# Resonance and mesomeric effect [closed]

Can you please explain to me why are A and E not relevant resonance structures of the molecule F? And is B aromatic?

## closed as off-topic by Mithoron, Mathew Mahindaratne, user55119, Jon Custer, andselisk♦Aug 12 at 8:38

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For A: The pyridine ring would be very strained. The lone pair on the nitrogen is also in the plane of the ring and would hardly be able to form a double bond this way.

For E: Putting a positive charge on an sp hybridized atom is never good. The more 's' character you have, the closer the charge is to the nucleus (since s orbitals are closer to the nucleus). This is higher in energy. On top of that, putting a negative charge on a 'sp3' carbon is not particularly favored either.

The relevant resonance structures of the given molecule are as follows.

• Structure A is shown below. NItrogen with 2 sigma bonds is linear as shown below. However , nitrogen in the ring is strained with an angle of $$\ce{120^o}$$ (figure 1).

In figure 2 ,an unstrained Nitrogen with $$\ce{120^o}$$ bond angle is shown for your reference. Therefore , this structure $$\ce{A}$$ is irrelevant resonance structure .

• Structure $$\ce{E}$$ is not a resonance structure of $$\ce{F}$$ or $$\ce{1}$$ in my structure. The carbanion in $$\ce{2}$$ is in conjugation with imine bond ($$\ce{ C=N }$$). This gives $$\ce{3}$$.

• Structure $$\ce{B}$$ in not aromatic .