The oxidation states of transition metal atoms in K Cr O , KMnO and K FeO , respectively, are x, y a — Redox Reactions and Volumetric Analysis Chemistry Question
Question
The oxidation states of transition metal atoms in K Cr O , KMnO and K FeO , respectively, are x, y and z. The sum of x, y and z is ……….. . 2 2 7 4 2 4
💡 Solution & Explanation
**Step 1: Determine oxidation state of Cr in K₂Cr₂O₇ (x)** Use the formula: Sum of oxidation states = charge of compound For K₂Cr₂O₇: - K: +1 (group 1 metal) - O: -2 (oxygen) - Cr: unknown (x) 2(+1) + 2(x) + 7(-2) = 0 2 + 2x - 14 = 0 2x = 12 **x = 6** **Step 2: Determine oxidation state of Mn in KMnO₄ (y)** For KMnO₄: - K: +1 - O: -2 - Mn: unknown (y) (+1) + y + 4(-2) = 0 1 + y - 8 = 0 **y = 7** **Step 3: Determine oxidation state of Fe in K₂FeO₄ (z)** For K₂FeO₄: - K: +1 - O: -2 - Fe: unknown (z) 2(+1) + z + 4(-2) = 0 2 + z - 8 = 0 **z = 6** **Step 4: Calculate the sum** x + y + z = 6 + 7 + 6 = **19** Therefore, the answer is 19.00.