Consider the following reactions
A+NaCl+H₂ SO₄ → CrO₂ Cl₂+ Side Products ; CrO₂ Cl_(2 (Vapour))+NaOH → B+NaCl+H₂ O ; B+H⁺ → C+H₂ O
The number of terminal ' O ' present in the compound ' C ' is__________

Solution

Chemistry ma’am · 5 steps

This is the chromyl chloride test: follow chromium to dichromate, then count which oxygens are not bridging.

1. Identify A. Red vapours of CrO₂Cl₂ form when a chloride is heated with conc. H₂SO₄ and a dichromate. So A is K₂Cr₂O₇: K₂Cr₂O₇ + 4NaCl + 6H₂SO₄ → 2CrO₂Cl₂ + 2KHSO₄ + 4NaHSO₄ + 3H₂O.
2. Identify B. Chromyl chloride vapour passed into NaOH gives yellow sodium chromate: CrO₂Cl₂ + 4NaOH → Na₂CrO₄ + 2NaCl + 2H₂O. So B is Na₂CrO₄.
3. Identify C. Acidifying chromate turns it orange as two chromate ions join and lose water: 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O. So C is Na₂Cr₂O₇, containing the dichromate ion.
4. Count terminal O. In Cr₂O₇²⁻ two CrO₄ tetrahedra share one corner. That one oxygen bridges the two Cr atoms; each Cr carries 3 more oxygens of its own, so 2 × 3 = 6 are terminal.
5. So. Compound C has 7 oxygens in total, of which 1 is bridging and 6 are terminal.

The common slip: 7. Counting all seven oxygens of Cr₂O₇²⁻ gives 7, but the shared Cr–O–Cr oxygen is bridging, not terminal.

The idea to keep. Chromate and dichromate switch with pH. Dichromate is two corner-sharing CrO₄ tetrahedra: 1 bridging O (Cr–O–Cr) and 6 terminal O.

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