Step 1: Understanding the Question:
We need to compare the VSEPR geometry of \( \text{ClF}_5 \) with the given Xenon compounds.
Step 3: Detailed Explanation:
Structure of \( \text{ClF}_5 \):
Cl has 7 valence electrons. It forms 5 bonds with F and has 1 lone pair.
Steric Number \( = 5 + 1 = 6 \). Geometry \( = \) Octahedral; Shape \( = \) Square Pyramidal.
Checking Xenon compounds:
(A) \( \text{XeF}_4 \): 4 bond pairs, 2 lone pairs. Shape \( = \) Square Planar.
(B) \( \text{XeF}_6 \): 6 bond pairs, 1 lone pair. Shape \( = \) Distorted Octahedral.
(C) \( \text{XeO}_3 \): 3 bond pairs, 1 lone pair. Shape \( = \) Pyramidal.
(D) \( \text{XeOF}_4 \): Xe has 8 valence electrons. It forms 1 double bond with O and 4 single bonds with F. It has 1 lone pair remaining.
Steric Number \( = 1 (\text{Xe=O}) + 4 (\text{Xe-F}) + 1 (\text{lp}) = 6 \). Shape \( = \) Square Pyramidal.
Thus, \( \text{XeOF}_4 \) and \( \text{ClF}_5 \) are structurally similar.
Step 4: Final Answer:
\( \text{XeOF}_4 \) has a square pyramidal structure similar to \( \text{ClF}_5 \).