iit jee 2012 paper 1 solutions

8/2/2019 Iit Jee 2012 Paper 1 Solutions

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MAHESH JANMANCHI IIT JEE 2012 PAPER 1

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PAPER-1

Maximum Marks: 70

Question paper format and Marking scheme:

1. In Section I ( Total Marks: 30), 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: 20), for each question you will be awarded 4 marks if you darken ALL the

bubble(s) corresponding to the correct answer(s) ONLY and zero marks other wise. There are no negative

marks in this section.

3. In Section III (Total Marks: 20), 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 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D)

out of which OLY ONE is correct.

21. Which ordering of compounds is according to the decreasing order of the oxidation state of

nitrogen?

(A) HNO ,NO,NH Cl,N3 4 2

(B) HNO , NO, N , NH Cl3 2 4

(C) HNO ,NH Cl, NO,N3 4 2

(D) NO,HNO ,NH Cl, N3 4 2

Sol. (B)

Oxidation states of nitrogen are as follows:

The correct order is HNO3( +5), NO(-2), N2(0), NH4Cl(-3)

22. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atoms is

[ 0a is Bohr radius]

(A)2

2 240

h

ma(B)

2

2 2160

h

ma

(C)2

2 2320

hma

(D)2

2 2640

hma

Sol. (C)

2

nhmvr

=

21

2KE mv=

0

2

2 2 (4 )

nh hv

mr m a = =

2 22

2 2 2 2

0 0

1 1

2 2 16 32

h hmv m

m a ma = =


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23. The number of aldol reaction(s) that occurs in the given transformation is

conc.aq.NaOHCH CHO+ 4HCHO3

OHOH

OH OH

(A) 1 (B) 2 (C) 3 (D) 4

Sol. (C)

Since it has 3 hydrogens, it undergoes 3 aldol reactions

CH CHO+HCHO3

OH1st aldol

condensation

CH CHO2

CH OH2

OH / HCHO2nd aldol

condensation

CH CHO

CH OH2

CH OH2

OH / HCHO3rd aldol

condensation

HOCH C CHO2

CH OH2

CH OH2

OH / HCHO

Cannizaroreaction

HOCH C CH OH2 2

CH OH2

CH OH2

24. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs. I/V plot is shownbelow. The value of the van der Waals constant a (atm. Litre

2mol

-2)

(A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0

Sol. (C)


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( )2

aP V b RT

v

+ =

aPV RT V

+ =

aPV RT

V= +

( )21.6 2 .........a RT i= +

( )20.1 3 .........a RT ii= +

Solving the equations , a = 1.5

25. In allene ( )C H3 4 , the type(s) of hybridization of the carbon atoms is (are)(A) 3spandsp (B) 2spandsp (C) only sp2 (D) 2 3sp andsp

Sol. (B)

Allene is propadiene

CH = C= CH2 2

sp 2sp

26. A compound M XP q

has cubic close packing (ccp) arrangement of X. Its unit cell structure us

shown below. The empirical formula of the compound is

(A) MX (B) MX2 (C) M2X (D) M5X14


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Sol. (B)

Number of ( )M 4 1/ 4 1 2= + =

( edge -1/4, centre 1)

Number of ( ) ( )X 8 1/ 8 6 1/ 2 4= + = (Corners 1/8, face centre )

M2X4

Therefore , MX2

27. The number of optically active products obtained from the complete ozonolysis of the given

compound is

CH CH CH C CH C CH CH CH3 3

CH = = =

CH3

CH3

H

H

(A) 0 (B) 1 (C) 2 (D) 4

Sol. (A)

The products of ozonolysis are:

None of the above is optically active since none of them have chiral carbon.

28. As per IUPAC nomenclature, the name of the complex ( ) ( )Co H O NH Cl2 3 34 2

is

(A) Tetraaquadiaminecobalt (III) chloride(B) Tetraaquadiamminecobalit (III) chloride

(C) Diaminetetraaquacobalt (III) chloride

(D) Diamminetetraaquacobalt (III) chloride

Sol. (D)

Diamminetetraaquacobalt (III) chloride


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29. The carboxyl functional group (-COOH) is present in

(A) Picric acid (B) barbituric acid (C) Ascorbic acid (D) aspirin

Sol. (D)

Aspirin

COOH

O C CH3

O

30. The colour of light absorbed by an aqueous solution of CuSO4

is

(A) orange-red (B) blue-green (C) yellow (D) violet

Sol. (A)

Orange-red is the complimentary colour of blue in which CuSO4 appears.

Section II

(Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out

of which ONE or MORE are correct.

