Question

Reaction of \(\mathrm{PtF_6}\) with oxygen \((\mathrm{O_2})\) gas results in the formation of an ionic compound, \(X^+Y^-\). Correct statement(s) is(are)

• A. The bond order of \(X^+\) is \(1.5\).
• B. Valence \(d\)-orbitals of the metal ion in \(X^+Y^-\) has \(5\) electrons.
• C. \(\mathrm{PtF_6}\) acts as an oxidant in this reaction.
• D. \(\mathrm{PtF_6}\) acts as a fluorinating agent in this reaction.

Hint:

Important reaction discovered by Neil Bartlett: \[ \mathrm{O_2 + PtF_6 \rightarrow O_2^+[PtF_6]^-} \] Key points:
• \( \mathrm{PtF_6} \) is a very strong oxidizing agent.
• \( \mathrm{O_2^+} \) is called the dioxygenyl ion.
• Oxidation state of Pt in \( [\mathrm{PtF_6}]^- \) is \(+5\).
• \( \mathrm{Pt^{5+}} \) has \(5d^5\) configuration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This relates to Neil Bartlett's famous discovery in 1962. He noticed that Platinum hexafluoride (\(\text{PtF}_6\)) was strong enough to oxidize oxygen to form dioxygenyl hexafluoroplatinate: \[ \text{O}_2 + \text{PtF}_6 \rightarrow \text{O}_2^+[\text{PtF}_6]^- \] Here, X\(^+\) is \(\text{O}_2^+\) and Y\(^-\) is \([\text{PtF}_6]^-\).
Step 3: Detailed Explanation:
(A) Bond Order of O\(_2^+\): Molecular orbital configuration of \(\text{O}_2\) has 2 unpaired electrons in \(\pi^*\) orbitals. Removing one electron increases the bond order. Bond Order = (10 - 5) / 2 = 2.5. Statement (A) says 1.5, so it is False. (C) Oxidant: In the reaction, \(\text{O}_2\) is oxidized from 0 to +1 (\(\text{O}_2 \rightarrow \text{O}_2^+\)). \(\text{PtF}_6\) accepts an electron to become \([\text{PtF}_6]^-\). Thus, \(\text{PtF}_6\) acts as an oxidizing agent (oxidant). Statement (C) is True. (D) Fluorinating agent: A fluorinating agent adds fluorine atoms to a substrate. Here, the fluorine atoms stay with Pt. Pt just changes oxidation state (Pt(VI) to Pt(V)). It is not acting as a fluorinating agent here. Statement (D) is False. (B) Valence d-orbitals of Metal: The metal is Pt. In \([\text{PtF}_6]^-\), the oxidation state of Pt is +5. Pt is in group 10 (Ni, Pd, Pt). Configuration: [Xe] 4f\(^{14}\) 5d\(^9\) 6s\(^1\). Pt\(^{5+}\): [Xe] 4f\(^{14}\) 5d\(^5\). Wait, the metal ion is Pt\(^{5+}\), which indeed has 5 d-electrons. However, check if Y\(^-\) is standardly defined. In many contexts, Statement (B) is considered true if we assume the standard Bartlett compound.
Step 4: Final Answer:
PtF\(_6\) is one of the strongest oxidants known. It oxidizes O\(_2\) to O\(_2^+\), leading to a bond order increase to 2.5.