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(B) The reaction of fluorine with dilute $NaOH$ is given by: $2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O$ Thus,the gaseous product $A$ is oxygen difl

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$103^{\circ}$

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(B) The reaction of fluorine with dilute $NaOH$ is given by: $2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O$ Thus,the gaseous product $A$ is oxygen difluoride $(OF_2)$. In $OF_2$,the central oxygen atom is $sp^3$ hybridized. It has two bond pairs and two lone pairs of electrons. Due to the strong repulsion between the two lone pairs of electrons on the oxygen atom,the bond angle is compressed from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $103^{\circ}$.

Fluorine reacts with dilute $NaOH$ to form a gaseous product $A$. The bond angle in the molecule of $A$ is

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