# Why does SO2 have a larger bond angle than H2O, H2S, and NH3

So we have to find which of the follwing compounds has the smallest bond angle: $\ce{H2O}$, $\ce{H2S}$, $\ce{NH3}$, $\ce{SO2}$.

So L $\propto \frac{1}{BA}$ where $L$ is the number of lone pairs and $BA$ is bond angle

So the number of lone pairs is:-

$\ce{H2O}$: 2

$\ce{H2S}$: 2

$\ce{NH3}$: 1

$\ce{SO2}$: 2

Since $\ce{NH3}$ has the fewest lone pairs, it must have the highest bond angle, so it is out.

Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom. The electronegativiy of the O atom in $\ce{H2O}$ is the highest, so it is out.

Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom. The electronegativiy of the H atom in $\ce{H2S}$ is the lowest of the remaining two. The only one left is $\ce{SO2}$ which must have the lowest bond angle but when I checked on google it had the highest bond angle. How is that possible?

Here are the bond angles for each molecule (data from wikipedia):

\begin{array}{|c|c|}\hline \mathrm{Molecule} & \mathrm{Bond \space Angle \space (^\circ)} \\ \hline \ce{H2S} & 92.1 \\ \hline \ce{H2O} & 104.5 \\ \hline \ce{NH3} & 107.8 \\ \hline \ce{SO2} & 119 \\ \hline \end{array}

So $L \propto \frac{1}{BA}$ where $L$ is the number of lone pairs and $BA$ is bond angle.

This is true but only in very specific situations; when dealing with molecules that have a central atom in the same period and outer atoms of the same element (e.g $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$). It breaks down as soon as you start comparing molecules with central atoms from different periods (e.g $\ce{PH3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle) or when you compare molecules with different outer atoms (e.g $\ce{NF3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle). For the reasons behind this I direct you to an excellent previous answer from @ron.

Additionally, I think you need to reconsider the number of lone pairs in $\ce{SO2}$, which can be described by these two resonance structures.

As you can see there is only one lone pair, but unfortunately this doesn't help us very much as we have no other molecule to compare it to.

Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom.

This is incorrect. In fact the reverse is true - that $BA \propto \frac{1}{ENC}$ - but only in the same situations as mentioned before. In this case the trend is only really pronounced in the second period; it is very slight in the third period and virtually non existent in the fourth. In fact, this 'rule' doesn't really have anything to do with electronegativity (that's just a coincidence) and it's essentially just a result of the first rule.

Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom.

This is sort of true - it's known as Bent's rule and it can be very useful but it's not really applicable here. It's been discussed many times on this site but here and here are some good introductions.

So how should you answer this question?

The first thing to note is that $\ce{SO2}$ only has three 'groups' on the central atom (sometimes called 'effective electron pairs' in VSEPR theory) - two bonds intermediate between a double and a single bond and a lone pair - whereas all the other molecules have three. Therefore we expect $\ce{SO2}$ to have the largest bond angle of the four molecules, and this is indeed the case. $\ce{H2O}$ and $\ce{NH3}$ are hydrides of the same period so we can use the first rule to determine that $\ce{H2O}$ has a smaller bond angle. Now we just have to decide whether $\ce{H2O}$ or $\ce{H2S}$ has a smaller bond angle. We can apply the hybridisation arguments given by @ron in the answer I linked earlier to determine that $\ce{H2S}$ has the smallest bond angle, and indeed we find that it is almost unhybridised with a bond angle very close to $\mathrm{90~^\circ}$.