The oxidation states of the most electronegative element in the products of the reaction between `BaO_2` with dilute `H_2SO_4` are

Options

A. `-1`
B. `+1`
C. `-2`
D. `0`

Explanation

To determine the oxidation states of the most electronegative element in the products of the reaction between barium peroxide (BaO₂) and dilute sulfuric acid (H₂SO₄), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are barium peroxide (BaO₂) and dilute sulfuric acid (H₂SO₄).

### Step 2: Write the Reaction Equation
When barium peroxide reacts with dilute sulfuric acid, the reaction can be represented as:
\[ \text{BaO}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + \text{H}_2\text{O} + \text{O}_2 \]

### Step 3: Identify the Products
The products of this reaction are barium sulfate (BaSO₄), water (H₂O), and oxygen gas (O₂).

### Step 4: Determine the Most Electronegative Element
In the products, we need to identify the most electronegative element. The most electronegative elements in the products are oxygen (O) and sulfur (S). However, oxygen is more electronegative than sulfur.

### Step 5: Assign Oxidation States
Now, let's assign oxidation states to oxygen in the products:
- In water (H₂O), the oxidation state of oxygen is -2.
- In barium sulfate (BaSO₄), the oxidation state of oxygen is also -2.
- In molecular oxygen (O₂), the oxidation state of oxygen is 0.

### Step 6: Conclusion
The oxidation state of the most electronegative element (oxygen) in the products is -2 in both BaSO₄ and H₂O, and 0 in O₂. Therefore, the oxidation states of the most electronegative element in the products of the reaction are -2 and 0.

### Final Answer
The oxidation states of the most electronegative element in the products of the reaction between BaO₂ and dilute H₂SO₄ are -2 (in BaSO₄ and H₂O) and 0 (in O₂).

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