Why is the binding angle of HOF lesser than that of H2O, despite the lone pairs and size of the flourine?
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.


Among the series H, F, Cl, hydrogen is the least electronegative and fluorine the most.
F is very electronegative and thus the electron density of the O-F bond near O is less than that in O-H.
If the electron density is farther away from the oxygen, then the 2 lone pairs of O squeeze the two bonding pairs closer together and the F-O-F bond angle will be less than the H-O-H angle.
Based on the electronegativity, you should expect that the Cl-O-Cl bond angle is less than H-O-H but more than F-O-F. This is not the case and the reason is that chlorine is big enough so that the steric repulsion prevails.

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 $F_2O$!



The reason for this unexpected small angle is that the bond length O-F and O-H are very different and their electron density(bond pairs) don't interact much and therefore they repel even less than in the F-O-F case(see picture above, where electron density is represented in blue).