Question

The electronic configuration of most electronegative element? is

• A. $ 1{{s}^{2}},2{{s}^{2}}2{{p}^{5}} $
• B. $ 1{{s}^{2}},2{{s}^{2}}2{{p}^{4}},3{{s}^{1}} $
• C. $ 1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{1}}3{{p}^{1}} $
• D. $ 1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{5}} $

Answer 1

Answer: (A)

(a) $1 s^{2}, 2 s^{2}, 2 p^{5}=2,7(\because$ it has capacity to accept electron therefore, it is electronegative.)
(b) $1 s^{2}, 2 s^{2}, 2 p^{4}, 3 s^{1}=2,6,1$ (configuration not correct $\left(2 p^{4}\right)$
(c) $1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{1}, 3 p^{5}=2,8,6$ (configuration not correct $3 s^{1}$ )
(d) $1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{5}=2,8,7 (\because$ it has capacity to accept electron therefore, it is electronegative) Smaller the size, greater will be electronegativity.
Since, element in choice (a) is smaller in size, it will be more electronegative than (d). In choice (a) the atomic number of element is $9$ , which is of fluorine and it is the most electronegative element of the periodic table.