Chemistry - 2012 Past Year Paper - IIT JAM

Correct option is (b

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

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)

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

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)

[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)

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)

Correct option is (d)

Correct option is (c)

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.

While in P there is OTs is sign to double bond, so reactivity will be less as compared to 'R'. Q is least reactive as there is no pi bond.

Rate of acetolysis R > P > Q

Correct option is (d)

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

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)

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)

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)

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)

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)

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)

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)

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

Correct option is (c)

The average speed,

H₂ = M = 2; O₂ = M = 16; N₂ = 14 = M

Correct option is (a)

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)

for n^th order reaction

When n = 0
t_1/2 ∝ a₀
Correct option is (c)

If molality is m.

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

Correct option is (a)

ΔH = ΔU + Δn_gRT
Formation of CF₂ClCF₂Cl(g)
2C(s) + 2F₂(g) + Cl₂ (g) = CF₂ClCF₂Cl(g)
Δn_g = gaseous mole of product - gaseous mole of reactant.
Δn_g = 1 - 3 = - 2; ΔH = ΔU + Δn_gRT
ΔU = ΔH - Δn_gRT = 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)

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)

Correct option is (d)

reduction

Overall,

Nernst equation,

Correct option is (d)

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)

(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.

(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

(a) K 3[Fe^III (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⁺

(a) Octahedral ligand field:

(b) Orbital overlaps in metal carbonyls:

(a) Complex is [Co(NH₃)₄SO₄]Cl

 No reaction. since SO₄^2– is present in the coordination sphere.

Cl^– is present in the ionic sphere

 [DMGH₂] → Dimethyl glyoximato  

[Ni(II)(DMGH)₂] complex → 

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

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

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

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

(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.

The rate of formation of BD is

... (1)

Applying SSA on CB^*,   

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

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

Therefore, (2) 

And at [CB] = 1 × 10^–5,

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

Dissociation of CH₂ClCOOH, CH₂ClCOOH →H⁺ + CH₂ClCOO^-

L_m^∞ = 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.