Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The total number of electrons in all bonding molecular orbitals of $O_2^{2-}$ is ........… . (Round off to the nearest integer)

Molecular Orbital Theory (MOT) is a powerful tool that allows us to understand the bonding, stability, and magnetic properties of molecules. In this problem, we are tasked with finding the total number of electrons residing in the bonding molecular orbitals of the peroxide ion, $O_2^{2-}$. Let's break down the process step-by-step.

The first and most crucial step in any MOT problem is determining the total number of electrons in the system.

A neutral oxygen atom ($O$) has an atomic number of 8, meaning it possesses 8 electrons. For a diatomic oxygen molecule ($O_2$), we have $8+8=16$ electrons. However, we are dealing with the peroxide ion, $O_2^{2-}$. The $2-$ charge indicates the presence of two additional electrons.

Total electrons = $16+2=18$ electrons.

Before we distribute these 18 electrons, we must establish the correct energy sequence of the molecular orbitals. For homonuclear diatomic molecules of elements like $O$ and $F$, the energy difference between the $2s$ and $2p$ atomic orbitals is large enough that they do not interact significantly (a phenomenon known as no s-p mixing).

Because there is no s-p mixing, the ${\sigma }_{2p_z}$ orbital, which forms from direct head-on overlap, drops lower in energy than the ${\pi }_{2p_x}$ and ${\pi }_{2p_y}$ orbitals, which form from sideways overlap. The energy order is:

Now, we apply the Aufbau principle, Pauli exclusion principle, and Hund's rule to fill the orbitals with our 18 electrons:

1. The 1s and 2s levels: The first 8 electrons completely fill the lower energy levels: ${\sigma }_{1s}^2$, \sigma^*_{1s}^2, ${\sigma }_{2s}^2$, and \sigma^*_{2s}^2. 2. The 2p level: We have 10 electrons remaining. - 2 electrons fill the ${\sigma }_{2p_z}$ orbital: ${\sigma }_{2p_z}^2$. - 4 electrons fill the degenerate bonding pi orbitals: ${\pi }_{2p_x}^2$ and ${\pi }_{2p_y}^2$. - The last 4 electrons completely fill the degenerate anti-bonding pi orbitals: \pi^*_{2p_x}^2 and \pi^*_{2p_y}^2.

The complete electronic configuration is: \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 question specifically asks for the number of electrons in bonding molecular orbitals. These are the orbitals that do not have an asterisk ($\ast$). Let's sum them up:

- From the 1s level: 2 electrons in ${\sigma }_{1s}$ - From the 2s level: 2 electrons in ${\sigma }_{2s}$ - From the 2p level: 2 electrons in ${\sigma }_{2p_z}$ and 4 electrons in ${\pi }_{2p_x},{\pi }_{2p_y}$

Total bonding electrons ($N_b$) = $2+2+2+4=10$.

As a bonus insight, if we calculate the bond order using the formula $\frac{N_b-N_a}{2}$, we get $\frac{10-8}{2}=1$. This confirms that the peroxide ion features a single $O-O$ bond. Furthermore, since all 18 electrons are paired, the ion is strictly diamagnetic.