Bond angle in ${P H}_{4}$ is higher than that in ${P H}_{3}$ . Why?

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Solution

Both are $s p^{3}$ hybridised. in $P H_{4}$ , all the four orbitals are bonded whereas in $P H_{3}$ there is a lone pair of electrons in P, which is responsible for lone pair-bond-pair-bond repulsion in $P H_{3}$ reducing the bond angle to less than $109^{0} 28^{-}$ .

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H - N - H

N H_{3}

H - P - H

P H_{3}

N H_{3}

N F_{3}

P H_{3}

P F_{3}

N H_{3}

P H_{3}

The bond angle in PH₃ is:

H - N - H

N H_{3}

H - P - H

P H_{3}

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