The oxidation state of Mn in `MnO_4^(2-)` is

Options

A. `+8`
B. `+6`
C. `+7`
D. `+5`

Answer

To determine the oxidation state of manganese (Mn) in the ion \( \text{MnO}_4^{2-} \), we can follow these steps:

### Step 1: Assign oxidation states
Let the oxidation state of manganese be \( x \). The oxidation state of oxygen is typically \(-2\).

### Step 2: Set up the equation
In the ion \( \text{MnO}_4^{2-} \), there are 4 oxygen atoms. Therefore, the total contribution from oxygen is:
\[ 4 \times (-2) = -8 \]

### Step 3: Write the equation based on the total charge
The total charge of the ion \( \text{MnO}_4^{2-} \) is \(-2\). We can set up the equation as follows:
\[ x + (-8) = -2 \]

### Step 4: Solve for \( x \)
Now, we can solve for \( x \):
\[ x - 8 = -2 \]
Adding 8 to both sides gives:
\[ x = -2 + 8 \]
\[ x = 6 \]

### Conclusion
The oxidation state of manganese in \( \text{MnO}_4^{2-} \) is \( +6 \).

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