An example of a disproportionation reaction is
(a) 2MnO4- + 10I- + 16H+ - 2Mn2+ + 5I2 + 8H2O
(b) 2N

Q: Why isn't option (c) a disproportionation reaction?

In option (c), $2\text{KMnO}_4 \longrightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2$, Manganese is reduced (from $+7$ to $+6$ and $+4$) and Oxygen is oxidized (from $-2$ to $0$). Since two different elements are undergoing oxidation and reduction, it is an intramolecular redox reaction, not disproportionation.

Q: Can a disproportionation reaction involve multiple reactants?

Yes, it can involve other reactants like water or ions (e.g., $\text{H}^+$ or $\text{OH}^-$) that provide the medium, but the key is that a *single element* from one of the reactants must be both oxidized and reduced.

Q: How do I quickly find the oxidation state of Copper in CuBr?

Halogens like Bromine typically have an oxidation state of $-1$ in halides. Since $\text{CuBr}$ is a neutral molecule, the sum of oxidation states must be zero. Therefore, Copper must be $+1$.

Q: Is disproportionation the opposite of a comproportionation reaction?

Exactly! In a comproportionation reaction, an element in two different oxidation states reacts to form a product where the element is in a single intermediate oxidation state.

Q: Why does Cu+ undergo disproportionation in aqueous solutions?

The $\text{Cu}^+$ ion is unstable in aqueous solutions because the hydration enthalpy of $\text{Cu}^{2+}$ is much higher than that of $\text{Cu}^+$, which compensates for the second ionization enthalpy of Copper. This drives the disproportionation of $\text{Cu}^+$ into $\text{Cu}^{2+}$ and solid $\text{Cu}$.

Question

An example of a disproportionation reaction is

Options:

(A) $2	ext{MnO}_4^- + 10	ext{I}^- + 16	ext{H}^+ \longrightarrow 2	ext{Mn}^{2+} + 5	ext{I}_2 + 8	ext{H}_2	ext{O}$

(B) $2	ext{NaBr} + 	ext{Cl}_2 \longrightarrow 2	ext{NaCl} + 	ext{Br}_2$

(C) $2	ext{KMnO}_4 \longrightarrow 	ext{K}_2	ext{MnO}_4 + 	ext{MnO}_2 + 	ext{O}_2$

(D) $2	ext{CuBr} \longrightarrow 	ext{CuBr}_2 + 	ext{Cu}$

Sigma Percentile · Accepted Answer

### The Dual Nature of Disproportionation

Have you ever seen a movie where the same actor plays both the hero and the villain? In the world of chemistry, we have a special type of reaction that does exactly this. It is called a **disproportionation reaction**.

In a standard redox reaction, one element gets oxidized (loses electrons) while a completely different element gets reduced (gains electrons). But in a disproportionation reaction, a **single element** in a specific oxidation state decides to split its personality. It simultaneously undergoes both oxidation and reduction!

To identify such a reaction, we need to act like detectives. We must assign oxidation states to every element in the reactants and products and look for the one element that goes both up and down in its oxidation number.

### Analyzing the Suspects

Let's investigate our first suspect, option (a): 
$$2	ext{MnO}_4^- + 10	ext{I}^- + 16	ext{H}^+ \longrightarrow 2	ext{Mn}^{2+} + 5	ext{I}_2 + 8	ext{H}_2	ext{O}$$

Here, Manganese ($	ext{Mn}$) starts at an oxidation state of $+7$ in the permanganate ion and reduces to $+2$. Meanwhile, Iodine ($	ext{I}$) goes from $-1$ to $0$, which is an oxidation. Since two different elements are involved, this is just a classic redox reaction.

Moving on to option (b): 
$$2	ext{NaBr} + 	ext{Cl}_2 \longrightarrow 2	ext{NaCl} + 	ext{Br}_2$$

In this case, Bromine ($	ext{Br}$) oxidizes from $-1$ to $0$, and Chlorine ($	ext{Cl}$) reduces from $0$ to $-1$. Again, different elements are changing their states. This is a standard displacement reaction where Chlorine kicks out Bromine.

Now, let's look at option (c): 
$$2	ext{KMnO}_4 \longrightarrow 	ext{K}_2	ext{MnO}_4 + 	ext{MnO}_2 + 	ext{O}_2$$

This is the thermal decomposition of potassium permanganate. Manganese reduces from $+7$ to both $+6$ and $+4$. Oxygen, on the other hand, oxidizes from $-2$ to $0$. Even though it's happening within the same molecule, the elements undergoing oxidation and reduction are different. So, it's an intramolecular redox reaction, not disproportionation.

### The True Disproportionation

Finally, we arrive at option (d): 
$$2	ext{CuBr} \longrightarrow 	ext{CuBr}_2 + 	ext{Cu}$$

Let's focus our magnifying glass on the Copper ($	ext{Cu}$) atom. In the reactant, Copper(I) bromide ($	ext{CuBr}$), Bromine has its typical $-1$ oxidation state. To balance this, Copper must be in a $+1$ oxidation state.

Now, let's look at the products. In Copper(II) bromide ($	ext{CuBr}_2$), Copper is bonded to two Bromine atoms, giving it an oxidation state of $+2$. Going from $+1$ to $+2$ means Copper has lost an electron. **This is oxidation!**

But wait, there's another product! We also have pure elemental Copper ($	ext{Cu}$). Any element in its pure, uncombined state has an oxidation state of $0$. Going from $+1$ to $0$ means Copper has gained an electron. **This is reduction!**

### The Final Verdict

We have found our culprit! The exact same element, Copper, starting from a $+1$ state, has simultaneously oxidized to $+2$ and reduced to $0$. This perfectly satisfies the condition for a disproportionation reaction.

Therefore, the correct answer is option (d). It is a beautiful example of how a single species can act as both an oxidizing and a reducing agent for itself!

Animated Solution for Chemistry - Redox Reactions: An example of a disproportionation reaction is

Select Answer:

Visualized Solution

Disproportionation Reaction

Analyzing Option (a)

Analyzing Option (b)

Analyzing Option (c)

Analyzing Option (d)

Oxidation of $Cu$

Reduction of $Cu$

Conclusion

The Sigma Insight: Oxidation and Reduction

The Dual Nature of Disproportionation

Analyzing the Suspects

The True Disproportionation

The Final Verdict

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