Question

What is meant by term bond order? Write bond orders for ${N_2}$ and ${O_2}$.

Answer 1

Hint: Bond order is the difference between the number of bonding and anti bonding electrons, divided by two. So first find out the total number of electrons present in ${N_2}$ and ${O_2}$ and write down its configuration. From this find out the number of bonding and antibonding electrons and use them in formula.
Formula used:
-Bond order: BO = $\dfrac{1}{2}$(Bonding electrons – Anti bonding electrons) (1)

Complete step by step answer:
-First of all we will see what bond order is.
Bond order is the number of electron pairs or bonds present between two atoms.
For example: let us take $N \equiv N$ molecule, it has a bond order of 3. Ethyne molecules ($H - C \equiv C - H$) also have a bond order of 3, etc.
-The Molecular Orbital Theory describes the bond order to be basically the difference between the number of bonding and anti bonding electrons, divided by two. Mathematically it can be written as:
                     BO = $\dfrac{1}{2}$(Bonding electrons – Anti bonding electrons) (1)
To find out the bond order we will first need to find out the number of bonding and anti bonding electrons. To do this we need to write down their configurations and the electrons are filled in these molecular orbitals according to their increasing order of energy as shown below:
       $\sigma 1s,{\sigma ^*}1s,\sigma 2s,{\sigma ^*}2s,\sigma 2{p_z},\pi 2{p_x} = \pi 2{p_y},{\pi ^*}2{p_x} = {\pi ^*}2{p_y},{\sigma ^*}2{p_z}$ and so on.
-Now let us begin by writing the electronic configuration:
- For ${N_2}$ : total number of electrons = 14
           Configuration: ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}{(\sigma 2{p_z})^2}$
Here we can see that: Number of bonding electrons = 10 and the number of antibonding electrons = 4. Calculate the bond order using equation (1):
                                                   BO = $\dfrac{1}{2}(10 - 4)$
                                                         = $\dfrac{1}{2} \times 6$
                                                        = 3
So, the bond order for ${N_2}$ is 3.
-Now for ${O_2}$: total number of electrons = 16
Configuration: ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\sigma 2{p_z})^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}{({\pi ^*}2{p_x})^1}{({\pi ^*}2{p_y})^1}$
Here we can see that: Number of bonding electrons = 10 and the number of antibonding electrons = 6. Calculate the bond order using equation (1):
                                                   BO = $\dfrac{1}{2}(10 - 6)$
                                                         = $\dfrac{1}{2} \times 4$
                                                        = 2
So, the bond order for ${O_2}$ is 2.

Hence the bond orders for ${N_2}$ is 3 and ${O_2}$ is 2.

Note: The electrons should always be filled according to the increasing energy level of the molecular orbitals. Also the MOT explains the existence of a molecule on the basis of bond order, but this concept is neither feasible nor appropriate to explain molecular existence of polyatomic molecules.