electronegativity - Why does NF3 have a smaller bond angle than NH3?

I've already read many answers about the reason why $\ce{NF3}$ has a smaller bond angle than $\ce{NH3}$ , but I can't seem to understand them. Here's my understanding of the situation:

1. $\ce{NH3}$: Here N is more electronegative than H so a large electron cloud is crowded over N. This will push the bond pairs $\ce{N-H}$ away from the central atom. So, the angle of $\ce{H-N-H}$ will decrease.


2. $\ce{NF3}$: Here F is more electronegative than N, so the lone pair cloud over N is scattered into the $\ce{N-F}$ bonds. Thus the smaller electron cloud over central atom is unable to push the $\ce{N-F}$ bonds away from itself as much as it did previously. So, bond angle should be higher than previous case.

Answer

The bond angle difference between $\ce{NH3}$ and $\ce{NF3}$ is not easily explained — but that is primarily because ammonia’s bond angles already violate the simple theories that work so well for phosphane, arsane and stibane.

From a nitrogen atom’s point of view, having its lone pair in an s-type orbital and forming bonds with the p orbitals alone would be energetically beneficial. p orbitals can overlap with other atoms much better because they are directional. However, if that was done the resulting ideal $90^\circ$ bond angles would bring the hydrogens far too close together. Therefore, s contribution is mixed into the bonding p orbitals to alleviate the steric stress until an observed ‘equilibrated bond angle’ of $107^\circ$.

The same thing occurs in $\ce{NF3}$. However, fluorine atoms also bond with p orbitals (rather than hydrogen’s s orbitals) and they are larger meaning that the bond lengths are also greater. Therefore, the fluorine atoms are already more spaced out as is and the bond angle needs to be expanded less far from the ideal $90^\circ$ to sufficiently prevent steric repulsion.