A) \[1{{s}^{2}},\]\[2{{s}^{2}},\]\[2{{p}^{6}},\]\[3{{s}^{1}}\]
B) \[1{{s}^{2}},\]\[2{{s}^{2}},\]\[2{{p}^{6}},\]\[3{{s}^{1}},\]\[3{{p}^{1}}\]
C) \[1{{s}^{2}},\]\[2{{s}^{2}},\]\[2{{p}^{6}},\]\[3{{s}^{2}}\]
D) \[1{{s}^{2}},\]\[2{{s}^{2}},\]\[2{{p}^{6}},\]\[3{{s}^{2}},\]\[3{{p}^{2}}\]
Correct Answer: C
Solution :
[c] On losing two electrons from \[1{{s}^{2}},\]\[2{{s}^{2}},\]\[2{{p}^{6}},\]\[3{{s}^{2}}\]electronic configuration, becomes \[1{{s}^{2}},\]\[2{{s}^{2}},\]\[2{{p}^{6}},\]\[3{{s}^{0}},\] it is more stable than \[1{{s}^{2}},\]\[2{{s}^{2}}2{{p}^{6}},\]\[3{{s}^{2}}.\] Since energy gap between \[2{{p}^{6}}\] and \[3{{s}^{2}}\] will be more due to changing of shell from 3s to 2p. Therefore, the difference in values of IInd ionization energy and IIIrd ionisation energy of this configuration is more. Hence, option [c] is correct.
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