Question

Give the correct order of bond length for $CO,C{O_2}$ and $C{O_3}^{2 - }$ (Carbonate) and why?

Answer 1

The chemical atoms or elements were separated by bonds in the chemical molecule, these chemical bonds will be different numbers in different molecules, the bond order is given by the number of bonds in a chemical molecule, this bond order is inversely proportional to bond length.

Complete answer: Given molecules are carbon monoxide $\left( {CO} \right)$ , carbon dioxide $\left( {C{O_2}} \right)$ and carbonate $\left( {C{O_3}^{2 - }} \right)$ .
The carbon monoxide has an element of carbon and hydrogen, these two atoms are separated by a triple bond.
Thus, the bond order of carbon monoxide $\left( {CO} \right)$ is $3$ . It has a triple bond character in the molecule.
The carbon dioxide has one carbon and two oxygen atoms, the bonds involved in carbon dioxide $\left( {C{O_2}} \right)$ are double bonds. There are two double bonds in this molecule.
Thus, the bond order of carbon dioxide $\left( {C{O_2}} \right)$ is 2. It has a double bond character in it.
The Carbonate ion has one carbon atom and three oxygen atoms, but there is a presence of both double bond and single bond in the molecule, these bonds are interchangeable due to resonance.
Thus, the bond order of carbonate lies between the single bond and double bond.
The bond length is defined as the distance between the two centres of covalently bonded atoms.
The higher the bond order, the shorter the bond length.
Thus, the correct order for bond length of given compounds are
$C{O_3}^{2 - } > C{O_2} > CO$ .

Note:
The bond order must be calculated based on the electrons entering into the shells.
The stronger is the bond i.e.., high bond order between the atoms, higher is the pulling force between the atoms, shortens the bond length, which results in the inversely proportional to the bond order.