In a mildly alkaline medium, thiosulphate ion is oxidized by MnO to “A”. The oxidation state of sulp — d and f Block Elements Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the oxidation state of sulfur in thiosulfate ion (S₂O₃²⁻)** In S₂O₃²⁻, oxygen is –2. Let x = oxidation state of S. 2x + 3(–2) = –2 2x = 4, so x = +2 (average oxidation state) **Step 2: Determine the oxidation state of Mn in MnO** In MnO, oxygen is –2, so Mn is +2. **Step 3: Analyze the redox reaction in mildly alkaline medium** MnO (+2) is oxidized to permanganate-like species or reduced to MnO₂/Mn(OH)₃. Since MnO is the oxidizing agent, it gets reduced. Thiosulfate (S in +2 state) must be oxidized. **Step 4: Apply electron transfer concept** In mildly alkaline conditions, thiosulfate is oxidized to sulfate (A = SO₄²⁻). In sulfate, the oxidation state of sulfur is: x + 4(–2) = –2 x = +6 **Step 5: Verify the oxidation state change** Sulfur in S₂O₃²⁻: +2 → Sulfur in SO₄²⁻: +6 Change: +6 – (+2) = +4 (loss of 4 electrons per S atom, confirming oxidation) Therefore, the answer is **6.00**.