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  4. Fluorine reacts with dilute NaOH and forms a gaseous product A. The bond angle in the molecule of A is

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Q. Fluorine reacts with dilute $NaOH$ and forms a gaseous product $A$. The bond angle in the molecule of $A$ is

VITEEEVITEEE 2009

Solution:

$2 F _{2}+\underset{\text{dilute}}{2 NaOH }\longrightarrow 2 NaF + \underset{(A)}{OF _{2} \uparrow}+ H _{2} O$
The structure of ' $A$ ' $\left( OF _{2}\right)$ is as
image
o bonds made by $O =2$
Lone pairs of electrons on $O =2$
$\therefore $ No. of orbitals used by $O$ for hybridisation $=2+2=4$
$\therefore $ Hybridisation of $O$ in $OF _{2}=s p^{3}$
Due to repulsion between two lone pairs of electrons, its shape gets distorted. Therefore, the bond angle in the molecule is $103^{\circ}$.