Question

Which of the following molecules represent the order of hybridization ${\text{s}}{{\text{p}}^2},{\text{s}}{{\text{p}}^2}{\text{, sp , sp}}$ from left to right carbon atoms?
A.${\text{C}}{{\text{H}}_2} = {\text{CH}} - {\text{CH}} = {\text{C}}{{\text{H}}_2}$
B.${\text{HC}} = {\text{C}} - {\text{CH}} \equiv {\text{CH}}$
C.${\text{C}}{{\text{H}}_3} - {\text{CH}} = {\text{CH}} - {\text{C}}{{\text{H}}_3}$
D.${\text{C}}{{\text{H}}_2} = {\text{CH}} - {\text{C}} \equiv {\text{CH}}$

Answer 1

The carbon is said to be ${\text{s}}{{\text{p}}^2}$ hybridized, when it is attached with double bond with the other group that may be carbon or any other atom that can be attached with double bond. A sp carbon is that carbon which is attached with a triple bond or is making 2 pi bonds.

Complete step by step answer:
Let us look at the hybridisation of carbon from left to right in each of the option:
${\text{C}}{{\text{H}}_2} = {\text{CH}} - {\text{CH}} = {\text{C}}{{\text{H}}_2}$
The first carbon is attached with a double bond with another carbon, hence it is ${\text{s}}{{\text{p}}^2}$ hybridized. The second carbon is also attached with a double bond, so this is also ${\text{s}}{{\text{p}}^2}$ hybridized. Here the second carbon is making only one double bond so it is ${\text{s}}{{\text{p}}^2}$ hybridized, if it would be making another double bond then it will have sp hybridization. The third carbon is attached with forth carbon with a double bond and hence is also ${\text{s}}{{\text{p}}^2}$ hybridized. The last carbon also is ${\text{s}}{{\text{p}}^2}$ hybridized; all the carbon atoms in this molecule are ${\text{s}}{{\text{p}}^2}$ hybridized.
${\text{HC}} = {\text{C}} - {\text{CH}} \equiv {\text{CH}}$
The above molecule is not correct because carbon always forms four bonds and the second carbon is making only 3 bonds, 2 sigma and 1 pi bond.
${\text{C}}{{\text{H}}_3} - {\text{CH}} = {\text{CH}} - {\text{C}}{{\text{H}}_3}$
The first carbon here is ${\text{s}}{{\text{p}}^3}$ hybridized, and this is not required in the question.
${\text{C}}{{\text{H}}_2} = {\text{CH}} - {\text{C}} \equiv {\text{CH}}$
In the above molecule the first carbon is forming only one double bond and hence is ${\text{s}}{{\text{p}}^2}$ hybridized. The second carbon is also ${\text{s}}{{\text{p}}^2}$ hybridized. If we look at the third carbon it is forming a triple bond and hence is sp hybridized and similarly the forth carbon is also sp hybridized. Hence the order of hybridization of carbon front left to right is ${\text{s}}{{\text{p}}^2},{\text{s}}{{\text{p}}^2}{\text{, sp , sp}}$.

Hence the correct option is D.

Note:
When an atom forms a single bond with another atom then it is termed as sigma bond. After the single bond, all other bonds formed are called pi bonds. Carbon belongs to group number 14 and has four electrons in its valence shell and hence always forms four covalent bonds to satisfy its valency.