Answer: The electrophilic SO 3 (from equilibrium H 2 S 2 O 7 ⇌ SO 3 + H 2 SO 4 ) attacks the aromatic ring.
- A The electrophilic SO<sub>3</sub> (from equilibrium H<sub>2</sub>S<sub>2</sub>O<sub>7</sub> ⇌ SO<sub>3</sub> + H<sub>2</sub>SO<sub>4</sub>) attacks the aromatic ring
- B H<sub>2</sub>SO<sub>4</sub> generally ionises largely into H+ and sulfate ions in solution overall in most cases
- C H<sub>2</sub>SO<sub>4</sub> acts here mainly as a reducing agent toward the aromatic ring under typical conditions
- D A free H+ ion directly attacks the aromatic ring as the electrophile according to standard textbooks
Correct answer: A. The electrophilic SO<sub>3</sub> (from equilibrium H<sub>2</sub>S<sub>2</sub>O<sub>7</sub> ⇌ SO<sub>3</sub> + H<sub>2</sub>SO<sub>4</sub>) attacks the aromatic ring
Explanation: In oleum, SO<sub>3</sub> is the active electrophile for sulfonation. In conc. H<sub>2</sub>SO<sub>4</sub>, the reaction is reversible and requires strong heating.
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.