Redox Reactions

Concept:Identify the ionic compound formed and evaluate each statement based on electronic configuration and redox behaviour.Explanation:The reaction is:$$\mathrm{O_2 + PtF_6 \rightarrow O_2^{+}[PtF_6]^{-}}$$Here, $\mathrm{X^+ = O_2^+}$ and $\mathrm{Y^- = [PtF_6]^{-}}$.Statement A: The bond order of $\mathrm{O_2^+}$ is calculated from its molecular orbital configuration:$$\sigma 1s^2 \ \sigma^{*}1s^2 \ \sigma 2s^2 \ \sigma^{*}2s^2 \ \sigma 2p_z^2 \ \pi 2p_x^2 = \pi 2p_y^2 \ \pi^{*}2p_x^1 = \pi^{*}2p_y^0$$Bond order = $\frac{10 - 5}{2} = 2.5$.Thus, statement A is incorrect (given bond order 1.5 is wrong).Statement B: In $\mathrm{[PtF_6]^{-}}$, the oxidation state of Pt is +5.The atomic configuration of Pt is $\mathrm{5d^9 6s^1}$. For $\mathrm{Pt^{+5}}$, it loses 5 electrons, leaving $\mathrm{5d^5}$.So, valence d‑orbitals have exactly 5 electrons. Hence, statement B is correct.Statement C: $\mathrm{PtF_6}$ oxidises $\mathrm{O_2}$ to $\mathrm{O_2^+}$ (oxidation increases from 0 to +1).Since $\mathrm{PtF_6}$ accepts electrons, it acts as an oxidising agent. Therefore, statement C is correct.Statement D: No fluorine atom is transferred from $\mathrm{PtF_6}$ to $\mathrm{O_2}$; the reaction is an electron transfer only.Thus, $\mathrm{PtF_6}$ does not act as a fluorinating agent. Statement D is incorrect.Answer:Statements B and C are correct.