Reaction of Platinum Hexafluoride with Oxygen Gas

When platinum hexafluoride ($\text{PtF}_6$) reacts with oxygen gas ($\text{O}_2$), an electron transfer reaction occurs to form the ionic compound dioxygenyl hexafluoroplatinate(V):

$\text{O}_2 + \text{PtF}_6 \longrightarrow \mathbf{X}^+\mathbf{Y}^- = \text{O}_2^+[\text{PtF}_6]^-$

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

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Step-by-Step Analysis of Options:

1. Option A: Bond Order of $\mathbf{X}^+$ ($\text{O}_2^+$)
   • Molecular orbital configuration for neutral $\text{O}_2$ ($16\text{ e}^-$): $\sigma_{1s}^2 \, \sigma_{1s}^{*2} \, \sigma_{2s}^2 \, \sigma_{2s}^{*2} \, \sigma_{2p_z}^2 \, (\pi_{2p_x}^2 = \pi_{2p_y}^2) \, (\pi_{2p_x}^{*1} = \pi_{2p_y}^{*1})$ $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 6}{2} = 2.0$
   • For the cation $\text{O}_2^+$ ($15\text{ e}^-$), an electron is removed from the antibonding $\pi^*$ orbital: $\text{Bond Order} = \frac{10 - 5}{2} = 2.5$
   • Therefore, the bond order of $\mathbf{X}^+$ is $2.5$ (not $1.5$).
   • Option A is incorrect.

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1. Option B: Valence $d$-electrons of the metal ion in $\mathbf{X}^+\mathbf{Y}^-$
   • In the anion $\mathbf{Y}^- = [\text{PtF}_6]^-$, each fluoride ligand has a charge of $-1$.
   • Let $x$ be the oxidation state of Platinum ($\text{Pt}$): $x + 6(-1) = -1 \implies x = +5$
   • The atomic number of Platinum ($\text{Pt}$) is $78$. Its ground state electronic configuration is: $\text{Pt}: [\text{Xe}] \, 4f^{14} \, 5d^9 \, 6s^1$
   • For the $\text{Pt}^{5+}$ ion, $5$ valence electrons are removed ($1$ from the $6s$ orbital and $4$ from the $5d$ orbitals): $\text{Pt}^{5+}: [\text{Xe}] \, 4f^{14} \, 5d^5$
   • Hence, the valence $d$-orbitals of the metal ion ($\text{Pt}^{5+}$) contain $5$ electrons.
   • Option B is correct.

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1. Option C: Role of $\text{PtF}_6$ as an Oxidant
   • Oxidation state of oxygen changes from $0$ in $\text{O}_2$ to $+\frac{1}{2}$ in $\text{O}_2^+$, indicating oxidation.
   • Oxidation state of platinum changes from $+6$ in $\text{PtF}_6$ to $+5$ in $[\text{PtF}_6]^-$, indicating reduction.
   • Since $\text{PtF}_6$ oxidizes $\text{O}_2$ while itself being reduced, it acts as an oxidant (oxidizing agent).
   • Option C is correct.

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1. Option D: Role of $\text{PtF}_6$ as a Fluorinating Agent
   • A fluorinating agent transfers fluorine atoms to form covalent bonds with the substrate. In this reaction, fluorine is not transferred to oxygen; instead, pure electron transfer takes place to form an ionic lattice.
   • Thus, $\text{PtF}_6$ acts as a powerful oxidizing agent, not a fluorinating agent.
   • Option D is incorrect.

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Conclusion

The correct statements are B and C.