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 compound \( Tl_n Ca_2 Ba_2 Cu_3 O_{10} \), we need to ensure that the overall charge of the compound is neutral. Let's break down the steps to calculate \( n \).

### Step-by-Step Solution:

1. **Identify the oxidation states of the elements:**
- Thallium (Tl) can have an oxidation state of +3.
- Calcium (Ca) has an oxidation state of +2.
- Barium (Ba) has an oxidation state of +2.
- Copper (Cu) typically has an oxidation state of +2 in this type of compound.
- Oxygen (O) has an oxidation state of -2.

2. **Write the oxidation state contributions:**
- For \( n \) thallium atoms: \( 3n \) (since each Tl contributes +3).
- For 2 calcium atoms: \( 2 \times 2 = 4 \).
- For 2 barium atoms: \( 2 \times 2 = 4 \).
- For 3 copper atoms: \( 3 \times 2 = 6 \) (since each Cu contributes +2).
- For 10 oxygen atoms: \( 10 \times (-2) = -20 \).

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

4. **Simplify the equation:**
Combine like terms:
\[ 3n + 14 - 20 = 0 \]
\[ 3n - 6 = 0 \]

5. **Solve for \( n \):**
\[ 3n = 6 \]
\[ n = \frac{6}{3} = 2 \]

Thus, the value of \( n \) is **2**.