Question

Consider the following ions: 

\[ \text{(I)}\ \mathrm{CH_3-CH_2^-} \qquad \text{(II)}\ \mathrm{CH_2=CH^-} \qquad \text{(III)}\ \mathrm{HC \equiv C^-} \]

The stability of the ions is in the order:

• A. \(\text{III}>\text{II}>\text{I}\)
• B. \(\text{II}>\text{III}>\text{I}\)
• C. \(\text{I}>\text{II}>\text{III}\)
• D. \(\text{I}>\text{III}>\text{II}\)

Hint:

For carbanion stability:

More s-character means carbon holds negative charge closer to nucleus.
sp-hybridized carbanions are the most stable.
Alkyl groups destabilize carbanions due to +I effect.

Answer 1

The Correct Option is A

Concept: 
Stability of carbanions mainly depends on:
Hybridization of the carbon atom
Greater the s-character, greater is the electronegativity of carbon.
Higher electronegativity stabilizes the negative charge.

Order of electronegativity based on hybridization: 
\[ \text{sp} > \text{sp}^2 > \text{sp}^3 \] 
Step 1: Identify the hybridization of each ion.

(I) \({CH3-CH2^-}\) : sp³-hybridized carbon
(II) \({CH2=CH^-}\) : sp²-hybridized carbon
(III) \({HC#C^-}\) : sp-hybridized carbon

Step 2: Compare stability.

Since: 
\[ \text{sp} > \text{sp}^2 > \text{sp}^3 \] 
Therefore: 
\[ \text{Stability: } {HC#C^-} > {CH2=CH^-} > {CH3-CH2^-} \] 
Conclusion:
\[ \boxed{\text{III} > \text{II} > \text{I}} \] 
Hence, the correct answer is (A).