Answer: H 3 PO 4 and PH 3.
- A H<sub>3</sub>PO<sub>2</sub> and P<sub>2</sub>O<sub>5</sub>
- B H<sub>4</sub>P<sub>2</sub>O<sub>7</sub> and H<sub>2</sub>O
- C P<sub>2</sub>O<sub>3</sub> and H<sub>2</sub>O
- D H<sub>3</sub>PO<sub>4</sub> and PH<sub>3</sub>
Correct answer: D. H<sub>3</sub>PO<sub>4</sub> and PH<sub>3</sub>
Explanation: H<sub>3</sub>PO<sub>3</sub> (P in +3) disproportionates on heating: 4H<sub>3</sub>PO<sub>3</sub> gives 3H<sub>3</sub>PO<sub>4</sub> (P +5) and PH<sub>3</sub> (P -3).
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.