Number of molecules/ions from the following in which the central atom is involved in sp3 hybridization is {NO_3^-,\…

🧠 $\text{sp}^3$ hybridisation needs exactly 4 electron groups (bonds + lone pairs) on the central atom. Count for each species.

| Species | Bonds | Lone pairs on central | Total | Hybridisation | |---|---|---|---|---| | $\ce{NO3-}$ | 3 (resonance) | 0 | 3 | sp² | | $\ce{BCl3}$ | 3 | 0 | 3 | sp² | | $\ce{ClO2-}$ | 2 | 2 | 4 | sp³ ✓ | | $\ce{ClO3-}$ | 3 | 1 | 4 | sp³ ✓ |

⚡ For $\ce{ClO2-}$: Cl has 7 valence e. 2 bonds use 2e from Cl; remaining 5e → 2 LP + 1 radical? No: with −1 charge, Cl has 8e available → 2 bonds (4e) + 2 LP (4e) = sp³.

$\boxed{\text{Answer: (c) 2}}$