Compound `YBa_(2)Cu_(3)O_(7)` is a super conductor The O.N. of the copper in the compound will be : [O.No. of Y = `+3`]

Options

A. `+7//3`
B. zero
C. `+2`
D. `+1`

Explanation

To find the oxidation number (O.N.) of copper in the compound \( YBa_2Cu_3O_7 \), we can follow these steps:

### Step 1: Assign known oxidation states
We are given that the oxidation number of \( Y \) (Yttrium) is \( +3 \). Barium (Ba) typically has an oxidation state of \( +2 \).

### Step 2: Set up the equation
Let the oxidation state of copper (Cu) be \( x \). The compound contains:
- 1 Yttrium atom: \( +3 \)
- 2 Barium atoms: \( 2 \times +2 = +4 \)
- 3 Copper atoms: \( 3x \)
- 7 Oxygen atoms: \( 7 \times -2 = -14 \)

### Step 3: Write the overall charge balance equation
The overall charge of the compound is neutral, so we can set up the equation:
\[ (+3) + (+4) + (3x) + (-14) = 0 \]

### Step 4: Simplify the equation
Combine the known values:
\[ 3 + 4 + 3x - 14 = 0 \]
\[ 3x - 7 = 0 \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ 3x = 7 \]
\[ x = \frac{7}{3} \]

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

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