Types of Redox Reactions: Disproportionation and Fractional Oxidation Numbers

In combination reactions, elements come together. In decomposition reactions, compounds fall apart. In displacement reactions, one element kicks another out. But what if a single substance plays both roles at once, acting as both the electron donor and the electron acceptor? That is exactly what happens in disproportionation, the most unusual category of redox reactions.

What Makes a Reaction a Disproportionation?

A disproportionation reaction is a special type of redox reaction in which one element, starting from a single oxidation state, is simultaneously oxidised (pushed to a higher state) and reduced (dropped to a lower state). Think of it as one element splitting its identity in two: part of it goes up on the oxidation-number scale, and part of it goes down.

For this to be possible, the element must be capable of existing in at least three different oxidation states. The element begins in the intermediate state and ends up in both a higher state and a lower state in the products.

Hydrogen Peroxide: The Classic Example

The decomposition of hydrogen peroxide ($H_2O_2$) in water is one of the most familiar disproportionation reactions:

$\overset{+1 \quad -1}{2H_2O_2} \text{ (aq)} \rightarrow \overset{+1 \quad -2}{2H_2O}\text{(l)} + \overset{0}{O_2}\text{(g)} \qquad \text{(7.45)}$

Let us trace what happens to the oxygen atoms:

• In $H_2O_2$, oxygen sits at the $-1$ oxidation state (this is the intermediate state).
• In the product $H_2O$, oxygen has dropped to $-2$. This is reduction (a decrease in oxidation number).
• In the product $O_2$, oxygen has risen to $0$. This is oxidation (an increase in oxidation number).

So the same element, oxygen, begins at one state ($-1$) and splits into two: one portion goes lower ($-2$) and another goes higher ($0$). That is disproportionation in action.

Disproportionation in Alkaline Solutions

Several non-metals undergo disproportionation when placed in an alkaline (basic) medium. Here are three important examples:

Phosphorus in Alkali

$\overset{0}{P_4}\text{(s)} + 3OH^-\text{(aq)} + 3H_2O\text{(l)} \rightarrow \overset{-3}{PH_3}\text{(g)} + 3\overset{+1}{H_2PO_2^-} \text{(aq)} \qquad \text{(7.46)}$

Phosphorus starts at $0$ in $P_4$. It gets reduced to $-3$ in phosphine ($PH_3$) and oxidised to $+1$ in the hypophosphite ion ($H_2PO_2^-$).

Sulphur in Alkali

$\overset{0}{S_8}\text{(s)} + 12 \text{ } OH^- \text{(aq)} \rightarrow \overset{-2}{4S^{2-}} \text{(aq)} + 2\overset{+2}{S_2O_3^{2-}}\text{(aq)} + 6H_2O\text{(l)} \qquad \text{(7.47)}$

Sulphur starts at $0$ in $S_8$. Some sulphur atoms are reduced to $-2$ (as $S^{2-}$) while others are oxidised to $+2$ (in the thiosulphate ion, $S_2O_3^{2-}$).

Chlorine in Alkali: Making Household Bleach

$\overset{0}{Cl_2} \text{(g)} + 2OH^- \text{(aq)} \rightarrow \overset{+1}{ClO^-} \text{(aq)} + \overset{-1}{Cl^-} \text{(aq)} + H_2O \text{(l)} \qquad \text{(7.48)}$

Chlorine begins at $0$ in $Cl_2$, then splits: one chlorine atom is oxidised to $+1$ in the hypochlorite ion ($ClO^-$), and the other is reduced to $-1$ as the chloride ion ($Cl^-$).

This reaction has everyday importance. The hypochlorite ion ($ClO^-$) formed here is a strong oxidising agent that breaks down colour-bearing stains in fabrics, turning them colourless. This is exactly how household bleaching agents work.

Bromine and iodine follow the same pattern as chlorine when they react with alkali. But fluorine is different.

Why Fluorine Does Not Disproportionate

When fluorine reacts with alkali, the products look different:

$2F_2\text{(g)} + 2OH^-\text{(aq)} \rightarrow 2F^-\text{(aq)} + OF_2\text{(g)} + H_2O\text{(l)} \qquad \text{(7.49)}$

(Note: fluorine will also attack the water in this mixture and release some oxygen.)

