Correct order of stability is

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(C) The stability of carbanions depends on the hybridization of the carbon atom bearing the negative charge. The order of electronegativity is $sp > s

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$CH_3COO^- > C_6H_5O^-$

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(C) The stability of carbanions depends on the hybridization of the carbon atom bearing the negative charge. The order of electronegativity is $sp > sp^2 > sp^3$. As the $s$-character increases,the electronegativity of the carbon atom increases,making it better at holding the negative charge. Thus,the stability order for carbanions is $CH \equiv \mathop {\ddot C}\limits^\Theta (sp) > CH_2 = \mathop {\ddot C}\limits^\Theta H (sp^2) > CH_3 - \mathop {\ddot C}\limits^\Theta H_2 (sp^3)$.Option $A$ shows the reverse of this order.Option $B$ refers to radicals,where stability increases with hyperconjugation and resonance,not just hybridization.Option $C$ compares the stability of acetate ion $(CH_3COO^-)$ and phenoxide ion $(C_6H_5O^-)$. In acetate,the negative charge is delocalized over two equivalent oxygen atoms,whereas in phenoxide,it is delocalized over one oxygen and the aromatic ring. The acetate ion is more stable due to equivalent resonance structures.Option $D$ compares a secondary carbocation with an oxygen-stabilized primary carbocation. The oxygen-stabilized carbocation is more stable due to the $+M$ effect of the oxygen atom.

Similar Questions

Carbocation as an intermediate is likely to be formed in the reaction:

Consider the following carbocations. What is the relative stability order of these carbocations?
$(1)$ $CH_3O-C_6H_4-CH_2^+$
$(2)$ $C_6H_5-CH_2^+$
$(3)$ $CH_3-C_6H_4-CH_2^+$

The species $A$,$B$,$C$,$D$ formed in the following bond cleavages respectively are:

Which of the following represents the incorrect order of stability?

The stability of the carbocations follows the order:

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