Question 7

O2 + PtF6 → O2⁺[PtF6]⁻ (dioxygenyl hexafluoroplatinate). X⁺ = O2⁺: 15 electrons, bond order = 2.5, so (A) is wrong. Y⁻ = [PtF6]⁻: Pt in +5 oxidation state. Pt is 5d⁹6s¹ (Z=78, 10 valence electrons); Pt(V) has 5 d-electrons → (B) correct. PtF6 oxidizes O2 to O2⁺, so PtF6 is an oxidant → (C) correct. No fluorine is transferred to oxygen; PtF6 is not acting as a fluorinating agent → (D) wrong.

**Solving chemical reaction** I need to tackle the reaction involving O2 and PtF6 to yield O2, PtF6-, and Pt5+ in a d5 group 10 context. I'm observing that there’s a false statement in option A and true statements in options B and C. I should focus on coming up with just the end JSON format for my answer, omitting anything unnecessary while ensuring it's concise and direct. Let's ensure we get that right!

**Evaluating fluorination reactions** I'm analyzing how PtF6 acts as a fluorinating agent in reactions. It seems PtF6 can oxidize, but does it fluorinate O2? I find that electron transfer occurs without fluorinating oxygen, which makes it inaccurate to label PtF6 as a fluorinating agent in this context. I also have to consider the formation of potential compounds like O2PtF6. I need to verify details thoroughly, possibly referencing official sources for the best final answer regarding the reactions involved. **Analyzing PtF6 and O2 reactions** I'm digging into the reaction of PtF6 with oxygen gas to see what ionic compounds form. It looks like the product might be something like X+Y−, possibly O2F+ and PtF6−. I’m considering bond orders, and if X+ equals O2+, then the bond order would be 2.5. I need to verify whether an ionic compound like O2+[PtF6]− actually exists, considering the oxidation state and valence electrons of platinum. There's quite a bit to verify! **Analyzing reaction equations** I'm looking at the reaction equation involving O2 and PtF6. There's no additional fluorine atom to form a bond with O2, which raises the question about PtF6 acting as a fluorinating agent. However, in a different context with xenon, PtF6 does act as both an oxidizing and fluorinating agent, yielding various products. In the case with O2, the products are O2 with a negative charge and PtF6, but not a fluorinated compound. So, if asked about the reaction with oxygen, it wouldn't be about forming oxygen fluoride.

This reaction forms O2+PtF6-, dioxygenyl hexafluoroplatinate, so X+ is O2+ and Y- is PtF6-. Removing an electron from O2's antibonding π* orbital raises its bond order from 2 to 2.5, so the claim of 1.5 is incorrect. Now I'm working out the oxidation state and d-electron configuration of platinum in the PtF6- anion. With overall charge -1 and six fluoride ions contributing -6, platinum must be +5. Starting from Pt's neutral valence configuration of 10 electrons (5d9 6s1), removing five gives a d5 count in Pt5+, confirming that the valence d-orbitals hold five electrons. PtF6 clearly acts as an oxidizer here since it converts O2 to O2+, but it isn't functioning as a fluorinating agent since the fluorines all remain bound to platinum in the product rather than transferring elsewhere. So the correct answers are B and C. Looking at this reaction, PtF₆ oxidizes O₂ to form the dioxygenyl cation O₂⁺ (X⁺) and PtF₆⁻ (Y⁻), i.e., O₂⁺PtF₆⁻ (analogous to the classic PtF₆ + O₂ reaction discovered by Bartlett). **Analysis:** - **X⁺ = O₂⁺**: O₂ has bond order 2 (σ2s², σ*2s², σ2pz², π2px²=π2py², π*2px¹=π*2py¹). Removing one electron from the antibonding π* orbital gives O₂⁺ a bond order of 2.5, not 1.5. **(A) is incorrect.** - **Y⁻ = PtF₆⁻**: Here Pt is in +5 oxidation state (6 F⁻ contribute −6 charge, overall ion charge −1, so Pt = +5). Neutral Pt has valence configuration 5d⁹6s¹ (10 valence electrons). Removing 5 electrons for Pt⁵⁺ leaves 5 d-electrons. **(B) is correct.** - PtF₆ oxidizes O₂ to O₂⁺ (removes an electron from O₂), consuming an electron itself (Pt goes from +6 to +5), confirming PtF₆ acts as an **oxidant**. **(C) is correct.** - Since fluorine remains bonded to Pt (forming PtF₆⁻) rather than being transferred to oxygen, PtF₆ does **not** act as a fluorinating agent here — it acts purely as an oxidizing agent. **(D) is incorrect.** ```json