Question

The mode of hybridization of carbon in $C{O_2}$ is;
A.$sp$
B.$s{p^2}$
C.$s{p^3}$
D.None of these

Answer 1

We have to know that carbon dioxide is a chemical compound having the chemical formula, $C{O_2}$ and it is an acidic colorless gas. The carbon dioxide molecules contain one carbon atom and two oxygen atoms. Here the carbon atom is covalently bonded with two oxygen atoms with double bonds. On the basis of octet rule, each oxygen atom is required to bond twice and the C atom is required to bond four times. And it contains two carbon – oxygen bonds and the dipoles point is in the opposite direction. Hence, it is polar molecules.

Complete answer:
We need to remember that the carbon atom is joined with two oxygen atoms with a double bond. Normally, the carbon atom contains two double bonds. But it is not enough to make the bonds with oxygen. Therefore, the electron present in the $2s$ orbital will jump to $2p$ orbital and there is a formation of two hybrid orbitals. And these hybrid sp orbitals present in the carbon atom will overlap with the $2p$ orbitals present in the oxygen atom and there is a formation of two sigma bonds. Thus, we can say that, the hybridization of carbon dioxide is sp. Hence, option (A) is correct.
The hybridization of carbon dioxide is not equal to $s{p^2}$. Hence, option (B) is incorrect.
The hybridization of carbon dioxide is not equal to $s{p^3}$. Hence, option (C) is incorrect.
The mode of hybridization of carbon in $C{O_2}$ is equal to sp. Hence, option (D) is incorrect.

Hence, option (A) is correct.

Note:
We must have to know that hybridization is the method of mixing two atomic orbitals having the same energy levels, and there is a formation of degenerated new molecular orbitals. It is explained on the basis of quantum mechanics. In the time of hybridization, the atomic orbitals with similar energy will be mixed. In the case of $s{p^2}$ –orbitals, one s-orbital and two p-orbitals are mixing.