Question

The compounds carbon dioxide (CO₂) and sulfur dioxide (SO₂) are formed by the burning of coal. Their apparently similar formulas mask underlying differences in molecular structure. Determine the shapes of these two types of molecules, identify the hybridization at the central atom of each, and compare the natures of their$\mathbf{\pi }$-bonds.

Short answer

The molecular formula of $C{O}_2$and$S{O}_2$ are same but their shapes are different. The shape of$C{O}_2$ is linear with sp hybridization whereas the shape of$S{O}_2$ is bent shaped with ${\mathrm{sp}}^2$hybridization.

This is because of presence of an extra p-orbital and lone pairs in sulphur atom. Sulphur is larger in size than carbon.

Step-by-step solution

Hybridization and shape of CO2 

Hybridization is evaluated by computing the number of sigma bonds and lone pairs around the central atom.

For carbon dioxide:

Structure of carbon dioxide

• Number of sigma bonds = 2
• Number of one pair on carbon = 0
• Steric number = 2+0 = 2
• Hybridization of C-atom = sp

The shape of the molecule is linear.

Hybridization and shape of SO2 

For sulphur dioxide:

Structure of sulphur dioxide

• Number of sigma bonds = 2
• Number of one pair on sulphur atom = 1
• Steric number is 2+1 = 3
• Hybridization of S-atom =${\mathrm{sp}}^2$

The shape of molecule is V-shaped or bent shaped.

Nature of π-bond

In $C{O}_2$, there is one σ and one π-bond formed between carbon and the two oxygen atoms and the formation of -bonds take place between $p\pi -p\pi$, that is, only p-orbital is involved in the formation of -bonds.

While in $S{O}_2$, there is one and one -bond formed between sulphur and the two oxygen atoms. The formation of π-bonds takes place between $d\pi -p\pi$, that is, one p-orbital overlaps with other d-orbital.