Question

Compare the bond lengths (x and y) in the following molecules:

• A. $x>y$
• B. $y>x$
• C. $x=y$
• D. None of these

Answer 1

The correct option is B $y>x$

The peripheral atoms in both cases are O and F, out of which F is more electronegative. F will want to combine with the orbital that has less s-character.

In the second molecule $\left(IO{F}_3\right)$, 3 flourine atoms are present. Therefore, the oxygen atom will combine with orbitals having more s-character.

In the first molecule $\left(I{O}_2{F}_2\right)$, 2 flourine atoms are present. Therefore, the two oxygens will combine with orbitals having less s-character than in $IO{F}_3$.
$\therefore$ $\text{bond length}\propto \frac{1}{\text{\%}scharacter}$

Thus, the bond length in $I{O}_2{F}_2$ is greater than that in $IO{F}_3$.

i.e $y>x$.