# Bond angles in NH3 and NCl3

The bond angle in a molecule is inversely proportional to the electronegativity of the surrounding atom if the central atom is same. This also happens with $\ce{NH3}$ and $\ce{NF3}$, as bond angle in $\ce{NH3 > NF3}$.

Why then is the bond angle in $\ce{NCl3 > NH3}$?

• We will need to look at Bent's rule as well as steric effects. In the case of NH3 and NCl3, it is clear that steric effects play the dominant role. This is due to chlorine atoms being significantly larger than hydrogen atoms and nitrogen being a small atom. – Tan Yong Boon Mar 14 '18 at 17:31
• Possible duplicate of Why does NF3 have a smaller bond angle than NH3? – Mithoron Mar 14 '18 at 22:58

The bond angle of a molecule depends on several factors. We have to look at all the factors and then decide the result according to them.

The difference in bond angles in $\ce{NF3}$ and $\ce{NH3}$ is only determined by the electronegativity difference between the central atom and the bonded atom. As $\ce{F}$ is more electronegative than $\ce{H}$, in $\ce{NH3}$, bonding electron pair shifts more towards $\ce{N}$ than they shift in $\ce{NF3}$. So, the bond angle increases. There is no other effect at all in this case.

But in the case of $\ce{NCl3}$, you can notice there is a lone pair on the central $\ce{N}$ atom. Most importantly, there is an introduction of vacant low lying $\ce{3d}$ orbital of $\ce{Cl}$ where the lone pair of $\ce{N}$ can delocalise. This effect is known as $\ce{p\pi-d\pi}$ backbonding. So, due to this backbonding, which is quite strong, the bond between $\ce{N}$ and $\ce{Cl}$ adopts a significant double bond character.

Although the electronegativity difference is decreased in the case of $\ce{NCl3}$ and bond pair is shifted away from central $\ce{N}$, this introduction of double-bond character increase the repulsion between the bond pairs even more, and that's why the bond angle increases.

So, in case of $\ce{NF3}$ and $\ce{NH3}$ there was no vacant and valence d orbitals either in $\ce{F}$ or in $\ce{H}$. So, this backbonding doesn't occur there. But in $\ce{NCl3}$, the case is different and more factors determine the bond angle. As the backbonding is much stronger, this effect overall dominates over the electronegativity factor and make the bond angle increase.

Experimental evidence of this double bond character: In $\ce{NCl3}$, $\ce{N-Cl}$ bond length is 1.759 angstroms but $\ce{N-Cl}$ single bond length is 1.91 angstroms. This indicates the presence of some partial double bond character which shrinks the bond length.

• Why 3d orbitals? There could be pi back-donation with the orbitals combinations from the usual valence sibshells. – Oscar Lanzi Mar 15 '18 at 8:31
• @Oscar Lanzi, I didn't get your point. Can you please specify which orbitals you are talking about?? – Soumik Das Mar 15 '18 at 10:48
• Overlap between "nonbonding" electrons on chlorine and antibonding sigma orbitals, which weakens the sigma bonding but gives more bond energy with pi bonding. It's a typical situation with molecular orbitals. – Oscar Lanzi Mar 15 '18 at 11:07

As Tan Yong Boon stated, it is Bent’s Rule with which we can explain the lower bond angle of $\ce{NF3}$ when compared to $\ce{NH3}$. The rule as stated by Henry Bent:

Atomic s character concentrates in orbitals directed towards electropositive substituents.

Fluorine is more electronegative that hydrogen, and the $\ce{N−F}$ bond would have greater p character than the $\ce{N−H}$ bond. And more s character leads to large bond angles. Thus, the bond angle is greater in $\ce{NH3}$ than in $\ce{NF3}$.
Now, consider $\ce{NCl3}$. Clearly, $\ce{Cl}$ atom islarger in size than the central atom, nitrogen. Hence the higher bond angle here is due to the steric crowding caused by $\ce{Cl}$ atoms.(More pronounced than the electrongativity of $\ce{Cl}$ atom). They repel eachother and hence bond angle increases. This is not the case with small $\ce{H}$ atoms around the nitrogen in $\ce{NH3}$. Therefore, the $\ce{Cl-N-Cl}$ bond angle of $\ce{NCl3}$ is greater than $\ce{H-N-H}$ bond angle of $\ce{NH3}$.