To determine the oxidation state of the most electronegative element in the products of the reaction between $ \text{BaO}_2 $ and dilute $ \text{H}_2\text{SO}_4 $, let's first write down the reaction:

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

In this reaction, $ \text{BaO}_2 $ reacts with $ \text{H}_2\text{SO}_4 $ to produce $ \text{BaSO}_4 $ and $ \text{H}_2\text{O}_2 $.

The most electronegative element present in the products is oxygen.

• In $ \text{BaSO}_4 $, the oxygen is bonded to sulfur. The oxygen here typically has an oxidation state of $ -2 $.


• In $ \text{H}_2\text{O}_2 $ (hydrogen peroxide), the oxygen also has a different oxidation state compared to when it's in water or oxides. In hydrogen peroxide, oxygen has an oxidation state of $ -1 $.

This indicates the oxidation states of the most electronegative element (oxygen) in the products are $ -2 $ and $ -1 $.

Thus, the correct answer is:

Option D: $ -2 $ and $ -1 $