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B
localized on every third C-atom.
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C
present in anti-bonding orbital
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D
delocalised forming extended π-bonding system.
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Solution
The correct option is D delocalised forming extended π-bonding system. Here each carbon atom is in sp2 hybridised state and is thus attached to three other carbon atoms by three sigma bonds forming a hexagonal planar structure. The fourth electron present in an unhybridised p−orbital of each carbon atom of a hexagonal unit thenoverlap with each other to form a pi bond.