The ratio of the number of hybrid and pure orbitals in ${\mathrm{C}}_6{\mathrm{H}}_6$ used for bonding is

Number of hybrid orbitals = number of carbons × number denoting the hybridization of carbon
Number of pure orbitals = number of hydrogen present + 2× (number of π−orbitals present)
In the given question, we need to find out the number of hybrid and pure orbitals of benzene.

In benzene, the carbon is sp² hybridized. The 2s and 2p orbitals of carbon undergo hybridization to form sp² hybrid orbitals, but the 1s and one 2p orbital does not undergo hybridization.
On the other hand, hydrogen has only one 1s orbital and a single orbital cannot undergo hybridization, so it remains in pure form.
Now, let’s find out the number of hybrid and pure orbitals in C₆H₆.
Number of hybrid orbitals == 6×3=18
Number of pure orbitals =6+2×3=12
Now, if we try to make it in the form of a ratio, we get 3:2