The correct stability order of the following resonance structures is $\mathrm{H}_{2} \mathrm{C}=\stackrel{+}{\mathrm{N}}=\stackrel{-}{\mathrm{N}}$ (I) $\mathrm{H}_{2} \stackrel{+}{\mathrm{C}}-\mathrm{N}=\stackrel{-}{\mathrm{N}}$ (II) $\mathrm{H}_{2} \stackrel{-}{\mathrm{C}}-\stackrel{+}{\mathrm{N}} \equiv \mathrm{N}$ (III) $\mathrm{H}_{2} \stackrel{-}{\mathrm{C}}-\mathrm{N}=\stackrel{+}{\mathrm{N}}$ (IV)

(B) (I) $>$ (III) $>$ (II) $>$ (IV)

(A) (I) $>$ (II) $>$ (IV) $>$ (III)

(C) (II) $>$ (I) $>$ (III) $>$ (IV)

(D) (III) $>$ (I) $>$ (IV) $>$ (II)

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · General Organic Chemistry

The correct stability order of the following resonance structures is

H_{2} C = N + = N -

(I)

H_{2} C + - N = N -

(II)

H_{2} C - - N + \equiv N

(III)

H_{2} C - - N = N +

(IV)

(A)(I) $>$ (II) $>$ (IV) $>$ (III)
(B)(I) $>$ (III) $>$ (II) $>$ (IV)✓ Correct
(C)(II) $>$ (I) $>$ (III) $>$ (IV)
(D)(III) $>$ (I) $>$ (IV) $>$ (II)

Answer: (B)

Solution

The stability of resonance structures depends on: (1) octet completion, (2) charge separation, and (3) electronegativity. Structure (I) has all octets complete with minimal charge separation. Structure (III) has a triple bond and complete octets. Structure (II) has charge separation with positive charge on less electronegative carbon. Structure (IV) has unfavorable charge distribution with negative charge on carbon and positive on nitrogen. Therefore, the order is (I) > (III) > (II) > (IV).

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Single correct option

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JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · General Organic Chemistry

Solution

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