What sorts of informations can you draw from the following reaction ? (CN)2(g) + 2OH–(aq) → CN–(aq) + CNO–(aq)

Exercises · 7.20

Q. What sorts of informations can you draw from the following reaction ? (CN)2(g) + 2OH-(aq) \to CN-(aq) + CNO-(aq) + H2O(l)

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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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This reaction shows the disproportionation of cyanogen $(CN)_2$ in a basic medium — it is simultaneously reduced to $CN^-$ and oxidised to $CNO^-$ . The oxidation number of carbon changes from $+3$ in $(CN)_2$ to $+2$ in $CN^-$ and $+4$ in $CNO^-$ , confirming a redox reaction where the same species acts as both oxidant and reductant.

Concept First — Why This Reaction is Interesting

At first glance, this looks like a simple neutralisation or substitution. But look closer: the same reactant, $(CN)_2$ , gives two different products containing cyanide. That's the hallmark of disproportionation — a reaction where one substance is both oxidised and reduced.

To see this, we track the oxidation numbers of carbon and nitrogen. Cyanogen $(CN)_2$ is like a dimer of cyanide: $N \equiv C - C \equiv N$ . Each carbon is bonded to a nitrogen (more electronegative) and to the other carbon. In $CN^-$ , carbon is bonded only to nitrogen. In $CNO^-$ (cyanate), carbon is bonded to both nitrogen and oxygen.

The key insight: the same element (carbon) changes oxidation state in opposite directions in the two products. That's the definition of disproportionation.

Step-by-Step Analysis

1. Assign oxidation numbers to carbon in each species

We use standard rules:

Oxygen is $-2$ (except in peroxides).
Nitrogen is more electronegative than carbon, so in each $C$ – $N$ bond nitrogen is assigned its usual value of $-3$ (as in nitriles/cyanides), and the oxidation number of carbon in each species follows from that.

Watch out

A common mistake: assume nitrogen always has $-3$ in cyanides. In $CN^-$ , the sum of oxidation numbers equals the charge $-1$ . If N is $-3$ , then C must be $+2$ to give $-1$ . But in $(CN)_2$ , the molecule is neutral, so each $CN$ unit must sum to $0$ . That forces N to be $-3$ and C to be $+3$ in $(CN)_2$ . Check: $+3 + (-3) = 0$ .

Let's verify systematically:

In $(CN)_2$ : molecule is neutral. Each $CN$ unit has net charge $0$ . Let $x$ = oxidation number of C, $y$ = oxidation number of N. For one $CN$ unit: $x + y = 0$ . Since N is more electronegative than C, $y = -3$ (usual for N in nitriles). Then $x = +3$ . So C in $(CN)_2$ is $+3$ .
In $CN^-$ : ion charge $-1$ . So $x + y = -1$ . With $y = -3$ , we get $x = +2$ . So C in $CN^-$ is $+2$ .
In $CNO^-$ (cyanate ion): structure is $[N \equiv C - O]^-$ . Let $x$ = oxidation number of C, $y$ = N, $z$ = O. Oxygen is $-2$ . The sum: $x + y + (-2) = -1$ (charge). So $x + y = +1$ . Nitrogen is more electronegative than C, so $y = -3$ . Then $x = +4$ . So C in $CNO^-$ is $+4$ .

(CN)_2

2. Identify the redox changes

From $(CN)_2$ to $CN^-$ : carbon goes from $+3$ to $+2$ — reduction (gain of electrons).

From $(CN)_2$ to $CNO^-$ : carbon goes from $+3$ to $+4$ — oxidation (loss of electrons).

So the same reactant is both reduced and oxidised. That's disproportionation (also called dismutation).

3. What about nitrogen?

In all three species, nitrogen stays at $-3$ (check: in $CNO^-$ , $y = -3$ as assumed; in $CN^-$ , $y = -3$ ; in $(CN)_2$ , $y = -3$ ). So nitrogen does not change oxidation state. The redox action is entirely on carbon.

4. Balance the half-reactions (optional but instructive)

Reduction half:

$(CN)_2 + 2e^- \rightarrow 2CN^-$

(Each C gains 1 electron, so two C gain 2 electrons.)

Oxidation half:

$(CN)_2 + 4OH^- \rightarrow 2CNO^- + 2H_2O + 2e^-$

(Each C loses 1 electron, so two C lose 2 electrons.)

Adding them:

$(CN)_2 + (CN)_2 + 4OH^- + 2e^- \rightarrow 2CN^- + 2CNO^- + 2H_2O + 2e^-$

Cancel the $2e^-$ and combine the two $(CN)_2$ :

$2(CN)_2 + 4OH^- \rightarrow 2CN^- + 2CNO^- + 2H_2O$ …

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Q.What sorts of informations can you draw from the following reaction ? (CN)2(g) + 2OH–(aq) → CN–(aq) + CNO–(aq) + H2O(l)

Concept First — Why This Reaction is Interesting

Step-by-Step Analysis

1. Assign oxidation numbers to carbon in each species

2. Identify the redox changes

3. What about nitrogen?

4. Balance the half-reactions (optional but instructive)

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