Which is true about ${\mathrm{OF}}_2$ and ${\mathrm{Cl}}_2\mathrm{O}$?

${\mathrm{OF}}_2$ and ${\mathrm{Cl}}_2\mathrm{O}$  have ${\mathrm{sp}}^3$ hybridized oxygen. Because the oxygen is ${\mathrm{sp}}^3$ hybridized, which implies it possesses four ${\mathrm{sp}}^3$ hybrid orbitals. To produce the $\mathrm{O}-\mathrm{H}$ sigma bonds, one of the ${\mathrm{sp}}^3$ hybridized orbitals overlaps with an orbital from a hydrogen. 

The$\mathrm{C}-\mathrm{O}$sigma bond is formed when one of the ${\mathrm{sp}}^3$ hybridized orbitals overlaps with a ${\mathrm{sp}}^3$ hybridized orbital from carbon.

Bond angles in ${\mathrm{OF}}_2$less than 109°28'. Because a lone pair of electrons requires more space than a bonding pair in oxyfluoride, the molecular geometry bends to provide the lone pair more room, resulting in bond angles that are less than those of the ideal geometry.

Bond angles in ${\mathrm{Cl}}_{2}0$ is greater than 109°28'. Because bonding electrons in '${\mathrm{Cl}}_{2}0$' are closer to O than to Cl, hence the bond angle is higher than 109°28'. Because the bonding electrons in $\text{'}{\mathrm{Br}}_2\mathrm{O}\text{'}$are closer to oxygen than in '${\mathrm{Cl}}_{2}0$' the bond angle is the greatest 116°.