The oxidation state of chromium in the final product formed in the reaction between KI and acidified $K_2Cr_2O_7$ solution is:

Question

The type of oxide formed by the element among Li, Na, Be, Mg, B and Al that has the least atomic radius is:

Answer 1

To solve this question, we need to determine the oxidation state of chromium in the final product formed from the reaction between potassium iodide (KI) and acidified potassium dichromate ($ K_2Cr_2O_7 $). The reaction occurs in an acidic condition, generally using sulfuric acid.

The balanced chemical equation for the reaction is:

\[ Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2 \]

In this reaction, the dichromate ion ($ Cr_2O_7^{2-} $) is reduced to chromium ions ($ Cr^{3+} $). Here, the relevant oxidation state changes occur:

In $ K_2Cr_2O_7 $, the oxidation state of chromium is +6. This is because the chemical formula of the dichromate ion, $ Cr_2O_7^{2-} $, can be analyzed: each $\text{O}^-$ is -2 and altogether they contribute -14, hence each Cr must be +6 to balance the -2 charge.
In the product, each chromium atom in $ Cr^{3+} $ has an oxidation state of +3.

Thus, the oxidation state of chromium in the final product is +3, which corresponds to the correct answer.

Let's analyze why the oxidation state is +3:

The dichromate ion is reduced, and simultaneously, the iodide ions ($ I^- $) are oxidized to iodine ($ I_2 $). The overall process is a redox reaction.
In a redox reaction, the oxidation state of the reduced species decreases. As $ Cr^{6+} $ is reduced to $ Cr^{3+} $, this indicates a decrease in the oxidation number, confirming the final oxidation state of chromium as +3.

Therefore, the oxidation state of chromium in the final product is +3.

Answer 2

To solve this question, we need to determine the oxidation state of chromium in the final product formed from the reaction between potassium iodide (KI) and acidified potassium dichromate ($ K_2Cr_2O_7 $). The reaction occurs in an acidic condition, generally using sulfuric acid.

The balanced chemical equation for the reaction is:

\[ Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2 \]

In this reaction, the dichromate ion ($ Cr_2O_7^{2-} $) is reduced to chromium ions ($ Cr^{3+} $). Here, the relevant oxidation state changes occur:

In $ K_2Cr_2O_7 $, the oxidation state of chromium is +6. This is because the chemical formula of the dichromate ion, $ Cr_2O_7^{2-} $, can be analyzed: each $\text{O}^-$ is -2 and altogether they contribute -14, hence each Cr must be +6 to balance the -2 charge.
In the product, each chromium atom in $ Cr^{3+} $ has an oxidation state of +3.

Thus, the oxidation state of chromium in the final product is +3, which corresponds to the correct answer.

Let's analyze why the oxidation state is +3:

The dichromate ion is reduced, and simultaneously, the iodide ions ($ I^- $) are oxidized to iodine ($ I_2 $). The overall process is a redox reaction.
In a redox reaction, the oxidation state of the reduced species decreases. As $ Cr^{6+} $ is reduced to $ Cr^{3+} $, this indicates a decrease in the oxidation number, confirming the final oxidation state of chromium as +3.

Therefore, the oxidation state of chromium in the final product is +3.

The oxidation state of chromium in the final product formed in the reaction between KI and acidified $K_2Cr_2O_7$ solution is:

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The Correct Option is A

Solution and Explanation

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