The oxidation state of molybdenum in `[Mo_(2)O_(4)(C_(2)H_(4))_(2)H_(2)O_(2)]^(-2)` is ____________

Answer

To determine the oxidation state of molybdenum in the complex \([Mo_{2}O_{4}(C_{2}H_{4})_{2}H_{2}O_{2}]^{-2}\), we can follow these steps:

### Step 1: Identify the components of the complex
The complex consists of:
- 2 Molybdenum (Mo) atoms
- 4 Oxygen (O) atoms (in the form of oxides)
- 2 Ethylene (C₂H₄) molecules
- 2 Hydrogen Peroxide (H₂O₂) molecules

### Step 2: Assign oxidation states to known components
- The oxidation state of oxygen in oxides is typically \(-2\).
- Ethylene (C₂H₄) is a neutral molecule, so its oxidation state is \(0\).
- Hydrogen peroxide (H₂O₂) is also a neutral molecule, so its oxidation state is \(0\).

### Step 3: Set up the equation for the oxidation states
Let the oxidation state of molybdenum be \(x\). The total contribution from the components can be expressed as follows:
- Contribution from 2 Mo: \(2x\)
- Contribution from 4 O: \(4 \times (-2) = -8\)
- Contribution from 2 C₂H₄: \(2 \times 0 = 0\)
- Contribution from 2 H₂O₂: \(2 \times 0 = 0\)

### Step 4: Write the equation based on the total charge
The total charge of the complex is \(-2\). Therefore, we can set up the equation:
\[ 2x - 8 + 0 + 0 = -2 \]

### Step 5: Solve for \(x\)
Rearranging the equation gives:
\[ 2x - 8 = -2 \]
Adding 8 to both sides:
\[ 2x = -2 + 8 \]
\[ 2x = 6 \]
Dividing both sides by 2:
\[ x = 3 \]

### Conclusion
The oxidation state of molybdenum in the complex \([Mo_{2}O_{4}(C_{2}H_{4})_{2}H_{2}O_{2}]^{-2}\) is \(+3\).

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