A) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}}\]
B) \[1{{s}^{2}},2{{s}^{2}}2{{p}^{5}}\]
C) \[1{{s}^{2}},2{{s}^{2}}\text{ }2{{p}^{6}},3{{s}^{2}}\text{ }3{{p}^{6}}\]
D) \[1{{s}^{2}},2{{s}^{2}}\text{ }2{{p}^{6}},3{{s}^{2}}\text{ }3{{p}^{6}}\text{ }3{{d}^{1}},4{{s}^{2}}\]
Correct Answer: B
Solution :
\[1{{s}^{2}}\,2{{s}^{2}}2{{p}^{5}}\]is the electronic configuration of fluorine. Electronegativity increases in period from left to right and decreases in group from up to down. Halogen element having high electronegativity in period. The electronegativity of fluorine is \[4.0\].
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