AIIMS Practice Paper Chemistry

1. When benzaldehyde reacts with acetaldehyde in presence of aq. NaOH, the product obtained is

(a) Crotonaldehyde

(b) Benzoin

(d) Paraldehyde

Answer: (c)

Solution

When benzaldehyde reacts with acetaldehyde, cinnamaldehyde is formed.

Identify the compound Z.

(a) Phenyl cyanide

(b) Benzoic acid

(d) Benzene diazonium chloride

Answer: (b)

Solution

3. The de-Broglie wavelength of an electron in the ground state of hydrogen atom is :

[K.E. = 13.6 eV; 1 eV = 1.602 × 10⁻¹⁹ J]

(a) 33.28 nm

(b) 3.328 nm

(d) 0.0332 nm

Answer: (c)

Solution

4. When R₃C−, R₂CH−, R−CH₂− groups are attached to an unsaturated group, the decreasing order of inductive effect is (R−alkyl group )

(a) R₃C− < R₂CH− > R−CH₂−

(b) R−CH₂− > R₂CH− > R₃C−

(d) R₃C− < R₂CH− < R−CH₂−

Answer: (c)

Solution

+I effect decreases in the order 3°>2^°>1^° due to presence of electron releasing alkyl group. The value also decreases due to presence of unsaturated through no bond resonance.

5. Which of the statements about anhydrous aluminium chloride is correct?

(a) It exists as a monomer molecule such as AlCl₃.

(b) It is not hydrolysed.

(d) It is obtained by passing dry chlorine over heated mixture of alumina and coke.

Answer: (d)

Solution

6. The composition of butter of tin is

(a) SnCl₂ ∙ 2H₂O

(b) SnCl₄ ∙ 5H₂O

(d) Sn + CH₃COOH

Answer: (b)

Solution

SnCl₄ ∙ 5H₂O is called butter of tin.

It is used as mordant in dyeing.

7. Which of the following is heated with silica at high temperature to form carborundum?

(a) Nitrogen

(b) Carbon

(d) None of these

Answer: (b)

Solution

Silicon carbide is known as carborundum is produced when silicondioxide is heated at about 2300 K with carbon.

8. Pick out the incorrect statement from the followings.

(a) sp² hybrid orbitals are equivalent and bond angle between any two of them is 120°.

(b) sp³d² hybrid orbitals are equivalent and are oriented towards corners of a regular octahedron.

(d) dsp² hybrid orbitals are equivalent with a bond angle of 90° between any two of them.

Answer: (d)

Solution

The shape of dsp² is square planar so the bond angle between any two hybrid orbitals will be either 90° or 180° as given below.

9. Calculate the percent dissociation of H₂S if 0.2 mole of H₂S is kept in a 800 ml vessel at 600° C for the reaction

. The equilibrium constant of the reaction is 8 × 10⁻⁶.

(a) 40%

(b) 4%

(d) 6%

Answer: (b)

Solution

10. Consider a 2^nd order reaction, 2A → Product.

If 40% of the reaction is completed in 400 seconds, how much time will it take for 80% completion?

(a) 2400

(b) 2600

(d) 2700

Answer: (a)

Solution

For a 2^nd order equation,

For 40% completion,

11. Calculate the free energy change of the reaction, when

If the heat of reaction is −216 kJ/mole, calculate the change in entropy of the reaction.

(a) 12.4 kJK^-1mol^-1

(b) 11.8 kJK^-1mol^-1

(d) 12.4 JK^-1mol^-1

Answer: (d)

Solution

12. Which of the following are arranged in the decreasing order of dipole moment?

(a) CH₃Cl, CH₃Br, CH₃F

(b) CH₃Cl, CH₃F, CH₃Br

(d) CH₃Br, CH₃F, CH₃Cl

Answer: (b)

Solution

Fluorine is most electronegative and Br is least electronegative. So CH₃ F should have highest dipole moment. But as C – F bond length is very small, inspite of greater polarity in CH₃F, it has less dipole moment than that of CH₃Cl.

13. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be

(a) 350

(b) 300

(d) 360

Answer: (a)

Solution

According to Raoult’s law:

p = p°_A X_A + p°_B X_B

290 = 200 × 0.4 + P°_B × 0.6

P°_B = 350

14. Aluminium oxide may be electrolyzed at 1000℃ to furnish aluminium metal(Atomic mass =27 u; 1 F= 96500 C). The cathode reaction is

Al³⁺ + 3e⁻ → Al⁰

To prepare 5.12 kg of aluminium metal by this method would require

(a) 5.49 ×10¹ C of electricity

(b) 5.49 ×10⁴ C of electricity

(d) 5.49 ×10⁷ C of electricity

Answer: (d)

Solution

Al³⁺ + 3e⁻ → Al

w = zQ

where w = amount of metal

= 5.12 kg = 5.12 × 10³g

z = electrochemical equivalent

15. A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct?

(a) MCl₂ is more volatile than MCl₄

(b) MCl₂ is more soluble in anhydrous ethanol than MCl₄

(d) MCl₂ is more easily hydrolyzed than MCl₄

Answer: (c)

Solution

In MCl₂, oxidation state of M = +2.

In MCl₄, oxidation state of M = +4

Higher the oxidation state, smaller the size, greater the polarizing power, greater the covalent characteristics.

Hence MCl₄ is more covalent and MCl₂ is more ionic.

16. The products obtained on heating LiNO₃ will be

(a) LiNO₂ + O₂

(b) Li₂O +NO₂ + O₂

(d) Li₂O + OH + O₂

Answer: (b)

Solution

LiNO₃ behaves differently from other alkali metal nitrates.

So second option is correct answer.

17. A metal M forms water soluble MSO₄ and inert MO. MO in aqueous solution forms insoluble M(OH)₂ that is soluble in NaOH. Metal M is

(a) Be

(b) Mg

(d) Si

Answer: (a)

Solution

On moving down water solubility of alkaline earth metals decreases.

Oxides and hydroxides of alkaline earth metals are basic, except Be.

Be gives amphoteric oxides and hydroxides. So metal M is Be.

18. Heating mixture of Cu₂O and Cu₂S will give

(a) Cu₂SO₃

(b) CuO + CuS

(d) Cu + SO₂

Answer: (d)

Solution

Following reaction takes place.

2Cu₂O + Cu₂S → 6Cu + SO₂

19. In context with the industrial preparation of hydrogen from water gas(CO + H₂), which of the following is the correct statement?

(a) CO and H₂ are fractionally separated using differences in their densities.

(b) CO is removed by absorption in aqueous Cu₂Cl₂ solution.

(d) CO is oxidized to CO₂ with steam in the presence of a catalyst followed by absorption of CO₂ in alkali.

Answer: (d)

Solution

CO is oxidized to CO₂ with steam in the presence of a catalyst followed by absorption of CO₂ in alkali.

20. The bond order in NO is 2.5 while that in NO⁺ is 3. Which of the following statements is true for these two species?

(a) Bond length in NO⁺ is greater than in NO

(b) Bond length in NO is greater than in NO⁺

(d) Bond length is unpredictable

Answer: (b)

Solution

Bond length is inversely proportional to bond order.

Bond order in NO⁺ = 3

Bond order in NO = 2.5

Thus bond length in NO > NO⁺

21. What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

(a) Cr³⁺ and Cr₂O₇²⁻ are formed

(b) Cr₂O₇²⁻ and H₂O are formed

(d) None of these

Answer: (b)

Solution

22. The equilibrium constants for the reactions X ⇌ 2Y and Z ⇌ P + Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal, then the ratio of total pressure at these equilibria is

(a) 1 : 36

(b) 1 : 3

(d) 1 : 9

Answer: (a)

Solution

23. Among Al₂O₃, SiO₂, P₂O₃ and SO₂ the correct order of acid strength is

(a) SO₂ < P₂O₃ < SiO₂ < Al₂O₃

(b) SiO₂ < SO₂ < Al₂O₃ < P₂O₃

(d) Al₂O₃ < SiO₂ < P₂O₃ < SO₂

Answer: (d)

Solution

While moving along a group from top to bottom, acidic nature of oxides decreases and along a period left to right, acidic nature increases.

