## IIT JAM Chemistry 2020

Previous Year Question Paper with Solution.

1. Treatment of formic acid with concentrated sulfuric acid gives

(a) CO+H_{2}O

(b) CO_{2} + H_{2}

(c) HCHO + ½ O_{2}

(d) no product (no reaction)

Ans. a

Sol.

2. The amino acid R configuration is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Hence, correct option is (b)

3. The nucleobase NOT found DNA is

(a) Thymine

(b) Uracil

(c) Guanine

(d) Adenine

Ans. 2

Sol. Uracil replaced by Thymine in DNA

Correct option is (b)

4. The graph that represents the temperature (T) – entropy (S) variation of a Carnot cycle is

(a)

(b)

(c)

(d)

Ans. 2

Sol.

Correct option is (b)

5. One of the products of the hydrolysis of calcium phosphide at 25ºC is

(a) phosphine

(b) phosophoric acid

(c) phosphorus pentoxide

(d) white phosphorus

Ans. a

Sol.

Correct, option is (a)

6. For the radical chain reaction below, the correct classification for step 2 and step 3 is.

(a) chain propagating, chain terminating

(b) chain branching, chain terminating

(c) chain propagating, chain propagating

(d) chain propagating, chain branching

Ans. c

Sol.

Step 2 is chain propagating step since a free radical leads to the formation of another free radical the same is with step 3.

So, correct, option is (c)

7. The salt bridge in a galvanic cell allows the flow of

(a) ions but NOT electrons

(b) BOTH ions and electrons

(c) electrons but NOT ions

(d) NEITHER ions NOR electrons

Ans. a

Sol. In a galvanic cell salt bridge allows the flow of ions but not of electrons.

Correct, option is (a)

8. The correct statement for the following structure is

(a) 1, 2 and 3 are resonance structures

(b) 1 and 2 are resonance structures, whereas 3 is an isomer of 1 and 2

(c) 1 and 3 are resonance structures, whereas 2 is an isomer of 1 and 3

(d) 1, 2 and 3 are constitutional isomers

Ans. d

Sol.

As in 1, 2 and 3 structures, connectivity of atoms is different. So, these are constitutional isomers.

Correct option is (d)

9. The d – orbitals involved in the hybridization to form square planar and trigonal bipyramidal geometries are, respectively.

(a)

(b)

(c)

(d)

Ans. c

Sol.

Correct option is (c)

10.

(a) II > I > III > IV

(b) II > III > I > IV

(c) I > III > IV > II

(d) I > IV > III > II

Ans. c

Sol. As in alcohols and ethers, boiling point of alcohols in higher because of H – bonding. As boiling point decreases with increase in branching.

So, Correct order of boiling point is I > III > IV > II

Correct option is (c)

11. The correct statement regarding the determinants (Det) of matrices R, S and T is

(a)

(b)

(c) Det(R) = Det(S) = Det(T)

(d) Det(R), Det(S), Det(T) are all different

Ans. b

Sol.

Det (R) = 3(5 × 8 – 3 × 7) – 2(4 × 8 –1 × 7) + 4 (4 × 3 – 1 × 5)

= Det. (R) = 3 (40 – 21) –2 (32 – 7) + 4 (12 – 5)

= 57 – 50 + 28 = 35

Det (S) = 2 (4 × 8 – 1 × 7) – 3(5 × 8 – 3 × 7) + 4(5 × 1 – 3 × 4)

= 2 (32 – 7) – 3(40 – 21) + 4 (5 – 12)

= 50 – 57 – 28 = – 35

Det (T) = 3 (5 × 8 – 3 × 7) – 4(2 × 8 – 4 × 3) + 1(2 × 7 – 5 × 4)

= 3 (40 – 21) – 4(16 – 12) + 1 (14 – 20)

= 57 – 16 – 6 = 35

Correct option is (b)

12. The order of the M – C bond strength in the following species is (Atomic number for Cr = 24, Mn = 25, Ti = 22, Co = 27)

