`XeF_(4)` on reaction with `H_(2)` gives

Options

A. `Xe`
B. `HF`
C. `XeF_(2)`
D. `XeF_(6)`

Explanation

To solve the question regarding the reaction of \( \text{XeF}_4 \) with \( \text{H}_2 \), we can follow these steps:

### Step 1: Write the Reactants
The reactants in this reaction are xenon tetrafluoride (\( \text{XeF}_4 \)) and hydrogen gas (\( \text{H}_2 \)).

### Step 2: Identify the Products
When \( \text{XeF}_4 \) reacts with \( \text{H}_2 \), it can produce xenon (\( \text{Xe} \)) and hydrofluoric acid (\( \text{HF} \)).

### Step 3: Write the Unbalanced Chemical Equation
The unbalanced equation for the reaction can be written as:
\[ \text{XeF}_4 + \text{H}_2 \rightarrow \text{Xe} + \text{HF} \]

### Step 4: Balance the Equation
Now, we need to balance the equation.
- On the left side, we have 4 fluorine atoms from \( \text{XeF}_4 \), so we need 4 \( \text{HF} \) on the right side to balance the fluorine.
- This means we will have 4 hydrogen atoms on the right side, so we need 2 \( \text{H}_2 \) on the left side to balance the hydrogen.

The balanced equation will be:
\[ \text{XeF}_4 + 2 \text{H}_2 \rightarrow \text{Xe} + 4 \text{HF} \]

### Step 5: Identify the Final Products
From the balanced equation, we can see that the products formed are xenon (\( \text{Xe} \)) and hydrofluoric acid (\( \text{HF} \)).

### Conclusion
Thus, the reaction of \( \text{XeF}_4 \) with \( \text{H}_2 \) gives xenon and hydrofluoric acid.

### Final Answer
The products formed are \( \text{Xe} \) and \( \text{HF} \).

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