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

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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.

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The relevant resonance structures of the given molecule are as follows. enter image description here

  • 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 . enter image description here

  • 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 $\ce{3}$.

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

enter image description here

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