The compound `Yba_(2)cUo_(7)` has copper in oxidation state Assume that the rare earth element Yttrium is in its usual oxidation state. The value of n is:

Options

A. `3//7`
B. `7//3`
C. 3
D. 7

Explanation

To find the oxidation state of copper in the compound \( YBa_2CuO_7 \), we can follow these steps:

### Step 1: Identify the oxidation states of the known elements
- Yttrium (Y) typically 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 2: Write down the formula and set up the equation
The compound is \( YBa_2CuO_7 \). We can express the total oxidation state of the compound as follows:

\[ \text{Total oxidation state} = \text{Oxidation state of Y} + 2 \times \text{Oxidation state of Ba} + \text{Oxidation state of Cu} + 7 \times \text{Oxidation state of O} \]

Since the compound is neutral, the total oxidation state must equal 0:

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

Where \( x \) is the oxidation state of copper.

### Step 3: Substitute the known values into the equation
Substituting the known oxidation states into the equation gives:

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

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

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

Combine like terms:

\[ x - 7 = 0 \]

### Step 5: Solve for the oxidation state of copper
Now, solve for \( x \):

\[ x = 7 \]

### Conclusion
Thus, the oxidation state of copper in \( YBa_2CuO_7 \) is +7.

### Final Answer
The value of \( n \) is 7.

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