Con NaOH

heat

KBr---->X----->Y+Z

H2S04

Oxidation numbers of halogen in the products Y and Z are

1)-1,+1 2)-1,+3

3)-1,+5 4)+1,+5

To tackle the question regarding the reaction of sodium hydroxide (NaOH) with potassium bromide (KBr) and sulfuric acid (H2SO4), we need to analyze the oxidation states of the halogens in the products formed. Let's break down the reaction step by step.

Understanding the Reaction

When NaOH reacts with KBr, it typically leads to the formation of sodium bromide (NaBr) and potassium hydroxide (KOH). However, in the presence of sulfuric acid, we can expect some oxidation-reduction reactions to occur. The key here is to identify the products Y and Z that are formed after the reaction.

Identifying Products Y and Z

In this scenario, we can anticipate that the bromide ion (Br^-) from KBr can be oxidized by sulfuric acid. A common oxidation product of bromide is bromine (Br₂), while the other product could be a bromate ion (BrO₃^-) or similar species, depending on the conditions of the reaction.

Assigning Oxidation States

Now, let’s assign oxidation states to the halogens in the products:

• In bromine (Br₂), each bromine atom has an oxidation state of 0.
• If we consider bromate (BrO₃^-), the oxidation state of bromine can be calculated as follows:

For BrO₃^-:

• Let the oxidation state of Br be x.
• The overall charge of the ion is -1.
• The three oxygen atoms contribute -6 (since each O has an oxidation state of -2).
• Setting up the equation: x + (-6) = -1.
• Solving for x gives us x = +5.

Analyzing the Options

Now, let’s summarize the oxidation states of bromine in the products:

• For product Y (assuming it is Br₂), the oxidation state is 0.
• For product Z (assuming it is BrO₃^-), the oxidation state is +5.

Given the options provided:

• 1) -1, +1
• 2) -1, +3
• 3) -1, +5
• 4) +1, +5

None of the options directly match our findings of 0 and +5. However, if we consider that the bromide ion (Br^-) is being oxidized, we can relate the oxidation states of the reactants to the products. The oxidation state of bromine in the reactant (Br^-) is -1, and in the product (BrO₃^-), it is +5.

Final Thoughts

Thus, the correct answer based on our analysis is option 3: -1 for the bromide ion and +5 for the bromate ion. This reflects the change in oxidation state during the reaction, showcasing the oxidation of bromine from -1 to +5. Understanding these oxidation states is crucial in redox chemistry, as it helps us track the transfer of electrons during chemical reactions.