Question:medium

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

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Calculate the lone pairs on Xe for each compound using (valence electrons of Xe minus bonding pairs) divided by 2, then add bonding pairs and lone pairs to get the total electron domains. A square planar shape needs exactly 4 bonding pairs and 2 lone pairs sitting opposite each other.
Updated On: Aug 17, 2026
  • \(XeF_2\)
  • \(XeF_4\)
  • \(XeF_6\)
  • None of these
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The Correct Option is B

Solution and Explanation

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.
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