1. Spin-only magnetic moments (in BM) of [NiCl₂(PPh₃)₂] and [Mn(NCS)₆]^4–, respectively, are

(a) 2.83 and 5.92

(b) 0.00 and 5.92

(c) 2.83 and 1.89

(d) 0.00 and 1.89

Ans. a

Sol.

Correct option is (a)

2. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

in case of Acetal hydrolysis non cyclic ester hydrolize first that means formation of (b) takes place.

Correct option is (b)

3. For Na⁺, Mg²⁺, Al³⁺ and F^–, the CORRECT order of ionic radii is

(a) Al³⁺> Mg²⁺> Na⁺> F^–

(b) Al³⁺> Na⁺> Mg²⁺> F^–

(c) Na⁺> F^–> Mg²⁺> Al³⁺

(d) F^–> Na⁺> Mg²⁺> Al³⁺

Ans. d

Sol. More the positive charge smaller will be the size.

Correct option is (d)

4. The CORRECT order of pK_a for the compounds I to IV in water at 298 K is

(a) I > III > II > IV

(b) IV > II > III > I

(c) IV > III > II > I

(d) I > II > III > IV

Ans. b

Sol. HCo(CO)₄ HCo(CO)₃(PPh₃) (HCo(Co)₃(P(OPh)₃) HCo(CO)₂(PPh₃)₂

Acidic strength acceptor strength of spectator ligands.

5. Among the following, the matrices with non-zero determinant are

(a) P, Q and R

(b) P, Q and S

(c) P, R and S

(d) Q, R and S

Ans. a

Sol. P, Q and R are triangular matrices so determinant will never zero. Hence, P, Q and R give non-zero determinant.

| P | = 1 × 1 × 1 × 1 = 1

| Q | = 1 × 2 × 3 × 4 = 24

| R | = 1 × 2 × 3 × 4 = 24

S in false because there are two columns are same and which give determinant zero.

Hence, P, Q and R correct options.

Correct option is (a)

6. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

7. A pure substance M has lesser density in solid state than in liquid state. The of M is +25 J K^–1mol^–1. The CORRECT representative Pressure-Temperature diagram for the fusion of M is

(a)

(b)

(c)

(d)

Ans. b

Sol. Since the pure substance 'M' has lesser density in solid state than in liquid state

Hence, the pure substance 'M has higher volume in solid state than in liquid state like in case of water.

So an pressure increases, temperature decreases & vice-versa

8. Two sets of quantum numbers with the same number of radial nodes are

(a) n = 3; l = 2; m_l = 0 and n = 2; l = 1; m_l = 0

(b) n = 3; l = 0; m_l = 0 and n = 2; l = 0; m_l = 0

(c) n = 3; l = 1; m_l = 1 and n = 2; l = 1; m_l = 0

(d) n = 3; l = 1; m_l = –1 and n = 2; l = 1; m_l = 0

Ans. a

Sol.

Radial node = n – l – 1

(a) (i) Radial node = n – l – 1

= 3 – 2 – 1 = 0

(ii) Radial node = 2 – 1 – 1 = 0

(b) (i) Radial node = 3 – 0 – 1 = 2

(ii) Radial node = 2 – 0 – 1 = 1

(c) (i) Radial node = 3 – 1 – 1 = 1

(ii) Radial node = 2 – 1 – 1 = 0

(d) (i) Radial node = 3 – 1 – 1 = 1

(ii) Radial node = 2 – 1 – 1 = 0

Correct option is (a)

9. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

Correct option is (d)

10. A compound shows ¹H NMR peaks at (in ppm) 7.31 (2H), 7.21 (2H), 4.5 (2H) and 2.3 (3H). The structure of the compound is

(a)

(b)

(c)

(d)

Ans. c

Sol.

11. For the consecutive reaction,

C_o is the initial concentration of X. The concentrations of X, Y and Z at time t are C_X, C_Y and C_Z, respectively. The expression for the concentration of Y at time t is

(a)

(b)

(c)

(d)

Ans. d

Sol.

This is a consecutive reaction and [Y] at any time 't' is given by

12. Half-life (t_1/2) of a chemical reaction varies with the initial concentration of reactant (A₀) as given below:

The order of the reaction is

(a) 1

(b) 3

(c) 2

(d) 0

Ans. c

Sol. For n^th order

13.

(a) 0

(b) 1

(c)

(d)

Ans. a

Sol.

If integrating factor is odd and limit is symmetric then integration will be zero.

So, option (a) is correct.

