The number of terminal oxygen atoms present in the product B obtained from the following reaction is — d and f Block Elements Chemistry Question

💡 Solution & Explanation

**Step 1: Balance the first reaction to identify product A** FeCrO₄ + Na₂CO₃ + O₂ → A + Fe₂O₃ + CO₂ The chromium is oxidized from CrO₄²⁻ to a higher oxidation state. With oxygen present, CrO₄²⁻ oxidizes to CrO₄²⁻ in sodium chromate (Na₂CrO₄). Product A = Na₂CrO₄. **Step 2: Balance the first equation** 2FeCrO₄ + 2Na₂CO₃ + ½O₂ → 2Na₂CrO₄ + Fe₂O₃ + 2CO₂ **Step 3: Identify the second reaction** A + H₂SO₄ → B + H₂O + Na₂SO₄ Substituting A = Na₂CrO₄: Na₂CrO₄ + H₂SO₄ → B + H₂O + Na₂SO₄ **Step 4: Determine product B** In acidic conditions, chromate converts to dichromate: 2Na₂CrO₄ + H₂SO₄ → Cr₂O₇²⁻ + H₂O + Na₂SO₄ Product B = Cr₂O₇²⁻ (dichromate ion) **Step 5: Count terminal oxygen atoms in Cr₂O₇²⁻** The dichromate ion structure has: - Two CrO₄ tetrahedra sharing one oxygen - Terminal oxygens (not bridging): 6 oxygen atoms bonded to Cr but not shared Therefore, the answer is **6.00**.