot aqueousNaOH

(B) Uis optically active

(C) Molecular formula ofWis10 18 4

C H O

(D) Vgives effervescence on treatment with aqueous 3NaHCO


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Sol. (A, C, D)

37. Which of the given statement(s) aboutN, O,P and Q with respect toMis (are) correct?

(A)MandNare nonmirror image stereoisomers

(B)Mand O are identical

(C)Mand P are enantiomers

(D)Mand Q are identical

Sol. (A, B, C)

HO

HO HO HO

HOOH

CH3CH3

CH3

ClCl

H

H H H

HH

M N O

Cl


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38. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown

in the figure. Which of the following statement(s) is (are) correct?

(A)T1 = T2 (B) T3 > T1

(B) isothermal adiabaticw > w (D) isothermal adiabaticU > U

Sol. (A, D)

(A) T1 = T2 (due to isothermal)(B) T3 > T1 (incorrect) cooling occurs in adiabatic expansion)

(C) Wisothermal > Wadiabatic { with sign, this is incorrect}

(D) Uisothermal = 0 > Uadiabatic = - ve


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39. The given graphs / data I, II, III and IV represent general trend observed for different

physisorption and chemisorption processes under mild conditions of temperature and pressure.

Which of the following choice(s) about I, II, III and IV is (are) correct ?

(A) I is physisorption and II is chemisorption

(B) I is physisorption and III is chemisorption

(C) IV is chemisorption and II is chemisorption

(D) IV is chemisorption and III is chemisorptions

Sol. (A, C)

I is physisorption because in physisorption on increasing temperature at constant pressure, adsorption

decreasesII is chemisorptions, because on increasing temperature adsorption increases at same pressure due to

activation energy

III is physical adsorption because , extent of adsorption is decreasing on increasing temperature,.IV is representing high enthalpy change during chemical adsorption (due to bond formation)

So, it is chemical adsorption.


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40. For the given aqueous reactions, which of the statement(s) is (are) true?

excess KI + ( )3 6K Fe CN 2 4

dilute H SO

brownish-yellow solution

4ZnSO

White precipitate + brownish-yellow filtrate

colourless solution

2 2 3Na S O

(A) The first reaction is a redox reaction.

(B) White precipitate is ( )3 6 2Zn Fe CN

(C) Addition of filtrate to starch solution gives blue colour.

(D) White precipitate is soluble inNaOHsolution.

Sol. (A, C, D)

KI(aq) + ( ) ( )3 6K Fe CN aq brownish-yellow

( )4+ ZnSO aq

White ppt

colourless

2 2 3Na S O

( ) ( ) ( )Kl aq K Fe CN aq63 4 +

( ) ( )K Zn Fe CN Kl aq62 3 32 +

( ) ( )2I aq S O aq4 6

+

OR

( ){ }K Zn Fe CN 62

(D) with NaOH

( ) ( ) ( ) ( ) ( )42

4 6K Zn Fe CN NaOH Zn OH aq Fe CN aq

62

+ + +

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Iit Jee 2012 Paper 2 Solutions