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Bond angle in PH_(4)^(+) is higher than in PH_(3). Why ? |
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Answer» Solution :In `PH_(3), P" is "sp^(3)` hybridised . Three ORBITALS are involved in BONDING with three hydrogenatoms and the fourth one contains a lonepair. As LONE PAIR - bond pair repulsion is stronger than bond pair - bond pair repulsion , the tetrahedralshape associated with `sp^(3)` bonding is CHANGED to pyramid. `PH_(3)` 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)^(+)` is higher than the bond angle in`PH_(3)`. 
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