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Why is the bond angle of $\ce{SF_2 (98.05^\circ) > SH_2 (92.11^\circ)}$?

Isn't this contradicting Bent's rule or otherwise electron repulsion rule ? Fluorine is more electronegative and hence it should have bonds with higher $\mathrm p$ character than those which hydrogen will form. Which means $\ce{F-S-F}$ bond angle should be smaller than $\ce{H-S-H}$ bond angle. Why is it not the case?

Source:

  1. $\ce{SH2}$ bond angle
  2. $\ce{SF2}$ bond angle
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  • $\begingroup$ Electrostatic repulsion? $\endgroup$ Commented Jun 11, 2021 at 11:37
  • $\begingroup$ Extent of hybridization is very low in these compounds and thus Bent's rule doesn't hold. $\endgroup$ Commented Jun 11, 2021 at 13:30
  • $\begingroup$ Moreover sulphur has a positive formal charge in $\ce{SF2}$ whereas in case of $\ce{SH2}$ it has a negative formal charge. $\endgroup$ Commented Jun 11, 2021 at 14:14
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    $\begingroup$ @NisargBhavsar Any reason for not correcting “flourine” and “diflouride”? $\endgroup$
    – Ed V
    Commented Jun 11, 2021 at 14:24
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    $\begingroup$ Please note that names of chemicals or chemical elements (like 'hydrogen') should not be capitalised, unless occurring at the beginning of a sentence. Additionally, IUPAC (and most people) use the spelling 'sulfur' nowadays. $\endgroup$ Commented Jun 11, 2021 at 14:43

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