Question

Find the bond order of the peroxide ion.

Answer 1

Hint: Think about the molecular formula of hydrogen peroxide (${{H}_{2}}{{O}_{2}}$) to make it easier to deduce what the peroxide ion is, and its charge. Bond order is calculated using the molecular orbital theory.

Complete step by step solution:
The peroxide ion has the formula ${{O}_{2}}^{2-}$. From this, we can calculate the bond order by identifying the number of electrons in the bonding (BMO) and anti-bonding (ABMO) orbitals of the ion. The formula used to calculate the bond order is:
\[\text{Bond order = }\dfrac{1}{2}\times (\text{No}\text{. of }{{\text{e}}^{-}}\text{ in BMO }-\text{No}\text{. of }{{\text{e}}^{-}}\text{ in ABMO})\]
While writing the electronic configuration, the anti-bonding molecular orbitals are denoted with a $*$.

The electrons in a bonding molecular orbital are present in the space between the 2 nuclei of the atoms that are bonded. The electrons in the antibonding molecular orbitals are present in any space that is not between the 2 nuclei.
Thus, when 2 orbitals of different atoms combine, one electron from the respective orbital of each atom enters the bonding molecular orbital and the other two enter the anti-bonding molecular orbital.
Now, we will calculate the number of electrons that are present in the ${{O}_{2}}^{2-}$ ion so that we can fill in the bonding and antibonding orbitals according to the electronic configuration.
Each oxygen atom has 8 electrons and the peroxide ion has a charge of 2-. Therefore, 2 more electrons will be added. Hence, the total number of electrons adds up to 18.
The electronic configuration will be:
${{O}_{2}}^{2-}=\sigma 1{{s}^{2}}\cdot \sigma *1{{s}^{2}}\cdot \sigma 2{{s}^{2}}\cdot \sigma *2{{s}^{2}}\cdot \sigma 2{{p}_{z}}^{2}\cdot \pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2}\cdot \pi *2{{p}_{x}}^{2}=\pi *2{{p}_{y}}^{2}\cdot \sigma *2{{p}_{z}}^{0}$
Here, the bonding orbitals fill up first since staying between 2 nuclei requires less energy than staying at the periphery. As lower energy orbitals fill up first, BMO fills up before ABMO.
Now, calculating the number of electrons in the bonding and antibonding molecular orbitals we can see that
No. of electrons in BMO = 10
No. of electrons in ABMO = 8
Plugging these values in the formula, we get:
\[\text{Bond order = }\dfrac{1}{2}\times (10-8)\]
Bond order = 1

Note: Please be careful while filling up the orbitals with electrons. In oxygen, the bonding $2{{p}_{x}}$ and $2{{p}_{y}}$ orbitals get filled up after the $2{{p}_{z}}$ orbital is filled. The exact opposite happens when the anti-bonding orbitals are filled. This order changes from atom to atom.