Question

Which of the following conditions should be met while writing the canonical structures of a resonance hybrid ?
A. The magnitude of resonance energy should be low and the hybrid will be more stable
B. The number of unpaired electrons should be same in all the canonical forms
C. Canonical forms where similar charges develop on vicinal atoms are highly significant forms
D. The positions of the nuclei of each canonical form and the hybrid must remain same
E. Only a pair of electrons should be shifted in writing the canonical forms
Choose the correct answer from the options given below :

• A. A, B, D only
• B. C, D, E only
• C. A, D, E only
• D. B, D, E only

Hint:

The absolute rules of resonance: 1. NEVER move atoms (nuclei). 2. NEVER exceed the octet rule for 2nd-row elements (C, N, O, F). 3. Conserve the total net charge and the number of unpaired electrons.

Answer 1

The Correct Option is D

Step 1: Concept:
The question asks to identify the fundamental rules governing the drawing and evaluation of resonance structures (canonical forms) in organic chemistry.

Step 2: Key Formula or Approach:
Review the rules of resonance:
- Resonance involves only the delocalization of electrons (specifically $\pi$ electrons and lone pairs). Nuclei (atoms) must never move.
- The overall spin state of the molecule cannot change; the number of unpaired electrons must be conserved across all structures.
- More stable resonance structures contribute more to the hybrid. Stability is lowered by separation of opposite charges, and severely lowered by placing like charges on adjacent (vicinal) atoms.
- A higher resonance energy (the energy difference between the hybrid and the most stable theoretical canonical form) implies greater stability, not lower.

Step 3: Step-by-step Explanation:

• A. Resonance energy should be low: False. Resonance energy is a measure of the extra stability gained by electron delocalization. A high magnitude of resonance energy means the hybrid is much more stable than any single canonical form (e.g., benzene has high resonance energy).

• B. Number of unpaired electrons should be same: True. Spin multiplicity must be conserved. You cannot draw a resonance form that converts a singlet state into a triplet state by breaking electron pairs arbitrarily.

• C. Similar charges on vicinal atoms are highly significant: False. Placing like charges (e.g., two positive charges) on adjacent atoms causes massive electrostatic repulsion. Such structures are extremely high in energy and represent highly insignificant contributors to the resonance hybrid.

• D. Positions of nuclei must remain same: True. This is the absolute golden rule of resonance. If atoms move, it is an isomerization reaction (like tautomerism), not resonance.

• E. Only a pair of electrons should be shifted: Generally true in the context of standard polar organic chemistry taught at this level. We use curved arrows to show the movement of pairs of electrons (lone pairs converting to $\pi$ bonds, or $\pi$ bonds breaking onto atoms as lone pairs). While radical resonance shifts single electrons, standard rules typically focus on electron pairs. Given the available options (since A and C are definitively false), E must be accepted as True in this context.
Therefore, the correct statements are B, D, and E.

Step 4: Final Answer:
The conditions that apply are B, D, and E, corresponding to option (D).