The oxidation states of the most electronegative elements in the products of the reaction between `BaO_(2)` and `H_(2)SO_(4)` are

Options

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

Explanation

To determine the oxidation states of the most electronegative elements in the products of the reaction between BaO₂ and H₂SO₄, we will follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
The reaction between barium peroxide (BaO₂) and sulfuric acid (H₂SO₄) produces barium sulfate (BaSO₄) and hydrogen peroxide (H₂O₂). The balanced equation can be written as:

\[ \text{BaO}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + \text{H}_2\text{O}_2 \]

### Step 2: Identify the products of the reaction.
The products of the reaction are:
1. Barium sulfate (BaSO₄)
2. Hydrogen peroxide (H₂O₂)

### Step 3: Determine the oxidation states of the most electronegative element in each product.
The most electronegative element in both products is oxygen (O).

- **For BaSO₄:**
- Barium (Ba) has an oxidation state of +2.
- Sulfate (SO₄) has a charge of -2.
- In sulfate, the oxidation state of sulfur (S) is +6 (since 4 oxygen atoms contribute -8, and the overall charge is -2).
- Therefore, the oxidation state of oxygen in BaSO₄ is -2.

- **For H₂O₂:**
- Hydrogen (H) has an oxidation state of +1.
- Since there are 2 hydrogen atoms, they contribute +2.
- The overall charge of hydrogen peroxide is neutral (0).
- Let the oxidation state of oxygen in H₂O₂ be x. The equation can be set up as:
\[ 2(+1) + 2x = 0 \]
\[ 2 + 2x = 0 \]
\[ 2x = -2 \]
\[ x = -1 \]
- Therefore, the oxidation state of oxygen in H₂O₂ is -1.

### Step 4: Summarize the oxidation states of oxygen in both products.
- In BaSO₄, the oxidation state of oxygen is -2.
- In H₂O₂, the oxidation state of oxygen is -1.

### Final Answer:
The oxidation states of the most electronegative element (oxygen) in the products of the reaction are:
- In BaSO₄: -2
- In H₂O₂: -1