The decreasing order of hydride affinity for following carbocations is:
(a)

(b)

(c)

(d)

Choose the correct answer from the options given below:

Hydride affinity is defined for the reaction
$$R^+ + H^- \;\rightarrow\; RH$$.
The larger the energy released (that is, the more negative the reaction enthalpy), the greater is the hydride affinity of the carbocation $$R^+$$.

A carbocation that is intrinsically less stable gains a larger amount of stabilisation on accepting a hydride ion, so it shows a higher hydride affinity.
Conversely, a more stable carbocation is already low in energy; it gains relatively little additional stabilisation on accepting $$H^-$$ and therefore exhibits a lower hydride affinity.
Hence,

$$\text{higher hydride affinity}\; \Longleftrightarrow\; \text{lower inherent stability of the carbocation}$$.

The methyl cation $$CH_3^+$$ possesses neither hyperconjugation nor resonance stabilisation. It is the least stabilised of the four; therefore it will have the highest hydride affinity.

The n-propyl (1°) carbocation $$CH_3CH_2CH_2^+$$ is stabilised by hyperconjugation with two α-hydrogen atoms. It is more stable than $$CH_3^+$$ but less stable than a 3° or resonance-stabilised cation. Its hydride affinity is therefore lower than that of (c) but higher than that of (b) and (d).

The tert-butyl (3°) carbocation $$(CH_3)_3C^+$$ enjoys extensive hyperconjugation (nine α-hydrogen atoms), making it markedly more stable than a 1° carbocation. Hence its hydride affinity is lower than those of (c) and (a).

The benzyl carbocation $$C_6H_5CH_2^+$$ is stabilised by resonance with the aromatic ring, distributing the positive charge over several atoms. This resonance stabilisation is stronger than the hyperconjugative stabilisation present in $$(CH_3)_3C^+$$. Therefore it is the most stable of the four and consequently has the lowest hydride affinity.

Combining the four cases, the order from highest to lowest hydride affinity is
$$\text{(c)} \; \gt \; \text{(a)} \; \gt \; \text{(b)} \; \gt \; \text{(d)}.$$

This corresponds to Option B: C, A, B, D.

Answer : Option B