In the compound `Yba_(2)Cu_(3)O_(7)` which shows superconductivity, what is the oxidation state of `Cu` ?
Assume that the rare earth element yttrium is in its usual `+3` oxidation state.

Options

A. `+(7)/(3)`
B. `-(7)/(3)`
C. `(5)/(3)`
D. `-(5)/(2)`

Explanation

To find the oxidation state of copper (Cu) in the compound \( YBa_2Cu_3O_7 \), we can follow these steps:

### Step 1: Understand the components of the compound
The compound consists of:
- Yttrium (Y)
- Barium (Ba)
- Copper (Cu)
- Oxygen (O)

### Step 2: Assign known oxidation states
We know from the problem that:
- Yttrium (Y) has an oxidation state of \( +3 \).
- Barium (Ba) typically has an oxidation state of \( +2 \).
- Oxygen (O) typically has an oxidation state of \( -2 \).

### Step 3: Set up the equation for the oxidation states
The overall charge of the compound must equal zero because it is neutral. We can express this as an equation:

\[ \text{Total oxidation state} = (\text{oxidation state of Y}) + 2 \times (\text{oxidation state of Ba}) + 3 \times (\text{oxidation state of Cu}) + 7 \times (\text{oxidation state of O}) = 0 \]

Substituting the known values:

\[ 3 + 2 \times 2 + 3X + 7 \times (-2) = 0 \]

### Step 4: Simplify the equation
Now we simplify the equation:

\[ 3 + 4 + 3X - 14 = 0 \]

Combine the constants:

\[ 3X - 7 = 0 \]

### Step 5: Solve for the oxidation state of Cu
Now, isolate \( X \):

\[ 3X = 7 \]
\[ X = \frac{7}{3} \]

### Conclusion
The oxidation state of copper (Cu) in the compound \( YBa_2Cu_3O_7 \) is \( +\frac{7}{3} \).

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