# Why is the binding angle of HOF lesser than that of H2O, despite the lone pairs and size of the flourine? [duplicate]

Why is the binding angle of HOF(101°) lesser than that of H2O?(104.5°) Since fluorine has unpaired electrons wouldn't it repel hydrogen more strongly than the corresponding hydrogen of H2O? Also, the size of fluorine is bigger than hydrogen, which also should have made the repulsion greater.

## marked as duplicate by Mithoron, Mathew Mahindaratne, Jon Custer, Buck Thorn, Todd MinehardtAug 10 at 21:36

In $$X_2O$$ molecules, such as $$H_2O$$, $$F_2O$$, $$Cl_2O$$, the oxygen is at the center of a tetrahedron and therefore theoretically the X-O-X bond angle should be 109.5°. However, this is never the case and it's always a balance between the repulsion of the two lone pairs of the oxygen(they repels more strongly than the bond pairs) and the steric repulsion of the X atoms.
In your case you are considering the molecule HOF: According to the logic above, you may expect that its bond angle is less than $$H_2O$$ but more than $$F_2O$$. Instead, H-O-F bond angle is 97° so even less than $$F_2O$$!