JEE Advanced2026ChemistryChemical Bonding and Molecular StructureActual
Consider the following species: SOCl ₂ , XeOF ₄ , ClF ₃ , ClF ₅ , XeF ₅^+ , SO ₃²⁻ , XeF ₃^+ , SF ₄ List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option. List-I List-II (P) See-saw (1) one (Q) T-Shaped (2) two (R) Trigonal Planar (3) t
Options
- AP 1; Q 2; R 5; S 3
- BP 5; Q 4; R 2; S 3
- CP 3; Q 2; R 1; S 4
- DP 1; Q 3; R 5; S 4
Correct answer
A. P 1; Q 2; R 5; S 3
Step-by-step solution
Let us determine the hybridization and shape of each given species by finding the number of bond pairs (bp) and lone pairs (lp) on the central atom: SOCl ₂ : Central atom S has 6 valence electrons. It forms 3 bonds (one with O, two with Cl) and has 1 lone pair. Steric number = 4 ( sp^3 ). Shape is Trigonal Pyramidal. XeOF ₄ : Central atom Xe has 8 valence electrons. It forms 5 bonds (one with O, four with F) and has 1 lone pair. Steric number = 6 ( sp^3d^2 ). Shape is Square Pyramidal. ClF ₃ : Central atom Cl has 7