The state of hybridisation of $\mathrm{C}_2, \mathrm{C}_3, \mathrm{C}_5$ and $\mathrm{C}_6$ of the hydrocarbon,
is in the following sequence

• Correct answer
  A

  $\mathrm{sp}, \mathrm{sp}^3, \mathrm{sp}^2$ and $\mathrm{sp}^3$
• B

  $s \mathrm{p}^3, \mathrm{sp}^2, \mathrm{sp}^2$ and $s \mathrm{p}$
• C

  $\mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^2$ and $\mathrm{sp}^3$
• D

  $\mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^3$ and $\mathrm{sp}^2$

Solution

Key Idea Count number of $\sigma$ bonds and then find hybridisation as follows. If number of $\sigma$ bonds $=2$; hybridisation is $\mathrm{sp}$, If number of $\sigma$ bonds $=3$; hybridisation is $\mathrm{sp}^2$, If number of $\sigma$ bonds $=4 ;$ hybridisation is $\mathrm{sp}^3$.
Double and triple bonds are not considered while finding hybridisation.

Hence, the correct answer is option: 'A'.