If the magnetic moment of a dioxygen species is $1.73\mathrm{B}.\mathrm{M}$, it may be :

$$\begin{gathered} \begin{array}{rl} & \mu =\sqrt{n(n+2)}=1.73\text{B.M}\\ & \therefore \sqrt{n(n+2)}=1.73\\ & \therefore (\sqrt{n(n+2)}{)}^2=(1.73{)}^2\\ & \therefore \mathrm{n}(\mathrm{n}+2)=3\\ & \mathrm{n}=1\end{array} \\ \therefore \text{One unpaired electron} \\ \text{In option}B\text{:} \\ {\mathrm{O}}_2^{-}\text{has}17\text{electrons.} \\ \sigma 1{\mathrm{s}}^2{\sigma }^{\ast }1{\mathrm{s}}^2\sigma 2{\mathrm{s}}^2{\sigma }^{\ast }2{\mathrm{s}}^2\sigma 2{{\mathrm{p}}_{\mathrm{z}}}^2\pi 2{{\mathrm{p}}_{\mathrm{x}}}^2=\pi 2{{\mathrm{p}}_{\mathrm{y}}}^2{\pi }^{\ast }2{\mathrm{px}}^2={\pi }^{\ast }2{{\mathrm{p}}_{\mathrm{y}}}^1 \\ \therefore \text{One unpaired electron} \\ {\mathrm{O}}_2^{+}\text{has}15\text{electrons.} \\ \sigma 1{\mathrm{s}}^2{\sigma }^{\ast }1{\mathrm{s}}^2\sigma 2{\mathrm{s}}^2{\sigma }^{\ast }2{\mathrm{s}}^2\sigma 2{{\mathrm{p}}_{\mathrm{z}}}^2\pi 2{{\mathrm{p}}_{\mathrm{x}}}^2=\pi 2{{\mathrm{p}}_{\mathrm{y}}}^2{\pi }^{\ast }2{\mathrm{px}}^1={\pi }^{\ast }2{{\mathrm{p}}_{\mathrm{y}}}^0 \\ \therefore \text{One unpaired electron} \\ \therefore {\mathrm{O}}_2^{-} \mathrm{and} {\mathrm{O}}_2^{+} \text{ both contain One unpaired electron. i.e, n=1 and } \\ \text{have a magnetic moment of 1.73 B.M.} \end{gathered}$$

Hence, the correct option is option (D) ${\mathrm{O}}_2^{-}\text{or }{\mathrm{O}}_2^{+}$.