JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
Match List-I with List-II :
Choose the correct answer from the options given below :
| List-I (Species) | List-II (Hybrid Orbitals) | ||
|---|---|---|---|
| (a) | SF₄ | (i) | sp³d² |
| (b) | IF₅ | (ii) | d²sp³ |
| (c) | NO₂ + | (iii) | sp³d |
| (d) | NH₄ + | (iv) | sp³ |
| (v) | sp |
- (a) (a)-(i), (b)-(ii), (c)-(v) and (d)-(iii)
- (b) (a)-(ii), (b)-(i), (c)-(iv) and (d)-(v)
- (c) (a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)
- (d) (a)-(iv), (b)-(iii), (c)-(ii) and (d)-(v)
Answer and solution
Answer (c) (a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)
Solution
Chemistry ma’am · 3 steps
For each centre, add sigma bonds and lone pairs. That count alone tells you the hybridisation.
- Count electron pairs. Steric number = sigma bonds + lone pairs on the central atom. Then 2 gives sp, 4 gives sp³, 5 gives sp³d and 6 gives sp³d². Pi bonds do not count.
Species Sigma bonds Lone pairs on centre Steric number Hybridisation SF₄ 4 1 (S has 6, uses 4) 5 sp³d, (iii) IF₅ 5 1 (I has 7, uses 5) 6 sp³d², (i) NO₂⁺ 2 0 (O=N=O, linear) 2 sp, (v) NH₄⁺ 4 0 (lone pair of N used for H⁺) 4 sp³, (iv) - Outer d orbitals. S and I use their outer nd orbitals, so their six-pair case is sp³d². The label d²sp³ means inner (n-1)d orbitals, used in transition metal complexes, so (ii) matches none of these.
- So. The match is (a)-(iii), (b)-(i), (c)-(v), (d)-(iv), which is option (c).
The idea to keep. Hybridisation of a central atom follows its steric number: sigma bonds plus lone pairs. Main-group atoms use outer d orbitals, so write sp³d², never d²sp³.
Answer from the NTA final answer key. Solution by PrepWiser.