Relation between various bond angles of Nitrogen compounds

Question

I was thinking what could be order of bond angles of NH₃, NF₃, N(CH₃)₃ and N(C₂H₅)₃.

Considering NF₃ and NH₃:

Since there is backbonding between 2p-2p orbitals of N and F, there will be a partial double bond character between N and F and therefore, since repulsion between lone pair and double bond is more than that between lone pair and single bond and thus bond angle in NF₃ will be less than NH₃

But, the size of F is very small, and F has lone pairs while H does not. If the N-F bond acquires partial double bond character, the lone pair of other F atoms should repel the bond making the F-N-F angle larger? Would the two effects cancel each other making the bond angle of NF₃ equal to NH₃

It is known that bond angle of NF₃ is less than NH₃. Why? How do I decide which factor will dominate over other?

Now since N(CH₃)₃ and N(C₂H₅)₃ have bulky groups, the groups will repel each other making the bond angle greater than NH₃.

But between N(CH₃)₃ and N(C₂H₅)₃ , which will have greater bond angle? I suppose C₂H₅ will repel other C₂H₅ groups more than CH₃ groups.

Again there is another factor to be considered here. There are more numbers of bonds in C₂H₅ and shape is tetrahedral which might (there is a possibility) bring the CH₃ connected to the C (which is connected to N) closer to N hence increasing repulsion between the group and N and thus decreasing C₂H₅-N-C₂H₅ bond angle.

Again, why does this not happen?

Can anyone explain the correct order between these molecules?

Answer 1

$\ce{NH3}$ and $\ce{NF3}$ are easier to analyse and predict since there is only one set of atoms around the nitrogen — not three groups of four or three groups of seven as for trimethyl- and triethylamine, respectively.

Note that you did not (but should!) include bond lengths in your analysis. It is important to remember them. For ease, here they are as taken from Wikipedia (along with the bonding angles):

$$\textbf{Table 1: Bond lengths and angles}\\\textbf{of nitrogen compounds}\\ \begin{array}{lcc} \hline \text{compound} & d(\ce{N-X}) & \angle (\ce{X-N-X})\\ \hline \ce{NH3} & 102~\mathrm{pm} & 107.8^\circ\\ \ce{NF3} & 137~\mathrm{pm} & 102.5^\circ\\ \ce{NMe3} & 146~\mathrm{pm} & 110.9^\circ\\ \hline \end{array}$$

Starting with the difference between $\ce{NH3}$ and $\ce{NF3}$ — oh wait, I already answered that once. The gist is both compounds would love a $90^\circ$ bond angle but cannot due to too small bond lengths — the outer atoms would clash into each other, figuratively speaking. Therefore, the bond angles are expanded (at a loss of energy) to reduce this steric strain (gain of energy) up to an optimal sweet spot. Since the $\ce{N-H}$ bond is shorter than $\ce{N-F}$, the fluorines are comfortable with a slightly smaller bond angle as they are further away from each other (even though they are larger).

The difference when going from fluorine atoms to methyl groups (or ethyl groups) — as highlighted above — is the additional set of atoms. These cause further crowding, even though the slightly larger carbon atom (with respect to fluorine) allows for an even longer bond length. The steric strain of the additional hydrogens means that the bond angle must be extended even further to accomodate for all these.

Not much changes when going from trimethyl to triethylamine — except that the steric strain is larger still, and that therefore the bond angle is probably increased slightly again. One source I found including a crystal structure including triethylamine did not, however, give notably different bond angles when compared to trimethylamine, so this point may be moot. Note, however, that the angles reported are already larger than the perfect tetrahedral angle, indicating that the lone pair has an even lower s-proportion which would destabilise the overall molecule. Only by reducing steric strain can this low s-proportion be explained in any way.

As I already commented on the question: Back-bonding will not happen since there are no accessable empty orbitals on nitrogen that could accept a back-bond. Try drawing a back-bond in a Lewis formula without exceeding nitrogen’s octet to see that it is impossible.