In which of the following pairs is the 2^{tex | vedclass

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(B) To determine the stability of carbanions,we look at the electron-withdrawing or electron-donating effects of the attached groups.$(A)$ $O_2N-\stac

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$\stackrel{\ominus}{C}F_3$ and $\stackrel{\ominus}{C}Cl_3$

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(B) To determine the stability of carbanions,we look at the electron-withdrawing or electron-donating effects of the attached groups.$(A)$ $O_2N-\stackrel{\ominus}{C}H_2$ vs $F-\stackrel{\ominus}{C}H_2$: The $-NO_2$ group is a stronger electron-withdrawing group ($-I$ and $-M$ effect) than the $-F$ atom ($-I$ effect). Thus,the $1^{	ext{st}}$ anion is more stable.$(B)$ $\stackrel{\ominus}{C}F_3$ vs $\stackrel{\ominus}{C}Cl_3$: $\stackrel{\ominus}{C}F_3$ is more stable due to the strong $-I$ effect of fluorine atoms compared to chlorine atoms.$(C)$ $F_3C-\stackrel{\ominus}{C}H_2$ vs $Cl_3C-\stackrel{\ominus}{C}H_2$: The $F_3C-$ group exerts a stronger $-I$ effect than $Cl_3C-$,making the $1^{	ext{st}}$ anion more stable.$(D)$ $CH_3-CO-\stackrel{\ominus}{C}H_2$ vs $H_2N-\stackrel{\ominus}{C}H_2$: In $CH_3-CO-\stackrel{\ominus}{C}H_2$,the negative charge is stabilized by resonance with the carbonyl group $(C=O)$. In $H_2N-\stackrel{\ominus}{C}H_2$,the $-NH_2$ group acts as an electron-donating group ($+M$ effect),which destabilizes the negative charge. Therefore,the $2^{	ext{nd}}$ anion $(H_2N-\stackrel{\ominus}{C}H_2)$ is less stable than the $1^{	ext{st}}$ anion. Wait,re-evaluating the question: The question asks where the $2^{	ext{nd}}$ is more stable. None of the options provided show the $2^{	ext{nd}}$ as more stable. However,if we compare the stability of $\stackrel{\ominus}{C}Cl_3$ vs $\stackrel{\ominus}{C}F_3$,$\stackrel{\ominus}{C}Cl_3$ is actually more stable due to back-bonding ($d$-orbital resonance) which is not possible in $\stackrel{\ominus}{C}F_3$. Thus,in option $(B)$,the $2^{	ext{nd}}$ anion is more stable.

In which of the following pairs is the $2^{\text{nd}}$ anion more stable than the $1^{\text{st}}$?

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