1. Identify the given carbanions

We need the decreasing order of stability of:

• $(CH_3)_3\overline{C}$ : tert-butyl carbanion
• $\overline{C}Cl_3$ : trichloromethyl carbanion
• $(CH_3)_2\overline{C}H$ : isopropyl carbanion
• $C_6H_5\overline{C}H_2$ : benzyl carbanion

2. General factors affecting carbanion stability

A carbanion is stabilized by:

• Electron-withdrawing groups through $-I$ effect
• Resonance delocalization of negative charge
• Less alkyl substitution, because alkyl groups show $+I$ effect and destabilize carbanions

So:

• More alkyl groups $\Rightarrow$ less stable carbanion
• Resonance stabilization greatly increases stability
• Strong $-I$ groups like $Cl$ also stabilize strongly

3. Compare each species

(i) $\overline{C}Cl_3$

The three chlorine atoms exert a very strong $-I$ effect, pulling electron density away from the negatively charged carbon. This strongly stabilizes the carbanion.

So this is highly stable.

(ii) $C_6H_5\overline{C}H_2$

This is a benzyl carbanion. The negative charge is delocalized into the benzene ring by resonance:

Hence it is also strongly stabilized.

Between benzyl and $CCl_3^-$, the strong stabilization by three chlorine atoms makes $\overline{C}Cl_3$ more stable in standard JEE ordering.

Thus:

(iii) $(CH_3)_2\overline{C}H$

This is a secondary carbanion. Two methyl groups show $+I$ effect, which push electron density toward the negatively charged carbon and destabilize it.

(iv) $(CH_3)_3\overline{C}$

This is a tertiary carbanion. Three methyl groups exert even stronger $+I$ effect, so it is the least stable.

Therefore:

4. Combine all comparisons

The decreasing stability order is:

5. Match with options

This matches Option B.

6. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So the answer agrees with the stored correct answer.