Question

Which of these species has a longer bond, \(\mathrm{B}_{2}\) or \(\mathrm{B}_{2}^{+} ?\) Explain in terms of molecular orbital theory.

Short answer

\(\mathrm{B}_{2}\) has a bond order of 3, whereas \(\mathrm{B}_{2^{+}}\) has a bond order of 2.5. Therefore, \(\mathrm{B}_{2^{+}}\) has a longer bond length compared to \(\mathrm{B}_{2}\).

Step-by-step solution

Write Electron Configurations

In order to perform our comparison, we must understand the electron configuration of each molecule. \(\mathrm{B}_{2}\) has a total of 10 electrons. The electron configuration, according to the MO theory, is \(\sigma_{1s}^{2}, \sigma_{1s}^{*2}, \sigma_{2s}^{2}, \sigma_{2s}^{*2}, \pi_{2p}^{2}\). \(\mathrm{B}_{2}^{+}\) has 9 electrons. Its electron configuration is therefore \(\sigma_{1s}^{2}, \sigma_{1s}^{*2}, \sigma_{2s}^{2}, \sigma_{2s}^{*2}, \pi_{2p}^{1}\).

Calculate Bond Order

Now, after obtaining the electron configuration for \(\mathrm{B}_{2}\) and \(\mathrm{B}_{2}^{+}\), calculate the bond order using the formula 'half the difference between number of bonding and antibonding electrons'. For \(\mathrm{B}_{2}\), the bonding electrons = 8 and the anti-bonding = 2, thus giving a bond order of 3. For \(\mathrm{B}_{2}^{+}\), the bonding electrons = 7 and the anti-bonding = 2, giving a bond order of 2.5.

Compare Bond Lengths

The bond length is inversely proportional to the bond order. Therefore, the molecule with the smaller bond order will have the longer bond length. \(\mathrm{B}_{2}^{+}\) has a lower bond order (2.5) than \(\mathrm{B}_{2}\) (3), and so it has a longer bond length.