PYQ 2018 - VedPrep

IIT JAM Chemistry 2018
Previous Year Question Paper with Solution.

1. NaF, KF, MgO and CaO are crystalline solids. They have NaCl structure. Their lattice energies vary in the order

(a) NaF < KF < MgO < CaO

(b) KF < NaF < CaO < MgO

(d) CaO < MgO < KF < NaF

Ans. b

Sol. Lattice energy of any ionic compound is a function of charge on constituent cations, anions and internuclear distance between ions.

In MgO, Mg²⁺ and O^2– ions have highest magnitude of charge product and smallest internuclear distance accounts for its largest lattice energy among other ionic compound

2. The number of crystal systems and the number of Bravais lattices are, respectively,

(a) 14 and 7

(b) 7 and 32

(d) 7 and 14

Ans. d

Sol. There are 7 crystal systems and 14 number of bravis lattices in solid state

(a) 0

(b)

(d) 2

Ans. a

Sol.

4. Carbonic anhydrase is an example of

(a) Hydrolysis enzyme

(b) Redox enzyme

(d) Heme protein

Ans. a

Sol. Carbonic anhydrase is a enzyme that catalyse rapid interconversion of carbon dioxide and water to bicarbonate.

It is a hydrolysis enzyme.

5. For adsorption of a gas on a solid suirface, the plot that represents Freundlich isotherm is (x = mass of gas, m = mass of adsorbent, P = pressure)

(a)

(b)

(c)

(d)

Ans. d

Sol.

6. The compound that contains the most acidic hydrogen is

(a) H₂C=CH₂

(b)

(c)

(d) H₃C–CH₃

Ans. b

Sol. Acidity of proton depends upon the hybrization of carbon to which it is attached. A sp hybridised carbon is very electronegative and can stablise easily negative charge on it

(a)

(b)

(c)

(d)

Ans. c

Sol.

8. The CORRECT order of melting points of group 15 trifluorides is

(a)

(b)

(c)

(d)

Ans. a

Sol. Melting point depends on strength of intermolecular forces of attractions between molecules.

As molecular mass increases, vander waal forces of attraction becomes higher and hence none energy is needed to convert into gaseous form from solid state therfore melting point increases

9. The C-2 epimer of D-glucose is

(a) D-Mannose

(b) D-Fructose

(d) D-Gulose

Ans. a

Sol. C–2 epimer mean configuration at C2 only is different and rest of the molecule is same.

C–2 epimer of D-glucose is D-Mannose.

10. On hydrolysis, aluminium carbide produces

(a) CH₄

(b) C₂H₆

(d) C₂H₂

Ans. a

Sol. Hydrolysis of aluminium carbide is given as

11. With respect to periodic properties, the CORRECT statement is

(a) Electron affinity order is F > O > Cl

(b) First ionisation energy order is Al > Mg > K

(d) Ionic radius order is K⁺ > Ca²⁺ > Mg²⁺

Ans. d

Sol. Ionic radius of any cation depends on its valance shell number, n and charge on cation

Ionic radius α principle quantum number, n

12. Among the following metal carbonyl species, the one with the highest metal-carbon back bonding is

(a)

(b)

(c)

(d)

Ans. a

Sol. (A)Strength of Metal-Carbonyl back bonding depends on availability of electron density on metal.

Ti is in –2 oxidation state and has highest donating tendency towards of carbonyl

Therefore highest metal-carbon back bonding is observed with

Correct option is (A)

13. With reference to the variation of molar conductivity with concentration for a strong electrolyte in an aqueous solution, the CORRECT statement is

(a) The asymmetry effect contributes to decrease whereas the electrophoretic effect contributes to increase

(b) The asymmetry effect contributes to increase whereas the electrophoretic effect contributes to decrease

(d) Both asymmetry effect and electrophoretic effect contribute to increase

Ans. c

Sol. Correct option is (c)

14. The CORRECT order of carbonyl stretching frequencies for the following compounds is

(a) II < I < III < IV

(b) I < III < II < IV

(d) III < IV < II < I

Ans. d

Sol. due to –I effect of Cl, carbonyl carbon becomes more deficient and hence bond is strengthened so absorb highest wavenumber.

lone pair of oxygen is in cross conjugation with CO group and phenyl sing and hence when lone pair donate to ring CO bond is strengthened (due to –I effect of =O⁺–) but when lone pair interact with CO, bond becomes weak therefore it absorbs tower wavenumber than (I)

Ring electron density interact with CO group and hence C=O bond becomes weak so absorb at lower wavenumber than (II)

due to +R effect of phenyl and +M effect of N lone pair, C=O has double bond character only, weakest among other.

