Question

The total number of electrons in all bonding molecular orbitals of O$_2^{2-}$ is ________.
(Round off to the Nearest Integer).

Hint:

To quickly find the bond order, use the formula: Bond Order = 1/2 (Number of bonding electrons - Number of antibonding electrons). For O$_2^{2-}$, this is 1/2 (10 - 8) = 1.

Answer 1

Correct Answer: 10

First, determine the total number of electrons in the peroxide ion, \( \mathrm{O_2^{2-}} \).
An oxygen atom (O) has 8 electrons.
An \( \mathrm{O_2} \) molecule has \( 8 \times 2 = 16 \) electrons.
The \( \mathrm{O_2^{2-}} \) ion has two additional electrons, so the total number of electrons is \( 16 + 2 = 18 \).
Now, write the molecular orbital (MO) configuration for an 18-electron species.
The order of MOs is:
\( \sigma_{1s} < \sigma^{*}_{1s} < \sigma_{2s} < \sigma^{*}_{2s} < \sigma_{2p_z} < (\pi_{2p_x} = \pi_{2p_y}) < (\pi^{*}_{2p_x} = \pi^{*}_{2p_y}) < \sigma^{*}_{2p_z} \)
Fill the orbitals with the 18 electrons:
\( (\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^{*}_{2p_x})^2 (\pi^{*}_{2p_y})^2 \)
The bonding molecular orbitals are those without an asterisk. These are \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y} \).
Count the number of electrons in these bonding orbitals:
- Electrons in \( \sigma_{1s} \): 2
- Electrons in \( \sigma_{2s} \): 2
- Electrons in \( \sigma_{2p_z} \): 2
- Electrons in \( \pi_{2p_x} \): 2
- Electrons in \( \pi_{2p_y} \): 2
Total number of electrons in bonding molecular orbitals \( = 2 + 2 + 2 + 2 + 2 = 10 \).