JEE Main · 2021 · 27 Jul 2021 · Shift 1 · General Organic Chemistry

In the benzyl cation the vacant $p$ orbital overlaps the ring $\pi$ system and the positive charge is spread over the ring carbons, so it is the most stable of the four.
In the ethyl cation the charged carbon is $s{p}^2$ hybridised and three $\alpha$ hydrogens of the methyl group release electron density into the vacant $p$ orbital by hyperconjugation.
In the vinyl cation $\mathrm{C}{\mathrm{H}}_2=\stackrel{+}{\mathrm{C}}\mathrm{H}$ and in the ethynyl cation $\mathrm{HC}\equiv \stackrel{+}{\mathrm{C}}$ the charged carbon is $sp$ hybridised. An $sp$ orbital has the highest $s$ character, so it holds its electrons closest to the nucleus and such a carbon resists giving up electron density to an adjacent electron-deficient centre; an $sp$ carbon therefore bears a positive charge far worse than an $s{p}^2$ carbon, and both lie below the ethyl cation. Of the two, the ethynyl cation is the poorer because its charged carbon is attached only to a triply bonded carbon that withdraws electron density inductively, while the vinyl cation still has a $\mathrm{C}{\mathrm{H}}_2$ group next to it.
The order of stability is benzyl $>$ ethyl $>$ vinyl $>$ ethynyl.