Which statements are NOT TRUE about XeO₂F₂$XeO_{2}F_{2}$? A. It has a see-saw shape. B. Xe has 5 electron pairs in its valence shell in XeO₂F₂$XeO_{2}F_{2}$. C. The O–Xe–O bond angle is close to 180°$180^{\circ}$. D. The F–Xe–F bond angle is close to 180°$180^{\circ}$. E. Xe has 16 valence electrons in XeO₂F₂$XeO_{2}F_{2}$. Choose the correct answer from the options given below.

Related Formula

Steric Number = Number of σ-bonds + Number of lone pairs on central atom$\text{Steric Number} = \text{Number of } \sigma\text{-bonds} + \text{Number of lone pairs on central atom}$

Core Logic

Evaluate the VSEPR structure of XeO₂F₂$XeO_2F_2$: Central atom Xe has 8 valence electrons. It forms 2 double bonds with Oxygen (2 × 2 = 4$2 \times 2 = 4$ electrons used) and 2 single bonds with Fluorine (2 × 1 = 2$2 \times 1 = 2$ electrons used). Total electrons used in bonding = 6$6$. Remaining valence electrons on Xe = 8 - 6 = 2$8 - 6 = 2$, which forms 1$1$ lone pair. Steric Number = 4 (bond pairs) + 1 (lone pair) = 5$\text{Steric Number} = 4 (\text{bond pairs}) + 1 (\text{lone pair}) = 5$ electron pairs. Hybridization is sp³d$sp^3d$. Geometry is trigonal bipyramidal. According to Bent's rule, the highly electronegative F atoms occupy axial positions, while the double-bonded O atoms and the lone pair occupy equatorial positions. This results in a "see-saw" shape.

Step 1: Check Statements

A. True. The molecule has a see-saw shape. B. False. Xe is surrounded by 4 σ$\sigma$-bonds, 2 π$\pi$-bonds, and 1 lone pair. In total, counting individual bonds, Xe has 7 electron pairs in its valence shell (4 σ$\sigma$, 2 π$\pi$, 1 lp) = 14 valence electrons. C. False. The O atoms occupy equatorial positions, so the O-Xe-O bond angle is close to 120°$120^{\circ}$, not 180°$180^{\circ}$. D. True. The F atoms are axial, so the F-Xe-F bond angle is close to 180°$180^{\circ}$. E. False. The central Xe atom has 14 valence electrons involved (8 from octet, plus sharing from ligands), not 16.

We need to find the INCORRECT statements, which are B, C, and E.

Pattern Recognition

In sp³d$sp^3d$ hybridization, bulky groups (double bonds, lone pairs) always occupy equatorial positions to minimize 90°$90^{\circ}$ repulsions. Highly electronegative groups (F) are forced into axial positions.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: p-Block Elements