covalent compounds - Why the bond angle of PH3 is lesser that that of PF3?

We can explain why the bond angle of $\ce{NF3}$ (102°29') is lesser than $\ce{NH3}$ (107°48') by the VSEPR theory, since lone pair lone pair repulsion is greater than lone pair bond pair repulsion. Then for $\ce{PH3}$ and $\ce{PF3}$, also, it is expected that the bond angle of $\ce{PF3}$ will be smaller. But I found that the thing is just reverse. Why is this so? Is it an exception?

Answer

The answer can not be determined using VSEPR theory. The number of lone pairs in both compounds is the same. The answer lies in electronegativity.

1. 
   

   In $\ce{NH3}$ nitrogen is more electronegative than hydrogen and therefore it will pull the electrons towards it. Therefore the electrons will be closer to each other hence more repulsion.

   
   


2. 
   
   

   But in $\ce{NF3}$ fluorine will pull electrons towards itself therefore the electrons are further comparatively and hence repulsion will be less and so is the bond angle.

   
   

But in case of $\ce{PF3}$ and $\ce{PH3}$ back bonding is possible in $\ce{PF3}$ therefore it has a larger bond angle .