So absorb lowest wavenumber

III < IV < II < I

Correct option is (d)

15. The reaction, A Products, follows first-order kinetics. If [A] represents the concentration of reactant at time t, the INCORRECT variation is shown in

(a)

(b)

(c)

(d)

Ans. b

Sol.

16. The behavior of Cl₂ is closest to ideal gas behavior at

(a) 100 ºC and 10.0 atm

(b) 0 ºC and 0.50 atm

(d) –100 ºC and 10.0 atm

Ans. c

Sol. Any gas can behave like ideal gas at extremely high temperature and low pressure, therefore for Cl₂ to behave ideally 200ºC and 0.50 atm is required.

Correct option is (c)

17. The decay modes of ¹⁴C and ¹⁴O are

(a)

(b) positron emission

(d) positron emission and decay, respectively

Ans. d

Sol.

18. is rotated through an angle and is also doubled in magnitude resulting in An acceptable value of x is

(a) 1

(b) 2

(d)

Ans. b

Sol.

19. The CORRECT expression that corresponds to reversible and adiabatic expansion of an ideal gas is

(a)

(b)

(c)

(d)

Ans. c

Sol.

In reversible adiabatic conditions dq_rev = 0.

Equation (iii) becomes

20. The major products Y and Z in the following reaction sequence are

(a)

(b)

(c)

(d)

Ans. b

Sol.

Undergoes decarboxylation on heat through cyclic mechanism to keto product.

Correct answer is (B)

21. Which plot represents a spectrophotometric titration, where the titrant alone absorbs light in the visible region?

(a)

(b)

(c)

(d)

Ans. a

Sol. When titrant is added from burelte into the conical flask; it reacts with titrend and gets eliminated from solution. Fill titrant reacts to titrand, there will be no change in absorbance, when all titrend is consumed.

Now adding titrant, there will be no reaction and hence in conical flask, there is titrant only and its concentrant increase with adding volume therefore absorbance rises after equivalence.

Correct option is (A)

22. The sequence of three steps involved in the following conversion is

(a) (i) Friedel-Crafts alkylation; (ii) Reduction; (iii) Friedel-Crafts acylation

(b) (i) Friedel-Crafts acylation; (ii) Friedel-Crafts alkylation; (iii) Reduction

(d) (i) Friedel-Crafts alkylation; (ii) Friedel-Crafts acylation; (iii) Reduction

Ans. b

Sol. Friedel craft acylation of benzene

23.

(a)

(b)

(c)

(d)

Ans. None

Sol.

24. Consider the following four xenon compounds: XeF₂, XeF₄, XeF₆ and XeO₃ The pair of xenon compounds expected to have non-zero dipole moment is

(a) XeF₄ and XeF₆

(b) XeF₂ and XeF₄

(d) XeF₆ and XeO₃

Ans.d

Sol. XeF₆ is capped octahedral with partial participation of lone pairs in hybridization which import distorted structure to XeF₆ therefore exhbit non zero dipole moment.

In XeO₃ there is one lone pair and 3 bond pairs possiessing sp³ hybridization and hence pyramidal shape with non zero dipole moment.

Correct option is (d)

25. The product X in the following reaction sequence is

(a)

(b)

(c)

(d)

Ans. d

Sol.

26. Among the dimethylcyclohexanes, which one can be obtained in enantiopure form?

(a)

(b)

(c)

(d)

Ans. d

Sol. Due to flunomial behaviour (A) will convert into axial-axial conformation but its equitorial-equitorial conformation is more stable. Hence can be obtained in enantiopure form.

27. The correct order of stability for the following carbocations is

(a) I < III < IV < II

(b) III < II < IV < I

(d) IV < III < I < II

Ans. d

Sol.

Bridge head carbocation are very unstable because these are too stained and can never become planar. It has high energy vacant sp³ orbital.

It has 9a hydrogen hyperconjugation stablization therefore more stable than

Which enjoys number hyper conjugation but resonance gives some sort of stablization to allylic carbocation.

28. The major product formed in the following reaction is

(a)

(b)

(c)

(d)

Ans. c

Sol.

29.

(a)

(b)

(c)

(d)

Ans. c

Sol.

According to Alder rule electron attracting activating group on dienophile will be towards developing bond in transition state due to secondary orbital stablizing interactions. so endo product is formed.

