JEE Main · 2021 · Shift-IImediumBOND-022

The hybridisations of the atomic orbitals of nitrogen in NO2-, NO2+ and NH4+ respectively are:

Chemical Bonding · Class 11 · JEE Main Previous Year Question

Question

The hybridisations of the atomic orbitals of nitrogen in NO2\mathrm{NO_2^-}, NO2+\mathrm{NO_2^+} and NH4+\mathrm{NH_4^+} respectively are:

Options
  1. a

    sp3,sp2\mathrm{sp^3, sp^2} and sp

  2. b

    sp,sp2\mathrm{sp, sp^2} and sp3\mathrm{sp^3}

  3. c

    sp3,sp\mathrm{sp^3, sp} and sp2\mathrm{sp^2}

  4. d

    sp2,sp\mathrm{sp^2, sp} and sp3\mathrm{sp^3}

Correct Answerd

sp2,sp\mathrm{sp^2, sp} and sp3\mathrm{sp^3}

Detailed Solution

🧠 Hybridisation = steric number (σ-bonds + lone pairs) on the central N. Count electron groups for each species.

| Species | σ-bonds | LP | Steric no. | Hybridisation | |---|---|---|---|---| | \ceNO2\ce{NO2-} | 2 | 1 | 3 | sp² | | \ceNO2+\ce{NO2+} | 2 | 0 | 2 | sp | | \ceNH4+\ce{NH4+} | 4 | 0 | 4 | sp³ |

🗺️ \ceNO2+\ce{NO2+} is isoelectronic with \ceCO2\ce{CO2}: N sits between two O with double bonds and no lone pair → linear → sp. \ceNO2\ce{NO2-} has an extra lone pair that reduces the geometry to sp² (bent shape). \ceNH4+\ce{NH4+} loses its lone pair on gaining H⁺ → fully sp³.

Answer: (d) sp2, sp, sp3\boxed{\text{Answer: (d) sp}^2\text{, sp, sp}^3}

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The hybridisations of the atomic orbitals of nitrogen in NO2-, NO2+ and NH4+ respectively are: (JEE Main 2021) | Canvas Classes