Fluorine reacts with dilute $\mathrm{NaOH}$ and forms a gaseous product $A$. The bond angle in the molecule…

Fluorine reacts with dilute $\mathrm{NaOH}$ and forms a gaseous product $A$. The bond angle in the molecule of $A$ is
  1. $104^{\circ} 40^{\prime}$
  2. $103^{\circ}$
  3. $107^{\circ}$
  4. $109^{\circ} 28^{\prime}$

Solution


The structure of ' $A$ ' $\left(\mathrm{OF}_2\right)$ is as
$\sigma$ bonds made by $\mathrm{O}=2$ Lone pairs of electrons on $\mathrm{O}=2$ $\therefore$ No. of orbitals used by $\mathrm{O}$ for hybridisation $=2+2=4$ $\therefore$ Hybridisation of $\mathrm{O}$ in $\mathrm{OF}_2=s p^3$ Due to repulsion between two lone pairs of electrons, its shape gets distorted. Therefore, the bond angle in the molecule is $103^{\circ}$.

Asked in: AP EAMCET 2009

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