Thus Al₂O₃ < SiO₂ < P₂O₃ < SO₂

24. The IUPAC name of the complex

(a) Triamminechromium(III) −μ−hydroxo− μ −hydroxo− μ −amido−chromium(III)sulphate

(b) Tetraamminebis(ethylene diamine) − μ −hydroxo− μ −imidodichromium(III)sulphate

(d) μ −hydroxo− μ −imido−tetraamminebis(ethylene diamine)dchromium(III)sulphate

Answer: (b)

Solution

First name of ligands then the bridging ligand followed by central metal atom and radical.

25. A compound of a metal ion (M^x+) has a spin only magnetic moment of √8 Bohr magnetons. The number of unpaired electrons in the compound are (Z = 28)

(a) 0

(b) 1

(d) 3

Answer: (c)

Solution

We know that

⇒ n(n + 2) = 8

⇒ n² + 2n – 8 = 0

⇒ (n + 4) (n – 2) = 0

⇒ n = +2, n ≠ – 4

26. Zinc granules are added in excess to 400 ml of 0.1 M nickel salt solution at 298 K until the equilibrium is reached. What will be the equilibrium constant at 25°C of

(a) 1.78 × 10¹⁵

(b) 1.78 × 10¹⁴

(d) 1.78 × 10¹⁷

Answer: (d)

Solution

27. What is the order of bond angle in the hydrides of the elements of nitrogen family?

(a) PH₃ > NH₃ > AsH₃ > SbH₃

(b) PH₃ > SbH₃ > AsH₃ > NH₃

(d) NH₃ > PH₃ > AsH₃ > SbH₃

Answer: (d)

Solution

As we more down the group, the bond angle decreases due to decrease in bp — bp repulsion.

So the bond angle in NH₃ = 107°

PH₃ = 93.5°

AsH₃ = 91.8°

SbH₃ = 91.3°

28. At what temperature, the kinetic energy of 6 gms of oxygen is equal to the kinetic energy of 0.5 mole of methane at 27℃?

(a) 600K

(b) 800℃

(d) 600℃

Answer: (c)

Solution

29. Calculate the pressure exerted by 32g of methane in a 500 ml container at 27℃ if it obeys Vander waal’s equation. Vander Waal’s constants are a = 2.253 atm L² mol⁻¹ and b = 0.0428 L mol⁻¹.

(a) 53.724 atm

(b) 65.402 atm

(d) 82.672 atm

Answer: (d)

Solution

Here n = 32/16 = 2 mole

Vander Waal’s equation

30. Which of the following reactions is not feasible?

(a) 2KI + Cl₂ → 2KCl + I₂

(b) 2KBr + F₂ → 2KF + Br₂

(d) 2KI + F₂ → 2KF + I₂

Answer: (c)

Solution

Bromine cannot oxidize Cl⁻ to Cl₂. The oxidizing nature follows the order

F₂ > Cl₂ > Br₂ > I₂

So Cl₂ can oxidize Br⁻ to Br₂ but Br₂ cannot oxidize Cl⁻ to Cl₂.

31. From the following conversion, identify the unknown compound “Z”.

(a)

(b)

(c)

(d)

Answer: (d)

Solution

32. A ketone reacts with methyl magnesium bromide followed by hydrolysis forms 2-methylbutan-2-ol. The ketone reacts with hydroxylamine followed reduction forms butan-2-amine. Name the ketone .

(a) Pentan-2-one

(b) Butan-2-one

(d) 2-methylbutanone

Answer: (b)

Solution

33. In aqueous solutions, amino acid mostly exist as

(a) NH₂−CHR−COOH

(b) NH₂−CHR−COO⁻

(c)

(d)

Answer: (d)

Solution

In aqueous solutions, amino acids mostly exist as zwitter ion.

34. Which of the following is wrong, when N₂ and O₂ are converted into mono anions?

(a) In oxygen mono anion, the oxygen-oxygen bond length increases.

(b) In Nitrogen monoanion, the N-N bond weaker.

(d) In N₂⁻, it becomes diamagnetic.

