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:
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:
Show Hint
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.
- \(\text{III}>\text{II}>\text{I}\)
- \(\text{II}>\text{III}>\text{I}\)
- \(\text{I}>\text{II}>\text{III}\)
- \(\text{I}>\text{III}>\text{II}\)
The Correct Option is A
Solution and Explanation
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).
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