JEE Advanced 2026 — Paper 1 — CHEM-II solution

JEE Advanced2026Paper 1CHEM-II

Reaction of PtF $_{6}$ with oxygen (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
PtF $_{6}$ acts as an oxidant in this reaction.
D
PtF $_{6}$ acts as a fluorinating agent in this reaction.

Solution

Identify the product. The classic Bartlett reaction: $PtF_{6} + O_{2} \to [O_{2}]^{+} [PtF_{6}]^{-}$ . So $X^{+} = O_{2}^{+}$ (dioxygenyl cation) and $Y^{-} = [PtF_{6}]^{-}$ .

(A) Bond order of O $_{2}^{+}$ . Using MO theory, O $_{2}$ has bond order 2; removing one electron from the antibonding $π^{*}$ orbital gives O $_{2}^{+}$ with bond order $\frac{8 - 3}{2} = 2.5$ , not 1.5. Incorrect.

(B) Electrons in valence d-orbitals of Pt in [PtF $_{6}$ ] $^{-}$ . Neutral Pt has configuration [Xe] $4 f^{14} 5 d^{9} 6 s^{1}$ . In [PtF $_{6}$ ] $^{-}$ the platinum is in the +5 oxidation state, removing the 6s electron and four d electrons, leaving $5 d^{5}$ — exactly 5 electrons in the valence d shell. Correct.

(C) Is PtF $_{6}$ an oxidant? Yes: it oxidises O $_{2}$ (oxidation state 0) to O $_{2}^{+}$ (oxidation state $+ \frac{1}{2}$ ), while platinum is reduced from +6 to +5. Correct.

(D) Is PtF $_{6}$ a fluorinating agent here? No fluorine is transferred to O $_{2}$ in this reaction; the product contains the intact [PtF $_{6}$ ] $^{-}$ anion. Incorrect.

Correct answers: (B), (C).

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