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Number of electrons transferred in each ...

Number of electrons transferred in each case when `KMnO_(4)` acts as an oxidising agent and converts into `MnO_(2), Mn^(2+), Mn(OH)_(3)` and `MnO_(4)^2(-)` are respectively 

A

3, 5, 4 and 1

B

4, 3, 1 and 5

C

1, 3, 4 and 5

D

5, 4, 3 and 1

Text Solution

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The correct Answer is:
To determine the number of electrons transferred when potassium permanganate (KMnO₄) acts as an oxidizing agent and converts into various manganese compounds, we need to analyze the oxidation states of manganese in each case. ### Step-by-Step Solution: 1. **Identify the oxidation state of manganese in KMnO₄:** - In KMnO₄, potassium (K) has an oxidation state of +1, and each oxygen (O) has an oxidation state of -2. - The overall charge of KMnO₄ is neutral (0). - Let the oxidation state of manganese (Mn) be x. - The equation can be set up as follows: \[ +1 + x + 4(-2) = 0 \] \[ +1 + x - 8 = 0 \implies x - 7 = 0 \implies x = +7 \] - Thus, the oxidation state of Mn in KMnO₄ is +7. 2. **Calculate the number of electrons transferred for each product:** a. **Conversion to MnO₂:** - In MnO₂, the oxidation state of Mn is +4. - The change in oxidation state from +7 to +4 involves: \[ 7 - 4 = 3 \text{ electrons} \] b. **Conversion to Mn²⁺:** - In Mn²⁺, the oxidation state of Mn is +2. - The change in oxidation state from +7 to +2 involves: \[ 7 - 2 = 5 \text{ electrons} \] c. **Conversion to Mn(OH)₃:** - In Mn(OH)₃, the oxidation state of Mn is +3. - The change in oxidation state from +7 to +3 involves: \[ 7 - 3 = 4 \text{ electrons} \] d. **Conversion to MnO₄²⁻:** - In MnO₄²⁻, the oxidation state of Mn is +6. - The change in oxidation state from +7 to +6 involves: \[ 7 - 6 = 1 \text{ electron} \] 3. **Summarize the results:** - KMnO₄ to MnO₂: 3 electrons - KMnO₄ to Mn²⁺: 5 electrons - KMnO₄ to Mn(OH)₃: 4 electrons - KMnO₄ to MnO₄²⁻: 1 electron ### Final Answer: - The number of electrons transferred in each case are: - KMnO₄ to MnO₂: **3** - KMnO₄ to Mn²⁺: **5** - KMnO₄ to Mn(OH)₃: **4** - KMnO₄ to MnO₄²⁻: **1**
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