Question: Assuming the rings are planar, which ions are aromatic?

According to Huckel’s rule, $\left(4n+2\right)\mathbf{}\mathbf{\pi }$electron system in which cyclic, planar, complete delocalization is present is aromatic, and $\mathbf{4}\mathit{n}\mathbf{\pi }$electron system in which cyclic, planar delocalization is present is anti-aromatic.

For the aromatic system, the number of $\mathrm{\pi }$electrons is $\left(4n+2\right)$, and for the anti-aromatic system, the number of $\mathrm{\pi }$electrons is $4n$ ,where$n=0,1,2,3........$

To be aromatic, the ions must have $\left(4n+2\right)\mathrm{\pi }$electrons. Ions in (b) and (c) do not have the right number of $\mathrm{\pi }$electrons, so they are not aromatic. For (a) and (d), the number of $\mathrm{\pi }$electrons is:

Representation of (a)

$$\begin{gathered} \left(4n+2\right)\mathrm{\pi } \\ \left(4\times 0+2\right)=2 \end{gathered}$$

$2\mathrm{\pi }$electrons are present; hence, it is aromatic.

Representation of (b)

$$\begin{gathered} \left(4n+2\right)\mathrm{\pi } \\ \left(4\times 2+2\right)=10 \end{gathered}$$

$10\mathrm{\pi }$ electrons are present; hence, it is aromatic.

Therefore, if the rings are planar, the aromatic ions are as follows:

Aromatic ions