Consider the following reactions
K₂ Cr₂ O₇ →(KOH)[ A] →(H₂ SO₄)[ B]+K₂ SO₄
The products [A] and [B], respectively are :

Solution

Chemistry ma’am · 4 steps

Base turns orange dichromate into yellow chromate, and acid turns it straight back. Chromium stays +6.

1. Add base. OH⁻ breaks the Cr-O-Cr bridge of dichromate and gives two chromate ions. So [A] is yellow K₂CrO₄.

   K₂Cr₂O₇ + 2KOH → 2K₂CrO₄ + H₂O

2. Add acid. H⁺ joins two chromate ions back together through an oxygen bridge and removes water. The potassium left over pairs with sulphate as K₂SO₄, so [B] is orange K₂Cr₂O₇ again: 2K₂CrO₄ + H₂SO₄ → K₂Cr₂O₇ + K₂SO₄ + H₂O.
3. No redox. Chromium is +6 in both CrO₄²⁻ and Cr₂O₇²⁻. Only the pH changes, so no lower-oxidation-state chromium product can form.
4. So. [A] is K₂CrO₄ and [B] is K₂Cr₂O₇, which is option (c).

The idea to keep. Chromate and dichromate interconvert with pH alone: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. Chromium stays +6, so if the only reagents are acid or base, look for no change in oxidation state.

Answer from the NTA final answer key. Solution by PrepWiser.