Q. The electronic configuration of P in $H_{3}PO_{4}$ is
A) $1s^{2},2s^{2},2p^{6},3s^{2},3p^{6}$
B) $1s^{2},2s^{2},2p^{6},3s^{2}$
C) $1s^{2},2s^{2},2p^{6}$
D) $1s^{2},2s^{2},2p^{6},3s^{2},3p^{6},4s^{1}$
Explanation
(c)
In $H_{3}PO_{4},$ P is present as $P^{5 +}$
$\ _{15}P = 1s^{2},2s^{2},2p^{6},3s^{2},3p^{3}$
$P^{5 +} = 1s^{2},2s^{2},2p^{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.