# Bond Order

The bond order (B.O.) in diatomic molecules is half of the difference between the total numbers of the bonding electron $(N_b$) and antibonding electrons $(N_a)$ therefore

$B.O. = \dfrac{N_b- N_a}{2}$
• If bond order is zero, the molecule does not exist.
• $\text{ Bond order }\propto \text{Bond angle} \propto \text{Bond energy} \\ \propto \dfrac{1}{\text{Bond length}}$

Molecular orbital configuration of some homonuclear diatomic molecules and ions with their bond order etc, are given here.

1. $H_2 : \sigma (1s)^2 \hspace{20mm} B.O. = \dfrac{2- 0}{2} = 1$

Bond energy of $H_2 \text{is} 436 kJ \text{mol}^{-1}$ and bond length is $0.751 \AA$

2. $H^+_2 : \sigma(1s))^1 \hspace{20mm} B.O. = \dfrac{1- 0}{2} = \dfrac{1}{2}$

Bond energy of $^+_2 \text{is} 269 kJ \text{mol}^{-1}$ and bond length is $1.04 \AA$

3. $He_2 : \sigma (1s)^2 \hspace{20mm} B.O. = \dfrac{2- 2}{2} = 0$

Since bond order of helium molecule $(He_2)$ is zero hence helium does not exist as $He_2$

4. $Li_2 : \sigma (1s)^2 \sigma^*(1s)^2 \sigma (2s)^2; \; \; B.O. = \dfrac{4- 2}{2} = 1$

The bond energy of $Li_2$ molecule is low $(105kJ \text{mol}^{-1})$ and its bond length is larger $2.67\AA$.

5. $Be_2 : \sigma (1s)^2 \sigma^* (1s)^2 \sigma (2s)^2 \sigma^* (2s)^2; \\ B.O. = \dfrac{4- 4}{2} = 0$

Since in this case, again bond order is zero hence beryllium does not exist as $Be_2$. It exists as Be.

6. $B_2 : \sigma (1s)^2 \sigma^* (1s)^2 \sigma (2s)^2 \pi )2p_x)^1 \pi(2p_y)^1; \\ B.O. = \dfrac{6- 4}{2} = 1$

The bond energy of $B_2$ molecule is $289kJ \text{mol}^{-1}$ and its bond length is $1.59 \AA$. This molecule is paramagnetic due to the presence of unpaired electrons.

7. $C_2 : \sigma (1s)^2 \sigma^* (1s)^2 \sigma (2s)^2 \sigma^* (2s)^2 \pi (2p_x)^2 \pi (2p_y)^2 \\ B.O. = \dfrac{8- 4}{2} = 2$

Its bond energy and bond length are found to be $606.7kJ \text{mol}^{-1}$ and $1.31 \AA$, respectively.

8. $N_2 \sigma (1s)^2 \sigma^* (1s)^2 \sigma (2s)^2 \sigma ^* (2s)^2 \pi (2p_x)^2 \pi (2p_y)^2 \sigma(2p_z)^2 \\ B.O. = \dfrac{10- 4}{2} = 3, i.e., N \equiv N$

Its bond energy and bond length are $945.6 kJ \text{mol}^{-1}$ and $1.10 \AA$ respectively.

This order of enrgy level is followed by H, He, Li, Be, B, C and N. this is due to the mixing of 2s and $2p_z$AOs. This mixing is not possible in the case of oxygen, fluorine etc. because in these case the difference between energies of 2s and $2p_z$ AOs is larger. Therefore the order of energy level of oxygen, fluorine etc. molecular orbitals is given below and shown in figure

$\sigma (1s) < \sigma^* (1s) < \sigma (2s) < \sigma^* (2s) < \sigma (2p_z) < \pi (2p_x) = \pi(2p_x) < \pi^* (2p_x) = \pi^* (2p_y) < \sigma^*(2p_z)$

9. $O_2 \sigma (1s)^2 \sigma ^* (1s)^2 \sigma (2s)^2 \sigma ^* (2s)^2 \sigma (2p_z)^2 \pi (2p_x)^2 \pi (2p_y)^2 \pi ^* (2p_x)^1 \pi ^* (2p_y)^1 \\ B.O. = \dfrac{10 - 6}{2} = 3, \\ i.e., O \equiv O$

Its bond energy is $494.6 kJ \text{mol}^{-1}$ and bond length is $1.21 \AA$ . Since there are more antibonding electrons than $N_2$ hence it is less stable than $N_2$. Due to the presence of unpaired electrons it is paramagnetic in nature.

10. $F_2 : \sigma (1s)^2 \sigma^* (1s)^2 \sigma (2s)^2 \sigma ^* (2s)^2 \sigma (2p_z)^2 \pi (2p_x)^2 \pi (2p_y)^2 \pi^* (2p_x)^2 \pi^* (2p_y)^2 \\ B.O. = \dfrac{10- 8}{2} = 1, i.e., e - F - F\equiv O$

Its bond energy and bond length are $155 kJ \text{mol}^{-1} \text{and} 1.42 \AA$ respectively. This is also more reactive than $N_2 \text{and} O_2$ due ot more antibonding electrons.

11. $Ne_2 : \sigma (1s)^2 \sigma^* (1s)^2 \sigma (2s)^2 \sigma ^* (2s)^2 \sigma (2p_z)^2 \pi (2p_x)^2 \pi (2p_y)^2 \pi^* (2p_x)^2 \pi^* (2p_y)^2 \pi^* (2p_z)^2 \\ B.O. = \dfrac{10- 10}{2} = 0$

Since in this case bond order is zero hence neon does not exist as $Ne_2$. It exists as Ne.

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