Arrange the following in order of increasing stability :
`CH_3 CH_2^(-) , CH -= C^(-) , CH_2 = CH^(-)`

Explanation

In acetylide ion `CH -= C^(-)`, the carbon atom carrying the -ve charge is sp-hybridized and has 50% s-character : in `CH_(2) = CH^(-)` ion, the carbon atom is `sp^(2)`-hybridized and has 33.3% s-character while in `CH_(3)CH_(2)^(-)` ion, the carbon atom bearing the -ve charge is `sp^(3)`-hybridized and has 25% s-character.
Since s-electrons, on the average, are closer to the nucleus than p-electrons, therefore, a carbon atom with greater s-character can accommodate or stabilize the negative charge better than a carbon atom with smaller s-character. In other words, the stability of the carbanion increases as the s-character of the carbon atom carrying the negative charge increases. Now since the s-character of the carbon decreases as we move from sp to `sp^(2)` to `sp^(3)`-carbon, therefore, the relative stability of the three carbanions follows the sequence: `CH -= C^(-) gt CH_(2) = CH^(-) gt CH_(3)CH_(2)^(-)`.