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 Multiple Choice QuestionsMultiple Choice Questions

61.

'Hydride Gap' is referred to which region of the Periodic Table ?

  • Groups 3, 4 and 5

  • Groups 5, 6 and 7

  • Groups 4, 5 and 6

  • Groups 7, 8 and 9


62.

The electronic configuration of the element with maximum electron affinity is

  • 1s2, 2s2,2p3

  • 1s2, 2s2,2p5

  • 1s2, 2s2,2p6, 3s2, 3p5

  • 1s2, 2s2,2p6, 3s2, 3p3


63.

The first ionisation energy of oxygen is less than that of nitrogen. Which of the following is the correct reason for this observation?

  • Lesser effective nuclear charge of oxygen than nitrogen.

  • Lesser atomic size of oxygen than nitrogen

  • Greater interelectron repulsion between two electrons in the same p-orbital counter balances the increase in effective nuclear charge on moving from nitrogen to oxygen.

  • Greater effective nuclear charge of oxygen than nitrogen.


64.

The correct match of contents in Column I with those in Column II is

  Column I   Column II
(A) He (i) High electron affinity
(B) Cl (ii) Most electropositive element
(C) Ca (iii) Strongest reducing agent
(D) Li (iv) Highest ionisation energy
  • A-iv, B-i, C-ii, D-iii

  • A-iii, B-i, C-ii, D-iv

  • A-iv, B-iii, C-ii, D-i

  • A-ii, B-iv, C-i, D-iii


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65.

The lattice enthalpy and hydration enthalpy of four compounds are given below

Compound Lattice enthalpy (in kJ mol-1) Hydration
enthalpy
(in kJ mol-1)
P +780 -920
Q +1012 -812
R +828 -878
S +632 -600

The pair of compound which is soluble in water is

  • P and R

  • Q and R

  • P and Q

  • Rand S


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66.

In which of the following the oxidation number of oxygen has been arranged in increasing order?

  • OF2 < KO2 < BaO2 < O3

  • BaO2 < KO2  < O< OF2 

  • BaO2  < O< OF2  < KO2 

  •  KO2  < OF2  < O< BaO2 


B.

BaO2 < KO2  < O< OF2 

Let the oxidation number of oxygen in following compound is x.

  In OF2

x+(-1) 2 = 0
          x = + 2

 In KO2

+ 1+( x 2)  = 0
             2x =-1
               x =-1/2

 In BaO2

+2+ (x × 2) = 0
             2x = -2
               x =-1

In O3, oxidation number of oxygen is zero because oxidation number of an element in free state or in any of its allotropic form is always zero. Thus, the increasing order of oxidation number is

            BaO2 < KO2  < O< OF2 

           -1        -1/2      0       +2

          


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67.

In which one of the following pairs the radius of the second species is greater than that of the first?

  • Na, Mg

  • O2-, N3-

  • Li+, Be2+

  • Ba2+, Sr2+


68.

The one electron species having ionization energy of 54.4 eV is :

  • H

  • He+

  • B4+

  • Li2+


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69.

Consider the ions : K+, S2-, Cl- and Ca2+. The radii of these ionic species follow the order

  • Ca2+ > K+ > Cl- > S2-

  • Cl- > S2- > K+ > Ca2+

  • S2- > Cl- > K+ > Ca2+

  • Ca2+ > S2- > Cl- > Ca2+


70.

Which one of the following oxides is amphoteric ?

  • ZnO

  • MgO

  • CaO

  • CO2


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