(a) II > I > IV > III

(b) I > III > II > IV

(c) III > IV > I > II

(d) III > I > II > IV

Ans. c

Sol. Higher the negative oxidation state of metal ion more will be the backbonding, greater will be the M – C bond strength

III > IV > I > II

Correct option is (c)

13. The forces constant for H^{35}Cl and D^{35}Cl are the same and both can be considered as harmonic oscillators. H^{35}Cl has a fundamental vibrational transition at 2886 cm^{–1}. The ratio of the zero – point energy of H^{35}Cl to that of D^{35}Cl is

(a) 0.515

(b) 0.717

(c) 1.395

(d) 1.946

Ans. c

Sol. As for harmonic oscillator,

For H^{35}Cl, x = 2886 cm^{–1} (given)

14. The rate of solvolysis of I – IV follows

(a) I > II > III > IV

(b) III > I > II > IV

(c) III > II > I > IV

(d) IV > I > II > III

Ans. c

Sol.

As more is the stability of carbocation, higher will be the rate of solvolysis. Stability of carbocation follows the order:-

15. The number of non – bonding electrons present in the frontier molecular orbitals of HF is

(a) 10

(b) 4

(c) 6

(d) 8

Ans. b

Sol.

So, there are 4 electrons in non - bonding MO of HF.

Correct option is (b)

16. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. a

Sol.

Correct option is (a)

17. The coordination number of aluminium ion and the number of bridging hydrogen atoms in [Al(BH_{4})_{4}]^{–} are respectively,

(a) 8 and 8

(b) 6 and 6

(c) 4 and 6

(d) 8 and 12

Ans. a

Sol.

Coordination number of Aluminium = 8

Number of Bridging Hydrogen = 8

Correct option (A)

18. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Correct option is (b)

19. The cage – type structure adopted by boron hydride [B_{5}H_{11}], is

(a) closo

(b) nido

(c) hypo

(d) arachno

Ans. d

Sol. B_{5}H_{11
}

_{ }[B_{5}H_{5}]^{6–
}

^{ }Archano

Correct option is (d)

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

(a)

(b)

(c)

(d)

Ans. b

Sol.

21. For the Diels – Alder reactions I – IV, the activation barriers follow the order

(a) II > I > III > IV

(b) I > III > IV > II

(c) III > IV > II > I

(d) IV > III > II > I

Ans. c

Sol. We know that

The more is the aromatic character the less will be the reactivity towards Diels - Alder reaction and hence more will be the activation barriers order of aromaticity.

Hence, the activation energy barriers follow the same order so Ans. is III > IV > II > I.

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

(a)

(b)

(c)

(d)

Ans. c

Sol.

23. The more stable species in each pair of conformers are

(a) II, IV and V

(b) I, IV and V

(c) II, III and V

(d) I, IV and VI

Ans. b

Sol.

This is comparatively more stable since t - Butyl group is on equvitorial position.

Since, the oxygen lone pair have small steric requirement, more over this structure is stabalised by hydrogen bonding.

So, the correct option is (b)

24. The number of d – d transition(s) expected for the complex [Cu(NH_{3})_{2}(H_{2}O)_{4}]^{2+} is

(a) 1

(b) 2

(c) 3

(d) 4

Ans. a

Sol. [Cu(H_{2}O)_{4}(NH_{3})_{2}]^{2+
}

^{ }Cu^{2+} = d^{9
}

^{
}

^{ }But due to Jahn Teller distortion ^{2}E_{g} splits into B_{1g} (lower energy) and A_{1g} (Higher energy) and ^{2}T_{2g} splits into B_{2g} and E_{g}.

Now transition occur from B_{1g} to the A_{1g}, B_{2g} or the E_{g} states.

Hence: If consider Jahn Teller distortion then expected d - d transition is 3.

If not consider Jahn Teller distortion then d - d transition is 1.

25. The plot showing the magnetic behavior of oxy – (solid line) and deoxy – heamoglobin (dashel line) is

(a)

(b)

(c)

(d)

Ans. a

Sol.

deoxy H_{b} is paramegnetic oxy H_{b} is diamegnetic.

