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

Step 1: Relate Magnetic Moment to Unpaired Electrons

The spin-only magnetic moment ($$\mu$$) of any chemical species is directly calculated from the number of unpaired electrons ($$n$$) using the formula:

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

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

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

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

$$3 \approx n^2 + 2n \implies n = 1$$

A magnetic moment of $$1.73\text{ B.M.}$$ precisely corresponds to the presence of exactly one unpaired electron.

Step 2: Electronic Configuration via Molecular Orbital (MO) Theory

According to Molecular Orbital Theory, the filling order for the valence shells of di-oxygen systems (containing 14 or more electrons) is:

$$\sigma_{1\text{s}}^2 \ \sigma_{1\text{s}}^{*2} \ \sigma_{2\text{s}}^2 \ \sigma_{2\text{s}}^{*2} \ \sigma_{2\text{p}_\text{z}}^2 \ (\pi_{2\text{p}_\text{x}}^2 = \pi_{2\text{p}_\text{y}}^2) \ (\pi_{2\text{p}_\text{x}}^{*1} = \pi_{2\text{p}_\text{y}}^{*1})$$

Let's count the total electrons and identify the number of unpaired electrons in the highest occupied antibonding pi orbitals ($$\pi^*_{2\text{p}_\text{x}}$$ and $$\pi^*_{2\text{p}_\text{y}}$$) for each choice:

1. Superoxide Ion ($$\text{O}_2^-$$):

   Total electrons = $$16 + 1 = 17$$.

   The valence configuration ends with: $$(\pi_{2\text{p}_\text{x}}^{*2} = \pi_{2\text{p}_\text{y}}^{*1})$$.

   It contains exactly 1 unpaired electron (n = 1).




2. Dioxygenyl Cation ($$\text{O}_2^+$$):

   Total electrons = $$16 - 1 = 15$$.

   The valence configuration ends with: $$(\pi_{2\text{p}_\text{x}}^{*1} = \pi_{2\text{p}_\text{y}}^{*0})$$.

   It contains exactly 1 unpaired electron (n = 1).

Why Other Options Are Incorrect:

• Neutral Oxygen molecule ($$\text{O}_2$$): Total electrons = 16. Its outermost configuration is $$(\pi_{2\text{p}_\text{x}}^{*1} = \pi_{2\text{p}_\text{y}}^{*1})$$, meaning it possesses 2 unpaired electrons. Its magnetic moment is $$\sqrt{2(2+2)} = \sqrt{8} \approx 2.83\text{ B.M.}$$

Conclusion:

Both $$\text{O}_2^-$$ and $$\text{O}_2^+$$ contain exactly one unpaired electron in their molecular orbital arrangement, yielding a spin-only magnetic moment value of $$1.73\text{ B.M.}$$

Answer: Option A — $$\text{O}_2^-$$ or $$\text{O}_2^+$$