MCQ
The radical, (image) is aromatic because it has


- A$7\, p-$orbitals and $7$ unpaired electrons
- B$6\, p-$orbitals and $7\,$ unpaired electrons
- ✓$6\, p-$orbitals and $6$ unpaired electrons
- D$7\, p-$orbitals and $6$ unpaired electrons.

formation of benzene ring (Aromatic nature).
$6 \pi$ electrons and $6 p$ orbitals
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$\left[\right.$ Given $, E_{C u^{2+} / C u}^{o}=0.34\, V , E _{ NO _{3}^{-} / NO_2 }^{\circ}=0.96\, V$ (Rounded-off to the nearest integer) $E _{ NO _{3} / NO _{2}}^{\circ}=0.79 \,V$ and at $298 \,K$ $\left.\frac{ RT }{ F }(2.303)=0.059\right]$
If $\Delta H^o = 25\,kcal/mol$ and $\Delta S^o = 50\,cal/K,$ at what temperature equilibrium will be established in the container. (Ignore variation of $\Delta H^o$ and $\Delta S^o$ with temperature.).......$K$
