JEE Challenger
More from Organic Chemistry - Basic Principles
ChemMedAdv 2009 • Paper 2

Stability Order of Diazomethane Resonance Structures

The correct stability order of the following resonance structures is

Diagram 0

Options

A

(I)>(II)>(IV)>(III)\text{(I)} > \text{(II)} > \text{(IV)} > \text{(III)}

B

(I)>(III)>(II)>(IV)\text{(I)} > \text{(III)} > \text{(II)} > \text{(IV)}

Correct
C

(II)>(I)>(III)>(IV)\text{(II)} > \text{(I)} > \text{(III)} > \text{(IV)}

D

(III)>(I)>(IV)>(II)\text{(III)} > \text{(I)} > \text{(IV)} > \text{(II)}

Topics & Concepts

Organic Chemistry - Basic PrinciplesResonance

Step-by-Step Solution

To determine the relative stability of the given resonance structures, we apply the standard rules governing resonance stability:

  1. Octet Rule (Number of Covalent Bonds):
    Structures in which all atoms have a complete valence shell (octet) are significantly more stable than structures with incomplete octets (electron-deficient atoms).

    • In (I) (H2C=N+=N−\mathrm{H_2C=\overset{+}{N}=\overset{-}{N}}):
      • C\mathrm{C} forms 44 bonds (complete octet).
      • Central N\mathrm{N} forms 44 bonds (complete octet).
      • Terminal N\mathrm{N} forms 22 bonds and has 22 lone pairs (complete octet).
    • In (III) (H2C−−N+≡N\mathrm{H_2\overset{-}{C}-\overset{+}{N}\equiv N}):
      • C\mathrm{C} forms 33 bonds and has 11 lone pair (complete octet).
      • Central N\mathrm{N} forms 44 bonds (complete octet).
      • Terminal N\mathrm{N} forms 33 bonds and has 11 lone pair (complete octet).

    Thus, both (I) and (III) have all octets complete and are more stable than (II) and (IV).

  2. Electronegativity and Location of Formal Charges (comparing I and III):

    • In (I), the negative formal charge resides on the more electronegative nitrogen atom (N−\mathrm{\overset{-}{N}}).
    • In (III), the negative formal charge resides on the less electronegative carbon atom (C−\mathrm{\overset{-}{C}}).

    Since a negative charge is more stable on a more electronegative atom:
    (I)>(III)\text{(I)} > \text{(III)}

  3. Comparison of Structures with Incomplete Octets (comparing II and IV):

    • In (II) (H2C+−N=N−\mathrm{H_2\overset{+}{C}-N=\overset{-}{N}}), carbon has an incomplete octet (66 valence electrons), while nitrogen carries the negative formal charge.
    • In (IV) (H2C−−N=N+\mathrm{H_2\overset{-}{C}-N=\overset{+}{N}}), the terminal nitrogen has an incomplete octet (66 valence electrons) and carries a positive formal charge, while the negative charge is located on the less electronegative carbon atom. Nitrogen is significantly more electronegative than carbon, so having an electron sextet and positive charge on nitrogen makes the structure extremely unstable.

    Therefore:
    (II)>(IV)\text{(II)} > \text{(IV)}

Combining these evaluations, the overall stability order is:
(I)>(III)>(II)>(IV)\text{(I)} > \text{(III)} > \text{(II)} > \text{(IV)}

Hence, the correct option is B.

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