Multiple Choice Questions

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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.


C.

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.

The electronic configuration of nitrogen is

                  7N = 1s2, 2s2, 2p2

                    

Due to presence of half filled p-orbital, (more stable) a large amount of energy is required to remove an electron from nitrogen. Hence, first ionisation energy of nitrogen is greater than that of oxygen.

The electronic configuration of oxygen is

            8O =1s2, 2s2, 2p4

              

The other reason for the greater IP of nitrogen is that in oxygen, there is a greater interelectronic repulsion between the electrons present in the same p-orbital which counter- balance the  increase in effective nuclear charge from nitrogen to oxygen.         


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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


The correct decreasing order of first ionisation enthalpies of five elements of the second period is

  • Be > B > C > N > F

  • N > F > C > B > Be

  • F > N > C > Be > B

  • N > F > B > C > Be


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+


'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


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

  • H

  • He+

  • B4+

  • Li2+


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 


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


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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The first () and second (ΔH2) ionisation enthalpies (in kJ mol-1) and the (ΔegH) electron gain enthalpy (in kJ mol-1) of the elements I, II, III, IV and V are given below

Element ΔH1 ΔH2 ΔegH
I 520 7300 -60
II 419 3051 -48
III 1681 3374 -328
IV 2372 5251 +48

The most reactive metal and the least reactive non-metal of these are respectively

 

  • I and IV

  • IV and II

  • II and IV

  • III and IV


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