Answer: P in +5 state with 3 P-OH (ionisable) and one P=O (not ionisable), making it triprotic.
- A P in +5 state with 3 P-OH (ionisable) and one P=O (not ionisable), making it triprotic
- B Mainly a single ionisable hydrogen among the four present in most cases under typical conditions
- C A direct P-H bond that does not contribute any ionisable proton according to standard textbooks
- D All four hydrogen atoms largely ionisable, making it tetraprotic in general practice
Correct answer: A. P in +5 state with 3 P-OH (ionisable) and one P=O (not ionisable), making it triprotic
Explanation: H<sub>3</sub>PO<sub>4</sub> has 3 P-OH groups (all ionisable, giving pKa1=2.1, pKa2=7.2, pKa3=12.4) and one P=O bond.
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.