AIIMS Practice Paper Chemistry – 1

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AIIMS Practice Paper Chemistry

 

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

(a)  Crotonaldehyde

(b) Benzoin

(c) Cinnamaldehyde

(d) Paraldehyde

Answer: (c)

Solution

When benzaldehyde reacts with acetaldehyde, cinnamaldehyde is formed.

2.

Identify the compound  Z.

(a) Phenyl cyanide

(b) Benzoic acid

(c) Nitrobenzene

(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

(c) 0.3328 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−

(c) R3C− > R2CH− > R−CH2

(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.

(c) It is a Lewis base.

(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

(c) SnO2

(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

(c) Carbon monoxide

(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.

(c) sp3d3 hybrid orbitals are not equivalent.

(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%

(c) 0.4%

(d) 6%

Answer: (b)

Solution

10. Consider a 2nd 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

(c) 1800

(d) 2700

Answer: (a)

Solution

For a 2nd 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

(c) 11.8 JK-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

(c) CH3Br, CH3Cl, CH3F

(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

(c) 700

(d) 360

Answer: (a)

Solution

According to Raoult’s law:

p = p°A XA + p°B XB

290 = 200 × 0.4 + P°B × 0.6

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

(c) 1.83 ×107 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

(c) MCl2 is more ionic 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

(c) Li3N + 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

(c) Ca

(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

(c) Cu + SO3

(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.

(c) H2 is removed through occlusion with Pd.

(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+

(c) Bond length in NO+ is equal to that 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

(c) CrO42 is reduced to +3 state of Cr

(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

(c) 1 : 1

(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

(c) Al2O3 < SiO < SO2 < 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

(c) Dichromium(III)tetraminebis(ethylene diamine) – μ− hydroxo− μ –inido−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 (Mx+) 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

(c) 2

(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

(c) 1.78 × 1016

(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

(c) SbH3 > AsH3 > PH3 > 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℃

(c) 800K

(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

(c) 95.225 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

(c) 2KCl + Br2 → 2KBr + Cl2

(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

(c) Petan-2-ol

(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.

(c) In O2, bond order decreases.

(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[Nb – Na]

= (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

(c) Covalent radius is bigger than Vander Waal’s radius

(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

(c) If the C.N of Ni is 6 then four H2O molecules should be inside 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

(c) None of these

(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.

(c) The concentration and basicity of solvent have no effect on rate of reaction of  SN1.

(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 C810, 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) CH− CH− C ≡ C – C  ≡ C – CH3

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

(c) CH3 – C ≡ C – CH2 – C ≡ C – 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

(c) (CH3)2C  = CHCH3

(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

(c) natural polymer occurring in plants with all trans configuration.

(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

(c) Solution of Mg(NO3)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.

(c) 80 litre of oxygen is necessary for the complete combustion of 20 lit. of propane.

(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

(c)     C3H8 + 5O2 → 3CO­2 + 4H2O

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

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

(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

(c) 2, 3-Dichloropropanol

(d) 1, 2, 3-Trichloropropane

Answer: (a)

Solution

46. Martius yellow is an example of

(a) Indirect dye

(b) Direct dye

(c) Mordant 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

(c) Coordination 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.

(c) Anglesite is PbCO3.

(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

(c) 11.1

(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

(c) CH3COCH3

(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.

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

(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.

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

(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.

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

(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.

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

(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.

(c) A is true, R is false.

(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.

(c) A is true, R is false.

(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.

(c) A is true, R is false.

(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.

(c) A is true, R is false.

(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.

(c) A is true, R is false.

(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*2Px and π*2Py 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.

(c) A is true, R is false.

(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.