14. The major products E and F formed in the following reactions are

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

15. Reaction of BCl₃ with NH₄Cl at 140ºC produces compound P. further, P reacts with NaBH₄ to give a colorless liquid Q. The reaction of Q with H₂O at 100ºC produces compound R and a diatomic gas S. Among the following, the CORRECT statement is

(a) S is Cl₂

(b) P is B₃N₃H₆

(c) R is [B(OH)NH]₃

(d) Q is [BClNH]₃

Ans. c

Sol.

Correct option is (c)

16. According to the crystal field theory, d–d transition observed in [Ti(H₂O)₆]³⁺ is

(a) Laporte forbidden and spin allowed

(b) Laporte allowed and spin forbidden

(c) Laporte forbidden and spin forbidden

(d) Laporte allowed and spin allowed

Ans. a

Sol.

Correct option is (a)

17. The CORRECT combination for metalloenzymes given in Column I with their catalytic reactions in Column II is

(a) (i)–(M); (ii)–(K); (iii)–(L); (iv)–(N)

(b) (i)–(N); (ii)–(K); (iii)–(L); (iv)–(M)

(c) (i)–(M); (ii)–(N); (iii)–(K); (iv)–(L)

(d) (i)–(N); (ii)–(L); (iii)–(K); (iv)–(M)

Ans. b

Sol. (i) – (N); (ii) – (K); (iii) – (L); (iv) – (M)

Correct option is (b)

18. The CORRECT statement regarding the molecules BF₃ and CH₄ is

(a) CH₄ is microwave active and infrared inactive

(b) Both BF₃ and CH₄ are infrared active

(c) BF₃ is microwave active and infrared active

(d) Both BF₃ and CH₄ are microwave active

Ans. b

Sol. For microwave active molecule should have permanent dipole moment & for IR active molecule must have dynamic dipole moment.

CH₄ = IR active

BF₃ = active

Correct option is (b)

19. Monochromatic X-rays having energy 2.8 × 10^–15 J diffracted (first order) from (200) plane of a cubic crystal at an angle 8.5º. The length of unit cell in Å of the crystal (rounded off to one decimal place) is

(Given: Planck's constant, h = 6.626 × 10^–34 J s; c = 3.0 × 10⁸ m s^–1)

(a) 9.8

(b) 4.8

(c) 2.4

(d) 3.4

Ans. b

Sol.

20. Hybridization of the central atoms in I₃^–, ClF₃ and SF₄, respectively, are

(a) sp³d, sp² and dsp²

(b) sp, sp² and sp³d

(c) sp, sp³d and dsp²

(d) sp³d, sp³d and sp³d

Ans. d

Sol.

Correct option is (d)

21. Reaction of [Ni(CN)₄]^2– with metallic potassium in liquid ammonia at –33ºC yields complex E. The geometry and magnetic behavior of E, respectively, are

(a) Octahedral and paramagnetic

(b) Tetrahedral and diamagnetic

(c) Square pyramidal and paramagnetic

(d) Square planar and diamagnetic

Ans. b

Sol.

22. The reaction that produces the following as a major product is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

23. The complex that does NOT obey the 18-electron rule is

(Given: Atomic numbers of Ti, Mn, Ta and Ir are 22, 25, 73 and 77, respectively)

(a)

(b) [TaCl₃(PEt₃)₂(CHCMe₃)]

(c) [Mn(SnPh₃)₂(CO)₄]^–

(d)

Ans. b

Sol.

= 5 + 4 + 8 + 1

= 18

(b) [TaCl₃(PEt₃)₂(CHCMe₃)] e^– count = 5 + 3 + 2 × 2 + 2

= 14

(c) [Mn(SnPh₃)₂(CO)₄]^–

electron count = 7+ 1 × 2 + 4 × 2 + 1

= 18

electron count = 5 + 9 + 2 + 2 = 18

b is the correct option.

24. The products P, Q, R and S formed in the following reactions are

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

25. The decreasing order of C=C bound length in the following complexes is

(a) II > III > IV > I

(b) IV > II > III > I

(c) II > IV > III > I

(d) IV > II > I > III

Ans. b

Sol.

Because O.S. of metal is same in all the cases, to determine the C = C bond length electron withdrawing capacity of ligand will be used.

