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Arrange the following componds in the increasing order of bond angle. $$\ce{HOF,~OF2,~SF2}$$

I know that bond angle of $\ce{SF2}$ will be less than the bond angle of $\ce{OF2}$ because of larger size of sulfur atom.

How can I compare $\ce{HOF}$ with the other two? Since $\ce{H}$ is less electronegative than $\ce{O}$, the bond pair of $\ce{O - H}$ bond will be more closer to $\ce{O}$. The opposite happens for $\ce{O - F}$ bond. In $\ce{SF2}$ both bond pairs are near $\ce{F}$ so the bonds can come closer to each other.

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  • $\begingroup$ This answer explains why $\ce{SF2}$ will have a bond angle close to $90^\circ$ and $\ce{OF2}$ will have a bond angle closer to $109.5^\circ$. For similar reasons, $\ce{HOF}$ will have a bond angle closer to $109.5^\circ$. However, comparing $\ce{HOF}$ and $\ce{OF2}$, you need to compare both the size difference of $\ce{H}$ and $\ce{F}$ and their electron-density (not just electronegativity). $\endgroup$ – Ben Norris Apr 13 '16 at 11:34
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    $\begingroup$ @BenNorris it given in Wikipedia that HOF bond angle is 97.2 and that in SF2 is 98. $\endgroup$ – Aditya Dev Apr 14 '16 at 2:23
  • $\begingroup$ Now, that I would not have expected. Interesting. $\endgroup$ – Ben Norris Apr 14 '16 at 2:42

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