The number of electrons exchanged when `KMnO_(4)` react with `H_(2)O_(2)` is equal to that of

Options

A. s electrons in Ca ( at. no. : 20)
B. p electrons in neon (at. no. : 10)
C. d electrons in chromium ( at. No. : 24)
D. f electron in lanthanum ( at. no. : 57)

Answer

To determine the number of electrons exchanged when KMnO₄ reacts with H₂O₂, we will analyze the reaction step by step.

### Step 1: Write the balanced chemical equation for the reaction in acidic medium.
The reaction of KMnO₄ with H₂O₂ in an acidic medium can be represented as:
\[ 2 \text{KMnO}_4 + 8 \text{H}_2\text{O}_2 + 3 \text{H}_2\text{SO}_4 \rightarrow 2 \text{MnSO}_4 + K_2\text{SO}_4 + 8 \text{H}_2\text{O} + 5 \text{O}_2 \]

### Step 2: Determine the oxidation states of manganese in KMnO₄ and MnSO₄.
- In KMnO₄, the oxidation state of manganese (Mn) can be calculated as follows:
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of Mn be \( x \).
- The equation becomes:
\[ +1 + x + 4(-2) = 0 \]
\[ +1 + x - 8 = 0 \]
\[ x - 7 = 0 \]
\[ x = +7 \]

- In MnSO₄, the oxidation state of Mn is:
- Sulfate (SO₄) has an oxidation state of -2.
- Let the oxidation state of Mn be \( y \).
- The equation becomes:
\[ y - 2 = 0 \]
\[ y = +2 \]

### Step 3: Calculate the change in oxidation state of manganese.
- The change in oxidation state of Mn from KMnO₄ to MnSO₄ is:
\[ +7 \text{ (in KMnO₄)} - +2 \text{ (in MnSO₄)} = 5 \]
Thus, 5 electrons are exchanged when KMnO₄ is reduced to MnSO₄.

### Step 4: Analyze the reaction in alkaline medium (optional).
In alkaline medium, the reaction can be represented as:
\[ 2 \text{KMnO}_4 + 3 \text{H}_2\text{O}_2 + 2 \text{KOH} \rightarrow 2 \text{MnO}_2 + 2 \text{KOH} + 3 \text{O}_2 + 4 \text{H}_2\text{O} \]
- Here, the oxidation state of Mn in MnO₂ is +4.
- The change in oxidation state is:
\[ +7 \text{ (in KMnO₄)} - +4 \text{ (in MnO₂)} = 3 \]
So, 3 electrons are exchanged in alkaline medium.

### Step 5: Compare the number of exchanged electrons with the options given.
- The question asks for the number of electrons exchanged when KMnO₄ reacts with H₂O₂. We found that in acidic medium, 5 electrons are exchanged.
- We compare this with the options:
- S electrons in calcium: 2 (not equal)
- P electrons in neon: 8 (not equal)
- D electrons in chromium: 6 (but chromium has 5 d electrons in its ground state)
- F electrons in lanthanum: 14 (not equal)

### Conclusion:
The number of electrons exchanged when KMnO₄ reacts with H₂O₂ in acidic medium is equal to the d electrons in chromium, which is 5.

**Final Answer:** The number of electrons exchanged when KMnO₄ reacts with H₂O₂ is equal to the d electrons in chromium.

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