For an element 'X' the process of oxidation is:
`X_(2)O_(4)^(-2) rarr` New compound
If `965 A` current when passed for `100` seconds discharged `0.1` of `X_(2)O_(4)^(-2)` find oxidation state of `X` in new compound?

Explanation

To solve the problem step by step, we will follow the outlined process to determine the oxidation state of element 'X' in the new compound formed after oxidation.

### Step 1: Understand the Reaction
The oxidation reaction given is:
\[ X_2O_4^{2-} \rightarrow \text{New Compound} \]
Here, we need to find the oxidation state of 'X' in the new compound.

### Step 2: Determine the Oxidation State of 'X' in \( X_2O_4^{2-} \)
The formula for the compound is \( X_2O_4^{2-} \). We can set up the equation for the oxidation state:
Let the oxidation state of 'X' be \( x \).
The oxidation state equation is:
\[ 2x + 4(-2) = -2 \]
\[ 2x - 8 = -2 \]
\[ 2x = 6 \]
\[ x = +3 \]

So, the oxidation state of 'X' in \( X_2O_4^{2-} \) is +3.

### Step 3: Identify the Oxidation Process
Since oxidation involves an increase in oxidation state, the oxidation state of 'X' in the new compound must be greater than +3. Let’s denote the oxidation state of 'X' in the new compound as \( m \).

### Step 4: Calculate the Number of Moles Discharged
According to the problem, a current of 965 A is passed for 100 seconds, discharging 0.1 moles of \( X_2O_4^{2-} \).
Using Faraday's law:
\[ \text{Weight} = \frac{I \cdot T}{F} \]
Where:
- \( I = 965 \, \text{A} \)
- \( T = 100 \, \text{s} \)
- \( F = 96500 \, \text{C/mol} \)

### Step 5: Calculate the Total Charge
The total charge (Q) can be calculated as:
\[ Q = I \cdot T = 965 \, \text{A} \times 100 \, \text{s} = 96500 \, \text{C} \]

### Step 6: Relate Charge to Moles and n-factor
Using the relationship:
\[ \text{Moles} = \frac{Q}{F} \]
We know that 0.1 moles of \( X_2O_4^{2-} \) are discharged. The n-factor can be calculated based on the change in oxidation state:
\[ n = \text{final oxidation state} - \text{initial oxidation state} \]
\[ n = 2m - 6 \]

### Step 7: Set Up the Equation
From Faraday's law:
\[ \text{Moles} = \frac{I \cdot T}{n \cdot F} \]
Substituting the values:
\[ 0.1 = \frac{96500}{(2m - 6) \cdot 96500} \]
This simplifies to:
\[ 0.1 = \frac{1}{2m - 6} \]

### Step 8: Solve for m
Cross-multiplying gives:
\[ 0.1(2m - 6) = 1 \]
\[ 0.2m - 0.6 = 1 \]
\[ 0.2m = 1.6 \]
\[ m = 8 \]

### Conclusion
The oxidation state of 'X' in the new compound is +8.