Question

The bond angle in \[\text{O}{{\text{F}}_{2}}\]out of \[\text{O}{{\text{F}}_{\text{2}}}\text{, C}{{\text{l}}_{\text{2}}}\text{O, B}{{\text{r}}_{\text{2}}}\text{O}\] is minimum. It is because in case of \[\text{O}{{\text{F}}_{2}}\]:
(A) Electrons are nearer to fluorine due to high electronegativity of F compared to Cl to Br
(B) Lone pair-lone pair repulsion decreases bond angle
(C) Both (a) and (b) is correct
(D) None of the above is correct

Answer 1

Hint: Lets us see what are the factors that affects the bond angle:
- Hybridization: Bond angle depends on the state of hybridization of the central atom
- Lone pair repulsion: Bond angle is affected by the presence of lone pair of electrons at the central atom.
- Electronegativity: If the electronegativity of the central atom decreases, bond angle decreases.With the help of these hints let's see the complete solution.

Complete step by step answer:
> Generally the bond angle of any compound depends on the hybridization of the central atom. For example $\text{C}{{\text{H}}_{\text{4}}}$has hybridization of $\text{s}{{\text{p}}^{\text{3}}}$with bond angle of 109.5 degree,$\text{BC}{{\text{l}}_{\text{3}}}$has hybridization of $\text{s}{{\text{p}}^{\text{2}}}$with bond angle of 120 degree.
> A lone pair of electrons at the central atom always tries to repel the shared pair of electrons, and it results in decreasing the bond angle.
> Sometimes electro negativity also comes into picture which plays a major role in determining the bond angle of any compound. If the electronegativity of the central atom decreases, the bond angle also decreases.
> In case of $\text{O}{{\text{F}}_{\text{2}}}$ the bond angle is minimum because the electrons are nearer to fluorine due to high electronegativity of F as compared to other halogens, and because of this repulsion between the electrons decreases the bond angle.
So, we can say that option C is the correct answer.

Note: Always remember these two points:
- Triple bonds repel other bonding-electrons more strongly than double bonds.
- Double bonds repel other bonding-electrons more strongly than single bonds.