Question 7

The reaction described is the discovery by Neil Bartlett: $O_2 + PtF_6 \rightarrow O_2^+[PtF_6]^-$ where $X^+ = O_2^+$ and $Y^- = [PtF_6]^-$.

Analysis of options: (A) Oxygen molecule ($O_2$) has a bond order of 2. $O_2^+$ is formed by removing an electron from an anti-bonding $\pi^*$ orbital. The bond order of $O_2^+$ is $\frac{10-5}{2} = 2.5$. Incorrect. (B) In $[PtF_6]^-$, the oxidation state of Platinum is $+5$. The ground state electronic configuration of $Pt$ ($Z=78$) is $[Xe] 4f^{14} 5d^9 6s^1$. For $Pt^{5+}$, the configuration is $[Xe] 4f^{14} 5d^5$. Thus, the valence $d$-orbitals have 5 electrons. Correct. (C) Oxygen is oxidized from state $0$ (in $O_2$) to $+1$ (in $O_2^+$). Platinum is reduced from state $+6$ (in $PtF_6$) to $+5$ (in $PtF_6^-$). Therefore, $PtF_6$ acts as an oxidizing agent (oxidant). Correct. (D) A fluorinating agent is a substance that introduces fluorine into another molecule. In this reaction, fluorine remains bonded to Platinum, and the primary process is electron transfer (oxidation-reduction), not fluorination of the oxygen molecule. Incorrect.