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?

  • $\begingroup$ I would assume that there is no back-bonding in $\ce{NF3}$. There are no accessable empty orbitals on nitrogen that you can back-bond into. $\endgroup$ – Jan Jan 28 '16 at 16:53

$\ce{NH3}$ 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.

  • 2
    $\begingroup$ You're right about no backbonding. My friend (whom I consider to be all knowing) told me that the reason would be backbonding so I wrote it without giving it much thought. I should've tried to think it out myself rather than believing him blindly. And that is a really good explanation and I forgot about bond lengths too. Thanks a lot. $\endgroup$ – Quark Jan 28 '16 at 19:30
  • $\begingroup$ You are forgetting about core electrons, when compare H and C/F $\endgroup$ – permeakra Jan 28 '16 at 22:21
  • $\begingroup$ @permeakra I’m curious: How do they affect the picture aside from what I already wrote? $\endgroup$ – Jan Jan 28 '16 at 22:55
  • $\begingroup$ @Jan No idea =). I suspect that exchange repulsion is a thing here, though. $\endgroup$ – permeakra Jan 28 '16 at 23:16

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