Q28 / 150 · MHT CET 2026 (18 April Shift 1) · Chemistry
Which of the following compound has square pyramidal structure?
- A$\text{XeF}_4$
- B$\text{XeF}_6$
- C$\text{XeO}_3$
- Correct answerD$\text{XeOF}_4$
Solution
The shape of the molecules can be determined using VSEPR theory by calculating the number of bond pairs (bp) and lone pairs (lp) on the central Xenon (Xe) atom, which has 8 valence electrons.
For $\text{XeF}_4$: Xe forms 4 single bonds with F atoms, using 4 electrons. Remaining electrons = 4 (2 lone pairs). With 4 bp and 2 lp, the hybridization is $sp^3d^2$ and the shape is square planar.
For $\text{XeF}_6$: Xe forms 6 single bonds with F atoms, using 6 electrons. Remaining electrons = 2 (1 lone pair). With 6 bp and 1 lp, the hybridization is $sp^3d^3$ and the shape is distorted octahedral.
For $\text{XeO}_3$: Xe forms 3 double bonds with O atoms, using 6 electrons. Remaining electrons = 2 (1 lone pair). With 3 bp ($\sigma$-bonds) and 1 lp, the hybridization is $sp^3$ and the shape is trigonal pyramidal.
For $\text{XeOF}_4$: Xe forms 1 double bond with O and 4 single bonds with F, using $2 + 4 = 6$ electrons. Remaining electrons = 2 (1 lone pair). With 5 bp ($\sigma$-bonds) and 1 lp, the hybridization is $sp^3d^2$ and the shape is square pyramidal.
Answer: $\text{XeOF}_4$
For $\text{XeF}_4$: Xe forms 4 single bonds with F atoms, using 4 electrons. Remaining electrons = 4 (2 lone pairs). With 4 bp and 2 lp, the hybridization is $sp^3d^2$ and the shape is square planar.
For $\text{XeF}_6$: Xe forms 6 single bonds with F atoms, using 6 electrons. Remaining electrons = 2 (1 lone pair). With 6 bp and 1 lp, the hybridization is $sp^3d^3$ and the shape is distorted octahedral.
For $\text{XeO}_3$: Xe forms 3 double bonds with O atoms, using 6 electrons. Remaining electrons = 2 (1 lone pair). With 3 bp ($\sigma$-bonds) and 1 lp, the hybridization is $sp^3$ and the shape is trigonal pyramidal.
For $\text{XeOF}_4$: Xe forms 1 double bond with O and 4 single bonds with F, using $2 + 4 = 6$ electrons. Remaining electrons = 2 (1 lone pair). With 5 bp ($\sigma$-bonds) and 1 lp, the hybridization is $sp^3d^2$ and the shape is square pyramidal.
Answer: $\text{XeOF}_4$
Hence, the correct answer is option: 'D'.