JEE Main · 2024 · Shift-ImediumBOND-006

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

Chemical Bonding · Class 11 · JEE Main Previous Year Question

Question

Number of molecules/ions from the following in which the central atom is involved in sp3\mathrm{sp^3} hybridization is

NO3, BCl3, ClO2, ClO3\mathrm{NO_3^-,\ BCl_3,\ ClO_2^-,\ ClO_3^-}

Options
  1. a

    4

  2. b

    3

  3. c

    2

  4. d

    1

Correct Answerc

2

Detailed Solution

🧠 sp3\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 | |---|---|---|---|---| | \ceNO3\ce{NO3-} | 3 (resonance) | 0 | 3 | sp² | | \ceBCl3\ce{BCl3} | 3 | 0 | 3 | sp² | | \ceClO2\ce{ClO2-} | 2 | 2 | 4 | sp³ ✓ | | \ceClO3\ce{ClO3-} | 3 | 1 | 4 | sp³ ✓ |

⚡ For \ceClO2\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.

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

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