Bond angles in NH3 and NF3 & VSEPR theory?

Question

I was trying to figure out if the bond angles in NF3 are larger than in NH3. My reasoning led me to the conclusion that they should be larger, though in reality the opposite is true (102 deg for NF3 and 106 deg for NH3). I can't pinpoint where I am mistaken.

Here is my reasoning:
According to VSEPR the repulsion for lone pair-bond e (electrons) is greater than bond e- bond e. Thus the bond angles will be determined by the strength of the lone-pair-bond e repulsion. F is more electronegative, thus it will pull closer to it the bond electrons, reduce the electron cloud density near N, reduce the repulsion lone-pair-bond e, thus decrease the lone pair-bond e angle and increase the bond e- bond e angle.

What am I missing?

Answer 1

Best Answer - Chosen by Asker

Electron cloud of N-F bonds will pull closer to N as you reasoned... However -- this will reduce the overall repulsion between the N-F bonds! Thus, the repulsion from the lone pair-bond e will (in essence) be greater)and will tend to push N-F bonds closer togehter -- thereby reducing the bond angle for NF3.

Asker's Rating:

Asker's Comment:

Obviously you mistyped N instead of F in the first sentence.This is the answer I found also at a uni website with a tutorial, so it is correct. Still, I find it not solid enough, but that has to do with VSEPR theory itself and not your reasoning.