For an element 'X' the process of oxidation is:
`X_(@)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, we need to determine the oxidation state of element 'X' in the new compound formed after the oxidation process. Here are the steps to find the solution:

### Step 1: Understand the given information
We have the compound \( X_2O_4^{2-} \) undergoing oxidation. We need to find the oxidation state of 'X' in the new compound formed after oxidation.

### Step 2: Calculate the total charge of the compound
The charge of the compound \( X_2O_4^{2-} \) is -2. The oxidation state of oxygen (O) is typically -2. Therefore, for 4 oxygen atoms:
\[ \text{Total charge from O} = 4 \times (-2) = -8 \]
Let the oxidation state of 'X' be \( n \). Since there are 2 'X' atoms, the total contribution from 'X' is \( 2n \).

### Step 3: Set up the equation for the total charge
The total charge of the compound can be expressed as:
\[ 2n - 8 = -2 \]
Now, we can solve for \( n \):
\[ 2n = -2 + 8 \]
\[ 2n = 6 \]
\[ n = 3 \]
Thus, the oxidation state of 'X' in \( X_2O_4^{2-} \) is +3.

### Step 4: Determine the charge passed during electrolysis
We are given that a current of 965 A is passed for 100 seconds. We can calculate the total charge (Q) using the formula:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 965 \, \text{A} \times 100 \, \text{s} = 96500 \, \text{C} \]

### Step 5: Calculate the amount of substance discharged
The charge required to discharge 0.1 moles of \( X_2O_4^{2-} \) can be calculated using Faraday's laws of electrolysis. The n-factor (number of electrons transferred per mole of substance) for the oxidation of 'X' can be expressed as:
\[ \text{n-factor} = 2(n - 3) \]
where \( n \) is the oxidation state of 'X' in the new compound.

### Step 6: Relate the charge to the n-factor
Using Faraday's law:
\[ Q = n \times \text{n-factor} \times F \]
where \( F \) (Faraday's constant) is approximately 96500 C/mol.

Substituting the values:
\[ 96500 = 0.1 \times 2(n - 3) \times 96500 \]
Dividing both sides by 96500:
\[ 1 = 0.1 \times 2(n - 3) \]
\[ 1 = 0.2(n - 3) \]
Now, solving for \( n \):
\[ n - 3 = \frac{1}{0.2} = 5 \]
\[ n = 5 + 3 = 8 \]

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