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Question
Bond angle in `PH_(4)^(+)` is higher than that in PH3. Why?
Solution 1
In PH3, P is sp3 hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with sp3 bonding is changed to pyramidal. PH3 combines with a proton to form `PH_4^(+)` in which the lone pair is absent. Due to the absence of lone pair in `PH_4^(+)`, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in PH3
Solution 2
P in PH3 is sp3-hybridized with 3 bond pairs and one lone pair around P. Due to stronger lp-bp repulsions than bp-bp repulsions, tetrahedral angle decreases from 109°28′ to 93.6°. As a result, PH3 is pyramidal. In PH4+, there are 4 bp’s and no lone pair. As a result, there are only identical bp-bp repulsions so that PH4+ assumes tetrahedral geometry and the bond angle is 109°28′.Hence, bond angle of PH4+ > bond angle of PH3
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