EWG F₂C = CF₂> N₂C = CN₂> F₂C = CH₂

More EWG Less will be e^– donation to metal less strength of M – C = C bond Also less strength of C = C bond

Order of bond length of C = C

Correct option is (b)

26. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

C – C bond length extent of back bonding acceptor strength of alkene.

acceptor strength of alkene e^– with drwaing strength of gp.

e^– with drawing strength

– CN > – F > – H

acceptor strength of alkene

C₂(CN)₄> C₂F₄> C₂H₂F₂> C₂H₄

C – C bond length : b > d > c > a

27. The major product formed in the following reaction sequence is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

28. The volume correction factor for a non-ideal gas in terms of critical pressure (p_c), critical molar volume (V_c), critical temperature (T_c) and gas constant (R) is

(a)

(b)

(c) 3p_cV_c²

(d)

Ans. b

Sol.

a Measure of strength of intermolecular force (pressure correction factor)

b volume correction factor

We know,

Substitute these expression into equation (i), we get

29. In the following reaction, compound Q is

(a)

(b)

(c)

(d)

Ans. b

Sol.

30. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. d

Sol.

31. The functional group (s) in reducing sugar that tests positive with Tollen's reagent is (are)

(a) Aldehyde

(b) Ketone

(c) Acetal

(d) Hemi – acetal

Ans. a, d

Sol. Aldehyde and Hemi-acetal reduced tollen's reagent.

Correct options are (a, d)

32. Hantzsch pyridine synthesis involves several steps. Some of those are

(a) Mannich reaction

(b) Michael addition

(c) Darzens reaction

(d) Aldol reaction

Ans. b, d

Sol. Hantzsch pyridine synthesis involves

Aldol reaction

Michael addition

Mech:

33. The product P and Q formed in the reaction are

(a)

(b)

(c)

(d)

Ans. d

Sol. Correct option is (d)

34. The correct statement(s) about the species is (are)

(a) CpMo(CO)₃ and CpW(CO)₃ are isoelectronic (where Cp is cyclopentadienyl)

(b) BH and CH are isolobal and isoelectronic

(c) CH₃ and Mn(CO)₅ are isolobal

(d) CH₂^– and NH₂ are isolobal and isoelectronic

Ans. d

Sol. (a) CpMo(CO)₃ and CpW(CO)₃

= 5 + 6 + 3 × 2 = 5 + 6 + 3 × 2

= 17 = 17

(b) BH and CH are not isolobal and isoelectronic

e^–count of BH = 3 + 1 = 4

e^–count of CH = 4 + 1 = 5

35. The Correct Maxwell relation(s) derived from the fundamental equations of thermodynamics is(are)

(a)

(b)

(c)

(d)

Ans. **

Sol. From thermodynamic magic square

Parallel walking along corners of magic square

36. Among the following, the anti – aromatic compound(s) is (are)

(a)

(b)

(c)

(d)

Ans. b, c, d

Sol.

37. The Correct statement(s) about sodium nitroprusside is(are)

(a) It contains nitrosyl ligand as NO⁺

(b) It is a paramagnetic complex

(c) It is used for the detection of S^2– in aqueous solution

(d) Nitroprusside ion is formed in the brown ring test for nitrates

Ans. a,c

Sol. Sodium Nitroprusside ion: Na₂ [Fe(CN)₅NO]

(a) Nitroprusside ion is not formed in Brown ring test.

(b) NO⁺ ligand

(c) detection of S^2–

(d) Diamagnetic

38. The complex (es) that show (s) Jahn – Teller distortion is(are)

(a) [Co(CN)₅(H₂O)]^3–

(b) [NiF₆]^2–

(c) [Co(en)₂F₂]⁺

(d) [Mn(CNMe)₆]²⁺

Ans. a,d

Sol. [Co(CN)₅(H₂O)]^3–

Therefore both (a) and (d) complex show Jahn Teller Distortion

39. The pigment responsible for red color in tomato has one functional group. The CORRECT statement(s) about this functional group is(are)

(a) It gives positive silver mirror test

(b) It gets cleaved on reaction with ozone

(c) It gives hydrazonederviative on reaction with 2, 4– dinitrophenylhydrazine

(d) It decolorizes bromine water

Ans. a, d

Sol. Tomato red Lycopena

(a) It decolorizes Br₂ water

(b) It gives positive silver mirror test.

(c) It hydrazone 2, 4 DNP

(d) It gets dissolved on reaction with ozone.

40. The compound(s), which gives) benzoic acid on oxidation with KMnO₄, is (are)

(a)

(b)

(c)

(d)

Ans. b, d

Sol.

Benzylic hydrogen required for the formation of acid via oxidation.

Correct options are b, d

41. The total number of microstates possible for a d⁸ electronic configuration is_______.

Ans. 45

Sol. No. of microstates for d⁸ electronic configuration

42. Among the following, the total number of terpenes (terpenoids) is ________.

Ans. 7

Sol.

43. The total number of optically active isomers of dichloridobis (glycinato)cobaltate(III) ion is _______.

Ans. 6

Sol.

