JEE Main 2024 · 30 January · evening · Single correct · +4 / −1
A and B formed in the following reactions are:
CrO₂ Cl₂+4 NaOH → A+2 NaCl+2 H₂ O, ; A+2 HCl+2 H₂ O₂ → B+3 H₂ O
- (a) A=Na₂ Cr₂ O₇, B=CrO₅
- (b) A=Na₂ CrO₄, B=CrO₅
- (c) A=Na₂ Cr₂ O₄, B=CrO₄
- (d) A=Na₂ Cr₂ O₇, B=CrO₃
Answer and solution
Answer (b) A=Na₂ CrO₄, B=CrO₅
Solution
Chemistry ma’am · 5 steps
One Cr in, one Cr out: A is chromate, and acidified H₂O₂ turns it into blue CrO₅.
- Find A. The left side has only one Cr, so A has one Cr, which rules out any dichromate. Alkali converts chromyl chloride into sodium chromate, with Cr staying at +6.
CrO₂Cl₂ + 4NaOH → Na₂CrO₄ + 2NaCl + 2H₂O
- Check the count. O: 2 + 4 = 4 + 2. H: 4 = 4. Na: 4 = 2 + 2. Cl: 2 = 2. Everything balances, so A = Na₂CrO₄.
- Find B. Acidified H₂O₂ converts chromate into chromium pentoxide, CrO₅, which gives a deep blue colour in ether. It has one oxo oxygen and two peroxo groups, so Cr is still +6.
- Check B. In Na₂CrO₄ + 2HCl + 2H₂O₂ → CrO₅ + 2NaCl + 3H₂O, O balances as 4 + 4 = 5 + 3 and H as 2 + 4 = 6. The 3 water molecules match the given equation.
- So. A = Na₂CrO₄ and B = CrO₅, which is option (b).
The idea to keep. Chromyl chloride in alkali gives yellow chromate. Chromate with H₂O₂ in acid gives blue CrO₅, a peroxo compound in which Cr is still +6.
Answer from the NTA final answer key. Solution by PrepWiser.