The bond angle in $O F_{2} is 103^{o}$ but in $O C l_{2} is 111^{o}$ though hybridization of oxygen is same $s p^{3}$ in both. The bond angle in $O C l_{2}$ is more because

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Fluorine is more electro negative than chlorine

Repulsion between two Cl atoms in $O C l_{2}$ is more than the repulsion between F atoms in $O F_{2}$

$O - F$ bond is shorter than $O - C l$ bond

Chlorine is larger atom than fluorine atom and also due to double bond character by back bonding.

answer is B.

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Detailed Solution

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$S i n \frac{ϑ}{2} = \frac{r_{x}}{d_{Δ Δ}}$

The $d_{O F} = 1.4 / A a n d r_{F} = 1.10 A^{o}, θ = 103^{o}$ for $O F_{2}$ and the $d_{o - C l} = 1.70 A^{o} and r_{c} = 1.4 A^{o}, θ = 111^{o} f o r O C l_{2} .$ Therefore the ligand close packing model correctly predicts $ϑ$ for $O C l_{2}$ that is larger than that in $O F_{2}$ and larger than the ideal tetrahedral angle. This indicates that larger size of $C l$ atoms makes the bond angle wider than in $O F_{2} .$ Also there is some double bond character between oxygen and chlorine due to back bonding from filled p-orbitals of O atom to vacant d-orbitals of $C l$ atoms which causes increases in bond character due to double bond character.