Concept: VSEPR Theory

VSEPR (Valence Shell Electron Pair Repulsion) theory is used to predict the geometry of molecules based on the repulsion between electron pairs in the valence shell of the central atom. The electron pairs (both bonding and lone pairs) arrange themselves to minimize repulsion, leading to a specific molecular geometry.

Why (C) is correct:

1. Determine the central atom: In ${\mathrm{XeOF}}_4$, Xenon (Xe) is the central atom.

2. Calculate the total number of valence electrons:
Xe: 8 valence electrons
O: 6 valence electrons (forms a double bond)
F: 7 valence electrons $\times$ 4 = 28 valence electrons
Total = 8 + 6 + 28 = 42 electrons. However, a simpler approach for VSEPR is to count the number of electron domains around the central atom.

3. Determine the number of electron domains (steric number) around Xe:
Number of bond pairs = 4 (Xe-F bonds) + 1 (Xe=O bond) = 5 bond pairs.
Number of lone pairs = $\frac{1}{2}$ [Valence electrons of Xe - (Number of single bonds $\times$ 1) - (Number of double bonds $\times$ 2)]
This formula is for calculating formal charge, not directly for VSEPR. A more direct way for VSEPR:
Valence electrons of Xe = 8.
Electrons used in bonding: 4 F atoms (single bonds) = 4 electrons. 1 O atom (double bond) = 2 electrons.
Total electrons used in bonding = 4 + 2 = 6 electrons.
Remaining electrons = 8 - 6 = 2 electrons.
Number of lone pairs = $\frac{2}{2}$ = 1 lone pair.

4. Steric number: Number of bond pairs + Number of lone pairs = 5 + 1 = 6.

5. Electron geometry: A steric number of 6 corresponds to an octahedral electron geometry.

6. Molecular geometry: With 5 bond pairs and 1 lone pair, the lone pair occupies one position in the octahedral arrangement. To minimize repulsion, the lone pair is usually placed opposite to the double bond or in an equatorial position if possible. In ${\mathrm{XeOF}}_4$, the lone pair is typically considered to be on one side, distorting the octahedral arrangement to a square pyramidal shape. The oxygen atom (double bond) and the four fluorine atoms form the base and apex of the pyramid, with the lone pair occupying the sixth position, pushing the other atoms into a pyramidal shape.

Option Analysis:

• A) tetrahedral: This geometry corresponds to a steric number of 4 with 4 bond pairs and 0 lone pairs (e.g., ${\mathrm{CH}}_4$).
• B) square planar: This geometry corresponds to a steric number of 6 with 4 bond pairs and 2 lone pairs (e.g., ${\mathrm{XeF}}_4$).
• C) square pyramidal: This is the correct geometry for ${\mathrm{XeOF}}_4$ with 5 bond pairs and 1 lone pair, derived from an octahedral electron geometry.
• D) octahedral: This is the electron geometry for a steric number of 6, but the molecular geometry is different due to the presence of a lone pair (e.g., ${\mathrm{SF}}_6$).

Correct Answer: (C)