Question

Shape of $\text{C}{{\text{O}}_{\text{2}}}$ is:
(A) Pyramidal
(B) Triangular
(C) Linear
(D) Spherical

Answer 1

The shape of any molecule is dependent on the hybridization of the central atom of that compound and the hybridization of the orbitals is defined as the intermixing of the atomic orbitals to give hybrid orbitals that have properties of the atomic orbitals involved.

Formula Used: $\text{H =}\dfrac{\text{V + X - C + A}}{\text{2}}$
Where H is the hybridization of the central element, V is the number of the electrons present in the valence shell of the element, x is the number of monovalent groups present in the molecule, C is the charge if the species is cationic and A is the charge if the species is anionic.

Complete step by step solution:
The central element in the carbon dioxide molecule is carbon which has 6 electrons in these shells among which 4 electrons are present in the valence shell.
For carbon dioxide, the number of electrons in the valence shell of carbon is 4, there are no monovalent groups, and the compound is electrically neutral, so there is no cationic or anionic charge. Therefore,
$\text{H =}\dfrac{\text{4 + 0 - 0 + 0}}{\text{2}}=\dfrac{4}{2}=2$
Hence there are two orbitals that are involved in the bonding and the hybridization is $\text{sp}$ . Hence the structure of carbon dioxide is “linear”.
So the correct option is C.

Note:
The Valence Shell Electron Pair Repulsion Theory of the VSEPR theory guides the bonding pattern and the shape of the molecules according to the number of the lone pairs and the bond pairs present in the molecule and this is because, the order of repulsion among the electron pairs is:
Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.