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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Given :
Molar mass $N =14\,g\,mol ^{-1} ; O =16\,g\,mol ^{-1} ; C =12\,g\,mol ^{-1} ; H =1\,g\,mol ^{-1}$;
$(I)\, SF_4$ $(II)\, OSF_4$ $(III)\, H_2CSF_4$
$S{O_{3(g)}} \rightleftharpoons S{O_{2(g)}} + 1/2\,{O_{2(g)}}$ is $4.9 \times 10^{-2}$ then find equilibrium constant for the reaction
$2S{O_{2(g)}} + {O_{2(g)}} \rightleftharpoons 2SO_3(g)$
