Chemistry - 2012 Past Year Paper


41 Questions MCQ Test IIT JAM Past Year Papers and Model Test Paper (All Branches) | Chemistry - 2012 Past Year Paper


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This mock test of Chemistry - 2012 Past Year Paper for IIT JAM helps you for every IIT JAM entrance exam. This contains 41 Multiple Choice Questions for IIT JAM Chemistry - 2012 Past Year Paper (mcq) to study with solutions a complete question bank. The solved questions answers in this Chemistry - 2012 Past Year Paper quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this Chemistry - 2012 Past Year Paper exercise for a better result in the exam. You can find other Chemistry - 2012 Past Year Paper extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.
QUESTION: 1

Molecular shape of SOCl2 is

Solution:

Correct option is (b

QUESTION: 2

Number of three-centre two-electron (3e-2e) bonds present in diborane is:

Solution:

2(3c – 2e) bonds are present in diborane Correct option is (a)

QUESTION: 3

The lattice energy of LiF calculated from Born-Lande equation – 1000 kJ mol–1. Assume that for both LiF and MgO the Madelung constants, interionic distances and Born exponents have the same value. The lattice energy of MgO in kJ mol–1 is:

Solution:

From Born-Lande equation

because ‘n’ is always greater than one.
LiF = 1000 kJ mole–1        ... (1)

MgO = x  =      ... (2)

x = 4000 kJ mole–1 Correct option is (d)

QUESTION: 4

The hydrogen-oxygen fuel cells have electrolyte solution of 2.5% __________

Solution:

The hydrogen-oxygen fuel cells have electrolyte solution of 2.5% of KOH. Through the anode hydrogen gas is bubbled.

QUESTION: 5

Number of moles of ions produced by complete dissociation of one mole of Mohr ’s salt in water is

Solution:

Iron (II) ammonium sulfate (NH4)2Fe(SO4).6H2O ions produced by complete dissociation of Mohr salt

(NH4)2 Fe(SO4 )2 6H2O → 2NH+4 + Fe2+ + 2SO4-2
Total ions = 5 Correct option is (c)

QUESTION: 6

The tetrachloro complexes of Ni(II) and Pd(II) respectively, are (atomic numbers of Ni and Pd are 28 and 46 respectively)

Solution:

[NiCl4]2– Paramagnetic


Pairing occurs in case of Pd sine it has greater nuclear charge leading to quater interaction to the ligands.
Correct option is (d)

QUESTION: 7

The total number of steps involved and number of beta particles emitted in the spontaneous decay of  Pb respectively, are

Solution:

Let ‘x’ = number of alpha-particles emitted y = No. of beta-particles emitted.
Conserving the number of nucleons
238 = 206 + 4x + y(0)
4x = 238 – 206 = 32 x = 8
Now, 92 = 82 + 8(2) + y(–1)
92 = 82 + 16 – y y = 98 – 92 = 6
Number of beta particles = 6
Total steps = 8 + 6 = 14
Correct option is (b)

QUESTION: 8

A filter paper moistioned with ammonical sodium nitroprusside solution turns violet on contact with a drop of alkaline Na2S solution. The violet colour is due to the formation of

Solution:

Correct option is (d)

QUESTION: 9

The species/compounds that are aromatic among the following are

Solution:
QUESTION: 10

The major product obtained in the reaction below is

Solution:


 

Correct option is (c)

QUESTION: 11

The rates of acetolysis for the following norbornyl derivatives are in the order

Solution:

Removal of the tosyl group (the rate determining step) is subject to strong anchimeric assistance by the double bond and forms non-classical carbocation. Hence acetolysis is fastest in case of R.
Among P and Q.

