The correct order of stability is:

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(B) Let us analyze each option based on the stability of reactive intermediates:$A$. Carbanion stability increases with the $s$-character of the carbo

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$CH_3-\dot{C}H_2 < CH_2=\dot{C}H < CH\equiv C^{\cdot}$

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(B) Let us analyze each option based on the stability of reactive intermediates:$A$. Carbanion stability increases with the $s$-character of the carbon atom. The $s$-character increases in the order $sp^3 $B$. Free radical stability increases with the $s$-character of the carbon atom. The stability order is $CH_3-\dot{C}H_2 $C$. The stability of anions is determined by the resonance effect. In $CH_3-COO^{\ominus}$,the negative charge is delocalized over two electronegative oxygen atoms. In $C_6H_5-O^{\ominus}$ (phenoxide ion),the negative charge is delocalized over the benzene ring. The acetate ion is more stable than the phenoxide ion due to the higher electronegativity of the oxygen atoms sharing the charge. The given order is correct.$D$. Carbocation stability is increased by the $+M$ effect of the oxygen atom. In $CH_3-O-\overset{\oplus}{C}H_2$,the oxygen atom donates electron density through resonance,making it more stable than the secondary carbocation $CH_3-\overset{\oplus}{C}H-CH_3$. The given order is correct.Note: Multiple options $(B, C, D)$ represent correct stability orders. However,in standard chemistry contexts,$B$ is a fundamental comparison of hybridization effects on radicals.

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