Arrange the carbanions (CH_(3))_(3)C^(-), Cl_(3)C^(-), (CH_(3))_(2)CH^(-), C_(6)H_(5)CH_(2)^(-) in order of their decreasing stability

Arrange the carbanions (CH_(3))_(3)C^(-), Cl_(3)C^(-), (CH_(3))_(2)CH^

1.	Arrange the carbanions (CH_(3))_(3)C^(-), Cl_(3)C^(-), (CH_(3))_(2)CH^(-), C_(6)H_(5)CH_(2)^(-) in order of their decreasing stability
Answer» `(CH_(3))_(2)CH^(-) gt Cl_(3)C^(-) gt C_(6)H_(5)CH_(2)^(-) gt (CH_(3))_(3)C^(-)` `Cl_(3)C^(-) gt C_(6)H_(5)CH_(2)^(-) gt (CH_(3))_(2)CH^(-) gt (CH_(3))_(3)C^(-)` `(CH_(3))_3)C^(-) gt (CH_(3))_(2)CH^(-) gt C_(6)H_(5)CH_(2)^(-) gt Cl_(2)C^(-)` `C_(6)H_(5)CH_(2)^(-) gt Cl_(3)C^(-) gt (CH_(3))_(3)C^(-) gt (CH_(3))_(2)CH^(-)` Solution :DUE to -I-EFFECT of the three Cl atoms, `Cl_(3)C^(-)` is the most stable. This is followed by `C_(6)H_(5)CH_(2)^(-)` which is stabilized by resonance. Out of `(CH_(3))_(3)C^(-)` and `(CH_(3))_(2)CH^(-), (CH_(3))_(2)CH^(-)` is more stable due to +I -effect of TWO rather than three `CH_(3)` groups. Thus option (b) is CORRECT.