Rate of acetolysis for P is faster than Q due to participation of p-electrons of two allylic 1-6, and 4-5 bonds. Rate of acetolysis R > P > Q

Correct option is (d)

QUESTION: 12

The Haworth projection for a-anomer of D-glucose is

Solution:

–OH which is right will be downward
–OH which is left will be upward.
Correct option is (c)

QUESTION: 13

The complementary DNA sequence of the given DNA 5’-G-A-A-T-T-C-3’ is:

Solution:

Complementary DNA sequence of 5´– G – A – A – T – T – C – 3´ → 3´ – C – T – T – A – A – G – 5´
This is called base pairing in DNA A (adenine) pair with T (thymine) G (guanine) pair with C (cytosine)
Correct option is (c)

QUESTION: 14

The order of nucleophilicity of the following anions in a SN2 reaction is:

Solution:

Among P, R and S conjugate base of the weakest acid is the strongest nucleophile


Between P and Q, is better nucleophilic than . (Large size of anion better nucleophile)
Correct option is (b)

QUESTION: 15

The pair of conformation that has maximum energy difference is:

Solution:

Chair form is the most stable and half chair is the least stable. hence maximum energy difference is between half chair and chair form.
Chair < twist boat < boat < half chair.
[Chair < half chair]
Correct option is (a)

QUESTION: 16

The major mono-sulfonation product of α-tetralone is

Solution:

E+ (SO3) can attack at different positions.

among I, II and III (I and III) are unstable so II is the more appropriate choice.
Correct option is (b)

QUESTION: 17

Electrophilic nitrations of the following compounds follow the trend

Solution:

Among P, Q, R and S, NHCOCH3 and –C2H5 groups are ring activators and –Cl, SO3H are ring deactivators.

Resonance effect dominates over hyper-conjugation hence 
 is the more powerful
activator. between –Cl and –SO3H, –SO3H is the powerful deactivator.
P > Q
Correct order → R > S > P > Q

Correct option is (b)

QUESTION: 18

The compounds those would not respond to tests of both nitrogen and sulfur with sodium fusion extracts are

Solution:

Organic compounds containing C, H, N, S give test with sodium fusion extracts. For test of Nitrogen atom presence of carbon is necessary, since it is the test for (CN) . (I) has no carbon atom hance it would not give test with sodium fusion extracts. (III) has no S atom.
So it would not give test of Sulfur. I and III is appropriate choice.
Correct option is (a)

QUESTION: 19

The correct epimeric pair of the following is

Solution:

Two diastereomers that differ in the configuration around one only stereogenic centre are called epimers.


So, the configuration is changed at the only C3 carbon. Hence it is called epimer and epimer is also called diastereomer.
Correct option is (a)

QUESTION: 20

α-Farnesene shown below is a

Solution:

d-farmesene has 3 isoprene unit hence it is a sesquiterpene. (Molecular formula = (C15H24))
Correct option is (d)

QUESTION: 21

For the equilibrium , the equilibrium constant, KP is expressed as

Solution:


Correct option is (c)

QUESTION: 22

The average speed of H2, N2 and O2 gas molecules is in the order

Solution:

The average speed,

H2 = M = 2; O2 = M = 16; N2 = 14 = M



Correct option is (a)

QUESTION: 23

The enthalpy of vaporization (ΔvapH) is zero at

Solution:

The heat of vaporization diminishes at the critical temperature because above the critical temperature the liquid and vapour phase no longer co-exist.
Correct option is (b)

QUESTION: 24

The half-life of any zero-order reaction is:

Solution:

for nth order reaction

When n = 0
t1/2 ∝ a0
Correct option is (c)

QUESTION: 25

The molality of (NH4)2SO4 solution that has the same ionic strength as 1 mol kg–1 solution of KCl is 

Solution:

If molality is m.

[I is equal to one same as KCl]
1 = 3m; m = 

Correct option is (a)

QUESTION: 26

The standard enthalpy of formation ( Δf U0300 ) at 1 bar and 300 K for the formation of CF2ClCF2Cl(g) from its constituent elements in the standard state is -900 kJ mol-1. Given R = 8.3 J K-1 mol-1, the standard internal energy of formation ( Δ​f U0300 ) at the same pressure and temperature is 

Solution:

ΔH = ΔU + ΔngRT
Formation of CF2ClCF2Cl(g)
2C(s) + 2F2(g) + Cl2 (g) = CF2ClCF2Cl(g)
Δng = gaseous mole of product - gaseous mole of reactant.
Δng = 1 - 3 = - 2; ΔH = ΔU + ΔngRT
ΔU = ΔH - ΔngRT = ΔH + 2RT
= -900 kJ / mole + 2 × 8.314 × 300 JK-1 mole-1 K
= -900 kJ / mole + 5kJ / mole = -895 kJ/mole
Correct option is (b)

