Question

The linear shape of ${\text{C}}{{\text{O}}_{\text{2}}}$ is:
A. ${\text{A s}}{{\text{p}}^{\text{3}}}$ hybridisation of carbon
B.${\text{B sp}}$ hybridisation of carbon
C. $C p\pi - p\pi$ bonding between carbon and oxygen
D. ${\text{D s}}{{\text{p}}^{\text{2}}}$ hybridisation of carbon

Answer 1

Hint: Linear shape is the shape in which atoms are deployed in straight line under ${\text{18}}{{\text{0}}^{\text{^o }}}$ angle. Therefore, molecules with linear electron pair geometries have ${\text{sp}}$ hybridisation at its centre of the atom. For example, ${\text{C}}{{\text{O}}_{\text{2}}}{\text{ and N}}{{\text{O}}_{\text{2}}}$ that is carbon dioxide and nitric oxide respectively.

Complete step by step answer: \[{\text{C}}{{\text{O}}_{\text{2}}}\] has linear shape so this is due to the VSEPR (valence shell electron repulsion) theory and the different number of electrons in each molecule. Clearly, this theory states that as electrons are negatively charged, their valence electrons in different atoms which are present in a molecule they repel each other. Whereas, lone pair electrons take up more space than bonding electrons, and as they are only attracted to the one atom rather than two. So, more bonding electrons are repelled by them. Therefore, if we talk about the order of repulsions between different types of electron pair that is:
Lone pair-lone pair ${\text{ > }}$ bonding pair-lone pair ${\text{ > }}$ bonding pair-bonding pair
Briefly, the total number of valence electrons in \[{\text{C}}{{\text{O}}_{\text{2}}}\] is $4$, we take this $4$ from \[{\text{C}}{{\text{O}}_{\text{2}}}\] and $6$ from each oxygen ${\text{ = 16}}$ . So as the carbon is in the centre as we know it very clearly because it has lower electronegativity. If we only take single bond from ${\text{C - O}}$ , then carbon does not form a stable octet of electrons so that’s why we need to form double bond ${\text{C = O = C}}$. The bonding electrons around the carbon repel equally so the molecule is linear. So from the question, the linear shape of \[{\text{C}}{{\text{O}}_{\text{2}}}\] is due to $p\pi - p\pi$ bonding between carbon and oxygen and${\text{ sp}}$hybridisation of carbon.
Hence, $\left( {\text{B}} \right){\text{and}}\left( {\text{C}} \right)$ are correct options.

Note: \[{\text{C}}{{\text{O}}_{\text{2}}}\] Have two regions of electron density and it has no lone pair present on the central atom${\text{'C'}}$. The shape of \[{\text{C}}{{\text{O}}_{\text{2}}}\] is not even distorted or bent and the s and p orbital of both atom overlap to form ${\text{ sp}}$hybrid and $p\pi - p\pi$ bonding.