Question
The correct order of $$C - O$$ bond length among $$CO,CO_3^{2 - },C{O_2}$$ is
A.
$$C{O_2} < CO_3^{2 - } < CO$$
B.
$$CO < CO_3^{2 - } < C{O_2}$$
C.
$$CO_3^{2 - } < C{O_2} < CO$$
D.
$$CO < C{O_2} < CO_3^{2 - }$$
Answer :
$$CO < C{O_2} < CO_3^{2 - }$$
Solution :
A bond length is the average distance between the centres of nuclei of two bonded atoms.
A multiple bond (double or triple bonds) is always shorter than the corresponding single bond.
The $$C - {\text{atom}}$$ in $$CO_3^{2 - }$$ is $$s{p^2}$$ hybridised as shown:
The $$C - {\text{atom}}$$ in $$C{O_2}$$ is $$sp$$ hybridised with bond distance carbon-oxygen is $$122\,pm.$$
The $$C - {\text{atom}}$$ in $$CO$$ is $$sp$$ hybridised with $$C - O$$ bond distance is $$110\,pm.$$
So, the correct order is
$$CO < C{O_2} < CO_3^{2 - }$$
A bond length is the average distance between the centres of nuclei of two bonded atoms.
A multiple bond (double or triple bonds) is always shorter than the corresponding single bond.
The $$C - {\text{atom}}$$ in $$CO_3^{2 - }$$ is $$s{p^2}$$ hybridised as shown:
The $$C - {\text{atom}}$$ in $$C{O_2}$$ is $$sp$$ hybridised with bond distance carbon-oxygen is $$122\,pm.$$
The $$C - {\text{atom}}$$ in $$CO$$ is $$sp$$ hybridised with $$C - O$$ bond distance is $$110\,pm.$$
So, the correct order is
$$CO < C{O_2} < CO_3^{2 - }$$