Question
Question: The radical given below is aromatic, because it has  pi electrons that should be in continuous delocalization or an uninterrupted cyclic pi electron cloud should be there. Here n= 0,1,2……
So if we put n=0 ,we get 2 pi electrons
If we put n=1, we get 6 pi electrons
If we put n=2, we get 10 pi electrons and so on
For example benzene, we all know are aware of the structure of benzene. It is planar, it is cyclic and it has 6 Pi electrons
In the above case, it is aromatic, because it is planar, cyclic and has 6 pi electrons which are in continuous delocalization. And hence satisfying the condition of Huckle’s rule of aromaticity.
But the free radical also has 1 unpaired electron and 1 p orbital. Here the free radical unpaired electron is also involved in the delocalization. So total we have 7 unpaired electrons 7 p- orbitals.
Hence the correct option will be B. total 7 unpaired electrons and 7 p- orbitals.
Note: Molecules which are cyclic, planar, and have 4n pi electrons which are in conjugation are called as anti-aromatic. For example: cyclobuta-1,3-diene. The compounds which do not follow huckel’s rule for aromaticity and anti-aromaticity are non-aromatic. For example: cyclooctatetraene.