Reaction of PtF_6 with oxygen ( O_2 ) gas results in the formation...

The reaction between $\text{O}_2$ and $\text{PtF}_6$ is:

$\text{O}_2 + \text{PtF}_6 \rightarrow \text{O}_2^+[\text{PtF}_6]^-$

Here, $X^+$ is $\text{O}_2^+$ and $Y^-$ is $[\text{PtF}_6]^-$.

For $\text{O}_2^+$, the total number of electrons is $15$. The molecular orbital configuration is $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1}$. The bond order is $\dfrac{10 - 5}{2} = 2.5$. Thus, the bond order is not $1.5$.

In the complex ion $[\text{PtF}_6]^-$, the oxidation state of $\text{Pt}$ is $+5$. The ground state electronic configuration of $\text{Pt}$ is $[\text{Xe}] 4f^{14} 5d^9 6s^1$. The configuration of $\text{Pt}^{5+}$ is $[\text{Xe}] 4f^{14} 5d^5$. Therefore, the valence $d$-orbitals of the metal ion contain $5$ electrons.

During the reaction, $\text{O}_2$ loses an electron to form $\text{O}_2^+$ and $\text{PtF}_6$ gains an electron to form $[\text{PtF}_6]^-$. Since $\text{PtF}_6$ accepts an electron, it undergoes reduction and acts as an oxidant.

Since no fluorine atoms are transferred to $\text{O}_2$, $\text{PtF}_6$ does not act as a fluorinating agent.

Answer: Valence $d$-orbitals of the metal ion in $\textbf{X}^+\textbf{Y}^-$ has $5$ electrons.; $\text{PtF}_6$ acts as an oxidant in this reaction.