The value of magnetic suspectibility for paramagnetic substances is greater the Diamegnetic Substance.

26. The geometries of the species [Br_{3}]^{+}. [Br_{3}]^{–} and [BrF_{3}] are, respectively

(a) linear, trigonal bipyramidal and trigonal bipyramidal

(b) linear, linear and trigonal planar

(c) tetrahedral, trigonal bipyramidal and trigonal bipyramidal

(d) tetrahedral, trigonal pyramidal and trigonal planar

Ans. c

Sol.

27. At constant pressure the diagram for a pure substance that sublimes is (s = solid, I = liquid and g = gas)

(a)

(b)

(c)

(d)

Ans. a

Sol.

Entropy of gas is maximum. So it will be (a)

Correct option is (a)

28. The solid state structure of HF is

(a)

(b)

(c)

(d)

Ans. b

Sol.

Solid HF consists of zig – zag chains of HF molecules.

Correct option is (b)

29. The complex which does NOT obey 18 – electron rule is

(atomic number for Mn = 25, Fe = 26, Co = 27, Ru = 44)

(a) [Co_{2}(CO)_{8}]

(b) [Fe(CO)_{4}]^{2–}

(c) [HMn(CO)_{5}]

(d)

Ans. d

Sol. [Co_{2}(CO)_{8}]

VEC = 9 × 2 + 2 × 8 + 2

= 18 + 16 + 2

= 36 e^{–}

(follows 18 e^{–} rule)

[Fe(CO)_{4}]^{2–}

VEC = 8 + 2 × 4 + 2

= 18 e^{–}

(follows 18 e^{–} rule)

[HMn(CO)_{5}]

1 + 7 + 2 × 5

= 18 e^{–}

(follows 18 e^{–} rule)

VEC = 5 + 8 + 1 + 2 + 3

= 19

(does not follows 18 e^{–} rule)

Correct option is (d)

30. The boyle temperature (T_{B}) is defined as the temperature at which the properties of a real gas coincide with those an ideal gas in the low pressure limit. The graph that shows the pressure dependence of the compression factor (Z) for a real gas at T_{B} is

(a)

(b)

(c)

(d)

Ans. b

Sol. The compression factors Z, approaches 1 at low pressures but does so with different slopes. For perfect gas, the slope, the slope is zero, but real gas may have either positive or negative sloper and the slope may vary with temperature, the slope is zero and the gas behaves perfectly over a wider range of conditions than at other temperatures.

Correct option is (b)

31. Achiral stereisomer(s) is(are) possible for

(a)

(b)

(c)

(d)

Ans. c, d

Sol. Achiral stereoisomers are those which are optically inactive but have the chiral centres within the same molecule

The molecule contains plane of symmetry hence is achiral but has 3 stereogenic centres (C_{2}, C_{3} and C_{4})

Similarly

dimethyl cyclopentane also contains plane of symmetry with 2 chiral centre hence is not optically active i.e. Achiral.

32. For the reaction shown in scheme 1, the concentration profiles of different species are provided.

Based on this graph the correct condition(s) regarding the rate constant is(are)

(a) k_{2} > k_{4}

(b) k_{3} > k_{1}

(c) k_{2} > k_{1}

(d) k_{1} = k_{2}

Ans. a, b

Sol. a, b are correct from the given data.

Correct options are (a, b)

33. The organometallic reagent(s) among the following is(are)

(a) Lithium divinylcuprate

(b) Lithium diisopropylamide

(c) Potassium tert - butoxide

(d) Isopropyl magnesiumidide

Ans. a, d

Sol.

Lithium divinylcuprate also called the gilman reagent is an organocopper reagent.

(d) Isopropyl magnesium iodide

This is an example of the organomagnesium reagent.

Hence, correct option is (a) and (d)

34. The compound(s) which will have only two signals in the ^{1}H NMR spectrum in 3 : 2 ratio is (are)

(a)

(b)

(c)

(d)

Ans. c, d

Sol. The compounds having only 2 signals are.