At first glance, you might wonder if this is disproportionation too. It is not. The reason comes down to a fundamental property: fluorine is the most electronegative element. It can never show a positive oxidation state. Since disproportionation requires the element to go to both a higher and a lower state, and fluorine has nowhere to go above zero, disproportionation is impossible for it.

In the reaction above, fluorine only goes down: from $0$ in $F_2$ to $-1$ in $F^-$. The $OF_2$ product has oxygen at $+2$ and fluorine at $-1$ (because fluorine is more electronegative than oxygen). Fluorine is reduced in every product, never oxidised. This makes fluorine the odd one out among the halogens when it comes to disproportionation.

Disproportionation of Chlorine Oxoanions

Chlorine forms a series of oxoanions (oxygen-containing ions), and whether each one can disproportionate depends on its oxidation state:

Oxoanion	Oxidation state of $Cl$	Can it disproportionate?
$ClO^-$ (hypochlorite)	$+1$	Yes
$ClO_2^-$ (chlorite)	$+3$	Yes
$ClO_3^-$ (chlorate)	$+5$	Yes
$ClO_4^-$ (perchlorate)	$+7$	No

$ClO_4^-$ cannot disproportionate because chlorine is already at its highest possible oxidation state ($+7$). There is no higher state to reach, so the “oxidation half” of disproportionation has no destination.

The disproportionation reactions for the other three are:

Hypochlorite ($ClO^-$, chlorine at $+1$):

$\overset{+1}{3ClO^-} \rightarrow \overset{-1}{2Cl^-} + \overset{+5}{ClO_3^-}$

Chlorine splits from $+1$ into $-1$ (reduced) and $+5$ (oxidised).

Chlorite ($ClO_2^-$, chlorine at $+3$):

$\overset{+3}{6 \text{ } ClO_2^-} \xrightarrow{h\nu} \overset{+5}{4ClO_3^-} + \overset{-1}{2Cl^-}$

This reaction is driven by light energy ($h\nu$). Chlorine moves from $+3$ to $+5$ (oxidised) and $-1$ (reduced).

Chlorate ($ClO_3^-$, chlorine at $+5$):

$\overset{+5}{4ClO_3^-} \rightarrow \overset{-1}{Cl^-} + \overset{+7}{3 \text{ } ClO_4^-}$

Chlorine shifts from $+5$ to $-1$ (reduced) and $+7$ (oxidised).

Classifying Redox Reactions: Putting It All Together

Now that we have studied all four types, let us practise classifying reactions. Consider these examples:

(a) $N_2 \text{ (g)} + O_2 \text{ (g)} \rightarrow 2 \text{ NO (g)}$

Two elemental substances combine to form a single compound (nitric oxide). This is a combination redox reaction.

(b) $2Pb(NO_3)_2\text{(s)} \rightarrow 2PbO\text{(s)} + 4 NO_2 \text{ (g)} + O_2 \text{ (g)}$

One compound (lead nitrate) breaks into three simpler substances. This is a decomposition redox reaction.

(c) $NaH\text{(s)} + H_2O\text{(l)} \rightarrow NaOH\text{(aq)} + H_2 \text{ (g)}$

The hydride ion ($H^-$) from sodium hydride pushes out the hydrogen from water, releasing it as $H_2$ gas. One hydrogen-containing species displaces hydrogen from another. This is a displacement redox reaction.

(d) $2NO_2\text{(g)} + 2OH^-\text{(aq)} \rightarrow NO_2^-\text{(aq)} + NO_3^- \text{(aq)} + H_2O\text{(l)}$

Nitrogen starts at $+4$ in $NO_2$, then splits: it drops to $+3$ in $NO_2^-$ (reduced) and climbs to $+5$ in $NO_3^-$ (oxidised). The same element from one state goes both higher and lower. This is disproportionation.

The Paradox of Fractional Oxidation Numbers

Sometimes, when you calculate the oxidation number of a particular element in a compound using the standard rules, you get a fraction. At first, this seems strange, because electrons are whole particles and cannot be shared or transferred in fractions.

