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Bond angle in PH_4^(+) is higher than that in PH_3. Why? |
Answer» Solution :P in `PH_3` is `sp^(3)`-hybridised. It has three bond pairs and one lone pair around P. Due to stronger lone pair-bond pair repulsions than bond pair-bond pair repulsions, the tetrahedral angle decreases from `109^(@)28. to 93.6^(@)`. As a result, `PH_3` is pyramidal. HOWEVER, when it REACTS with a PROTON, it forms `PH_4^(+)` ion which has four bond pairs and no lone pair. Now, there are no lone pair-bond pair repulsions. Only four identical bond pair-bond pair interactions exist. `PH_4^(+)` therefore assumes tetrahedral geometry with a bond angle of `109^(@)28.`. This explains why the bond angle in `PH_4^(+)` is higher than in `PH_3` 
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