Question

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

• A. $104^{\circ }40^{\prime }$
• B. $103^{\circ }$
• C. $107^{\circ }$
• D. $109^{\circ }28^{\prime }$

Answer 1

Answer: (B)

$2F_2+\underset{\text{dilute}}{2NaOH}⟶2NaF+\underset{(A)}{O{F}_2\uparrow }+H_{2}O$
The structure of ' $A$ ' $\left(O{F}_2\right)$ is as

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 $O{F}_2=s{p}^3$
Due to repulsion between two lone pairs of electrons, its shape gets distorted. Therefore, the bond angle in the molecule is $103^{\circ }$.