Question:
Which of the following is disproportionation reaction?
Which of the following is disproportionation reaction?
Show Hint
$\text{XeF}_2$ undergoing hydrolysis behaves as a reducing agent, $\text{XeF}_6$ undergoes non-redox reactions, and $\text{XeF}_4$ is the only one that disproportionates.
This is an important distinguishing reaction of xenon tetrafluoride.
This is an important distinguishing reaction of xenon tetrafluoride.
Updated On: Jul 22, 2026
- Complete hydrolysis of $\text{XeF}_6$
- Complete hydrolysis of $\text{XeF}_4$
- Complete hydrolysis of $\text{XeF}_2$
- Partial hydrolysis of $\text{XeF}_6$
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The question is from noble gas chemistry.
We need to determine which of the given Xenon fluoride hydrolysis reactions is a disproportionation reaction.
Step 2: Key Formula or Approach:
A disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized (its oxidation state increases) and reduced (its oxidation state decreases).
Step 3: Detailed Explanation:
• Let us write and analyze the chemical equations for the options:
• (A) Complete hydrolysis of $\text{XeF}_6$:
\[ \text{XeF}_6 + 3\text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6\text{HF} \] Oxidation state of Xe in $\text{XeF}_6$ is +6, and in $\text{XeO}_3$ it is also +6.
No redox process occurs here; it is a simple non-redox hydrolysis.
• (B) Complete hydrolysis of $\text{XeF}_4$:
\[ 6\text{XeF}_4 + 12\text{H}_2\text{O} \rightarrow 4\text{Xe} + 2\text{XeO}_3 + 24\text{HF} + 3\text{O}_2 \] Oxidation state of Xe in reactant $\text{XeF}_4$ is +4.
In the products:
For elemental Xe, the oxidation state is 0 (reduction: +4 to 0).
For $\text{XeO}_3$, the oxidation state of Xe is +6 (oxidation: +4 to +6).
Since the same Xenon atom is both reduced to $\text{Xe}(0)$ and oxidized to $\text{Xe}(+6)$, this is a disproportionation reaction.
• (C) Complete hydrolysis of $\text{XeF}_2$:
\[ 2\text{XeF}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Xe} + 4\text{HF} + \text{O}_2 \] Oxidation state of Xe decreases from +2 (in $\text{XeF}_2$) to 0 (in Xe).
Oxidation state of Oxygen increases from -2 (in $\text{H}_2\text{O}$) to 0 (in $\text{O}_2$).
This is a redox reaction, but not a disproportionation, as different elements undergo oxidation and reduction.
• (D) Partial hydrolysis of $\text{XeF}_6$:
\[ \text{XeF}_6 + \text{H}_2\text{O} \rightarrow \text{XeOF}_4 + 2\text{HF} \] The oxidation state of Xe remains +6 throughout. This is a non-redox reaction.
Step 4: Final Answer:
The complete hydrolysis of $\text{XeF}_4$ is a disproportionation reaction.
The question is from noble gas chemistry.
We need to determine which of the given Xenon fluoride hydrolysis reactions is a disproportionation reaction.
Step 2: Key Formula or Approach:
A disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized (its oxidation state increases) and reduced (its oxidation state decreases).
Step 3: Detailed Explanation:
• Let us write and analyze the chemical equations for the options:
• (A) Complete hydrolysis of $\text{XeF}_6$:
\[ \text{XeF}_6 + 3\text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6\text{HF} \] Oxidation state of Xe in $\text{XeF}_6$ is +6, and in $\text{XeO}_3$ it is also +6.
No redox process occurs here; it is a simple non-redox hydrolysis.
• (B) Complete hydrolysis of $\text{XeF}_4$:
\[ 6\text{XeF}_4 + 12\text{H}_2\text{O} \rightarrow 4\text{Xe} + 2\text{XeO}_3 + 24\text{HF} + 3\text{O}_2 \] Oxidation state of Xe in reactant $\text{XeF}_4$ is +4.
In the products:
For elemental Xe, the oxidation state is 0 (reduction: +4 to 0).
For $\text{XeO}_3$, the oxidation state of Xe is +6 (oxidation: +4 to +6).
Since the same Xenon atom is both reduced to $\text{Xe}(0)$ and oxidized to $\text{Xe}(+6)$, this is a disproportionation reaction.
• (C) Complete hydrolysis of $\text{XeF}_2$:
\[ 2\text{XeF}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Xe} + 4\text{HF} + \text{O}_2 \] Oxidation state of Xe decreases from +2 (in $\text{XeF}_2$) to 0 (in Xe).
Oxidation state of Oxygen increases from -2 (in $\text{H}_2\text{O}$) to 0 (in $\text{O}_2$).
This is a redox reaction, but not a disproportionation, as different elements undergo oxidation and reduction.
• (D) Partial hydrolysis of $\text{XeF}_6$:
\[ \text{XeF}_6 + \text{H}_2\text{O} \rightarrow \text{XeOF}_4 + 2\text{HF} \] The oxidation state of Xe remains +6 throughout. This is a non-redox reaction.
Step 4: Final Answer:
The complete hydrolysis of $\text{XeF}_4$ is a disproportionation reaction.
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