The protons on Benzene are equivalent since plane of symmetry exists.

4 Ha hydrogens given a signal.

"6 H" of methyl groups gives one signal.

The hydrogens on C_{3} and C_{4} are equivalent and that on C_{1} and C_{6} are also equivalent since the plane of symmetry exists.

"4 H" on C_{3} and C_{4} are equivalent gives one signal.

"6 H" on C_{1} and C_{6} are equivalent gives one signal.

So correct option is (c) and (d).

35. The boron adduct(s), which show(s) three signals in ^{1}H NMR spectrum with the intensity ratio 1 : 2 : 3 is (are)

(a) (CH_{3})_{3}B:N(CH_{3})_{3}

(b) (CH_{3}CH_{2})_{3}B : N(CH_{2}CH_{3})_{3}

(c) H_{3}B : N(CH_{2}CH_{3})_{3}

(d) (CH_{3}CH_{2})_{3}B : NH_{3}

Ans. c, d

Sol.

3 Ha [Hydrogenal NH_{3} group]

6 Hb [Hydrgen of CH_{2} group]

9 Hc [Hydrogen of CH_{3} group]

3Ha : 6 Hb: 9Hc

Intensity 1 : 2 : 3

3 Ha [Hydrogenal BH_{3} group]

6 Hb [Hydrogen of OH_{2} group]

9 Hc [Hydrogen of CH_{3} group]

3Ha : 6 Hb: 9Hc

Intensity 1 : 2 : 3

Hence, correct option is c & d

36. The transition metal complexe(s) with zero magnetic moment, zero dipole moment and CFSE of 2.4 is (are)

(a) [Mn(CO)_{5}(CH_{3})]

(b) [trans - Ni(ethylene diamine)_{2}Cl_{2}]

(c) [trans-Co(CN)_{4}(H_{2}O)_{2}]^{–}

(d) [trans - Fe(CN)_{4}Cl_{2}]^{4–}

Ans. c, d

Sol. (c) trans – [Co(CN)_{4}(H_{2}O)_{2}]^{–
}

^{
}

^{ }dipole moment = 0

oxidation state of Co is

37. In water, the enthalpy of a protein in its folded state(H_{F}) is lower than that in its unfolded state (H_{UF}). The entropies of the folded and unfolded states are S_{F} and S_{UF} respectively. The condition(s) under which this protein spontaneouly folds in water at a temperature T, is (are)

(a) S_{UF} > S_{F}

(b) S_{UF} = 0

(c) S_{UF} = S_{F}

(d) (S_{F} – S_{UF})>(H_{F} – H_{UF})/T

Ans. a, c, d

Sol.

Then also, value of folding process would be negative so above condition can lead to spontaneous folding of protein in water.

= S_{UF} can never be zero.

option a, c, d are correct.

38. The soft Lewis base(s) is(are)

(a) I^{–}

(b) CO

(c) H^{–}

(d) CH_{3}NC

Ans. a, b, c, d

Sol. I^{–}, CO, H^{–}, CH_{3}CN

All the molecules are the examples of soft Lewis bases.

Soft Lewis bases are generally the ones that contains larger, relatively polarizable donor atoms.

39. describes the wavefunction of a particle. The probability of finding the particle between x and x + dx, y and y + dy, z and z + dz, can be expressed as

(a)

(b)

(c)

(d)

Ans. b,c

Sol. If wave function is denoted by Then probability of finding a particle between x and x + dx, y and y + dy, z and z + dz is given by =

or if function is real function, then,

40. The correct sequence of reaction for the synthesis of the following molecule is(are)

(a)

(b)

(c)

(d)

Ans. a, b, c

Sol.

41. The number of lone pairs present in phosphonic acid (phosphorus acid) is______

Ans. 6

Sol. Number of lone pairs in phosphonic acid (Phosphorous Acid) is

Correct answer is 6

42. Total number of constitutional isomers possible for trimethyl cyclohexane is _____.

Ans. 6

Sol. Trimethyl cyclohexane is

All the substituted product i.e. conformational isomers are shown and hence the answer is 6.

