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 determine the oxidation state of copper (Cu) in the compound YBa2CuO7, we can follow these steps:

### Step 1: Identify the oxidation states of the other 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 the oxidation state equation
The compound YBa2CuO7 can be broken down into its components:
- The total contribution from Yttrium (Y) is +3.
- The total contribution from Barium (Ba) is 2 × (+2) = +4 (since there are 2 Ba atoms).
- The total contribution from Oxygen (O) is 7 × (-2) = -14 (since there are 7 O atoms).

### Step 3: Set up the equation
The overall charge of the compound must equal zero. Therefore, we can set up the equation as follows:

\[ \text{Total oxidation state} = \text{Oxidation state of Y} + \text{Oxidation state of Ba} + \text{Oxidation state of Cu} + \text{Oxidation state of O} = 0 \]

Substituting the known values into the equation:

\[ (+3) + (+4) + n + (-14) = 0 \]

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

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

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

Combine the constants:

\[ n - 7 = 0 \]

### Step 5: Solve for n
Now, we can solve for \( n \):

\[ n = 7 \]

### Conclusion
The oxidation state of copper (Cu) in the compound YBa2CuO7 is +7.

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