Question

The electronic configuration of $P$ in $H_{3}PO_{4}$ is

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

Answer 1

Answer: (A)

The electronic configuration of $P$ in $H _{3} PO _{4}$ is $1 s ^{2} 2 s ^{2} 2 p ^{6} 3 s ^{2} 3 p ^{6}$
$P$ atom has 5 valence electrons. It shares 3 with 3 - $OH$ groups to form $3 P -$ $OH$ bonds. It shares 2 with one $O$ atom to form $P = O$ bond.