The electronic configuration of P in ${{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}$ is

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

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

JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

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Understanding +4 | -1

The electronic configuration of P in

{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}

View answer and explanation

Correct option: C

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

The correct answer is C (

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

) because in

{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}

, phosphorus is in the +5 oxidation state (

{\mathrm{P}^{5 + }}

), having lost its 3s and 3p electrons.

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n^{\text {th }}

\mathrm{A}

\times 10^{12}

\frac{1}{{ }^{\prime} \mathrm{B}^{\prime}}

\ell = 1,\,\,m = 0

1.25 \times {10^{ - 13}} \times {A^{1/3}}\mathrm{cm}.