43. Fullerene (C_{60}) crystallizes in an FCC unit cell (edge length = 14.14 Å) with one C_{60} centered at each lattice point. The smallest distance (in Å) between the centers of two C_{60} molecules is _____ (Round off to two decimal places)

Ans. 10.00

Sol. Edge length = 14.14 Å

given : one C_{60} centered at each lattice point.

We know that

To find: The smallest distance (in Å) between the centres of two C_{60} molecule i.e. 2r

44. The degree of unsaturation (double bond equivalent) for a compound with molecular formula C_{14}H_{12}O_{2} is ______.

Ans. 9

Sol. Degree of unsaturation double bond equivalent

(C_{14}H_{12}O_{2})

45. The value of 'n' in [P_{n}O_{18}]^{6–} is______

Ans. 6

Sol. [P_{n}O_{18}]^{6–
}

^{ }Metaphosphoric acid with empirical formula HPO_{3} basic unit is [PO_{3}]^{–1},

Given formula [P_{n}O_{18}]^{6–
}

^{ }Here number of oxygen atom is 18.

So, multiply basic unit by 6 to make number of oxygenation equal to 18.

[PO_{3}]^{1–} × 6 = [P_{6}O_{18}]^{6–
}

^{ }

46. A film of stearic acid partially covers the water surface in a container. The work needed to decrease this coverage by 1 cm^{2} is 25.0 × 10^{–7} J. The surface tension (in N/m) of the film is _____ (Round off to three decimal places)

(Surface tension of pure water is 0.072 N/m)

Ans. 0.047

Sol.

47. The dihedral (torsional) angle (in degrees) between the two methyl groups in the most stable conformation of n – butane is ______ (Round off to nearest integer)

Ans. 180º

Sol. Most stable conformation of n – butane is.

The angel between the 2 methyl groups is 180º in the most stable conformation.

Correct Answer is 180º.

48. The longest wavelength of light absorbed by a hydrogen – like atoms is 2.48 nm. The nuclear charge (Z) of the atom is ______ (Round off to nearest integer)

(Rydberg constant R_{H} = 109700 cm^{–1}).

Ans. 7

Sol. The longest wave length can be get when transition from n = 1 to n = 2

On solving it will be 7.

Correct answer is 7.

49. The function x^{4}e^{–2x/3} (for x > 0) has a maximum at a value of x equal to _______.

Round off to two decimal places)

Ans. 6

Sol. y = x^{4}.e^{–2x/3
}

^{ }

50. The total number of all possible isomers of [Co(H_{2}NCH_{2}CH_{2}NH_{2})_{2}Cl_{2}]^{+ }and [Co(H_{2}NCH_{2}CH_{2}NH_{2})_{3}]^{3+}

Ans. 5

Sol. [Co(NH_{2} – CH_{2} – CH_{2} – NH_{2})Cl_{2}]^{+
}

^{ }Possible geometrical isomer

[Co(en)_{3}]^{3+} is chiral. Complex can exist in two entaiomeric forms. [d and *l*]

Hence, total isomer which possible for both compounds are 3 + 2 = 5

Correct option answer is 5

51. The heat of formation of MgO at 300 K and 1 bar pressure is –600.60 kJ mol^{–1}. The free energy (in kJ mol^{–1}) of formation of MgO at 280 K is _____ (Round off to nearest integer)

Given: In the range 280 – 300 K, the constant pressure heat capacities (C_{P}) and molar entropies (S_{m}) are:

Ans. –572

Sol.

= – 600600 – 280 (–102.6)

= –600600 + 28728

= –571,872 J mol^{–1
}

^{ }or –572 kJ mol^{–1}

52. At a certain wavelength, liquid P transmits 70%, whereas liquid Q transmits 30% of the incident light when separately placed in a spectrophotometric cell (path length = 1 cm). In a binary mixture of liquids P and Q (assume non - interacting liquids), the absorbance in the same cell is 0.25. The volume fraction of liquid P in the binary mixture is ______ (Round off to two decimal places)

Ans. 0.74

Sol.

