Refer to the periodic table given in your book and now answer the following questions: (a) Select the possible

Exercises · 7.24

Q. Refer to the periodic table given in your book and now answer the following questions:

(a) Select the possible non metals that can show disproportionation reaction.

(b) Select three metals that can show disproportionation reaction.

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Concept understanding — Disproportionation Reactions

Disproportionation Reactions: The Self-Oxidation-Reduction

Imagine you have a group of friends who are all equally tall. Now imagine that, for no external reason, some of them suddenly grow taller while others shrink shorter — all starting from the same height. That sounds strange, right? But that's exactly what happens in a disproportionation reaction: the same element, in the same starting oxidation state, simultaneously gets oxidised (loses electrons) and reduced (gains electrons).

In other words, one part of the substance acts as the oxidising agent, and another part acts as the reducing agent — on itself.

The Intuition First

Think of a chemical element like chlorine. In its elemental form ( $Cl_2$ ), each chlorine atom has an oxidation state of 0. Now, if you put chlorine gas into water, something odd happens:

Some chlorine atoms gain an electron (reduction) and become $Cl^-$ (oxidation state -1).
Other chlorine atoms lose an electron (oxidation) and become $ClO^-$ (oxidation state +1).

The same starting material ( $Cl_2$ ) produces two different products — one more reduced, one more oxidised. That's disproportionation.

Note

The word "disproportionation" literally means "not in proportion" — the original uniform state splits into two different states.

The Precise Definition

Important

A disproportionation reaction is a redox reaction in which a single substance (element or compound) in an intermediate oxidation state simultaneously undergoes both oxidation and reduction, producing two different products — one with a higher oxidation state and one with a lower oxidation state.

The key conditions are:

One reactant — the same species is both oxidised and reduced.
Intermediate oxidation state — the starting element must be able to go both higher and lower.
Two products — one more oxidised, one more reduced.

The Classic Example: Chlorine in Water

Cl_2 + H_2O \rightarrow HCl + HOCl

Let's track the oxidation states:

Species	Oxidation state of Cl
$Cl_2$	0
$HCl$	-1 (reduced)
$HOCl$	+1 (oxidised)

The chlorine in $Cl_2$ (oxidation state 0) goes to -1 (gain of electron = reduction) and to +1 (loss of electron = oxidation). The same element, same starting state, two different directions.

Watch out

A common mistake is to think that because $Cl_2$ is a single molecule, it must be either oxidised or reduced. But each chlorine atom in $Cl_2$ can behave differently — one gets oxidised, the other gets reduced. The reaction as a whole is a disproportionation.

How to Spot a Disproportionation Reaction

Look for these three clues:

One reactant, two products — especially if the reactant contains an element that can exist in multiple oxidation states.
The same element appears in two different oxidation states in the products — one higher, one lower than the starting state.
No external oxidising or reducing agent — the substance does it to itself.

More Examples

Hydrogen peroxide decomposition:

2H_2O_2 \rightarrow 2H_2O + O_2

Oxygen in $H_2O_2$ has oxidation state -1. In $H_2O$ , oxygen is -2 (reduced). In $O_2$ , oxygen is 0 (oxidised). The -1 state is intermediate between -2 and 0.

Copper(I) chloride in aqueous solution:

2CuCl \rightarrow Cu + CuCl_2

Copper in $CuCl$ is +1. It goes to 0 (reduced) and +2 (oxidised).

Ammonium nitrite decomposition:

NH_4NO_2 \rightarrow N_2 + 2H_2O

Nitrogen in $NH_4^+$ is -3, in $NO_2^-$ is +3. Both go to 0 in $N_2$ — this is actually the reverse (comproportionation), but it shows how the same element can meet in the middle.

The Opposite: Comproportionation …

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Disproportionation requires an element in an intermediate oxidation state that can both increase and decrease its oxidation number. For non-metals, possible candidates are Cl, S, P, N, and C; for metals, Cu, Hg, and Tl are three that show this behaviour.

The Concept: Why Disproportionation Happens

A disproportionation reaction is a special type of redox reaction where the same element simultaneously undergoes both oxidation (increase in oxidation number) and reduction (decrease in oxidation number). This is only possible if the element exists in an intermediate oxidation state — one that is neither the lowest nor the highest possible for that element.

Think of it like a seesaw: the element in the middle can "give away" electrons to become more positive (oxidised) and also "accept" electrons to become more negative (reduced). The driving force is often the greater stability of the two extreme oxidation states compared to the intermediate one.

+x

Step-by-Step Analysis

1. Understanding the Periodic Table Context

The periodic table in your textbook organises elements by their electron configurations. The key to spotting disproportionation is knowing the common oxidation states of elements. Elements that show multiple stable oxidation states are the ones to watch.

For non-metals, these are typically found in the p-block (Groups 14–17). For metals, transition metals and some post-transition metals (Groups 11–13) are the usual suspects.

2. Identifying Non-Metals That Can Disproportionate

Let's scan the non-metals systematically:

Chlorine (Cl): Has oxidation states from $-1$ to $+7$ . The intermediate states like $+1$ (in $\ce{HClO}$ ), $+3$ (in $\ce{HClO2}$ ), and $+5$ (in $\ce{HClO3}$ ) readily disproportionate. For example:

3\ce{ClO-} \rightarrow 2\ce{Cl-} + \ce{ClO3-}

Here Cl goes from $+1$ to $-1$ (reduction) and $+5$ (oxidation).

Sulphur (S): Shows states from $-2$ to $+6$ . Elemental sulphur ( $0$ ) disproportionates in hot alkali:

3\ce{S} + 6\ce{OH-} \rightarrow 2\ce{S^{2-}} + \ce{SO3^{2-}} + 3\ce{H2O}

Sulphur goes from $0$ to $-2$ and $+4$ .

Phosphorus (P): White phosphorus ( $0$ ) disproportionates in alkali:

\ce{P4} + 3\ce{OH-} + 3\ce{H2O} \rightarrow \ce{PH3} + 3\ce{H2PO2-}

P goes from $0$ to $-3$ and $+1$ .

Nitrogen (N): Has states from $-3$ to $+5$ . $\ce{NO2}$ ( $+4$ ) disproportionates in water:

2\ce{NO2} + \ce{H2O} \rightarrow \ce{HNO3} + \ce{HNO2}

N goes from $+4$ to $+5$ and $+3$ .

Carbon (C): Though less common, carbon in $+2$ (as in CO) can disproportionate:

2\ce{CO} \rightarrow \ce{C} + \ce{CO2}

C goes from $+2$ to $0$ and $+4$ .

Watch out

A common mistake is to think that all non-metals with multiple oxidation states can disproportionate. For example, fluorine (F) has only $-1$ and $0$ — no intermediate state — so it cannot disproportionate. Similarly, oxygen has $-2$ , $-1$ , and $0$ , but the $-1$ state (peroxides) can disproportionate, while elemental oxygen ( $0$ ) cannot.

3. Identifying Metals That Can Disproportionate …

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Q.Refer to the periodic table given in your book and now answer the following questions:

The Concept: Why Disproportionation Happens

Step-by-Step Analysis

1. Understanding the Periodic Table Context

2. Identifying Non-Metals That Can Disproportionate

3. Identifying Metals That Can Disproportionate …

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