# Why is the bond angle of sulphur difluoride greater than that of hydrogen sulphide?

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:

• Electrostatic repulsion? Jun 11 at 11:37
• Extent of hybridization is very low in these compounds and thus Bent's rule doesn't hold. Jun 11 at 13:30
• Moreover sulphur has a positive formal charge in $\ce{SF2}$ whereas in case of $\ce{SH2}$ it has a negative formal charge. Jun 11 at 14:14
• @NisargBhavsar Any reason for not correcting “flourine” and “diflouride”?
– Ed V
Jun 11 at 14:24
• 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. Jun 11 at 14:43