Answer: (d)

Solution

In O₂⁻, the no. of electrons = 16 + 1 = 17

B. O= (½)[10 – 7] = 1.5

Thus B.O decreases and bond length increases.

In N₂⁻, the no. of electrons = 14 + 1 = 15

B.O = 1/2[N_b – N_a]

= (½)[10 – 5]

= 2.5.

So N-N bond will be weaker.

But due to presence of unpaired electron in antibonding orbital, N₂⁻ is paramagnetic not diamagnetic.

35. Which of the following statement is correct?

(a) Electron gain enthalpy of inert gas is positive.

(b) Fluorine has highest electron affinity

(d) Increase in effective nuclear charge increases the ionization energy.

Answer: (d)

Solution

The electron affinity of inert gases is equal to zero due to stable electronic configuration. Chlorine shows the highest value of electron affinity. Due to the distance between non bonded atoms, Vander Waal’s radius is greater than covalent radius.

Nuclear charge ∝ IP due to strong attraction between outer most electron and nucleus.

36. When excess of AgNO₃ is added to a solution of one mole of Ni(H₂O)_x(Br)₃, only one mole of AgBr gets precipitated, so

(a) two Br⁻ ions are outside the co-ordination sphere

(b) three Br⁻ ions are outside the co-ordination sphere

(d) the oxidation state of Ni in the molecule should be +1.

Answer: (c)

Solution

Since it gives only one AgBr thus only one Br⁻ ion is outside the co-ordination sphere. The coordination number of Ni is 6. So the formula of the complex is [Ni(H₂O)₄Br₂]Br.

Thus four H₂O ligands are inside the co-ordination sphere.

37. Magnesia mixture solution is used to test

(a) As³⁺

(b) PO₄³⁻

(d) Both (a) & (b)

Answer: (d)

Solution

Both ions form a white ppt with magnesia mixture (MgCl₂ + NH₄Cl + NH₄OH).

38. Which of the following is a wrong statement?

(a) The order of reactivity of alkyl halide in SN¹ reaction is 3° > 2° > 1° > CH₃−X.

(b) E₁ reaction is two step reaction.

(d) The concentration of reagent does not depend on the rate of SN² reaction.

Answer: (d)

Solution

SN² is a single step reaction in which both substrate and reagent are used.

Thus rate ∝ [Substrate][Reagent]

e.g.

CH₃Cl + aq.KOH → CH₃OH + KCl

Here rate ∝ [CH₃Cl] [KOH]

39. A hydrocarbon (A) of formula C₈H₁₀, on ozonolysis gives compound (B), C₄H₆O₂ only. The compound (B) can also be obtained from alkyl bromide (C), C₃H₅Br upon treatment with magnesium in dry ether, followed by carbon dioxide and acidification. The compound A is

(a) CH₃− CH₂− C ≡ C – C ≡ C – CH₃

(b) CH₃ – C ≡ C – CH₂ – C ≡C – CH₂ – CH₃

(d) C₃H₅ – C ≡ C – C₃H₅

Answer: (d)

Solution

so (B) can be formed only from C₃H₅ – C ≡ C – C₃H₅

40. Which of the following alkenes is the most stable?

(a) CH₃CH = CHCH₃

(b) (CH₃)₂ C = CH₂

(d) (CH₃)₂C = C(CH₃)₂

Answer: (d)

Solution

(CH₃)₂C = C(CH₃)₂ is the most stable. It is due to maximum hyperconjugation. It has 12 hyperconjugate structures.

41. Gutta percha is a

(a) synthetic polymer obtained from isoprene by free radical polymerization.

(b) synthetic polymer obtained from isoprene by co-ordination polymerization

(d) natural polymer occurring in plants with a mixture of cis and trans configuration.

Answer: (c)

Solution

Gutta percha is natural polymer occurring in plants with all trans-configuration.

42. Which of the following is called milk of magnesia?

(a) Suspension of MgSO₄

(b) Suspension of Mg(OH)₂

(d) Solution of MgCO₃

Answer: (b)

Solution

Suspension of Mg(OH)₂ in water is used as a medicine as an antacid under the name milk of magnesia.

