Find the oxidation state of Mn in Mn{{O}_{4}}^{-}.

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

1. **Identify Known Oxidation States**:
- Oxygen (O) generally has an oxidation state of \(-2\).

2. **Set Up the Equation**:
- In \(\text{MnO}_4^{-}\), there are four oxygen atoms. Therefore, the total contribution from the oxygen atoms is:
\[
4 \times (-2) = -8
\]
- Let \(x\) be the oxidation state of manganese (Mn). Since the overall charge of the permanganate ion is \(-1\), we can set up the following equation:
\[
x + (-8) = -1
\]

3. **Solve for \(x\)**:
\[
x - 8 = -1
\]
\[
x = -1 + 8
\]
\[
x = +7
\]

Thus, the oxidation state of manganese (Mn) in \(\text{MnO}_4^{-}\) is \(+7\).

### Summary
- **Oxidation State of Mn**: \(+7\)
- This means that in the permanganate ion, manganese is in a high oxidation state, which is characteristic of its strong oxidizing properties.