Answer: 4NH 3 + 5O 2 → 4NO + 6H 2 O (catalysed by Pt/Rh).
- A 4NH<sub>3</sub> + 5O<sub>2</sub> → 4NO + 6H<sub>2</sub>O (catalysed by Pt/Rh)
- B N<sub>2</sub> + O<sub>2</sub> → 2NO occurring directly without a catalyst at this stage
- C NO + O<sub>2</sub> → NO<sub>2</sub>, the subsequent atmospheric oxidation step
- D 3NO<sub>2</sub> + H<sub>2</sub>O → 2HNO<sub>3</sub> + NO, the final absorption step in water
Correct answer: A. 4NH<sub>3</sub> + 5O<sub>2</sub> → 4NO + 6H<sub>2</sub>O (catalysed by Pt/Rh)
Explanation: The first and key step in Ostwald's process is catalytic oxidation of NH<sub>3</sub> over Pt/Rh gauze at 850-900°C.
Xenon tetrafluoride (XeF4) has six electron domains around xenon — four Xe–F bonds plus two lone pairs. The lone pairs occupy opposite axial positions, leaving the four fluorine atoms in a square-planar shape (Xe–F ≈ 194 pm). Image: ChemSim, Public Domain, via Wikimedia Commons.
Concept context
Covers the nitrogen family (Group 15), oxygen family (Group 16), halogens (Group 17), and noble gases (Group 18). One of the highest-weightage inorganic chapters in NEET and JEE - includes preparation and properties of HNO₃, H₂SO₄, interhalogens, and xenon compounds.