QUESTION: 27

The percent transmittance of a solution having absorbance (optical density) 1.0 is

Solution:

A = –log T; –A = log T, 10–A = T
T = 10–A = 10–1 = 1/2 = 0.01
∴%T = 0.1 × 100 = 10
Correct option is (b)

QUESTION: 28

The matrix which transforms 

Solution:

Correct option is (d)

QUESTION: 29

A concentration cell with two hydro gen electro des at t wo differnet pressures is depicted as

The potential (Ecell) of the cell is

Solution:

 

reduction

Overall,

Nernst equation,

Correct option is (d)

QUESTION: 30

An aqueous solution containing 1 g L–1 of a polymer exerts osmotic pressure of 4 torr at 300 K.Given R = 0.082 L atm, the molar mass (g mol–1) of the polymer is

Solution:

Given that mass of solute, w = 1.0 g L Volume, V = 1 ltr.
Osmotic pressure, P = 4 Torr T = 300 K

∴ Molar mass = 

(∵ 1 atm = 760 torr)


= 4674 g mole–1

Correct option is (c)

QUESTION: 31

(a) identify the most acidic compound from the following: CH3 – CH3, CH2 = CH2 and CH ≡ CH, and justify your answer. Draw overlap of the orbitals to show bonding in the most acidic compound using the concept of hybridization.
(b) Write a balanced chemical equation to represent acid-base reaction of orthoboric acid in water.
Addition of ethylene glycol to aqueous orthoboric acid enhances its acidity. Explain the above statement using appropriate chemical equation.


Solution:

(a) CH ≡ CH is the most acidic compound among ethane and ethylene.

  • The acidity of H–A increases as the percent S-character of the A increases.
  • The higher the percent S-character of the hybrid orbital, the closer the lone pair is held to the nucleus, and the more stable the conjugate base.

Acetylene → H - C ≡ C - H

(b) orthoboric acid in water:

Addition of ethylene glycol of aqueous orthoboric acid forms chelated complex compound and generates H+ ions.

QUESTION: 32

(a) Draw the unit cell structure of NaCl. Calculate the limiting radius ratio of any ionic solid having NaCl like structure.
(b) Give molecular formula and structure of the compound formed by reaction of Be(OH)2 with acetic acid.


Solution:

(a) Unit cell of NaCl Cl = F.C.C arrangement Na+ = All octahedral voids.

Limiting Radius Ratio:
Let a = Edge length





 = 1.414 -1 =  0.414

QUESTION: 33

(a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 µB and 1.73 µB, respectively.
Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.

(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of O–N–O bond angles.


Solution:

(a) K 3[FeIII (oxalate)3 ]

 

= 5.91 BM

Ligand field electronic configuration = 

Ligand field electronic configuration =

In both the complexes ligand is same but one complex is high spin and another complex is low spin. Fe and Ru belongs to same group. On moving down the group crystal field splitting energy (Δ0) increases. hence Ru complex is a low spin complex.

increasing order of ONO bond angles NO2- < NO2 < NO2+

QUESTION: 34

(a) Show with labels the splitting of d-orbitals in an octahedral ligand field. Calculate the CFSE of (i) high spin d6 and (ii) low spin d6 metal ions in octahedral field. (b) Schematically represents orbital overlaps in metal carbonyls. Show the correct signs of the lobes


Solution:

(a) Octahedral ligand field:


(b) Orbital overlaps in metal carbonyls:

QUESTION: 35

(a) A coordination compound is composed of one Co(III), one chloride, one sulfate and four molecules of ammonia. The aqueous solution of the compound gives no precipitate when combined with aqueous BaCl2, while a white precipitate is formed with aqueous AgNO3 solution. Draw its structure and explain the observations with chemical equations.

(b) Draw the structure of dimethylglyoxime (DMGH2) and its Ni(II) complex formed in aqueous ammonia.


