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

To solve this problem, we analyze the chemical structure and geometry of the compound XeO₂F₂. Each statement is examined using VSEPR theory and bonding considerations.

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1. The compound XeO₂F₂ has xenon (Xe) as the central atom. Xenon is bonded to two oxygen atoms and two fluorine atoms.
2. According to VSEPR theory, the geometry depends on the number of bonding and lone electron pairs around the central atom. Xenon has the electronic configuration:
   [Kr] 4d¹⁰ 5s² 5p⁶.
   In XeO₂F₂, xenon forms four σ-bonds and has one lone pair, leading to sp³d hybridization.
3. For an sp³d hybridized atom with one lone pair, the molecular shape is see-saw. Hence, statement (A) is correct.
4. In a see-saw geometry, the larger atoms prefer equatorial positions, while fluorine atoms occupy the axial positions. Therefore, the F–Xe–F bond angle is approximately 180°. Hence, statement (B) is correct.
5. The O–Xe–O bond angle does not need to be close to 180° in a see-saw geometry. Therefore, statement (C) is incorrect.
6. Xenon has 8 valence electrons. In XeO₂F₂, xenon shares electrons with four atoms but still retains a lone pair. Thus, its valence electron count is not reduced to 5. Hence, statement (D) is incorrect.

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Conclusion:

The correct set of statements is: A and B only.