The oxidising state of molybdenum in its oxo complex species
`[Mo_(2)O_(4)(C_(2)H_(4))_(2)(H_(2)O)]^(2-)` is

Options

A. `+2`
B. `+3`
C. `+4`
D. `+5`

Explanation

To find the oxidation state of molybdenum in the oxo complex species \([Mo_2O_4(C_2H_4)_2(H_2O)]^{2-}\), we can follow these steps:

### Step 1: Assign oxidation states
Let the oxidation state of molybdenum (Mo) be \(X\). Since there are two molybdenum atoms, we will represent their total oxidation state as \(2X\).

### Step 2: Determine oxidation states of other elements
- The oxidation state of oxygen (O) is \(-2\). Since there are 4 oxygen atoms, their total contribution is:
\[ 4 \times (-2) = -8 \]
- The oxidation state of carbon (C) in \(C_2H_4\) is \(0\) because it is a neutral molecule. Therefore, the contribution from \(C_2H_4\) is \(0\).
- The oxidation state of hydrogen (H) in \(H_2O\) is also \(0\) because it is a neutral molecule. Therefore, the contribution from \(H_2O\) is \(0\).

### Step 3: Set up the equation
The overall charge of the complex is \(-2\). Therefore, we can set up the equation as follows:
\[ 2X + (-8) + 0 + 0 = -2 \]

### Step 4: Solve for \(X\)
Now, we simplify the equation:
\[ 2X - 8 = -2 \]
Adding 8 to both sides gives:
\[ 2X = 6 \]
Dividing both sides by 2 gives:
\[ X = 3 \]

### Conclusion
The oxidation state of molybdenum in the complex \([Mo_2O_4(C_2H_4)_2(H_2O)]^{2-}\) is \(+3\).