Solution:

(a) Complex is [Co(NH3)4SO4]Cl

 No reaction. since SO42– is present in the coordination sphere.

 

Cl is present in the ionic sphere

 [DMGH2] → Dimethyl glyoximato  

[Ni(II)(DMGH)2] complex → 

QUESTION: 36

(a) Write the structure of E, F and G in the following scheme of reactions.

(b) Identify the structure of H and I in the following synthetic transformation


Solution:

(a) Chemical transformation involved in above chemical reaction can be illustrated as




QUESTION: 37

(a) Complete the following reaction sequence with appropriate structures of J, K and L.

(b) Identify the structures of M and N in the following synthetic transformation


Solution:

(a) Chemical transformation involved in above chemical reaction can be illustrated as



 

(b) Chemical transformation involved in above chemical reaction can be illustrated as


QUESTION: 38

(a) In the following reaction scheme, write the structure of O, P and Q

(b) Given below are structure of some natural products. Identify them as vitamin A, B6, C and D and classify them according to their classes (isoprenoid, alkaloid, carbohydrate and steroid)


Solution:

(a) Chemical transformation involved in above chemical reaction can be illustrated as










 

QUESTION: 39

(a) Write the appropriate structures for R, S and T in the following scheme.


                                                                 

b) Choose the correct stereoisomer between U and V that would furnish W on controlled hydrolysis. Write the stable conformation of W


Solution:

(a) Chemical transformation involved in above chemical reaction can be illustrated as







 

In both U and V, V is the correct choice. Since in V both groups are anti to each other. (Suitable condition for SN2)

Stable conformation of W → due to presence of three membered ring. The cyclohexane ring is deformed chair.

QUESTION: 40

The mechanism of isomerization of cyclobutene (CB) to 1, 3-butadiene (BD) is as follows.

(a) Show that the rate law is 

(b) The apparent first-order rate constant,    At the CB concentration of 1 × 10–5 mol dm–3, the value of kapp reaches 50% of its limiting value obtained at very high concentrations of CB. Evaluate the ratio


Solution:


(a) The rate of formation of BD is

... (1)

Applying SSA on CB*,   

Therefore, (1) 
Hence proved.
(b) Given, ... (2)

At very high concentration of CB, k -1[CB] >> k 2

Therefore, (2) 

And at [CB] = 1 × 10–5,

QUESTION: 41

(a) The molar conductance of 0.012 mol dm–3 aqueous solution of chloroacetic acid is 100 Ω-1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 Ω-1 cm2 mol-1 and 300 Ω-1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.
(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.


Solution:

(a) Concentration of chloroacetic acid = 0.012 mol dm–3

Dissociation of CH2ClCOOH, CH2ClCOOH →H+ + CH2ClCOO-

Lm = conductance of H+ + conductance of CH2ClCOO-

= 50 + 300 = 350 Ω-1 cm2 mol-1




(b) (1) Weak acid with strong base: CH3COOH with NaOH

(i) Weak acid (CH3COOH) has low conductance than its salt (CH3COONa). Due to its lower dissociation w.r.t. salt. So, addition of base increase conductance (region ab in graph)

 

(ii) The reson for initial decrease in conductance (region (Oa)) is because small addition of base leads to common anion i.e. CH3COO ion formation which represses the dissociation of CH3COOH which gives highly conducting H+ ions. (iii) After the end point there is higher rate of increases in conductance because Na+ + OH has higher conductance than Na+ + CH3COO(region bc in graph).

(2) Weak acid with weak base: CH3COOH + NH4OH →CH3COO- + NH4+ + H2O
(i) Weak acid (CH3COOH) has low conductance than its salt (CH3COONa). Due to its lower dissociation w.r.t. salt. So, addition of base increase conductance (region ab in graph). (ii) The reason for initial decrease in conductance (region(Oa)) is because small addition of base leads to common anion i.e. CH3COO ion formation which represses the dissociation of CH3COOH which gives highly conducting H+ ions.

(iii) The region b´c´ of graph is due to weak nature of base. So, conductance remains almost constant with the addition of weak base after end point.