If the magnetic moment of a di-oxygen species is 1.73 B.M., it may be:

We start with the spin‐only magnetic-moment formula that is used for molecules and transition-metal ions:

$$\mu = \sqrt{n(n+2)}\;\text{B.M.}$$

Here $$n$$ is the number of unpaired electrons and $$\mu$$ is given in Bohr Magnetons (B.M.).

The question tells us that the observed magnetic moment is $$\mu = 1.73\;\text{B.M.}$$

Now we substitute this value into the formula and solve for $$n$$:

$$1.73 = \sqrt{n(n+2)}$$

Squaring both sides gives

$$1.73^2 = n(n+2).$$

Since $$1.73^2 \approx 3.00,$$ we have

$$n(n+2) = 3.$$

Because $$n$$ must be a whole number (one cannot have a fractional unpaired electron), we test small integral values of $$n$$:

For $$n = 0:\;0(0+2)=0\ne3$$

For $$n = 1:\;1(1+2)=1\times3=3$$

The equality is satisfied when $$n = 1$$. So the species we are looking for must possess exactly one unpaired electron.

Next we examine the molecular orbital configurations of the given dioxygen species using the well-accepted ordering for $$\mathrm{O_2}$$ and its ions (after $$\sigma_{2p_z}$$ the $$\pi_{2p_x}$$ and $$\pi_{2p_y}$$ come before $$\pi_{2p_x}^\*,\;\pi_{2p_y}^\*$$):

Neutral $$\mathrm{O_2}$$ has 16 electrons. Its valence-orbital filling ends as

$$\pi_{2p_x}^2\;\pi_{2p_y}^2\;\pi_{2p_x}^\*{}^1\;\pi_{2p_y}^\*{}^1$$

This leaves two unpaired electrons (one in each $$\pi^\*$$), so $$n = 2$$ and $$\mu \approx 2.83\;\text{B.M.}$$, not 1.73.

The cation $$\mathrm{O_2^+}$$ has one electron fewer (15 electrons). Removing one of the antibonding electrons gives

$$\pi_{2p_x}^2\;\pi_{2p_y}^2\;\pi_{2p_x}^\*{}^1\;\pi_{2p_y}^\*{}^0$$

Now only one antibonding orbital contains a single electron, so $$n = 1$$ and $$\mu = 1.73\;\text{B.M.}$$ exactly as required.

The anion $$\mathrm{O_2^-}$$ has one electron more than neutral $$\mathrm{O_2}$$ (17 electrons). The extra electron enters one of the antibonding $$\pi^\*$$ orbitals:

$$\pi_{2p_x}^2\;\pi_{2p_y}^2\;\pi_{2p_x}^\*{}^2\;\pi_{2p_y}^\*{}^1$$

Again there is only one unpaired electron (now in the other $$\pi^\*$$), so $$n = 1$$ and $$\mu = 1.73\;\text{B.M.}$$

Finally, $$\mathrm{O_2^{2-}}$$ (peroxide) would have even numbers of electrons in all orbitals, giving $$n = 0$$, so it is diamagnetic and not listed among the options anyway.

Putting everything together, the species that exhibit a magnetic moment of $$1.73\;\text{B.M.}$$ are those containing exactly one unpaired electron, namely $$\mathrm{O_2^-}$$ and $$\mathrm{O_2^+}$$.

Hence, the correct answer is Option A.