Total number of optical active isomers = 6 (six)

44. For the following fusion reaction,

the Q-value (energy of the reaction) in MeV (rounded off to one decimal place) is ______.

(Given: Mass of ¹H nucleus is 1.007825 u and mass of ⁴He nucleus is 4.002604 u)

Ans. 26.7

Sol. Q-value = × 931.5 MeV.

= (m_R – m_P) × 931.5

= (4 × 1.007825 – 1 × 4.002604) × 931.5

= 0.0287 × 931.5

= 26.7 MeV.

45. The dissociation constant of a weak monoprotic acid is 1.6 × 10^–5 and its molar conductance at infinite dilution is 360.5 × 10^–4 mho m²mol^–1. For 0.01 M solution of this acid, the specific conductance is n × 10^–2 mho m^–1. The value of n (rounded off to two decimal places) is___________.

Ans. 1.442

Sol. k_a = 1.6 × 10^–5

k = 14.42 × 10^–5Scm^–1

k = 14.42 S cm^–1 × 10^–5

k = 14.42 S (10²) m^–1 × 10^–5

k = 14.42 × 10^–3 S m^–1

k = 1.442 × 10^–2 S m^–1

k = n × 10^–2 S m^–1

n = 1.442

46. Calcium crystallizes in fcc lattice of unit cell length 5.56 Å and density 1.4848 g cm^–3. The percentage of Schottky defects (rounded off to one decimal place) in the crystal is ________.

(Given: Atomic mass of Ca is 40 g mol^–1; N_A = 6.022 × 10²³mol^–1)

Ans. 3.94

Sol. Ca²⁺fcc 2= 4

= 3.94%

47. If the root mean square speed of hydrogen gas at a particular temperature is 1900 m s^–1, then the root mean square speed of nitrogen gas at the same temperature, in m s^–1 (rounded off to the nearest integer), is _________.

(Given: atomic mass of H is 1 g mol^–1; atomic mass of N is 14 g mol^–1)

Ans. 507.8

Sol. We know,

48. Adsorption of a toxic gas on 1.0 g activated charcoal is 0.75 cm³ both at 2.5 atm, 140 K and at 30.0 atm, 280 K. The isosteric enthalpy for adsorption of the gas in kJ mol^–1 (rounded off to two decimal places) is _________.

(Given: R = 8.314 J K^–1mol^–1)

Ans. –5.784.6

Sol. P₁ = 2.5 atm P₂ = 30 atm

T₁ = 140 k T₂ = 280 k

Relate by

49. A buffer solution is prepared by mixing 0.3 M NH₃ and 0.1 M NH₄NO₃. If K_b of NH₃ is 1.6 × 10^–5 at 25ºC, then the pH (rounded off to one decimal place) of the buffer solution at 25ºC is_________.

Ans. 9.7

Sol. (NH₃) b = 0.3M

S (NH₄NO₃) = 0.1

k_b = 1.6 × 10^–5

pOH = 4.322

pH = 14 – 4.322

pH = 9.67

50. MgO crystallizes as rock salt structure with unit cell length 2.12 Å. From electrostatic model, the calculated lattice energy in kJ mol^–1 (rounded off to the nearest integer) is _________.

(Given: N_A = 6.022 × 10²³mol^–1; Madelung constant = 1.748; = 8.854 × 10^–12 J^–1 C² m^–1; charge of an electron = 1.602 × 10^–19 C)

Ans. – 7880

Sol. Rock salt type structure

51. The separation of energy levels in the rotational spectrum of CO is 3.8626 cm^–1. The bond length (assume it does not change during rotation) of CO in Å (rounded off to two decimal places) is ______.

(Given: Planck's constant h = 6.626 × 10^–34 J s; N_A = 6.022 × 10²³mol^–1; atomic mass of C is 12 g mol^–1; atomic mass of O is 16 g mol^–1; c = 3 × 10⁸ m s^–1)

Ans. 1.14

Sol. 2B = Spacing between energy line

2B = 3.8626 cm^–1

B = 1.931 cm^–1

= 0.00130 × 10^–17

r² = 1.3 × 10^–20 m²

= 1.14 × 10^–10 m

= 1.14Å

52. A dilute solution prepared by dissolving a nonvolatile solute in one liter water shows a depression in freezing point of 0.186 K. This solute neither dissociates nor associates in water. The boiling point of the solution in K (rounded off to three decimal places) is )_________.

