Question

Statement-I : Among $\text{BF}_4^-$, $\text{SiF}_4$, $\text{SF}_4$, and $\text{XeF}_4$, the bond lengths are not identical in two of these molecules.
Statement-II : Among $\text{O}_2^+$, $\text{O}_2$, $\text{O}_2^{2-}$ and $\text{F}_2$ the highest bond order is found in $\text{O}_2^-$.

• A. Statement I is true and statement II is false.
• B. Statement I is false and statement II is true.
• C. Both statement are true.
• D. Both statement are false.

Hint:

Check VSEPR geometries. Only structures with non-equivalent positions (like trigonal bipyramidal $sp^3d$ derived) have different bond lengths.

Answer 1

The Correct Option is D

Statement I:
$\text{BF}_4^-$ (Tetrahedral) - All bonds identical.
$\text{SiF}_4$ (Tetrahedral) - All bonds identical.
$\text{XeF}_4$ (Square Planar) - All bonds identical.
$\text{SF}_4$ (See-saw) - Axial and equatorial bonds are different lengths.
So, only ONE molecule ($\text{SF}_4$) has non-identical bonds. The statement claims "two of these molecules". Thus, Statement I is False.
Statement II:
Bond Orders:
$\text{O}_2^+ = 2.5$.
$\text{O}_2 = 2.0$.
$\text{O}_2^{2-} = 1.0$.
$\text{F}_2 = 1.0$.
$\text{O}_2^- = 1.5$.
The highest bond order is in $\text{O}_2^+$ (2.5), not $\text{O}_2^-$. Thus, Statement II is False.