30. The correct order of (the octahedral crystal field splitting of d orbitals) values for the following anionic metal complexes is

(a)

(b)

(c)

(d)

Ans. b

Sol. Ir⁺³ is a 5d⁶ metal ion having highest Z_eff.complexes with strong acceptor CN^– to provide largest splilting of d orbital in OH environment

Rh⁺³ is 4d⁶ metal ion shows larger splilting with acceptor CN^– is compared to doner I^– ligands.

For CO³⁺ which is 3d⁶ Is^– is weaker field ligand therefore is smallest in

31. Consider the following six solid binary oxides: CaO, Al₂O₃, PbO, Cs₂O, SiO₂ and Sb₂O₃. The pair(s) of ionic oxides is(are)

(a) CaO and Al₂O₃

(b) CaO and PbO

(d) SiO₂ and Sb₂O₅

Ans. a,b,c

Sol. CaO, Al₂O₃, PbO, CS₂O are ionic oxides as consituents metal ions are very electro positive.

32. Which of the following metal(s) is(are) extracted from its(their) sulfide ore(s) by self-reduction air reduction method?

(a) Cu

(b) Al

(d) Pb

Ans. a, d

Sol. Hg, Ph and Cu are extracted from their PhS and CuS over through air reduction method.

Correct options are (A and D)

33. The correct statement(s) about carbene is(are)

(a) Carbene is a neutral species

(b) Carbene is an intermediate in the Curtius rearrangement

(d) Carbene is generated from amines on reaction with nitrous acid

Ans. a,c

Sol. (A) Carbene is neutral species

(B) Nitrene is intermediate in curtius rearrangement

(C) Carbene shows N–H/O–H and C–H insertions and also cyclopropanation ring formation with alkenes so can insert in both bonds.

(D) No carbene can never be generated from amine on reaction with nitrons acid, it gives carbocation, not carbene.

Correct options are (A and C)

34. Choose the correct answer(s) with respect to the magnesium-EDTA titration carried out in the pH range 7 – 10.5, using solochrome black as indicator

(a) Magnesium-indicator complex is more stable than the magnesium-EDTA complex

(b) At the end point, the colour changes from red to blue

(d) pH range of 7 – 10.5 is necessary for observing the specific colour change

Ans. b,c,d

Sol. Correct options are (B, C and D)

35. The correct expression(s) for isothermal expansion of 1 mol of an ideal gas is(are)

(a)

(b)

(c)

(d)

Ans. a,b,d

Sol.

In isothermal expansion of ideal gas, T is constant and as

So, option (D) is incorrect

Correct options are (A, B and D)

36. Tetrapeptide(s) that gives (give) the following product on reaction with Sanger's reagent followed by hydrolysis is(are)

(a) Ala-Gly-Leu-Phe

(b) Asp-Phe-Leu-Pro

(d) Ala-Phe-Tyr-Pro

Ans. a,d

Sol. To react with sanger reagent nucleophilic nitrogen is required. Tertiary and amide nitrogens are not nucleophilic. Hence only A and D will produced the above product.

37. The compound(s) that shows (show) positive haloform test is(are)

(a)

(b)

(c)

(d)

Ans. a,b

Sol.

Product of iodoform reaction.

Correct options are (A and B)

38. Which of the following set(s) of quantum numbers is (are) NOT allowed?

(a) n = 3, l = 2, m₁ = –1

(b) n = 4, l = 0, m₁ = –1

(d) n = 5, l = 3, m₁ = +2

Ans. b,c

Sol. (B) n = 4 principle quantum number

l = 0 means S orbital and for S orbital is not valid or allowed.

Radial Nodes = n – l – 1 can never be negative

= 3 – 3 – 1

= 3 – 4

= – 1

So (C) is not allowed.

39. In a saturated calomel electrode, the saturation is with respect to

(a) KCl

(b) Hg₂Cl₂

(d) AgCl

Ans. a,b

Sol. Saturated calomel electrode is composed of saturated solution of KCl and paste of Hg₂Cl₂. Sparingly soluble salt.

Correct options are (A and B)

40. On reaction with NaNO₂ and HCl, which of the following amino alcohol(s) will yield compound P?

(a)

(b)

(c)

(d)

Ans. b,c

Sol.

bond is Anti periplanar to bond therefore easily / smoothly donates this electron density into vacant so ring contraction assisted by

Correct options are (B and C)

41. Among the following hydrocarbon(s), how many of them would give rise to three groups of proton NMR peaks with 2:2:3 integration ratio?

Ans. 2

Sol.

So, 2 : 2 : 3 integration ratio in NMR.

Correct answer is (2)

42. The nuclear spin quantum number (I) of a nucleus is When placed in an external magnetic field. the number of possible spin energy states it can occupy is __________.

