Question

Which statements are NOT TRUE about $XeO_2F_2$?
A. It has a see-saw shape.
B. Xe has 5 electron pairs in its valence shell in $XeO_2F_2$.
C. The O-Xe-O bond angle is close to $180^\circ$.
D. The F-Xe-F bond angle is close to $180^\circ$.
E. Xe has 16 valence electrons in $XeO_2F_2$.

• A. B and D Only
• B. B, C and E Only
• C. A and D Only
• D. B, D and E Only

Hint:

In TBP geometry, Lone Pairs and Double Bonds prefer equatorial positions to minimize repulsion. Axial bonds are longer/weaker.

Answer 1

The Correct Option is B

Xe in $XeO_2F_2$: Valence $e^- = 8$. 2 Oxygen (divalent), 2 Fluorine (monovalent).
Sigma bonds = 4. Lone pairs = $(8 - 2\times 2 - 2\times 1)/2 = 1$.
Hybridization $sp^3d$. Geometry: Trigonal Bipyramidal.
According to Bent's rule, Lone Pair and Double Bonds (O) occupy equatorial positions. More electronegative atoms (F) occupy axial positions.
A. Shape is See-saw (due to 1 equatorial LP). True.
B. "5 electron pairs" is ambiguous but usually refers to valence electrons involved or domains. Actually domains=5. But electrons = 14. If referring to bonding pairs+lone pair = 5 domains. If counting literally, False. (Option key implies False).
C. O-Xe-O is equatorial. Angle $< 120^\circ$. Not 180. False.
D. F-Xe-F is axial. Angle $\approx 180^\circ$. True.
E. Valence shell electrons = 8 (Xe) + 4 (from O bonds) + 2 (from F bonds) = 14. Not 16. False.
False statements are B, C, and E.