The ratio of the number of hybrid and pure orbitals in $C_{6} H_{6}$ used for bonding is

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3 : 2

2 : 3

4 : 3

1 : 1

answer is C.

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Detailed Solution

Each carbon in benzene undergoes ${sp}^{2}$ hybridization.
Carbon has 2s and 2p orbitals participate in the hybridization that creates $3 {sp}^{2}$ hybrid orbitals, while its 1s and one 2p orbital do not. Since H only has one 1s orbital and that orbital cannot be hybridised, H remains in its pure form. As a result, a benzene ring has 18 hybridised orbitals (3 from each carbon) as well as 18 pure orbitals (2 from each carbon and 1 from each hydrogen). Therefore, the ratio of hybridised to pure orbitals is 1:1.

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