Reaction of $\mathrm{PtF}_{6}$ with oxygen $\left(\mathrm{O}_{2}\right)$ gas results in the formation of an ionic compound, $\mathbf{X}^{+} \mathbf{Y}^{-}$. Correct statement(s) is (are)

(B) Valence d-orbitals of the metal ion in $\mathbf{X}^{+} \mathbf{Y}^{-}$has 5 electrons.; (C) $\mathrm{PtF}_{6}$ acts as an oxidant in this reaction.

(A) The bond order of $\mathbf{X}^{+}$is 1.5.

(D) $\mathrm{PtF}_{6}$ acts as a fluorinating agent in this reaction.

Chemical Bonding and Molecular Structure — JEE Advanced · 2026 · 17 May 2026 · Paper 1 Question with Solution

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.✓ Correct
(C) $PtF_{6}$ acts as an oxidant in this reaction.✓ Correct
(D) $PtF_{6}$ acts as a fluorinating agent in this reaction.

Answer: (B), (C)

Solution

The reaction is:

PtF_{6} + O_{2} \to O_{2} [PtF_{6}]

X^{+} = O_{2}^{+}

and

Y^{-} = [PtF_{6}]^{-}

(A) Bond order of

O_{2}^{+}

: Removing one electron from

O_{2}

(which has bond order 2) removes it from an antibonding orbital, giving bond order = 2.5. Incorrect.

(B) Electronic configuration of Pt: [Xe]4f¹⁴5d⁹6s¹
In

[PtF_{6}]^{-}

, Pt is in +5 oxidation state: [Xe]4f¹⁴5d⁵
Valence d-orbitals have 5 electrons. Correct.

(C)

PtF_{6}

oxidizes

O_{2}

O_{2}^{+}

, so it acts as an oxidant. Correct.

(D) No fluorination occurs in this reaction;

PtF_{6}

acts as an oxidant, not a fluorinating agent. Incorrect.

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · One or more correct options

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Chemical Bonding and Molecular Structure

Solution

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