Q. Consider the following reactions $\mathrm{K_2Cr_2O_7}\xrightarrow[-H_2O]{KOH}[A]\xrightarrow[-H_2O]{H_2SO_4}[B]+\mathrm{K_2SO_4}$ The products $[A]$ and $[B]$, respectively are :
A) K₂CrO₄ and CrO
B) K₂CrO₄ and Cr₂O₃
C) K₂CrO₄ and K₂Cr₂O₇
D) K₂Cr(OH)₆ and Cr₂O₃
Explanation
[Solution by Dr. Abubacker] Dichromate with KOH gives yellow chromate: K₂Cr₂O₇ + 2KOH → 2K₂CrO₄ + H₂O [A]. Acidifying with H₂SO₄ regenerates orange dichromate: 2K₂CrO₄ + H₂SO₄ → K₂Cr₂O₇ + K₂SO₄ + H₂O [B]. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.