Consider the following reactions :
NaCl + K₂Cr₂O₇ + H₂SO₄_((conc.)) → (A) + side products
(A) + NaOH → (B) + Side products
( B ) + H₂SO₄_((dilute)) + H₂O₂ → (C) + Side products
The sum of the total number of atoms in one molecule each of (A) and (B) and (C) is

Solution

Chemistry ma’am · 4 steps

This is the chromyl chloride test, step by step: red vapour, then yellow chromate, then blue CrO₅.

1. Find (A). A chloride heated with K₂Cr₂O₇ and conc. H₂SO₄ gives red-orange vapours of chromyl chloride: 4NaCl + K₂Cr₂O₇ + 6H₂SO₄ → 2CrO₂Cl₂ + 2KHSO₄ + 4NaHSO₄ + 3H₂O. So (A) is CrO₂Cl₂.
2. Find (B). Passing these vapours into NaOH gives a yellow solution of sodium chromate: CrO₂Cl₂ + 4NaOH → Na₂CrO₄ + 2NaCl + 2H₂O. So (B) is Na₂CrO₄.
3. Find (C). Chromate with H₂O₂ in dilute acid gives blue chromium pentoxide: Na₂CrO₄ + 2H₂O₂ + H₂SO₄ → CrO₅ + Na₂SO₄ + 3H₂O. So (C) is CrO₅, which is Cr(O)(O₂)₂ with Cr still in the +6 state.

   Compound	Formula	Atoms
   (A)	CrO₂Cl₂	1 + 2 + 2 = 5
   (B)	Na₂CrO₄	2 + 1 + 4 = 7
   (C)	CrO₅	1 + 5 = 6

4. So. The total number of atoms is 5 + 7 + 6 = 18.

The common slip: 29. Taking (C) as Cr₂(SO₄)₃, as if H₂O₂ reduced chromium to +3, counts 17 atoms for (C) and gives 29. The blue product is CrO₅, with Cr still +6.

The idea to keep. Remember the chain for chloride: red CrO₂Cl₂, then yellow Na₂CrO₄ in NaOH, then blue CrO₅ with H₂O₂ in acid. Chromium stays +6 throughout.

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