Q. The electronic configuration of most electronegative elements is

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

(a)
$1s^{2},2s^{2},2p^{5} = 2,\ 7$
($\because$ It has capacity to accept electron therefore, it is electronegative.)
(b) $1s^{2},2s^{2},2p^{4},3s^{1} = 2,\ 6,\ 1$
(configuration not correct $(2p^{4})$)
(c)$\ 1s^{2},2s^{2},2p^{6},3s^{1},3p^{5} = 2,\ 8,\ 6$
(configuration not correct $3s^{1}$)
(d) $1s^{2},2s^{2},2p^{6},3s^{2},3p^{5} = 2,\ 8,\ 7$
($\because$ 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.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.