Question

For $\text{XeO}_2\text{F}_2$, select the correct statements : (A) It has see-saw shape (B) $\angle FXeF \approx 180^\circ$ (C) $\angle OXeO \approx 180^\circ$ (D) Number of valence electron on $\text{Xe} = 5$

• A. A, B, C and D
• B. A and B only
• C. B and D only
• D. A & B only

Hint:

For $sp^3d$ hybridization ($\text{SN}=5$), the electronic geometry is TBP. Lone pairs and atoms forming multiple bonds typically occupy equatorial positions to minimize repulsion.

Answer 1

The Correct Option is B

Xenon is the central atom, having 8 valence electrons.
$\text{XeO}_2\text{F}_2$ has 2 double bonds ($\text{O}$) and 2 single bonds ($\text{F}$).
Total bonding electron pairs = 4. Remaining electrons $= 8 - (2\times 2 + 2\times 1) = 2$. This forms 1 lone pair ($\text{LP}$).
Steric Number $(\text{SN}) = 4 \text{ bonding pairs} + 1 \text{ LP} = 5$. Hybridization is $sp^3d$.
(A) The geometry is $\text{AX}_4\text{E}_1$, which results in a see-saw molecular shape. (A) is Correct.
(B) The $\text{F}$ atoms occupy the axial positions in the TBP arrangement. The $\angle FXeF$ angle is close to $180^\circ$. (B) is Correct.
(C) The $\text{O}$ atoms occupy the equatorial positions, along with the LP. The $\angle OXeO$ is roughly $120^\circ$ (equatorial angle), not $180^\circ$. (C) is Incorrect.
(D) Xenon is a Group 18 element and has 8 valence electrons, not 5. (D) is Incorrect.
Statements (A) and (B) are correct.