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

Topic: Chemical Bonding and Molecular Structure
Step 1: Understanding the Question:
We need to determine which Xenon fluoride compound exhibits a "Square Planar" molecular geometry based on electron pair arrangements.
Step 2: Key Formula or Approach:
We use the VSEPR (Valence Shell Electron Pair Repulsion) theory.
Steric Number (SN) = (Number of valence electrons on central atom + Number of monovalent atoms - charge) / 2.
Step 3: Detailed Explanation:
For \(XeF_4\):
Central atom \(Xe\) has 8 valence electrons.
Number of fluorine atoms (monovalent) = 4.
SN = \(\frac{8 + 4}{2} = 6\).
A Steric Number of 6 corresponds to \(sp^3d^2\) hybridization (Octahedral geometry).
Out of 6 electron pairs, 4 are Bonding Pairs (BPs) and 2 are Lone Pairs (LPs) since there are 4 Fluorine atoms.
According to VSEPR theory, to minimize repulsion, the 2 lone pairs occupy axial positions (opposite to each other).
The 4 fluorine atoms occupy the equatorial positions, resulting in a Square Planar molecular shape.
Step 4: Final Answer:
\(XeF_4\) has a square planar geometry.