`[HXeO_(4)]^(-)+Ohrarr[X]+[Y]+O_(2)+H_(2)O`
The products `[X]` and `[Y]` in unbalanced reaction are:

Options

A. `[XeO_(6)]^(4-)` & `Xe`
B. `[XeO_(6)]^(4-)` & `XeO_(3)`
C. `XeO_(3)` & `Xe`
D. `H_(2)XeO_(4)` & `Xe`

Answer

To solve the reaction `[HXeO_(4)]^(-) + OH^(-) → [X] + [Y] + O_(2) + H_(2)O`, we need to determine the products `[X]` and `[Y]`. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Reactants
The reactants in the equation are:
- `[HXeO_(4)]^(-)`: This is the anion of xenon oxoacid, specifically the xenate ion.
- `OH^(-)`: This is the hydroxide ion.

### Step 2: Analyze the Reaction
When `[HXeO_(4)]^(-)` reacts with `OH^(-)`, it is likely to undergo a redox reaction where the oxidation states of xenon change.

### Step 3: Determine the Products
From the reaction, we can predict the products:
- The xenate ion `[HXeO_(4)]^(-)` can be oxidized further.
- The hydroxide ion can facilitate the release of oxygen gas and water.

### Step 4: Write the Products
Based on the reaction:
1. The xenate ion can be converted into `[XeO_(6)]^(4-)`, which is a higher oxidation state of xenon.
2. Xenon can also be released as a product, which is simply `Xe`.
3. The other products include `O2` (oxygen gas) and `H2O` (water).

Thus, we can summarize the products as:
- `[X] = [XeO_(6)]^(4-)`
- `[Y] = Xe`

### Final Answer
Therefore, the products `[X]` and `[Y]` are:
- **[X] = [XeO_(6)]^(4-)**
- **[Y] = Xe**