The electronic configuration of $P$ in $H_{3} {P O}_{4}$ is:

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}

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1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}

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1 s^{2} 2 s^{2} 2 p^{6}

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1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}

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Solution

The correct option is B $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$
The option (A) is the correct answer.
The electronic configuration of P in $H_{3} P O_{4}$ 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 $- O H$ groups to form 3 $P - O H$ bonds. It shares 2 with one O atom to form $P = O$ bond.
Thus, in $H_{3} P O_{4}$ , P atom has 8 valence electrons.

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The correct electronic configuration of $V^{2 +}$ is:

1 s^{2} 2 s^{2} 2 p^{6}

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5} 4 s^{2}

= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}

= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}

= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} 4 s^{2} 3 d^{1}

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