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CORRECT order of stability for the following is \({\text{CH}}_{2}={\text{CH}}^{−},{\text{CH}}_{3}−{\text{CH}}_{2}^{−},\t…

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CORRECT order of stability for the following is

\({\text{CH}}_{2}={\text{CH}}^{−},{\text{CH}}_{3}−{\text{CH}}_{2}^{−},\text{CH}\equiv {\text{C}}^{−}\)

[JEE Main 2026, 28 Jan (Shift 2)]

a

\(\mathrm{CH}\equiv {\mathrm{C}}^{−}>{\mathrm{CH}}_{2}={\mathrm{CH}}^{−}>{\mathrm{CH}}_{3}−{\mathrm{CH}}_{2}^{−}\)

b

\({\text{CH}}_{3}−{\text{CH}}_{2}^{−}>{\text{CH}}_{2}={\text{CH}}^{−}>\text{CH}\equiv {\text{C}}^{−}\)

c

\({\text{CH}}_{2}={\text{CH}}^{−}>\text{CH}\equiv {\text{C}}^{−}>{\text{CH}}_{3}−{\text{CH}}_{2}^{−}\)

d

\(\text{CH}\equiv {\text{C}}^{−}>{\text{CH}}_{3}−{\text{CH}}_{2}^{−}>{\text{CH}}_{2}={\text{CH}}^{−}\)

✓ Correct answer: a)

\(\mathrm{CH}\equiv {\mathrm{C}}^{−}>{\mathrm{CH}}_{2}={\mathrm{CH}}^{−}>{\mathrm{CH}}_{3}−{\mathrm{CH}}_{2}^{−}\)

Explanation

These are carbanions, so stability depends mainly on hybridization.

Rule: Greater s-character stabilizes the negative charge more.

sp → 50% s-character
sp² → 33% s-character
sp³ → 25% s-character

So:

CH≡C⁻ (sp) → most stable
CH₂=CH⁻ (sp²) → moderately stable
CH₃–CH₂⁻ (sp³) → least stable

Final order:

CH≡C⁻ > CH₂=CH⁻ > CH₃–CH₂⁻

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