CORRECT order of stability for the following is
$$CH_{2}=CH^{-},CH_{3}-CH_{2}^{-},CH\equiv C^{-}$$

We need to arrange three carbanions in decreasing order of stability.

Key Concept: Stability of carbanions depends on the hybridisation of the carbon bearing the negative charge. The greater the s-character of the orbital holding the lone pair, the closer the electrons are held to the nucleus, and the more stable the carbanion. The s-character of different hybridisations is:

- $$sp$$ hybrid: 50% s-character

- $$sp^2$$ hybrid: 33.3% s-character

- $$sp^3$$ hybrid: 25% s-character

The carbanion $$CH \equiv C^-$$ (acetylide ion) features an $$sp$$-hybridised carbon bearing the negative charge. With 50% s-character, its lone pair is held closest to the nucleus, making it the most stable of the three.

In the case of $$CH_2 = CH^-$$ (vinyl carbanion), the negatively charged carbon is $$sp^2$$ hybridised. With 33.3% s-character, this carbanion has intermediate stability.

The carbanion $$CH_3 - CH_2^-$$ (ethyl carbanion) is $$sp^3$$ hybridised, possessing only 25% s-character, so its lone pair is held furthest from the nucleus. Additionally, the methyl group exerts a +I (electron-donating inductive) effect that increases electron density on the negatively charged carbon, further destabilising it. This is the least stable carbanion.

Therefore, the decreasing order of stability is $$CH \equiv C^- > CH_2 = CH^- > CH_3 - CH_2^-$$.

The correct answer is Option 2: $$CH \equiv C^- > CH_2 = CH^- > CH_3 - CH_2^-$$.