JEE (Advanced) 2026, Paper 1, Chemistry, Section 2 — one or more options correct: +4 only if all the correct options are chosen, partial marks +3 / +2 / +1 for a correct subset, 0 if unanswered, −1 otherwise.
Q.7 Reaction of PtF6 with oxygen (O2) gas results in the formation of an ionic compound, X+Y–. Correct statement(s) is(are)
Four stations and an analysis board. Press START EXPERIMENT — the whole investigation runs by itself: mix → electron jump → oxidation states → MO diagram → bond order → platinum's d-electrons → fluorine audit → verdict on each option.
One gas, one hexafluoride, one ionic solid. The question's conditions are loaded — press START EXPERIMENT.
Ionisation energies are measured values (NIST). The ~12.2 eV limit for PtF6 is Bartlett's empirical observation: O2 (12.07 eV) and Xe (12.13 eV) react, N2 (15.58 eV) does not.
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Build any second-row diatomic and fill its molecular orbitals yourself. Pick a molecule, then add or remove electrons — the diagram, bond order and magnetism update at once. Below it, strip electrons from platinum and watch its 5d count.
Run the experiment — the solution opens with the answer.
The experiment has not finished yet. Press START and let it run to the end — or use Reveal anyway in the header if you are revising.
PtF6 has one of the highest electron affinities known (≈ 7 eV). It pulls one electron straight off O2.
O2+ has 15 electrons; the one removed came from an antibonding π* orbital, so the bond gets stronger.
In [PtF6]− platinum is +5. Pt is 5d96s1; take away 5 electrons (6s first).
Only an electron moves: Pt +6 → +5 (reduced), O2 → O2+ (oxidised). All six F stay on Pt — nothing is fluorinated.
The first ionisation energy of O2 (12.07 eV) is low enough for PtF6 to take one electron. X+ = O2+ (dioxygenyl), Y− = [PtF6]− (hexafluoridoplatinate(V)).
Bonding electrons Nb = 10, antibonding Na = 5, bond order = (10 − 5)/2 = 2.5 — not 1.5. (1.5 is the bond order of the superoxide ion O2−, a common trap.) The bond also shortens, 121 → 112 pm. (A) ✗
In [PtF6]−: x + 6(−1) = −1, so Pt is +5. Pt (Z = 78) is [Xe]4f145d96s1; removing 6s1 and four 5d electrons leaves 5d5 (octahedral, low spin t2g5). (B) ✓
Pt goes from +6 in PtF6 to +5 in [PtF6]−: PtF6 gains an electron, it is reduced, so it is the oxidant. (C) ✓
A fluorinating agent hands F atoms to the other reactant (e.g. U + 3ClF3 → UF6 + 3ClF). Here all six F atoms stay on Pt and no O–F bond forms. (D) ✗
For a student opening it for the first time. The whole investigation runs by itself after one press.
O2 and PtF6 give an ionic compound X+Y−. Which of the four statements are correct? More than one can be.
The question's reagents are loaded — O2 and PtF6. You do not need to change anything.
Swap the gas for NO or N2, or replace PtF6 with F2. The lab tells you honestly when the question's pathway no longer applies. ↺ RESTORE QUESTION CONDITIONS puts everything back.
Press ▶ START EXPERIMENT. There is nothing else to click (about a minute; 2× makes it faster).
An electron jumps from O2 to PtF6, oxidation numbers change, the MO diagram fills and loses one π* electron, platinum's d-orbitals fill, and every fluorine atom is counted.
When all four statements are decided the lab says ✓ EXPERIMENT COMPLETE and reveals the answer. Badges unlock on the way.
In the Electron explorer add or remove electrons from O2, N2 or NO and watch the bond order change; step platinum through its oxidation states. Then read the Detailed solution.
Press ▶ REPLAY EXPERIMENT, or change a condition and press ⚙ RUN WITH CUSTOM VALUES. CLASSROOM MODE enlarges everything for a projector.
| Exam | JEE Advanced 2026, Paper 1, Chemistry, Section 2, Q.7 |
|---|---|
| Chapter | Chemical Bonding and Molecular Structure |
| Topic | Molecular orbital theory |
| Difficulty | Easy-Moderate |
| Answer | — run the experiment |