– log T = A

– log 0.7

A_{1} = 0.155

– log T = A

– log 0.3 = A

0.523 = A_{2
}

_{ }A_{T} = A_{1}C + A_{2} (1 – C)

Let the volume fraction of liquid P in the binary mixture be C

A_{T} = A_{1}C + A_{2} (1 – C)

0.25 = 0.155 C + 0.523 (1 – C)

0.25 = 0.155 C + 0.523 – 0.523

0.25 – 0.523 = 0.155 C – 0.523

0.273 = 0.368 C

53. The total number of head to tail isoprene linkages in the following molecule is _______

Ans. 4

Sol. Isoprene is a hemiterpene having formula

In the given molecule,

As in this there are 4 head to tail isoprene linkage.

Correct option is (4)

54. Titanium tetrachloride (TiCl_{4}) reacts with THF to form an octahedral complex X under inert atmosphere at 25ºC. If 5.0 g of TiCl_{4} is used and the yields is 80%, the amount of X (in grams) formed is _______ (Round off to one decimal place)

(Use atomic weights: Ti = 48, Cl = 35.5 Q = 16, C = 12 and H = 1)

Ans. 6.96

Sol. As TiCl_{4} reacts with THF to form TiCl_{4}(THF)_{2
}

_{
}

_{
}

_{ }amount of TiCl_{4}(THF)_{2} formed is 6.96 g.

55. The mean ionic activity coefficient for a 0.01 M aqueous solution of Ca_{3}(PO_{4})_{2} is _______ (Round off to three decimal places)

Ans. 0.066

Sol.

56. The Maxwell distribution of speeds of a gas at 300 K is given below.

The molar mass (in g mol^{–1}) of this gas is _____ (Round off to one decimal place)

(R = 8.3 J mol^{–1} K^{–1})

Ans. 19.9

Sol. As from the graph, it is observed that, most probable speed is 500 ms^{–1}.

Temperature = 300 K

57. A backterial colony grows via cell division where each mother bacterium independently produces two daughter cells in 20 minutes. If the concentration of bacteria is 10^{4} cm^{–3}, the colony becomes harmful. Starting from a colony with an initial concentration of 5 cm^{–3}. the time taken (in minutes) for the colony to become harmful is ______ (Round off to nearest integer)

Ans. 219.33

Sol.

We know each cell division generation 2, 2^{2}, 2^{3} and so an daughter cells and to reach required number of daughter cells colony forming units we assume there are n divisions occured therefore concentration and number of division are related as

C = 2^{n} × initial concentration.

10^{4} = 2^{n} × 5

2 × 10^{3} = 2^{n
}

^{ }taking log both sides, we get log (2 × 10^{3}) = n log^{2
}

^{
}

^{ }For 1^{st} cell division process time taken is 20 min., so for n cell divisions time consumed would be t = 20 n

t = 20 × 10.97

58. Sea water containing 1 M NaCl has to be desalinated at 300 K using a membrane permeable only to water. The minimum pressure (in bars) required on the sea – water side of the membrane is _____ (Round off to one decimal place)

(R = 8.3 J mol^{–1} K^{–1}, 1 bar = 10^{5} N/m^{2})

Ans. 49.91

Sol. For NaCl i = 2 as it gives 2 ions in solution.

59. For the reaction, CuSO_{4}(aq) + Zn(s) ZnSO_{4}(aq) + Cu(s), the value of (in kJ mol^{–1}) is ____ (Round off to the nearest integer)

(Reduction potential: Cu^{2+} (aq)/Cu(s) = + 0.34 V: Zn^{2+} (aq)/Zn(s) = –0.76 V)

(Faraday constant = 96485 C mol^{–1})

Ans. –212

Sol. Cathode

60. The total number of tautomers possible for I and II together is _________.

Ans. 6

Sol. Total number of possible tautomers

Structure I