The oxide `Tl_n Ca_2 Ba_2 Cu_3 O_10` is found to be superconductor at 125 K. The value of n is

Explanation

To find the value of \( n \) in the oxide \( Tl_n Ca_2 Ba_2 Cu_3 O_{10} \), we need to determine the oxidation states of each element in the compound and ensure that the overall charge balances to zero, as it is a neutral compound.

### Step-by-Step Solution:

1. **Identify the oxidation states of each element:**
- **Thallium (Tl)**: Thallium is in Group 13 and typically has an oxidation state of +3. Therefore, for \( n \) thallium atoms, the contribution to the charge is \( +3n \).
- **Calcium (Ca)**: Calcium is in Group 2 and has an oxidation state of +2. Since there are 2 calcium atoms, the contribution is \( 2 \times +2 = +4 \).
- **Barium (Ba)**: Barium is also in Group 2 and has an oxidation state of +2. For 2 barium atoms, the contribution is \( 2 \times +2 = +4 \).
- **Copper (Cu)**: Copper typically has an oxidation state of +2. With 3 copper atoms, the contribution is \( 3 \times +2 = +6 \).
- **Oxygen (O)**: Oxygen typically has an oxidation state of -2. With 10 oxygen atoms, the contribution is \( 10 \times -2 = -20 \).

2. **Set up the equation for the overall charge:**
The sum of the oxidation states must equal zero for the compound to be neutral:
\[ +3n + 4 + 4 + 6 - 20 = 0 \]

3. **Simplify the equation:**
Combine the constants:
\[ +3n + 14 - 20 = 0 \]
This simplifies to:
\[ 3n - 6 = 0 \]

4. **Solve for \( n \):**
Rearranging gives:
\[ 3n = 6 \]
Dividing both sides by 3:
\[ n = \frac{6}{3} = 2 \]

5. **Conclusion:**
The value of \( n \) is 2.

### Final Answer:
The value of \( n \) is 2.