43. Which of the following is a correct statement?

(a) The no. of electrons present in 1.6 gm of CH₄ is 6.023 × 10²².

(b) The normality of mixture obtained by mixing 100 ml of 0.2 M H₂SO₄ and 100 ml of 0.2 M NaOH is 0.3 N.

(d) 0.14 mole of Al₂(SO₄)₃ would be in 50 g of the substance.

Answer: (d)

Solution

16 g of CH₄ contains 6.023 ×10²³ molecules

1.6 of CH₄ contains = 6.023 × 10²² molecules

so 1.6 g of CH₄ contains 6.023 × 10²³ electrons

(b) No. of milimoles = 100 × 0.4 – 100 × 0.2

= 20

1vol. 5vol.

1 vol. of C₃H₈ = 20 l

5 vol. of O₂ = 5 × 20

= 100 litr.

(d) Mol. mass of Al₂(SO₄)₃ = 342

342 g ≡ 1 mole

1 g= (1/342) mole

50 g = 50/342

= 0.14 mole

44. Match the following :

List I List II

A- Excluded molar volume (I) All gases behave ideally

B- Higher compressibility factor (II) b

C – At Boyles temperature (III) Difficult is the liqueification of gas

D – Only a small fraction of (IV) Maxwell distribution law.

molecules have either very low

or high velocity

(a) A-III B-II C-I D-IV

(b) A-II B-III C-IV D-I

(d) A-II B-III C-I D-IV

Answer: (d)

Solution

Self Explanatory

45. Addition of HOCl to allyl alcohol gives

(a) 2-Chloropropane-1, 3-diol

(b) 3-Chloropropane-1, 2-diol

(d) 1, 2, 3-Trichloropropane

Answer: (a)

Solution

46. Martius yellow is an example of

(a) Indirect dye

(b) Direct dye

(d) Basic dye

Answer: (b)

Solution

It is directly applied to fabric from an aqueous solution.

47. K₃[Fe(C₂O₄)₃] exhibits

(a) Geometrical isomerism

(b) Optical isomerism

(d) Ionization isomerism

Answer: (b)

Solution

48. Which of the following is a correct statement?

(a) Tin stone is magnetic in nature.

(b) Wolframite is magnetic in nature.

(d) Cassiterite is sulphide ore.

Answer: (b)

Solution

Anglesite is PbSO₄.

Cassiterite is SnO₂.

Wolframite is Fe, Mn, & WO₄.

49. Mass percentage of deuterium in heavy water is

(a) same as that of protium in water

(b) 20

(d) unpredictable

Answer: (b)

Solution

Molar mass of D₂O in 20 g mol⁻¹

50. A certain compound has a formula C₃H₆O. It combines with hydroxylamine to form two compounds which are geometrical isomers of each other. What is the compound?

(a) HCHO

(b) CH₃CH₂CHO

(d) CH₂ = CHCH₂OH

Answer: (b)

Solution

51. Assertion: A solution of FeCl₃ in water produces precipitate on standing.

Reason: Hydrolysis of FeCl₃ takes place in water.

(a) If both the assertion and reason are true and reason is the correct explanation of the assertion.

(b) If both the assertion and reason are true but the reason is not the correct explanation of the assertion.

(d) If the assertion is false but the reason is true.

Answer: (a)

Solution

Aqueous solution of FeCl₃ on standing produce brown ppt. due to hydrolysis of FeCl₃ to Fe(OH)₃ which is brown in color.

52. Assertion: Benzene is reactive while inorganic benzene is unreactive compound.

Reason: Inorganic benzene is borazine, B₃N₃H₆.

(a) If both the assertion and reason are true and reason is the correct explanation of the assertion.

(b) If both the assertion and reason are true but the reason is not the correct explanation of the assertion.

(d) If the assertion is false but the reason is true.

Answer: (d)

Solution

The assertion is false as benzene is less reactive than inorganic benzene because benzene is stabilized due to resonance. But reason that inorganic benzene is borazine B₃N₃H₆ is true.

53. Assertion : The O—O bond length in H₂O₂ is shorter than that of O₂F₂.

Reason : H₂O₂ is an ionic compound.

