Which compound given below has ${\mathrm{sp}}^3, {\mathrm{sp}}^2$ and sp orbitals in the ratio of 6:3: 2?

There are three types of hybridized carbon atoms:  ${\mathrm{sp}}^3, {\mathrm{sp}}^2$, and sp.

Therefore, the number of ${\mathrm{sp}}^3$ hybrid orbitals is 4+4, the number of ${\mathrm{sp}}^2$ hybrid orbitals is 3+3, and the number of sp hybrid orbitals is 2+2+2.

Therefore, the $\mathrm{sp}:{\mathrm{sp}}^2:{\mathrm{sp}}^3$ ratio is 6:3:2 

Hence the correct answer is D) ${\mathrm{CH}}_3 – \mathrm{CH} = \mathrm{CH} – \mathrm{C} \equiv \mathrm{CH}$.