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🧪 Chemistry  ·  p-Block Elements (Groups 15 to 18)  ·  NEET & JEE

In the reaction between Cl<sub>2</sub> and hot concentrated NaOH, the product is:

Answer: NaClO 3 (sodium chlorate) and NaCl.

  • A NaClO<sub>3</sub> (sodium chlorate) and NaCl
  • B NaOCl and NaCl (cold dilute NaOH gives hypochlorite)
  • C NaCl only
  • D NaClO<sub>4</sub>

Correct answer: A. NaClO<sub>3</sub> (sodium chlorate) and NaCl

Explanation: Cold dilute NaOH + Cl<sub>2</sub> → NaOCl + NaCl (bleach); hot concentrated NaOH + 3Cl<sub>2</sub> → NaClO<sub>3</sub> + 5NaCl + 3H<sub>2</sub>O (disproportionation).

Xenon tetrafluoride XeF4: a central xenon bonded to four fluorine atoms in a square-planar arrangement with two lone pairs above and below, Xe-F bond length 194 pm.

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.

Read the full p-Block Elements (Groups 15 to 18) notes →