A) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{5}}\]
B) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{1}}\]
C) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}\]
D) \[I{{s}^{2}},\text{ }2{{s}^{2}}\text{ }2{{p}^{5}}\]
Correct Answer: D
Solution :
Fluorine is the most electronegative element. Its atomic number is nine and hence its configuration is as\[F(9)=1\,{{s}^{2}},2{{s}^{2}},2{{p}^{5}}\] Due to its configuration, it has maximum effective nuclear charge, hence it shows maximum electro negativity.
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