Question

Which of the following noble gas compounds has a square planar geometry: \(XeF_2\), \(XeF_4\), or \(XeF_6\)?

• A. \(XeF_2\)
• B. \(XeF_4\)
• C. \(XeF_6\)
• D. None of these

Hint:

For \(sp^3d^2\) hybridization with two lone pairs (as in \(XeF_4\)), the lone pairs occupy opposite positions, leaving four atoms arranged in a square plane.

Answer 1

The Correct Option is B

Concept: The geometry of molecules can be predicted using VSEPR theory, which states that electron pairs around the central atom arrange themselves to minimize repulsion.

Step 1: Count the valence electrons of Xenon. Xenon has 8 valence electrons.

Step 2: Determine bonding and lone pairs in \(XeF_4\). In \(XeF_4\): \[ \text{Number of Xe-F bonds} = 4 \] After forming four bonds, two lone pairs remain on Xenon. Thus, \[ \text{Total electron pairs} = 6 \] Hybridization: \[ sp^3d^2 \]

Step 3: Determine the molecular geometry. The six electron pairs arrange octahedrally, and the two lone pairs occupy opposite positions to minimize repulsion. Therefore, the four fluorine atoms lie in one plane forming a: \[ \text{Square planar geometry} \] Hence, the compound with square planar geometry is: \[ \boxed{XeF_4} \]