The number of $P - H$ bond(s) in $H_{3} P O_{2}$ , $H_{3} P O_{3}$ and $H_{3} P O_{4}$ respectively, is:

2, 0, 1

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1, 1, 1

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2, 0, 0

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2, 1, 0

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Solution

The correct option is D $2, 1, 0$
The number of P−H bonds in $H_{3} P O_{2}$ , $H_{3} P O_{3}$ and $H_{3} P O_{4}$ is $2, 1, 0$ respectively.

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