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JEE Main 2024 · Chemistry · Chemical Bonding and Molecular Structure · medium

The correct increasing order of the bond angle in NO2^-, O3, NO2 and NO2^+ is

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The correct increasing order of the bond angle in NO2−\mathrm{NO_2^-}, O3\mathrm{O_3}, NO2\mathrm{NO_2} and NO2+\mathrm{NO_2^+} is
  1. A. NO2−<O3<NO2<NO2+\mathrm{NO_2^-}<\mathrm{O_3}<\mathrm{NO_2}<\mathrm{NO_2^+}
  2. B. O3<NO2−<NO2<NO2+\mathrm{O_3}<\mathrm{NO_2^-}<\mathrm{NO_2}<\mathrm{NO_2^+}
  3. C. NO2+<NO2<O3<NO2−\mathrm{NO_2^+}<\mathrm{NO_2}<\mathrm{O_3}<\mathrm{NO_2^-}
  4. D. NO2−<NO2<O3<NO2+\mathrm{NO_2^-}<\mathrm{NO_2}<\mathrm{O_3}<\mathrm{NO_2^+}
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Answer: (A) NO2−<O3<NO2<NO2+\mathrm{NO_2^-}<\mathrm{O_3}<\mathrm{NO_2}<\mathrm{NO_2^+}

NO2+\mathrm{NO_2^+} has no lone pair on N and is linear (180∘180^\circ, sp). NO2\mathrm{NO_2} has a single unpaired electron on N, which repels the bonds less than a lone pair, so the angle is ≈134∘\approx134^\circ. NO2−\mathrm{NO_2^-} has a lone pair on N (sp2^2) with angle ≈115∘\approx115^\circ, and O3\mathrm{O_3} has one lone pair on the central O with angle ≈117∘\approx117^\circ. Order: NO2−(115∘)<O3(117∘)<NO2(134∘)<NO2+(180∘)\mathrm{NO_2^-}(115^\circ)<\mathrm{O_3}(117^\circ)<\mathrm{NO_2}(134^\circ)<\mathrm{NO_2^+}(180^\circ).
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