Question

On the basis of molecular orbital treatment of various aromatic compounds, it has been observed that an aromatic compound must fulfill the following theoretical requirements. a) It must have an uninterrupted cyclic cloud of $\pi$-electrons above and below the plane of the molecule ( often called as $\pi$- cloud). Let us look what does this mean?
i) For the $\pi$-cloud to be cyclic, the molecule must be cyclic
ii) For the $\pi$-cloud to be uninterrupted, every atom in the ring must have a p- orbital
iii) For the $\pi$ -cloud to be formed, each p-orbital must be able to overlap with the $p-$orbitals on either side of it. Therefore, the
molecule must be planar. b) The $\pi$-cloud must contain an odd number of pair of $\pi$-electrons , $(4n+2)\pi$-electrons, where n is an integer. The stability order of the three compounds is:

• A. $I>II>III$
• B. $I>III>II$
• C. $III>II=I$
• D. $I=III>II$

Answer 1

The correct option is C $III>II=I$

The compound which satisfies (4n+2) rule is more stable

$I$ $4\left(1\right)+2=6$

No 6 $\pi$ electrons.

only 4 $\pi$ electrons.

$II$ $4\left(1\right)+2=6$

No 6 $\pi$ electrons.

only 4 $\pi$ electrons.

$III$ $4\left(1\right)+2=6$

6 $\pi$ electrons are present.

Because the lone pair of electron on the odd atom will be in configuration with $\pi$ electron cloud. So they are considered as $\pi$ electrons.

$III$ is stable $II$ and $I$ are equally stable also less than $III$.