The state of hybridisation of $\mathrm{C}_2, \mathrm{C}_3, \mathrm{C}_5$ and $\mathrm{C}_6$ of the…

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
  1. $\mathrm{sp}, \mathrm{sp}^3, \mathrm{sp}^2$ and $\mathrm{sp}^3$
  2. $s \mathrm{p}^3, \mathrm{sp}^2, \mathrm{sp}^2$ and $s \mathrm{p}$
  3. $\mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^2$ and $\mathrm{sp}^3$
  4. $\mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^3$ 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.

Asked in: NEET 2009 (Screening)

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