Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: If the magnetic moment of a dioxygen species is $1.73$ B.M, it may be.

When we talk about the magnetic properties of molecules, we are essentially playing a game of counting unpaired electrons. The spin-only magnetic moment, denoted by $\mu$, is a direct window into the quantum mechanical soul of a molecule.

Let's start by recalling the master formula for the spin-only magnetic moment:

Here, $n$ represents the number of unpaired electrons. The problem states that our mystery dioxygen species has a magnetic moment of $1.73\text{ B.M.}$ If we square $1.73$, we get approximately $3$.

Solving this simple quadratic equation reveals that $n$ must be exactly $1$. Our mission is now clear: we need to find which of the given dioxygen species ($O_2$, $O_2^{+}$, or $O_2^{-}$) possesses exactly one unpaired electron.

Let's analyze the neutral oxygen molecule first. It has a total of $16$ electrons. According to Molecular Orbital Theory, the filling of electrons proceeds smoothly until we reach the highest energy levels. The last two electrons must enter the degenerate antibonding orbitals: ${\pi }^{\ast }2p_x$ and ${\pi }^{\ast }2p_y$.

Following Hund's rule of maximum multiplicity, these two electrons will occupy the degenerate orbitals singly with parallel spins.

As we can see, there are two unpaired electrons ($n=2$) in $O_2$. This would give a magnetic moment of $\sqrt{2(4)}=2.82\text{ B.M.}$, which doesn't match our target of $1.73\text{ B.M.}$

Now consider the $O_2^{+}$ ion. It has $15$ electrons, one less than neutral $O_2$. To form this cation, we must remove one electron from the highest occupied molecular orbital (HOMO), which is one of the ${\pi }^{\ast }$ orbitals.

This leaves us with exactly one unpaired electron ($n=1$) in the ${\pi }^{\ast }2p_x$ orbital. The magnetic moment is $\sqrt{1(3)}=1.73\text{ B.M.}$ This is a perfect match!

Finally, let's check the superoxide ion, $O_2^{-}$. It has $17$ electrons, one more than neutral $O_2$. This extra electron must enter the ${\pi }^{\ast }$ orbitals, pairing up with one of the existing single electrons.

Even after pairing, one electron remains unpaired in the ${\pi }^{\ast }2p_y$ orbital. So, $n$ is again $1$. The magnetic moment is $\sqrt{1(3)}=1.73\text{ B.M.}$ This is also a match!

Both the $O_2^{+}$ and $O_2^{-}$ ions have exactly one unpaired electron, giving them both a magnetic moment of $1.73\text{ B.M.}$ Therefore, the correct option is the one that includes both of these species.