The oxidation number of an element in a compound decides its nature to acts as oxidant or reductant. Oxidation number is defined and the residual charge which an atom has or appears to have in a molecule whenal other atoms are removed from the molecule as ions. Oxidation number is frequently used interchangebly with oxidation state. The stock notations of oxidation number are based on the periodic property-electronegativity. An atom in a molecule can be assigned positive, negative or zero oxidation number by considering its environment. In few cases, oxidation number can evenn be fractional.
Oxidation number of Y in `YBa_(2)Cu_(2)O_(7)` is +3, then oxidation number of Cu is:

Options

A. `+7//3`
B. `+5//3`
C. `+2`
D. `+1`

Answer

To find the oxidation number of Cu in the compound \( YBa_2Cu_2O_7 \), we can follow these steps:

### Step 1: Identify the known oxidation states
We know the oxidation state of Y is +3 and that of Ba (Barium) is +2. The oxidation state of oxygen (O) is typically -2.

### Step 2: Write the formula for the total oxidation states
The compound has the following components:
- 1 Y atom: +3
- 2 Ba atoms: \( 2 \times +2 = +4 \)
- 2 Cu atoms: Let the oxidation state of Cu be \( x \).
- 7 O atoms: \( 7 \times -2 = -14 \)

### Step 3: Set up the equation
The sum of the oxidation states in the compound must equal 0 because the compound is neutral. Therefore, we can write the equation:
\[ (+3) + (+4) + (2x) + (-14) = 0 \]

### Step 4: Simplify the equation
Combine the constants:
\[ 3 + 4 - 14 + 2x = 0 \]
This simplifies to:
\[ -7 + 2x = 0 \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ 2x = 7 \]
Dividing both sides by 2:
\[ x = \frac{7}{2} = 3.5 \]

### Conclusion
The oxidation number of Cu in \( YBa_2Cu_2O_7 \) is +3.5.

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