Question

How bond order is calculated and why ${\mathrm{CO}}^{+}$ has bond order 3.5?

Answer 1

Step 1: Calculate the total no of electrons:

The number of valence electrons in the Carbon atom with electronic configuration$1s^2,2s^2,2p^2$ is $6$

And in the Oxygen atom (atomic number=8)$1s^2,2s^2,2p^4$, it is $8$.

As the total number of electrons in, $\mathrm{CO}=6+8=14$

So total number of electrons in ${\mathrm{CO}}^{+}=14-1=13$

Step 2: Calculate no of bonding antibonding electrons by considering the electronic configuration of $C{O}^{+}$ :

In the case of ${\mathrm{CO}}^{+}$, the $2\mathrm{s}$ atomic orbital of Oxygen is much lower than the energy than the $2\mathrm{s}$ atomic orbital of Carbon.

This unusual trend of energy allows the $\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}\&\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}$ BMO to lower the energy than $\mathrm{sigma}*2\mathrm{s}$ MO in the MO diagram of CO which can be shown as:

Hence, the molecular orbital configuration of $C{O}^{+}$ is given as:
${\left({\mathrm{\sigma }}^{\mathrm{a}}1\mathrm{s}\right)}^2<{\left({\mathrm{\sigma }}^{\mathrm{b}}1\mathrm{s}\right)}^2<{\left({\mathrm{\sigma }}^{\mathrm{b}}2\mathrm{s}\right)}^2<{\left({{\mathrm{\pi }}^{\mathrm{b}}}_{\mathrm{x}}\right)}^2={\left({{\mathrm{\pi }}^{\mathrm{b}}}_{\mathrm{y}}\right)}^2<{\left({\mathrm{\sigma }}^{\mathrm{b}}2{\mathrm{p}}_{\mathrm{z}}\right)}^2<{\left({\mathrm{\sigma }}^{\mathrm{a}}2\mathrm{s}\right)}^1$

Thus from the above configuration of ${\mathrm{CO}}^{+}$, we can calculate the total electrons of bonding molecular orbitals $=10$

and electrons in its antibonding molecular orbital$=3$

Step 3 : formula of bond order:

The bond order defines the number of chemical bonding present between a pair of atoms, for which the formula can be given as;

$\mathrm{Bond}\mathrm{order}:\frac{1}{2}\left[{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}\right]$ (Where N_b=number of nonbonding electrons, N_a =number of bonding electrons)

Step 4: calculation of Bond order of ${\mathrm{CO}}^{+}$:

Hence, applying the Bond order formula for ${\mathrm{CO}}^{+}$;

$$\begin{gathered} =\frac{1}{2}[{\mathrm{N}}_{\mathrm{b}}–{\mathrm{N}}_{\mathrm{a}}] \\ =\frac{1}{2}[10–3] \\ =3.5 \end{gathered}$$

$\because$Hence, the bond order of ${\mathrm{CO}}^{+}\mathrm{is}3.5$ has been explained by using the Bond order and MOT concept.