A) \[1{{s}^{2}}\,2{{s}^{2}}\,2{{p}^{5}}\]
B) \[1{{s}^{2\,}}\,2{{s}^{2}}\,2{{p}^{3}}\]
C) \[1{{s}^{2}}\,2{{s}^{2}}\,2{{p}^{5}}\,3{{s}^{1}}\]
D) \[1{{s}^{2}}\,2{{s}^{2}}\,2{{p}^{6}}\]
Correct Answer: C
Solution :
The electronic configuration \[1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}3{{s}^{1}}\] shows lowest ionisation energy because this configuration is unstable due to the presence of one electron in s-orbital. Hence, less energy is required to remove the electron.
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