31. For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. Thefinal state Z can be reached by either of the two paths shown in the figure. Which of the following

choice(s) is (are) correct? [Take S as change in entropy and w as work done]

(A) S S Sx z x zy y

= +

(B) W W Wx z x zy y

= +

(C) W Wx z xy y

=

(D) S Sx z xy y

=


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Sol. (A, C)

Entropy is a state function. Area under the curve is identical for both

Wx zy

and Wx y

as Wy z

is isochoric.

32. Which of the following molecules, in pure form, is(are) unstable at room temperature?

(A) (B) (C)

O

(D)

Sol. (B, C)B is antiaromatic

C is antiaromatic in polar form

33. Identify the binary mixture(s) that can be separated into individual compounds, by differential

extraction, as shown in the given scheme.

Binary mixture containing

Compound 1 and compound 2

NaOH (aq)

( )3NaHCO aq

Compound 1

Compound 1

Compound 2

Compound 2

(A)6 5 6 5C H OH and C H COOH (B) 6 5 6 5 2C H COOH and C H CH OH

(C)6 5 2 6 5C H CH OH and C H OH (D) 6 5 2 6 5 2C H CH OH and C H CH OH

Sol. (B, D)

Acid reacts with NaOH & NaHCO3Phenol reacts with NaOH but not NaHCO3

Alcohol reacts with none of these.

34. Choose the correct reason(s) for the stability of the lyophobic colloidal particles.

(A) Preferential adsorption ions on their surface from the solution

(B) Preferential adsorption of solvent on their surface from the solution

(C) Attraction between different particles having opposite charges on their surface

(D) Potential difference between the fixed layer and the diffused layer of opposite charges around the

colloidal particles

Sol. (A, D)Facts about colloidal solutions.


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35. Which of the following hydrogen halides react(s) with ( )AgNO aq3

to give a precipitate that

dissolves in ( )Na S O aq2 2 3

?

(A) HCl (B) HF (C) HBr (D) HI

Sol. (A, C, D)

AgNO HBr HNO3 3

AgBr+ +

( )AgBr 2Na S O Na Ag S O +NaBr2 2 3 3 2 3 2

+ Similar reactions occur with HCl & HI

Section III

(Integer Answer Type)

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to9 (both inclusive)

36. An organic compound undergoes first-order decomposition. The time taken for its decomposition

to 1/8 and 1/10 of its initial concentration are 1/8t and 1/10t respectively. What is the value of

[ ]

[ ]1/8

1/10

t10?

t (take 10log 2 0.3= )

Sol. (9)

1/8

1/10

1/8

1/10

1

log 1. log8 3log 28 0.9

1. log10 1log

1

10

109

k t

k t

t

t

= = = =

=

37. When the following aldohexose exists in its D-configuration, the total number of stereoisomers in

its pyranose form is

CHO|

CH2

|CHOH

|

CHOH

|

CHOH

|

CH2OH


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Sol. (8)

2CH OH

H

OH

H

HH

HCHO

OH

1

23

4

5

6

This can exist in two forms and .

Arrangement at position 5 remains same, therefore eight stereoisomers are possible

4 each for and -form.

38. The substituents R1 and R2 for nine peptides are listed in the table given below. How many of

these peptides are positively charged at pH = 7.0?

H N CH CO NH CH CO NH CH CO NH CH COO3

H1R 2R H

Peptide R1

R2

I H H

II H 3CH

III2

CH COOH H

IV2 2

CH CONH ( )2 24CH NH

V2 2

CH CONH 2 2

CH CONH

VI ( )2 24CH NH ( )2 24CH NH

VII2CH COOH 2 2CH CONH

VIII2CH OH ( )2 24CH NH

IX ( )2 24CH NH 3CH

Sol. (4)

The peptide must be basic in nature.

IV,VI,VIII,IX


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39. The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons,

undergoes a nuclear reaction yielding element X as shown below. To which group element X

belongs in the periodic table?63 1 1 1

29 1 0 1Cu+ H 6 n + 2 H + X +

Sol. (8)

63 1 1 4 1 52

29 1 0 2 1 26Cu + H 6 n + He + 2 H + Fe

So, X = Fe

group number of Fe is 8.

40. 29.2% (w/w) HCl stock solution has a density of 1.25 g mL-1. The molecular weight of HCl in 36.5g mol

-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4M HCl is

Sol. (8)

M1V1=M2V2

29.2 1.25200 0.4

3.65V

=

200 0.4 3.658

29.2 1.25V

= =

iit jee 2012 paper 1 solutions

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