JEE Main 2025 · 2 April · morning · Single correct · +4 / −1
Among SO₂, NF₃, NH₃, XeF₂, ClF₃ and SF₄, the hybridization of the molecule with nonzero dipole moment and highest number of lone-pairs of electrons on the central atom is
- (a) sp³
- (b) dsp²
- (c) sp³ d²
- (d) sp³ d
Answer and solution
Answer (d) sp³ d
Solution
Chemistry ma’am · 3 steps
First drop the molecules with zero dipole, then pick the one with the most lone pairs on the centre.
- Count electron domains. For each molecule, count bond pairs and lone pairs on the central atom. Their total sets the hybridisation: 3 is sp², 4 is sp³, 5 is sp³d.
Molecule Bond pairs + lone pairs Hybridisation and shape Dipole moment SO₂ 2 + 1 sp², bent nonzero NF₃ 3 + 1 sp³, pyramidal nonzero NH₃ 3 + 1 sp³, pyramidal nonzero XeF₂ 2 + 3 sp³d, linear zero ClF₃ 3 + 2 sp³d, T-shaped nonzero SF₄ 4 + 1 sp³d, see-saw nonzero - Remove XeF2. XeF₂ has the most lone pairs (3), but its two Xe–F bonds point in opposite directions and cancel, so its dipole moment is zero. It is ruled out.
- So. Among the polar molecules, ClF₃ has the most lone pairs on the central atom (2 on Cl). Its hybridisation is sp³d, option (d).
The idea to keep. Apply the conditions in order: check symmetry for the dipole first, then compare lone pairs. The molecule with the most lone pairs is often the symmetric one that gets excluded.
Answer from the NTA final answer key. Solution by PrepWiser.