Question

The oxidation states of the most electronegative element in the products of the reaction, \(\mathrm{BaO}_{2}\) with diluted \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are: (a) 0 and \(-1\) (b) \(-1\) and \(-2\) (c) \(-2\) and 0 (d) \(-2\) and \(+1\)

Short answer

The oxidation states are -1 and -2.

Step-by-step solution

Understand the Reaction

Write down the reaction of barium peroxide \(\textrm{BaO}_2\) with diluted sulfuric acid \(\textrm{H}_2\textrm{SO}_4\). The typical reaction occurs as:\[ \textrm{BaO}_2 + \textrm{H}_2\textrm{SO}_4 \rightarrow \textrm{BaSO}_4 + \textrm{H}_2\textrm{O}_2 \] Upon reaction, \(\textrm{BaO}_2\) with \(\textrm{H}_2\textrm{SO}_4\) forms barium sulfate \(\textrm{BaSO}_4\) and hydrogen peroxide \(\textrm{H}_2\textrm{O}_2\). We need to analyze the products to identify the oxidation states of their elements.

Identify Most Electronegative Element

In the reaction products \( \textrm{BaSO}_4 \) and \( \textrm{H}_2\textrm{O}_2 \), find the most electronegative element. Oxygen (O) is the most electronegative element in both \( \textrm{BaSO}_4 \) and \( \textrm{H}_2\textrm{O}_2 \).

Determine the Oxidation States of Oxygen in Products

- In \(\textrm{H}_2\textrm{O}_2\) (hydrogen peroxide), each oxygen has an oxidation state of \(-1\) because the overall molecule is neutral and the oxidation state of hydrogen is \(+1\). Thus, \(2(+1) + 2(O) = 0 \Rightarrow 2O = -2 \Rightarrow O = -1\).- In \(\textrm{BaSO}_4\), the sulfate ion (SO₄²⁻) typically has oxygen in its standard oxidation state of \(-2\), since barium has a \(+2\) charge. \Therefore, oxygen in \(\textrm{BaSO}_4\) has an oxidation state of \(-2\).

Compare and Conclude

Based on the analysis:- In \(\textrm{H}_2\textrm{O}_2\), oxygen has an oxidation state of \(-1\).- In \(\textrm{BaSO}_4\), oxygen has an oxidation state of \(-2\).Thus, the correct option in the given choices where the oxidation states of oxygen are \(-1\) and \(-2\) is option (b).

Key concepts

Electronegativity

Electronegativity is a measure of how strongly an atom attracts electrons in a chemical bond. In simple terms, it's like the atom's desire to "grab" the electrons when it is involved in a bond with another atom.
Oxygen is one of the most electronegative elements, which means it has a strong tendency to pull electrons towards itself in chemical compounds. This property is essential in determining the behavior of oxygen in various chemical reactions, including those we analyze for their oxidation states.
When comparing elements like sulfur, barium, and oxygen in compounds, oxygen ends up being the most electronegative. This characteristic leads oxygen to commonly exist in negative oxidation states, as seen in the products of the barium peroxide and sulfuric acid reaction.

Chemical Reactions Analysis

To analyze chemical reactions, we need to understand the transformation of reactants into products. The process involves breaking old bonds and forming new ones, resulting in the generation of new substances, some of which may have different properties from the original reactants.
Analyzing the reaction of barium peroxide (\(\textrm{BaO}_2\)) with diluted sulfuric acid (\(\textrm{H}_2\textrm{SO}_4\)), we observe that it results in barium sulfate (\(\textrm{BaSO}_4\)) and hydrogen peroxide (\(\textrm{H}_2\textrm{O}_2\)).

• First, identify the chemical species before and after the reaction.
• Determine the changes in oxidation states, especially for key elements like oxygen.
• Understanding these changes helps in predicting how reactions proceed and the stability of the new compounds formed.

Barium Peroxide Reactions

Barium peroxide (\(\textrm{BaO}_2\)) is an interesting compound often used in fireworks for its ability to release oxygen. It reacts with various acids, such as sulfuric acid, to produce different substances, a process useful in understanding and determining oxidation states.
When barium peroxide reacts with sulfuric acid, the barium separates to form barium sulfate (\(\textrm{BaSO}_4\)) while the remaining peroxide portion forms hydrogen peroxide (\(\textrm{H}_2\textrm{O}_2\)).
This behavior highlights the phenomenon where compounds containing peroxide ions (\(\textrm{O}_2^{2-}\)) undergo such reactions, resulting in a product capable of further reactions due to the presence of unstable bonds like those in hydrogen peroxide.

Oxidation State Calculation

Oxidation state calculations help us understand the electron distribution among atoms in a compound by assigning a numerical value to each element. This value indicates the number of electrons lost or gained by an atom.
In the reaction between barium peroxide (\(\textrm{BaO}_2\)) and sulfuric acid (\(\textrm{H}_2\textrm{SO}_4\)), determining the oxidation states of oxygen in the products is crucial for analyzing the chemical's behavior.

• In hydrogen peroxide (\(\textrm{H}_2\textrm{O}_2\)), oxygen has an oxidation state of \(-1\), balancing the \(+1\) charge of hydrogen.
• In barium sulfate (\(\textrm{BaSO}_4\)), oxygen's more familiar oxidation state of \(-2\) is present, as it balances barium's \(+2\) charge coupled with the sulfate ion.
• These calculations help us recognize the electron transfer occurring in the reaction and confirm the transformation of reactants into their respective products.

Understanding this concept provides clarity on the movement of electrons during reactions and the stabilization of resulting compounds.