. $OCl _2$ has unexpectedly greater angle due to bulky surrounding group $Cl$.
Angle of $OF _2$ is smaller than that of $H _2 O$ as bond angle is inversely to electronegativity.
All these are $sp ^3$ hybridized. And have $2$ lone pairs but $ClO _2$ has only $3$ free electrons. Therefore, has highest bond angle.
Therefore, the bond angle order is: $ClO _2\,>\, OCl _2\, >\, H _2 O \, >\, OF _2$.