Predict the shape of XeOF4 on the basis of VSEPR theory
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Square pyramidal
By VSEPR theory the shape of a molecule is decided by the number of electron pairs (bond pairs + lone pairs) around the central atom. The pairs take up the arrangement that keeps them farthest apart, and the shape is described by the positions of the atoms only.
In \( \ce{XeOF4} \), xenon has 8 valence electrons. Four go into the four Xe–F bonds and two into the Xe=O bond, leaving one lone pair:
\( \displaystyle \text{bond pairs (directions)} = 4\ (\ce{Xe-F}) + 1\ (\ce{Xe=O}) = 5 \)
\( \displaystyle \text{electron pairs} = 5 + 1\ \text{lone pair} = 6 \;\Rightarrow\; sp^3d^2,\ \text{octahedral arrangement} \)
One corner of the octahedron is taken by the lone pair, so the five atoms form a square pyramid: four F atoms at the base and the O atom at the apex.
Why the other options are incorrect
Option (1) Linear — Incorrect. This is the shape of \( \ce{XeF2} \) (2 bond pairs + 3 lone pairs, trigonal bipyramidal arrangement).
Option (3) Pyramidal — Incorrect. This is the shape of \( \ce{XeO3} \) (3 bond directions + 1 lone pair, tetrahedral arrangement).
Option (4) Square planar — Incorrect. This is the shape of \( \ce{XeF4} \) (4 bond pairs + 2 lone pairs, octahedral arrangement with the lone pairs opposite each other).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Xenon-oxygen Compounds – “… XeO3 is a colourless explosive solid and has a pyramidal molecular structure (Fig. 7.9). XeOF4 is a colourless volatile liquid and has a square pyramidal molecular structure (Fig.7.9).”
