# How to calculate bond angle of simple molecules quantitatively?

Hi I'm new to this forum so spare me if I've made mistakes.

Our high school chemistry teacher explained that bond angles of different molecules varies slightly, e.g. 107 degrees in ammonia and 104.5 degrees in water, which could be predicted by VSEPR theory qualitatively.

I have checked literature either through google or stack-exchange, but none point me to any formula/algorithm/computationally inexpensive methods to find non-ideal bond angles by calculation.

• I was taught that for each unbonded electron pair, the usual bond angle for a certain configuration is reduced by ≈2º. – ringo Oct 9 '15 at 14:58
• Your first mistake: This is not a forum! D= No, don't worry, joking ;) Welcome to chemistry.stackexchange.com ^^ – Jan Oct 9 '15 at 14:59
• @ringo never heard of the approximation - could you point me to the reasoning behind? – MarcoXerox Oct 11 '15 at 3:48

## 1 Answer

This has a simple reason: To the best of my knowledge there is no quantitative way to predict bond angles without resorting to computational chemistry.

This is because the entire subject of predicting bond angles is a lot more tricky than it was displayed to you. Take phosphane and hydrogen sulphide, the heavy analogues to ammonia and water. Their bond angles are nowhere near $109^\circ$ but much closer to $90^\circ$.

If you dig deeper into the matter, you find that ammonia and water are more the exception than the rule, that the VSEPR model fails spectacularly on the long run, and that their geometries are actually finely balanced between an ideal $90^\circ$ angle and the hydrogen-hydrogen repulsion.

In fact, not even with computers can you calculate the bond angle directly with a certain set of axiomic assumptions. Rather, computational chemistry would empirically calculate energies for different angles and deem the angle that has the lowest energy to be ‘correct’ and ‘calculated’.

So no simple quantitative rule exists nor will it exist.

Note: There are, of course, semi-quantitative rules like the one mentioned by Ringo in a comment. But I don’t consider them quantitative, because they still contain approximations.

• Thank you for the explanation. But then, does the smaller angle between H2S and PH3 compared to H2O and NH3 related to their hybridization, as S & P are in the same group as O & N? – MarcoXerox Oct 11 '15 at 3:38
• @MarcoXerox You can kind of say that. $\ce{S}$ and $\ce{P}$ mainly use the p-orbitals for bonding to hydrogen while $\ce{N}$ and $\ce{O}$ use somehow hybridised orbitals. But from a theoretical perspective, it’s rather that $\ce{N}$ and $\ce{O}$ would use only their p-obitals if they weren’s damn small. The hybridisation and their corresponding angles are obtimised by nature. – Jan Oct 11 '15 at 15:06