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.6 Correct statement(s) about the compounds X, Y and Z is(are)
Three laboratory stations and an analysis board. Press START EXPERIMENT — the whole investigation runs by itself: clue → reaction → observation → identification → structure → property → verdict on each option.
Three unknowns, three reactions. The question's conditions are loaded — press START EXPERIMENT.
Temperature, flow and progress are visualisation values — the question gives only 573 K.
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Nothing investigated yet.
Pick a molecule. Drag to rotate, use the slider or the mouse wheel to zoom, arrow keys also rotate.
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.
MnO2 oxidises chloride from concentrated HCl. The only greenish-yellow gas that can form is chlorine.
With chlorine in excess, all three N–H hydrogens of NH3 are replaced by Cl, one after another.
Chlorine with excess fluorine at 573 K gives the interhalogen ClF3 (equal volumes at a lower temperature give ClF).
(A) Cl2 disinfects water ✓ · (B) NCl3 is pyramidal ✗ · (C) ClF3 makes UF6 for 235U enrichment ✓ · (D) NCl3 is a weaker base than NH3 ✗
MnO2 (Mn +4) is reduced to Mn2+; chloride (−1) is oxidised to Cl2 (0). Chlorine is the greenish-yellow gas. X = Cl2.
Y = NCl3. Nitrogen has 3 bond pairs and 1 lone pair: electron geometry tetrahedral, molecular geometry trigonal pyramidal — N sits above the plane of the three Cl atoms. Not planar.
(With excess NH3 instead, the products are N2 and NH4Cl: 8NH3 + 3Cl2 → N2 + 6NH4Cl — which is why the question says "excess".)
Z = ClF3. Chlorine has 3 bond pairs and 2 lone pairs: trigonal-bipyramidal electron geometry with both lone pairs equatorial, so the molecule is T-shaped.
(A) ✓ Cl2 is widely used as a disinfectant for drinking water.
(B) ✗ NCl3 is trigonal pyramidal, not planar.
(C) ✓ ClF3 is a powerful fluorinating agent used to make volatile UF6 — the form in which uranium is handled for 235U enrichment: U + 3ClF3 → UF6 + 3ClF.
(D) ✗ The electronegative Cl atoms pull electron density away from nitrogen (−I effect), so its lone pair is less available: NCl3 is a weaker Lewis base than NH3.
For a student opening it for the first time. The whole investigation runs by itself after one press.
Three reactions hide three compounds X, Y and Z. Which of the four statements about them are correct? More than one can be.
The question's conditions are loaded — conc. HCl, excess Cl2, excess F2, 573 K. You do not need to change anything.
Change a condition — dilute HCl, excess NH3, equal volumes of 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).
Gas bubbles off and fills the jar, molecules collide and swap atoms, the temperature climbs to 573 K, each product is built in 3D, and each statement is tested on the analysis board.
When all four statements are decided the lab says ✓ EXPERIMENT COMPLETE and reveals the answer. Badges unlock on the way.
Rotate Cl2, NCl3, ClF3 and NH3 in the Molecular explorer, 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.6 |
|---|---|
| Chapter | The p-Block Elements (Group 17) |
| Topic | Chlorine, nitrogen trichloride and interhalogens - preparation, structure and uses |
| Difficulty | Easy-Moderate |
| Answer | — run the experiment |