A) \[Ne[3{{s}^{2}}3{{p}^{1}}]\]
B) \[Ne[3{{s}^{2}}3{{p}^{3}}]\]
C) \[Ne[3{{s}^{2}}3{{p}^{2}}]\]
D) \[Ar[3{{d}^{10}}4{{s}^{2}}4{{p}^{3}}]\]
Correct Answer: B
Solution :
[b] and [d] have exactly half-filled 3p orbital, but is smaller in size than . Hence has highest ionization enthalpy.
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