JEE Main · 2020 · Shift-IIeasyBOND-089

The compound that has the largest H–M–H bond angle (M = N, O, S, C) is:

Chemical Bonding · Class 11 · JEE Main Previous Year Question

Question

The compound that has the largest H–M–H bond angle (M = N, O, S, C) is:

Options
  1. a

    H2O\mathrm{H_2O}

  2. b

    NH3\mathrm{NH_3}

  3. c

    H2S\mathrm{H_2S}

  4. d

    CH4\mathrm{CH_4}

Correct Answerd

CH4\mathrm{CH_4}

Detailed Solution

🧠 Lone pairs compress bond angles. More lone pairs → smaller angle. Compare the LP count and hybridisation of each central atom.

| Molecule | Central | Hybridisation | LP | H–M–H angle | |---|---|---|---|---| | \ceCH4\ce{CH4} | C | sp³ | 0 | 109.5° | | \ceNH3\ce{NH3} | N | sp³ | 1 | ~107° | | \ceH2O\ce{H2O} | O | sp³ | 2 | ~104.5° | | \ceH2S\ce{H2S} | S | ~p (little hybridisation) | 2 | ~92° |

With no lone pairs, \ceCH4\ce{CH4} achieves the ideal tetrahedral angle. \ceH2S\ce{H2S} is smallest because S is large and uses nearly unhybridised 3p orbitals (near 90°).

Answer: (d) \ceCH4\boxed{\text{Answer: (d) } \ce{CH4}}

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