Chemistry Question 64 – NEET-UG 2025

For each compound, calculate the steric number (number of atoms bonded to central atom + number of lone pairs on central atom) to determine hybridization and then deduce the molecular geometry.

A. ${\text{XeO}}_3$: Xe has 8 valence electrons. 3 oxygen atoms form double bonds (6 electrons used). Remaining 2 electrons form 1 lone pair. Steric number = 3 (bonding pairs) + 1 (lone pair) = 4. Hybridization is $s{p}^3$. Geometry is Pyramidal.

B. ${\text{XeF}}_2$: Xe has 8 valence electrons. 2 fluorine atoms form single bonds (2 electrons used). Remaining 6 electrons form 3 lone pairs. Steric number = 2 (bonding pairs) + 3 (lone pairs) = 5. Hybridization is $s{p}^{3}d$. Geometry is Linear (lone pairs occupy equatorial positions).

C. ${\text{XeOF}}_4$: Xe has 8 valence electrons. 1 oxygen forms a double bond (2 electrons) and 4 fluorine atoms form single bonds (4 electrons). Total 6 electrons used. Remaining 2 electrons form 1 lone pair. Steric number = 5 (bonding pairs) + 1 (lone pair) = 6. Hybridization is $s{p}^3{d}^2$. Geometry is Square Pyramidal.

D. ${\text{XeF}}_6$: Xe has 8 valence electrons. 6 fluorine atoms form single bonds (6 electrons used). Remaining 2 electrons form 1 lone pair. Steric number = 6 (bonding pairs) + 1 (lone pair) = 7. Hybridization is $s{p}^3{d}^3$. Geometry is Distorted Octahedral.