Ans. 4

Sol. No. of possible nuclear energy state for a nuclei whose nuclear quantum number, is given by 2I + 1

This nuclei will have 4 nuclear spin energy levels which are apparently available for occupation.

So answer is 4.

43. The number of possible isomers for is _______ (py is pyridine)

Ans. 3

Sol.

3 isomers are possible for [Pt(Py)BrCl].

44. The number of stereoisomers possible for the following compound is __________.

Ans. 3

Sol. No. of stereoisomer for molecule having internal plane of symmetry / meso compound.

n = number of chiral centers = 2

no. of stereoisomer = 2ⁿ – 1

= 2² – 1

= 4 – 1

= 3.

45. Assuming ideal gas behavior, the density of O₂ gas at 300 K and 1.0 atm is ______ gL^–1 (rounded up to two decimal places).

[R = 0.082 L atm mol^–1 K^–1, molar mass of O₂ = 32]

Ans. 1.3

Sol.

Correct answer is (1.3)

46. The time for 50% completion of a zero order reaction is 30 min. Time for 80% completion of this reaction is _______ min.

Ans. 48

Sol. For zero order reaction, concentration – time relation

47. The volume of 0.3 M ferrous ammonium sulphate solution required for the completion of redox titration with 20 mL of 0.1 M potassium dichromate solution is _______ mL.

Ans. 40

Sol.

No. of g equivalents of Fe²⁺ = No. of g equivalents of Cr₂O₇^2–

48. The value of C_v for 1 mol of N₂ gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is _______ J K^–1 mol^–1 (rounded up to two decimal places).

[R = 8.3 JK^–1 mol^–1]

Ans. 20.75

Sol. According to low of equipartition of energy each coordinate axis is associated with energy for rotational and transitional degrees of freedom for N₂ linear molecule

There is 2 rotational degree of freedom 3 translational degree of freedom

49. The number of hydrogen bond(s) present in a guanine-cytosine base pair is _______.

Ans. 3

Sol.

50.

The value of for the reaction at 300 K is –281.8 kJ mol^–1. The value of at same temperature is ________ kJ mol^–1 (rounded up to the first decimal place).

[R = 8.3 J K^–1 mol^–1]

Ans. -283.0

Sol.

51. the following information is given T = 300 K

Ans. 788

Sol.

52. How many of the following interhalogen species have 2 lone pairs of electrons on the central atom?

Ans. 2

Sol.

53. The solubility of PbI₂ in 0.10 M KI(aq) is ______ × 10^–7 M (rounded up to two decimal places).

[The solubility product, ]

Ans. 7.1

Sol.

54. The number of compounds having S-configuration among the following is ______.

Ans. 4

Sol.

Since least priority is at horizontal, therefore Anticlockwise S configuration will revert to R configuration.

Correct answer is (4)

55. The magnitude of crystal field stabilization energy (CFSE) of octahedral complex is 7680 cm^–1. The wavelength at the maximum absorptionof this complex is ______ nm (rounded up to the nearest integer).

Ans. 1

Sol. Electronic contant Ti³⁺ 3d'45º

56. The electron of a hydrogen atom is in its n^th Bohr orbit having de Broglie wavelength of 13.4 Å. The value of n is _________ (rounded up to the nearest integer).

[Radius of n^th Bohr orbit = 0.53 n² Å, = 3.14]

Ans. 4

Sol.

57. Elemental analysis of an organic compound contaming C, H and O gives percentage composition: C : 39.9% and H : 6.7%. If the molecular weight of the compound is 180, the number of carbon atoms present in the molecule is ________.

Ans. 6

Sol. In given organic compound % of C is 39.9 therefore fraction of C is 0.399.

So, weight of C in given compound = 0.399 × Molecular weight of compound

= 0.399 × 180

58. The emf of a standard cadmium cell is 1.02 V at 300 K. The temperature coefficient of the cell is . The value of for the cell is _______ kJ mol^–1 (rounded up to two decimal places).

[1 F = 96500 C mol^–1]

Ans. –199.75

Sol.

Eº = 1.02 V;T = 300 K

For cadmium cell, number of electron exchanged n = 2

= –2 × F × 1.02 + 2F × 300 (–5 × 10^–5)

= –2.04 F – 0.03 F

= –2.07 F

= –2.07 × 96500

= –199755 J mol^–1

= –199.75 kJ mol^–1

59.For H₂ molecule, the fundamental vibrational frequency in wavenumbers can be taken as . the zero-point energy of the molecule is _______ kJ mol^–1 (rounded up to two decimal places).