Saved Bookmarks
| 1. |
Bond angle in PH_(4)^(+) is higher than that in PH_(3). Why ? |
Answer» Solution :P in `PH_(3)` is `SP^(3)`-hybridized. 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. Due to the absence of lone pair-bond pair repulsions and presence of four identical bond pair-bond pair interactions, `PH_(4)^(+)` 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)`. 
|
|