Q.What is meant by 'disproportionation' of an oxidation state? Give an example.

Concept: Disproportionation Reaction

A disproportionation reaction occurs when a particular oxidation state of an element becomes less stable relative to two OTHER oxidation states of the same element — one higher, one lower — so it simultaneously undergoes oxidation and reduction.

Reasoning

1. Identify an oxidation state that is genuinely less stable than states on either side of it.
2. Confirm that the same element ends up in a higher (oxidised) state and a lower (reduced) state in the products.
3. NCERT's own example: manganese in the $+6$ state (manganate, $\text{MnO}_4^{2-}$) is unstable relative to manganese in $+7$ (permanganate, $\text{MnO}_4^-$) and $+4$ (manganese dioxide, $\text{MnO}_2$) in acidic solution:

Two Mn($+6$) are oxidised to Mn($+7$); one Mn($+6$) is reduced to Mn($+4$). …

The Core Idea — Why an Element Would "Attack Itself"

A disproportionation reaction is one of the most elegant concepts in redox chemistry. It happens when an element in one oxidation state splits into two different oxidation states — one higher, one lower. The same species acts as both the oxidising agent and the reducing agent.

Think of it as a chemical "civil war": the intermediate oxidation state is unstable under the given conditions, so it rearranges itself into more stable forms. The driving force is always thermodynamic — the free energy change for the overall reaction must be negative.

Step-by-Step Reasoning

1. Recognise the pattern.

In disproportionation, the same element appears in three oxidation states in the reaction: the starting state (intermediate), a higher state, and a lower state. The starting state must be thermodynamically unstable with respect to the other two.

2. Check the condition for spontaneity.

For a reaction to disproportionate, the standard electrode potential for the reduction of the intermediate to the lower state must be more positive than the potential for its oxidation to the higher state. In other words, the intermediate is both a stronger oxidising agent (gets reduced) and a stronger reducing agent (gets oxidised) than its neighbours — which sounds contradictory, but happens when the intermediate is "out of place" on the stability scale.

3. The classic example: Copper(I) in aqueous solution.

Copper has common oxidation states $0$, $+1$, and $+2$. In aqueous acidic solution, $\ce{Cu+}$ is unstable. It disproportionates:

Let's verify:

• One $\ce{Cu+}$ is oxidised to $\ce{Cu^2+}$ (loses an electron).
• The other $\ce{Cu+}$ is reduced to $\ce{Cu}$ (gains an electron).

The standard potentials tell the story:

For the disproportionation, the overall cell potential is $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.521 - 0.153 = +0.368\ \text{V}$. A positive $E^\circ$ means the reaction is spontaneous. …

Disproportionation Reaction — Concept & Method

What is Disproportionation?

Disproportionation is a redox reaction where the same element in one oxidation state simultaneously undergoes both oxidation and reduction — forming two different products with higher and lower oxidation states.

Key idea: One oxidation state splits into two — one goes up, one goes down.

Method: Oxidation Number Tracking Method

Step 1 — Identify the element that changes oxidation state.

Step 2 — Write the initial oxidation state.

Step 3 — Write the oxidation states of the products.

Step 4 — Check: one product has a higher oxidation state (oxidation), the other has a lower oxidation state (reduction).

Step 5 — Confirm the element is the same in reactant and both products.

Example: Disproportionation of Copper(I) in Aqueous Solution

Reaction:

Step-by-step using the method:

Here is a breakdown of the common mistakes students make on the concept of Disproportionation Reactions, along with how to avoid each one.

1. Confusing the Definition (The "Why" is Key)

The Mistake: Students often memorize the definition as "a reaction where the same element undergoes both oxidation and reduction," but they fail to understand why this happens. They then apply the term to any reaction where an element appears on both sides.

How to Avoid It: Understand the core condition: The element must exist in a single oxidation state on the reactant side and split into two different oxidation states (one higher, one lower) on the product side.

• Think of it as "self-oxidation-reduction." The same species is both the oxidizing agent and the reducing agent.
• The "why" is usually because the intermediate oxidation state is unstable in that specific medium (e.g., aqueous solution).

Example: In $3 \text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}$, the Mn in $\text{MnO}_4^{2-}$ (oxidation state +6) is the only reactant. It goes to +7 (in $\text{MnO}_4^-$) and +4 (in $\text{MnO}_2$). This is disproportionation.

2. Forgetting the "Same Element" Rule

The Mistake: Students see a reaction like $\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O}$ and correctly identify it as disproportionation. But then they incorrectly label a reaction like $\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$ as disproportionation because "oxidation and reduction are happening."

How to Avoid It: Check the element. In the Zn/Cu reaction, Zn is oxidized (0 → +2) and Cu is reduced (+2 → 0). These are two different elements. Disproportionation requires the same element to undergo both changes.

• Quick Check: Ask yourself: "Is the same atom (e.g., Cl, Mn, Cu) appearing in two different oxidation states in the products, starting from one oxidation state in the reactants?" If yes, it's disproportionation.

3. Misidentifying the Oxidation States (The Math Error)

The Mistake: Students calculate oxidation states incorrectly, especially in polyatomic ions like $\text{MnO}_4^{2-}$ or $\text{Cu}_2\text{O}$. They might think Mn in $\text{MnO}_4^{2-}$ is +7 (like in permanganate) instead of +6.

How to Avoid It: Always calculate the oxidation state using the standard rules.

• Rule: Oxygen is usually -2. Hydrogen is usually +1. The sum of oxidation states in a neutral compound is 0; in an ion, it equals the charge.
• Example: For $\text{MnO}_4^{2-}$:
  • Let Mn = $x$.
  • $x + 4(-2) = -2$
  • $x - 8 = -2$
  • $x = +6$ (This is the key intermediate state).
• Practice: Do this calculation for every element in the reaction before deciding if it's disproportionation.

4. Giving a Wrong or Incomplete Example

The Mistake: Students give an example that is not a disproportionation reaction (e.g., $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$) or they give a correct example but fail to show the oxidation states.

How to Avoid It: Memorize one or two classic, foolproof examples and write them with the oxidation states clearly labeled.

• Best Example: Decomposition of Hydrogen Peroxide:

- **O in $\text{H}_2\text{O}_2$:** $-1$ (the intermediate state)
- **O in $\text{H}_2\text{O}$:** $-2$ (reduced)
- **O in $\text{O}_2$:** $0$ (oxidized)
- **Why it works:** The -1 state of oxygen is unstable.

• Another Good Example: Disproportionation of Copper(I): …