JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

The reaction between O _2 and PtF _6 is: O _2 + PtF _6 O _2^+[ PtF _6]^- Here, X^+ is O _2^+ and Y^- is [ PtF _6]^-. For O _2^+, the total number of electrons is 15. The molecular orbital configuration is _ 1s ^2 _ 1s ^ *2 _ 2s ^2 _ 2s ^ *2 _ 2p_z ^2 _ 2p_x ^2 _ 2p_y ^2 _ 2p_x ^ *1 . The bond order is 10 - 5 2 = 2.5. Thus, the bond order is not 1.5. In the complex ion [ PtF _6]^-, the oxidation state of Pt is +5. The ground state electronic configuration of Pt is [ Xe ] 4f^ 14 5d^9 6s^1. The configuration of Pt ^ 5+ is [ Xe ] 4f^ 14 5d^5. Therefore, the valence d-orbitals of the metal ion contain 5 electrons. During the reaction, O _2 loses an electron to form O _2^+ and PtF _6 gains an electron to form [ PtF _6]^-. Since PtF _6 accepts an electron, it undergoes reduction and acts as an oxidant. Since no fluorine atoms are transferred to O _2, PtF _6 does not act as a fluorinating agent. Answer: Valence d-orbitals of the metal ion in X ^+ Y ^- has 5 electrons.; PtF _6 acts as an oxidant in this reaction.