Question

What is meant by the term bond order? Calculate the bond order of: $N_2$ , $O_2$ , $O_2^{+}$ and $O_2^{-}$ .

Answer 1

Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.

If $N_a$ is equal to the number of electrons in an anti-bonding orbital, then $N_b$ is equal to the number of electrons in a bonding orbital.

Bond order = $\frac{1}{2}(N_b-N_a)$

If $N_b>N_a$ , then the molecule is said be stable. However, if $N_b\le N_a$, then the molecule is considered to be unstable.

Bond order of $N_2$ can be calculated from its electronic configuration as:

$\left[\sigma \right(1s{)}^2\left]\right[{\sigma }^{\ast }\left(1s{)}^2\right]\left[\sigma \right(2s{)}^2\left]\right[{\sigma }^{\ast }\left(2s{)}^2\right]\left[\pi \right(2p_x\left){]}^2\right[\pi \left(2p_y\right){]}^2\left[\sigma \right(2p_z){]}^2$

Number of bonding electrons, $N_b$ = 10

Number of anti-bonding electrons, $N_a$= 4

Bond order of nitrogen molecule = $\frac{1}{2}(10-4)$

=3

There are 16 electrons in a dioxygen molecule, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

$\left[\sigma \right(1s{)}^2\left]\right[{\sigma }^{\ast }\left(1s{)}^2\right]\left[\sigma \right(2s{)}^2\left]\right[{\sigma }^{\ast }\left(2s{)}^2\right]\left[\sigma \right(1p_z\left){]}^2\right[\pi \left(2p_x\right){]}^2\left[\pi \right(2p_y\left){]}^2\right[{\pi }^{\ast }\left(2p_x\right){]}^2\left[{\pi }^{\ast }\right(2p_y){]}^1$

Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = $N_b$ and the number of anti-bonding electrons = 4 = $N_a$.

Bond order = $\frac{1}{2}(N_b-N_a)$

= $\frac{1}{2}(8-4)$

=2

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration of $O_2^{+}$can be written as:

$KK\left[\sigma \right(2s{)}^2\left]\right[{\sigma }^{\ast }\left(2s{)}^2\right]\left[{\sigma }^{\ast }\right(2p_z{)}^2\left]\right[\pi \left(2p_x\right){]}^2\left[\pi \right(2p_y\left){]}^2\right[{\pi }^{\ast }\left(2p_x\right){]}^1$

$N_b$ = 8 $N_a$

=3

Bond order of = $O_2^{+}=\frac{1}{2}(8-3)$

=2.5

Thus, the bond order $O_2^{-}$ of is 2.5.

$KK\left[\sigma \right(2s{)}^2\left]\right[{\sigma }^{\ast }\left(2s{)}^2\right]\left[{\sigma }^{\ast }\right(2p_z{)}^2\left]\right[\pi \left(2p_x\right){]}^2\left[\pi \right(2p_y\left){]}^2\right[{\pi }^{\ast }\left(2p_x\right){]}^1\left[{\pi }^{\ast }\right(2p_y){]}^1$

$N_b$ = 8 $N_a$

=5

Bond order of $O_2^{-}=\frac{1}{2}(8-5)$

=1.5

Thus, the bond order of $O_2^{-}$ ion is 1.5.