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

Given, magnetic moment of di-oxygen species$=1.73 \mathrm{B}.\mathrm{M}.$

We know, magnetic moment $=\sqrt{\text{n}\left(\text{n}+2\right)}$, where $\mathrm{n}=$number of unpaired electrons

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

$\text{n}=1$
So, species should contain one unpaired electron.

Let us check by electronic configuration of each species

$$\begin{gathered} {\mathrm{O}}_2=\mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{{\mathrm{p}}^2}_{\mathrm{z}} \mathrm{\pi }2{{\mathrm{p}}^2}_{\mathrm{x}}=\mathrm{\pi }2{{\mathrm{p}}^2}_{\mathrm{y}} {\mathrm{\pi }}^{*}2{{\mathrm{p}}^1}_{\mathrm{x}}={\mathrm{\pi }}^{*}2{{\mathrm{p}}^1}_{\mathrm{y}} \\ {{\mathrm{O}}_2}^{+}=\mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{{\mathrm{p}}^2}_{\mathrm{z}} \mathrm{\pi }2{{\mathrm{p}}^2}_{\mathrm{x}}=\mathrm{\pi }2{{\mathrm{p}}^2}_{\mathrm{y}} {\mathrm{\pi }}^{*}2{{\mathrm{p}}^1}_{\mathrm{x}}={\mathrm{\pi }}^{*}2{{\mathrm{p}}^0}_{\mathrm{y}} \\ {{\mathrm{O}}_2}^{-}=\mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{{\mathrm{p}}^2}_{\mathrm{z}} \mathrm{\pi }2{{\mathrm{p}}^2}_{\mathrm{x}}=\mathrm{\pi }2{{\mathrm{p}}^2}_{\mathrm{y}} {\mathrm{\pi }}^{*}2{{\mathrm{p}}^2}_{\mathrm{x}}={\mathrm{\pi }}^{*}2{{\mathrm{p}}^1}_{\mathrm{y}} \end{gathered}$$
 

As clear by above electronic configurations that, ${\mathrm{O}}_2$ has two unpaired electrons and ${{\mathrm{O}}_2}^{-} \& {{\mathrm{O}}_2}^{+}$  Each has only one unpaired electron; hence magnetic moment of  ${{\mathrm{O}}_2}^{+} \& {{\mathrm{O}}_2}^{-}$ is $1.73 \mathrm{B}.\mathrm{M}.$