(Given: For pure water, boiling point = 373.15 K; cryoscopic constant = 1.86 K (mol kg^–1)^–1; ebullioscopic constant = 0.51 K (mol kg^–1)^–1)

Ans. 371.201

Sol. V : 1 litre

Freezing point of pure water = 0ºC

0.186 = 1.86 ºC mol^–1 kg × m

m = 0.1 m

Aqueous solution is 1 molal

Eleration in boiling point

Boiling point of water = 371.15 k

(T_b) solution – (T_b) solvent = 0.051

(T_b) solution = 0.051 + 371.15

(T_b) solution = 371.201 K

53. A salt mixture (1.0 g) contains 25 wt% of MgSO₄ and 75 wt% of M₂SO₄. Aqueous solution of this salt mixture on treating with excess BaCl₂ solution results in th precipitation of 1.49 g of BaSO₄. The atomic mass of M in g mol^–1 (rounded off to two decimal places) is ________.

(Given: the atomic masses of Mg, S, O, Ba and Cl are 24.31, 32.06, 16.00, 137.33 and 35.45 g mol^–1, respectively)

Ans. 24.31

Sol. Mass of BaSO₄ precipitated = 1.49 g

Molar mass of BaSO₄ = 233.38 g mol

Moles of BaSO₄ precipitated

Percentage of MgSO₄ in salt mixture = 25%

Mass of salt mixture = 1.0 g

Mass of MgSO₄ in salt mixture

= 0.25 × 1.0 = 2.5 g

Molar mass of MgSO₄ = 120.37 g mol

Moles of MgSO₄ in salt mixture

Calculate moles of M₂SO₄ in salt mixture percentage of M₂SO₄ in salt mixture = 75%

Mass of salt mixture = 1.0 g

Mass of M₂SO₄ in salt mixture = 0.75 × 1.0 = 0.75 g

Molar mass of M₂SO₄ = 2 × Molar mass of M + Molar mass of s + 4 × molar mass of 0.

Molar mass of M₂SO₄ = 2M + 32 + 64

= 2M + 96

Reaction

BaCl₂ + M₂SO₄BaSO₄ + 2MCl

1 mole of M₂SO₄ react with 1 mole of BaCl₂ to produce 1 mole of BaSO₄

Substituting values of moles of M₂SO₄ in equation moles of

M = 24.31 g mol

54. For the molecule,

CH₃–CH=CH–CH(OH)–CH=CH–CH=C(CH₃)₂

the number of all possible stereoisomers is_______.

Ans. 8

Sol.

total n = 3

Possible stereoisomers = 2ⁿ = 2³ = 8

55. For the reaction,

k₁ = 2.5 × 10⁵ L mol^–1 s^–1, k_–1 = 1.0 × 10⁴ s^–1 and k₂ = 10 s^–1. Under steady state approximation, the rate constant for the overall reaction in L mol^–1 s^–1 (rounded off to the nearest integer) is _______.

Ans. 250

Sol.

= 250

56. The thermodynamic data at 298 K for the decomposition reaction of limestone at equilibrium is given below

The partial pressure of CO₂(g) in atm evolved on heating limestone (rounded off to two decimal places) at 1200 K is ___________.

(Given: R = 8.314 J K^–1mol^–1)

Ans. 4.313

Sol.

57. The mean ionic activity coefficient of 0.004 molalCaCl₂ in water at 298 K (rounded off to three decimal places) is ___________.

(Given: Debye-Huckel constant for an aqueous solution at 298 K is 0.509 kg^1/2mol^–1/2)

Ans. 0.773

Sol. Given, 0.004 molalCaCl₂

using Debye-Huckel limiting law.

58. The intensity of a monochromatic visible light is reduced by 90% due to absorption on passing through a 5.0 mM solution of a compound. If the path length is 4 cm, then the molar extinction coefficient of the compound in M^–1 cm^–1 is ________

Ans. 50

Sol. Since intensity reduced by 90%

59. The surface tension of a solution, prepared by mixing 0.02 mol of an organic acid in 1 L of pure water, is represented as

is the surface tension of pure water, A = 0.03 N m^–1, B = 50 mol^–1 L and c is concentration in mol L^–1. The excess concentration of the organic acid at the surface of the liquid, determined by Gibbs adsorption equation at 300 K is n × 10^–6mol m^–2. The value of n (rounded off to two decimal places) is ____________.

(Given: R = 8.314 J K^–1mol^–1)

Ans. 3.6

Sol.

60. If the crystal field splitting energy of [Co(NH₃)₄]²⁺ is 5900 cm^–1, then the magnitude of its crystal field stabilization energy, in kJ mol^–1 (rounded off to one decimal place), is __________.

Ans. – 84.58

Sol.