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.

4 Answers 4


This can be argued on the basis of Bent's rule; concisely stated

Atomic s character concentrates in orbitals directed toward electropositive substituents

What follows below is a crude explanation. Before that, I'll note that we concern ourselves with the hybridisation of the orbitals at the central atom.

Since s orbitals are lower in energy than p orbitals, they are better at stabilising electrons. Which is why molecules like to use them (s orbitals) in bonds where there is "more" electron density to stabilise.

Thus, Bent's rule can just as easily be formulated as follows: p character concentrates in orbitals directed towards electronegative substituents; (the central atom doesn't have to waste its low energy s orbitals, the orbitals on the electronegative atom, can (possibly) take care of it.

What has all of this got to do with bond angles? Well, more s character leads to larger bond angles. You can think of how s orbitals are spherical in shape, and a large s character would lead to more "spherical" hybrid orbitals, and that would lead to larger bond angles.

Going back to your question, we are supposed to compare $\ce{NF3}$ and $\ce{NH3}$.

Fluorine is more electronegative that hydrogen, and the $\ce{N-F}$ bond would have greater p character than the $\ce{N-H}$ bond, (or $\ce{N-H}$ bond has greater s character, whichever you prefer). Thus $\ce{H-N-H}$ angle is greater in $\ce{NH3}$ than it is in $\ce{NF3}$

P.S I'm sure you can learn more about what I just tried to explain (rather poorly) by reading the Wikipedia article, and/or a textbook and/or using the search function on this site to find related questions/answers.


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.

  • $\begingroup$ This explanation is just your own conjecture right? Although undoubtedly, the N-F bond has a greater length than the N-H bond, the fluorine atom is also a significantly larger atom than the hydrogen atom. In fact, fluorine has approximately twice the covalent radius of hydrogen according to one Internet source I looked at. Thus, your steric repulsion argument may not be so convincing. $\endgroup$ Jul 11, 2018 at 11:39
  • $\begingroup$ @TanYongBoon No, it’s actually what I was taught by Professor Klüfers in the introductory and inorganic chemistry lecture in Munich in 2006/7. $\endgroup$
    – Jan
    Oct 16, 2018 at 13:30

When you're comparing bond angles between $\mathrm{NH_{3}}$ and $\mathrm{NF_{3}}$, you'd want to take the electronegativities the of hydrogen and fluorine into consideration.

Fluorine hits a 3.98 on the Pauling Scale for electronegativity, while hydrogen does a 2.2 on the same scale.

Hydrogen, with its said electronegativity of 2.2 does a fairly decent job of pulling the N-H electron bond pairs towards it. As you would know, adjacent electron bond pairs repel, and it's this repulsion that gives ammonia an H-N-H bond angle of 107°.

Now fluorine, with an electronegativity of 3.98, does an even better job of pulling the N-F electron bond pairs towards it. Since its got a much greater electronegativity than hydrogen, fluorine's able to pull out the electron pairs even further. The consequence of this is that the N-F bonds can be pushed closer than N-H bonds before they're stopped by the repulsion of their electron bond pairs. This is what gives $\mathrm{NF_{3}}$ a smaller bond angle of 102°.


Both $\ce{NH3}$ and $\ce{NF3}$ molecules formed as the shape of pyramid (bonds pairs-3 lone pairs-1) But the angles between those molecules different to the each other because of the electronegativity of the each atom.

Electronegativity of each atom




Get the molecule $\ce{NH3}$. Actually according to the Folin's law nitrogen's electonegativity is more than the Hydrogen. Therefore the bond electron cloud tries to attract to the direction of the nitrogen atom. There for the whole electron cloud attract to the central atom. Then the each electron pair come closer to the each and because of that electrons tries to repeal.(When electrons come closer to the each other they try to repeal) And then the the angles been increased because of the repultion.

Then get the molecule $\ce{NF3}$. It also making a pyramidal shape. The electronagitivity of Florin is more than the Nitrogen. Therefore the bond pairs attract to the florin atoms, that means they attract to the terminal atoms. Then the distance between electon pairs increase, and not try to repulsion. Because of thar the angles be smaller.

*If the repulsion is very large electrons try to go far. Then the angle increase. *If the repulsion is small electrons not try to go far. Then the angle is try to be bended.

Therefore the bond angle of $\ce{NH3}$ is more than $\ce{NF3}$.

Now think about the angle of $\ce{NCl3}$ also.


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