Question

$\mathrm{S}-\mathrm{O}$ bond length is max in:

• A. ${\mathrm{SOCl}}_2$
• B. ${\mathrm{SOBr}}_2$
• C. ${\mathrm{SOF}}_2$
• D. All have same length

Answer 1

The correct option is B

${\mathrm{SOBr}}_2$

The correct option is (B):

Explanation for the correct answer:

Bent rule:

1. It basically describes the relationship in between the orbital hybridizations of the atoms which are there in the center position as well as the electronegativity of the substituents.
2. So, bond length generally depends on the bond order of the electrons.
3. As when more electrons participate in the bond formation the bond length will become shorter.
4. Basically the Bond length is just inversely proportional to the bond strength as well as the bond dissociation energy.
5. And the bond length also depends upon the bond angle, as larger the bond angle the shorter would be the bond length.
6. The greater the size of the Halogen, the greater would be the angle between $\mathrm{S}-\mathrm{X}$ and shorter the $\mathrm{S}-\mathrm{O}$ bond angle.
7. So as the size of the halogen atom increases in order$\mathrm{F}<\mathrm{Cl}<\mathrm{Br}$ thus the order of the bond angle $\mathrm{X}-\mathrm{S}-\mathrm{X}$ increases in the order${\mathrm{SOF}}_2<{\mathrm{SOCl}}_2<{\mathrm{SOBr}}_2$.
8. Thus making the order of the bond angle of $\mathrm{S}-\mathrm{O}$ increases in the opposite order i.e.${\mathrm{SOBr}}_2<{\mathrm{SOCl}}_2<{\mathrm{SOF}}_2$.
9. Hence the order of the bond length of $\mathrm{S}-\mathrm{O}$ appears to be in the order ${\mathrm{SOF}}_2<{\mathrm{SOCl}}_2<{\mathrm{SOBr}}_2$

Therefore, the correct option is (B): ${\mathrm{SOBr}}_2$