Use the spin-only magnetic moment formula

For a species with $n$ unpaired electrons, $\mu = \sqrt{n(n+2)}\ \text{B.M.}$

Given: $\mu = 1.73\ \text{B.M.}$

Find the number of unpaired electrons

Check values:

• If $n=1$, $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\ \text{B.M.}$
• If $n=2$, $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\ \text{B.M.}$

So the species must have 1 unpaired electron.

Examine dioxygen species using MO theory

For oxygen and its ions, the relevant antibonding orbitals are $\pi^{*}_{2p}$.

• $O_2$ has configuration ending as: $\pi^{*}_{2p_x}{}^1\,\pi^{*}_{2p_y}{}^1$ Hence it has 2 unpaired electrons. So, $\mu \approx 2.83\ \text{B.M.}$ Not suitable.

• $O_2^-$ has one extra electron. One of the $\pi^*$ orbitals gets paired, leaving 1 unpaired electron. So, $\mu \approx 1.73\ \text{B.M.}$

• $O_2^+$ has one electron removed from $O_2$. Then only one electron remains unpaired in the $\pi^*$ set, so it also has 1 unpaired electron. Thus, $\mu \approx 1.73\ \text{B.M.}$

Match with options

Species with magnetic moment $1.73\ \text{B.M.}$ are: $O_2^- \text{ and } O_2^+$

Therefore, the correct option is A.