Question

Arrange the following in the order of their stability:
A. $CH={{C}^{-}}$
B. $C{{H}_{2}}=C{{H}^{-}}$
C. $Ph{{C}^{-}}$
D. $CH_{3}^{-}$

Answer 1

Hint: In chemistry, resonance has been defined as a way of describing the bonding patterns in certain molecules or ions that do not have a single stable structure but a number of resonating or canonical structures represent the molecules. It involves the dispersion of charges.

Complete step by step answer:
The stability of the resonating structures which in a combined way give rise to the resonance hybrid which has the property of all the resonating structures and has the lowest energy among all the resonating structures, depends on a number of factors. Firstly, the presence of opposite charges on adjacent atoms makes the structure very unstable.
Higher the number of covalent bonds in a molecule, higher will be the stability. The positive charge on a carbon atom is satisfied more by a secondary carbon atom and a primary one due to the increased inductive effect of the alkyl groups in the molecules. Taking all these steps into consideration we see that the stability of the carbocations has the following order:
$\left( Ph{{C}^{-}} \right)>\left( C{{H}_{2}}=C{{H}^{-}} \right)>\left( CH={{C}^{-}} \right)>\left( CH_{3}^{-} \right).$
So, the carbocation with the highest stability is the one that is in the second option while the least amount of stability is there in option number four.
Therefore, correct order of their stability is $c>b>a>d.$

Note: A carbocation is a positively charged carbon species that has carbon in the hybridized state and adopts a triangular planar structure. Due to the presence of the positive charge the order of stability of the carbocations is: tertiary carbocation $>$ secondary carbocation $>$ primary carbocations.