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.

(d) If both assertion and reason are false.

Answer: (d)

Solution

The O―O bond length is shorter in O₂F₂ than in H₂O₂ due to higher electronegativity of F atom.

H₂O₂ is a covalent compound.

54. Assertion : Effusion rate of oxygen is smaller than nitrogen.

Reason : Molecular size of nitrogen is smaller than oxygen.

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.

(d) If both assertion and reason are false.

Answer: (c)

Solution

Rate of effusion ∝ 1/√M

[M = Molecular mass]

as molecular mass of O₂ = 32 g

molecular mass of N₂ = 28 g

molecular mass of O₂ is more than N₂

∴ rate of effusion of O₂ is less than that of N₂

because r ∝ 1/√M

Molecular size of nitrogen is greater than oxygen as size decrease in a period from left to right.

55. A: The angular momentum of electron in any Bohr orbit is given by the integral multiple of h/2π where ‘n’ is the principal quantum number.

R : The integral multiple of wavelength of electron is integral multiple of 2 πr.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not correct explanation of A.

(d) A is false, R is true.

Answer: (a)

Solution

From the reason; we have 2πr = nλ.

From de-Broglie equation we know that,

Thus angular momentum (mvr) is the integral multiple of h/2π. So both are correct and R is the correct explanation of A.

56. A : Primary benzylic halides are more reactive than primary alkyl halides towards SN¹ reaction.

R : Reactivity depends upon the nature of the nucleophile and the solvent.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not correct explanation of A.

(d) A is false, R is true.

Answer: (b)

Solution

Primary alkyl halide undergoes SN¹ reaction mechanism. Alkyl halide follows carbocation intermediate. Benzylic carbocation is more stable than primary alkyl halide due to resonance

The reaction depends on the base (Nucelophile) and the medium (Solvent).

57. A : If 1 Faraday of electricity is passed through acidified water, the volume of O₂ liberated at NTP will be 5.6 litres.

R : As 16 g of oxygen at NTP liberates 22.4l.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not correct explanation of A.

(d) A is false, R is true.

Answer: (c)

Solution

1 Faraday deposts/liberates 1 gm eqv of the electrolyte

So 1 Faraday liberates 8 gms of oxygen at NTP.

as Eq. wt of oxygen = 8

We know that;

32 g of oxygen occupy 22.4 l at NTP

So 8 of oxygen occupy

Thus A is correct and R is wrong.

58. A : Tetra cyano Nickelate (II) ion is square planar and diamagnetic in nature.

R : It has dsp² hybridized and no unpaired electrons due to presence of strong ligand.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not correct explanation of A.

(d) A is false, R is true.

Answer: (a)

Solution

The e.c of Ni²⁸ → 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸

The e.c of Ni²⁺ → 1s²2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁸

The structure of ion is [Ni(CN)₄]²⁻, Ni²⁺ contain the half filled “d” orbitals. Due to presence of strong CN ligand, the half filled orbitals paired up to form dsp² hybridization as

The structure of dsp² is square planar. Due to absence of unpaired electron it is diamagnetic in nature.

Hence option (a) is correct.

59. A : Caesium is the most electropositive non radioactive element in the periodic table.

R : The first ionization potential of Cs is less than that of Potassium.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not the correct explanation of A.

(d) A is false, R is true.

Answer: (b)

Solution

Due to bigger atomic size of Cs, the force of attraction between the last electron and nucleus decreases. So it shows lowest ionization energy in the periodic table. So the R is correct. Due to low IP value, the last electron is easily removed making more electropositive. Thus A is correct.

60. A : The molecule of oxygen is paramagnetic in nature.

R : This property is due to the presence of unpaired electron in π^*2P_x and π^*2P_y antibonding orbital.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not the correct explanation of A.

(d) A is false, R is true.

Answer: (a)

Solution

Oxygen molecule contains 16 electrons. The electronic configuration of oxygen molecule = σ 1s² < σ^*1s² < σ2s² < σ^*2s²

Due to the presence of unpaired electron it has certain magnetic moment. So it is paramagnetic in nature.

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