The first ionization energy of Na, NO, Xe and O₂ follows the order $\mathrm{Na}<\mathrm{NO} <\mathrm{Xe}={\mathrm{O}}_2.$ O₂ reacts with powerful oxidizing agent, ${\mathrm{PtF}}_6$ and yields ${\mathrm{O}}_2^{+}{[{\mathrm{PtF}}_6 ]}^{-}.$  If ${\mathrm{PtF}}_6$ is allowed to react with other mentioned species then the product is/are

The given trend of ionization energies of Na, NO, O₂ and Xe states that ionization of O₂ is very high, still ${\mathrm{PtF}}_6$ successfully removed electrons from O₂. This indicates that ${\mathrm{PtF}}_6$ is a powerful oxidizing agent and forms compounds with Na, NO, and Xe.

$$\begin{gathered} \mathrm{Na} + {\mathrm{PtF}}_6 \to {\mathrm{Na}}^{+} {\left[{\mathrm{PtF}}_6\right]}^{-} \\ \mathrm{NO} + {\mathrm{PtF}}_6 \to {\mathrm{NO}}^{+} {\left[{\mathrm{PtF}}_6\right]}^{-} \\ \mathrm{Xe} + {\mathrm{PtF}}_6 \to {\mathrm{Xe}}^{+} {\left[{\mathrm{PtF}}_6\right]}^{-} \end{gathered}$$