Organic Chemistry Basics 2 Question 22
22. The correct stability order of the following resonance structure is
(I) $\mathrm{H}{2} \mathrm{C}=\stackrel{+}{\mathrm{N}}=\overline{\mathrm{N}}$ (III) $\mathrm{H}{2} \overline{\mathrm{C}}-\stackrel{\mathrm{N}}{ } \equiv \mathrm{N}$
(a) (I) $>$ (II) $>$ (IV) $>$ (III)
(b) (I) $>$ (III) $>$ (II) $>$ (IV)
(c) (II) $>$ (I) $>$ (III) $>$ (IV)
(d) (III) $>$ (I) $>$ (IV) $>$ (II)
(II) $\mathrm{H}_{2} \stackrel{+}{\mathrm{C}}-\mathrm{N}=\stackrel{-}{\mathrm{N}}$
(IV) $\mathrm{H}_{2} \stackrel{\rightharpoonup}{\mathrm{C}}-\mathrm{N}=\stackrel{+}{\mathrm{N}}$
(2009)
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Solution:
- I is most stable because it has more covalent bonds and negative charge on electronegative nitrogen. III is more stable than II and IV due to greater number of covalent bonds. Between II and IV, II is more stable since, it has negative charge on electronegative atom and positive charge on electropositive atom. Hence, overall stability order is
$$ \text { I }>\text { III }>\text { II }>\text { IV } $$
