Arrange the following carbocations in the increasing order of their...

The stability of carbocations is determined by the electronic effects of the attached groups.

In (IV), the $\mathrm{NO}_2$ group exerts a strong $-\mathrm{I}$ and $-\mathrm{M}$ effect, which highly destabilizes the positive charge. Thus, it is the least stable.

In (III), the ethyl carbocation is stabilized by the $+\mathrm{I}$ effect and hyperconjugation from the three $\alpha$-hydrogens of the methyl group.

In (I), the allyl carbocation is stabilized by resonance with the adjacent $\pi$ bond. Resonance stabilization is generally more effective than hyperconjugation, making (I) more stable than (III).

In (II), the oxygen atom has lone pairs that stabilize the adjacent positive charge through a strong $+\mathrm{M}$ effect. The resulting resonance structure ($\mathrm{CH}_3-\stackrel{+}{\mathrm{O}}=\mathrm{CH}_2$) has complete octets for all atoms, which provides exceptional stability. Therefore, (II) is more stable than (I).

The increasing order of stability is (IV) $<$ (III) $<$ (I) $<$ (II).

Answer: (IV) $<$ (III) $<$ (I) $<$ (II)