Here are three compounds that produce fractional values:

• $C_3O_2$ (carbon suboxide): average oxidation number of carbon = $\frac{4}{3}$
• $Br_3O_8$ (tribromine octaoxide): average oxidation number of bromine = $\frac{16}{3}$
• $Na_2S_4O_6$ (sodium tetrathionate): average oxidation number of sulphur = $2.5$

The Resolution: Look at the Structure

The fractional number is just an average across all atoms of that element in the molecule. The actual, real oxidation states are always whole numbers, but different atoms of the same element sit at different states. The molecular structure reveals the truth.

Carbon suboxide ($C_3O_2$):

The molecule has a linear structure: $O{=}C{=}C{=}C{=}O$. The two terminal carbon atoms are each bonded to an oxygen atom and carry oxidation state $+2$. The middle carbon sits between the other two carbons and carries oxidation state $0$.

The average: $(+2 + 2 + 0) \div 3 = \frac{4}{3}$. But the real picture is $+2$, $+2$, and $0$.

Tribromine octaoxide ($Br_3O_8$):

The structure has three bromine atoms in a chain. Each of the two terminal bromines bonds to three oxygen atoms, giving them an oxidation state of $+6$. The central bromine bonds to two oxygens, putting it at $+4$.

The average: $(+6 + 6 + 4) \div 3 = \frac{16}{3}$. But the reality is $+6$, $+4$, and $+6$.

Tetrathionate ion ($S_4O_6^{2-}$):

The ion has four sulphur atoms in a row. The two outer sulphurs are each bonded to three oxygen atoms and carry oxidation state $+5$. The two inner sulphurs bond only to other sulphur atoms and carry oxidation state $0$.

The average: $(+5 + 0 + 0 + 5) \div 4 = 2.5$. But the actual states are $+5$, $0$, $0$, and $+5$.

The General Rule

Whenever you encounter a fractional oxidation state for any element in any compound, it means:

1. The fractional value is just a mathematical average.
2. The element is actually present in two or more different whole-number oxidation states within the molecule.
3. Only the molecular structure reveals which atoms are in which state.

Other examples of compounds with fractional averages include the mixed oxides: $Fe_3O_4$, $Mn_3O_4$, and $Pb_3O_4$. In each case, the metal atoms sit in two distinct oxidation states, and the average comes out as a fraction.

There is one exception worth noting: in molecular ions like $O_2^+$ and $O_2^-$, the oxidation state per atom is genuinely $+\frac{1}{2}$ and $-\frac{1}{2}$ respectively, because both oxygen atoms in these symmetric species are truly equivalent.

How $Pb_3O_4$ Behaves Differently with Different Acids

$Pb_3O_4$ (red lead) is not a simple oxide. It is actually a stoichiometric (fixed-ratio) mixture of 2 mol of $PbO$ and 1 mol of $PbO_2$. In $PbO$, lead sits at $+2$ (a stable state), while in $PbO_2$, lead sits at $+4$ (a less stable, oxidising state).

Reaction with Hydrochloric Acid ($HCl$)

$Pb_3O_4 + 8HCl \rightarrow 3PbCl_2 + Cl_2 + 4H_2O$

This reaction can be split into two parts:

Acid-base part:

$2PbO + 4HCl \rightarrow 2PbCl_2 + 2H_2O$

The basic oxide $PbO$ simply reacts with the acid. No redox here.

Redox part:

$\overset{+4}{PbO_2} + \overset{-1}{4HCl} \rightarrow \overset{+2}{PbCl_2} + \overset{0}{Cl_2} + 2H_2O$

The $PbO_2$ component acts as an oxidising agent, pulling electrons from $Cl^-$ ions ($-1 \rightarrow 0$) and converting them into chlorine gas. Lead itself gets reduced from $+4$ to $+2$.

Reaction with Nitric Acid ($HNO_3$)

$Pb_3O_4 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + PbO_2 + 2H_2O$

Here, only the $PbO$ portion reacts:

$2PbO + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + 2H_2O$

This is just an acid-base reaction. The $PbO_2$ sits there unreacted and appears as a separate product.

Why the difference? Nitric acid ($HNO_3$) is itself a powerful oxidising agent. Since $PbO_2$ can only act as an oxidising agent (it wants to gain electrons, not lose them), and $HNO_3$ will not give up electrons either, there is no redox partner for $PbO_2$ to react with. With $HCl$, the chloride ion ($Cl^-$) is happy to donate electrons, so the redox reaction proceeds. This is a neat example of how the nature of the acid, not just its acidity, determines the outcome of the reaction.