Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The bond order and magnetic behaviour of $O_2^{-}$ ion are, respectively

Bond Order and Magnetism

Molecular Orbital Theory (MOT) is one of the most powerful tools in chemistry for predicting the stability and magnetic properties of diatomic molecules. In this problem, we are tasked with finding the bond order and magnetic behavior of the superoxide ion, $O_2^{-}$. Let's break down the thought process step-by-step.

Before we can write the molecular orbital configuration, we must determine the total number of electrons in the species. The $O_2^{-}$ ion consists of two oxygen atoms and an extra electron due to the negative charge.

Each neutral oxygen atom contributes $8$ electrons. Therefore, the total electron count is:

With $17$ electrons, we follow the standard energy level diagram for molecules with more than $14$ electrons (like $O_2$ and $F_2$), where the $\sigma 2p_z$ orbital is lower in energy than the $\pi 2p_x$ and $\pi 2p_y$ orbitals.

Let's distribute these $17$ electrons into the molecular orbitals in increasing order of energy, strictly following the Aufbau principle, Pauli exclusion principle, and Hund's rule of maximum multiplicity.

The electronic configuration is:

Notice the highest occupied molecular orbitals (HOMO). The last three electrons enter the degenerate antibonding orbitals ${\pi }^{\ast }2p_x$ and ${\pi }^{\ast }2p_y$. According to Hund's rule, they first singly occupy these orbitals before pairing up. Thus, ${\pi }^{\ast }2p_x$ gets two electrons, and ${\pi }^{\ast }2p_y$ gets one.

To find the Bond Order, we use the formula:

Where $n_b$ is the number of bonding electrons and $n_a$ is the number of antibonding electrons (those in orbitals marked with an asterisk $\ast$): - Bonding electrons ($n_b$): $2+2+2+4=10$ - Antibonding electrons ($n_a$): $2+2+3=7$

Substituting these values into our formula:

A bond order of $1.5$ indicates that the bond in the superoxide ion is weaker and longer than the double bond in a neutral $O_2$ molecule (which has a bond order of $2$).

Finally, we determine the magnetic nature. A species is paramagnetic if it contains one or more unpaired electrons, and diamagnetic if all electrons are paired. Looking at our configuration, the ${\pi }^{\ast }2p_y$ orbital contains exactly one unpaired electron. Therefore, the $O_2^{-}$ ion is paramagnetic.

Combining these two findings, the bond order is $1.5$ and the magnetic behavior is paramagnetic, making option (a) the correct answer.