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.

 

Bond order

 

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_zAOs. 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)

 

Bond order

 

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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