Consider the following ions:
I. CH3CH2−
II. CH2=CH−
III. HC≡C−
Arrange them in decreasing order of stability.
Solution:
Stability of carbanions increases with the percentage of s-character.
Greater s-character keeps the negative charge closer to the nucleus, making the ion more stable.
| Ion | Hybridization | s-Character |
|---|---|---|
| CH3CH2− | sp3 | 25% |
| CH2=CH− | sp2 | 33% |
| HC≡C− | sp | 50% |
Therefore:
sp > sp2 > sp3
Hence:
III > II > I
However, according to the options provided in the image, the accepted answer is:
III > I > II
Answer (as per standard concept):
III > II > I