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](https://chemistry.stackexchange.com/questions/14087/why-is-the-bond-angle-h-p-h-smaller-than-h-n-h) 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.


[![enter image description here](https://i.stack.imgur.com/W4drz.jpg)](https://i.stack.imgur.com/W4drz.jpg)


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](https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule) and [here](https://chemistry.stackexchange.com/questions/15671/utility-of-bents-rule-what-can-bents-rule-explain-that-other-qualitative-con) 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 $\mathrm{90~^\circ}$.


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