A) \[3s{{p}^{3}},\,\,3s{{p}^{2}}\]
B) \[3s{{p}^{3}},\,\,3sp\]
C) All \[6sp\]
D) All \[6s{{p}^{2}}\]
Correct Answer: D
Solution :
In benzene all 6 carbons are \[s{{p}^{2}}\] hybridised.
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