Question

In the compound ${}^{1}C{H}_2$ = ${}^{2}C-\stackrel{H}{\stackrel{|}{{}^3\underset{\stackrel{|}{H}}{C}}}-\underset{H}{\underset{|}{{}^4\stackrel{\underset{|}{H}}{C}}}{-}^{5}C$ ≡ ${}^{6}CH$, the hybridization of $C_3$ is:

• A. ${sp}^2$
• B. $sp$
• C. ${dsp}^2$
• D. ${sp}^3$

Answer 1

The correct option is D

${sp}^3$

Lets draw the structure again,

${}^{1}C{H}_2$ = ${}^{2}C-\stackrel{H}{\stackrel{|}{{}^3\underset{\stackrel{|}{H}}{C}}}-\underset{H}{\underset{|}{{}^4\stackrel{\underset{|}{H}}{C}}}{-}^{5}C$ ≡ ${}^{6}CH$

Now , Number of hybrid oribitals = number of bonded pairs + number of lone pairs

$\Rightarrow$ Number of hybrid orbitals = 4

$\Rightarrow$ ${sp}^3$ hybridization.

Therefore,the hybridization of $C_3$ is ${sp}^3$.