Question

In $\stackrel{1}{C{H}_2}=\stackrel{2}{C}=\stackrel{3}{CH}-\stackrel{4}{C{H}_3}$ molecule, the hybridization of carbon 1, 2, 3 and 4 respectively are

• A. $s{p}^2,sp,s{p}^2,s{p}^3$
• B. $s{p}^2,s{p}^2,s{p}^2,s{p}^3$
• C. $s{p}^2,s{p}^3,s{p}^2,s{p}^3$
• D. $s{p}^3,sp,s{p}^3,s{p}^3$

Answer 1

The correct option is A $s{p}^2,sp,s{p}^2,s{p}^3$

Hybridization is the combining of atomic orbitals to produce new orbitals of varying energies and shapes than the initial orbitals. Hybridization occurs as atomic orbitals combine to create a new atomic orbital. The new orbital will contain the same total electron number as the old electrons.
In double bond, one is formed by sigma bond and other is by pi bond. Pi bond are formed by lateral overlap of p orbitals.

$\underset{s{p}^2}{C{H}_2}=\underset{sp}{C}=\underset{s{p}^2}{CH}-\underset{s{p}^3}{C{H}_3}$

Correct answer is (a).