# Oxidizing Ability of the Group 17 Elements

Consider a reaction between one halogen—​chlorine, for example—and the ions of another—iodide, in this case. The iodide ions are dissolved from a salt such as sodium iodide or potassium iodide. The sodium or potassium ions are spectator ions and therefore irrelevant to the reaction, which proceeds as follows:

$Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$

• The iodide ions lose electrons to form iodine molecules; they are oxidized.
• The chlorine molecules gain electrons to form chloride ions. They are reduced.

This is a redox reaction in which chlorine is acting as an oxidizing agent. The driving force force this reaction is straightforward to identify from the table of Standard Reduction Potentials (Table P2).

### Fluorine

Fluorine must be excluded from this section, because it is too strong an oxidizing agent. Fluorine oxidizes water to oxygen (as shown in the equation below), making it impossible to carry out simple solution reactions with it.

$2F_2 + 2H_2O \rightarrow 4HF + O_2$

### Chlorine, bromine and iodine

In each case, a halogen higher in the group can oxidize the ions of one lower down. For example, chlorine can oxidize bromide to bromine:

$Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2$

The bromine appears as an orange solution. As observed above, chlorine can also oxidize iodide ions to iodine:

$Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$

Iodine appears either as a red solution if a small amount of chlorine is added, or as a dark gray precipitate if the chlorine is in excess. The explanation for the red solution is that iodine dissolves in potassium iodide (or other soluble iodides) by reacting to give a red ion, I3-. If chlorine is in excess, there is nothing left to react with the iodine, and so it remains as the gray precipitate.

Bromine can oxidize iodide ions to iodine, as in the reaction shown below; it is not a strong enough oxidizing agent to convert chloride ions into chlorine (the reverse of the previous reaction). A red solution of iodine is formed (see the note above) unless the bromine is in excess. In that case, a dark gray precipitate is formed.

$Br_2 + 2I^- \rightarrow 2Br^- + I_2$

Iodine cannot oxidize any of the other halide ions (except the extremely radioactive and rare astatide ion).

### Summary

• Each of the halogens can potentially oxidize another species—they are oxidizing agents.
• Fluorine is such a powerful oxidizing agent that it cannot be used in solution reactions.
• Chlorine oxidizes both bromide ions and iodide ions. Bromine and iodine cannot reclaim those electrons from the chloride ions formed—chlorine is a more powerful oxidizing agent than either bromine or iodine.
• Similarly, bromine is a more powerful oxidizing agent than iodine: it can remove electrons irreversibly from iodide ions to give iodine.

In other words, oxidizing ability decreases down the group.

### Explaining the trend

Whenever a halogen is acts as an oxidant in solution, the it converts to a halide ion which is solvated by water molecules. Looking at all four of the common halogens:

From fluorine to iodine, the ease with which these hydrated ions are formed decreases; therefore, oxidizing ability decreases down the group. The reason for this effect is a complicated mixture of several factors. Unfortunately, this is often over-simplified to a faulty and misleading explanation, which is explained below. The correct explanation will be given thereafter.

### The faulty explanation

This is the explanation normally given for the trend in oxidizing ability of chlorine, bromine and iodine:

How easily the element forms its ions depends on how strongly the new electrons are attracted. As the atoms increase in size, the new electrons are farther from the nucleus, and increasingly screened by the inner electrons (offsetting the effect of the greater nuclear charge). The bigger atoms are therefore less good at attracting new electrons and forming ions.

This appears to be a reasonable explanation, but it is flawed. Consider the trend in electron affinity from chlorine to bromine to iodine; it decreases down the group, consistent with the above explanation. However, problems arise if the argument is expanded to include fluorine: fluorine has a much higher tendency to form its hydrated ion than chlorine does, but the electron affinity of the fluorine atom is less than that of chlorine. This inconsistency calls the accuracy of the argument into question.

The problem stems from examining one step in a complicated process. The prior argument about atoms accepting electrons applies to isolated atoms in the gas state picking up electrons to make isolated ions—also in the gas state. This is a very different situation. The process to examine is the following:

• The halogen starts as a diatomic molecule, $$X_2^-$$ which may be gas, liquid or solid, depending on the halogen.
• This molecule is broken into two individual halogen atoms.
• Each atom gains an electron (this is the sole step considered in the flawed explanation).
• The isolated ions are solvated to form hydrated ions.

### The proper explanation

The table below examines the amount of energy involved in each of these changes. The quantities of interest are defined below:

### Atomization energy

Atomization energy is the energy required to produce 1 mole of isolated gaseous atoms from an element in its standard state (gas for chlorine and liquid for bromine, for example, both diatomic). For chlorine, this is half of the bond enthalpy (breaking a Cl-Cl bond produces 2 chlorine atoms).

$1/2 \,Cl_{2\, (g)} \rightarrow Cl^._{(g)}$

For a liquid like bromine or a solid like iodine, the atomization energy also includes the energy required to convert these species into gases (evaporation or sublimation, respectively).

### Electron affinity

The first electron affinity is the energy released when one mole of gaseous atoms acquires an electron to form one mole of gaseous ions, according the the equation below:

$X_{(g)} + e^- \rightarrow X^-_{(g)}$

### Hydration enthalpy (hydration energy)

This is the energy released when 1 mole of gaseous ions dissolves in water to produce hydrated ions, illustrated by the equation below:

$X^-_{(g)} \rightarrow X^-_{(aq)}$

The amount of heat evolved decreases dramatically from the top to the bottom of the group, with the largest drop from fluorine to chlorine. Fluorine produces a large quantity of heat in the formation of the hydrated ion, chlorine less so, and so on down the group.

Atomization energy
(kJ mol-1)
Electron affinity
(kJ mol-1)
Hydration enthalpy
(kJ mol-1)
Overall
(kJ mol-1)
F +79 -328 -506 -755
Cl +121 -349 -364 -592
Br +112 -324 -335 -547
I +107 -295 -293 -481

Table 1: Energetics underlying the Reduction Potential of Halogens. The final column of the table shows the overall heat evolved in the process. It is calculated by adding the figures in the previous 3 columns.

Note:  It is important to remember that only half of a redox reaction is considered in each case. There are other energy terms involving the reductant; these changes are endothermic overall. For example, if chlorine oxidizes iodide to iodine, that half of the total reaction requires +481 kJ mol-1, producing an overall reaction enthalpy change of (-592 + 481) = -111 kJ per mole of I- oxidized.

### The oxidative strength of fluorine

There are two main factors responsible for the highly negative enthalpy change for the reduction of fluorine to fluoride. The  first is that atomization energy of fluorine is abnormally low, reflecting the low bond enthalpy of fluorine. The reason for fluorine's low bond enthalpy is described elsewhere. The main reason, however, is the high hydration enthalpy of the fluoride ion. The ion is very small and experiences a strong attraction to water molecules. The stronger the attraction, the more heat is evolved when the hydrated ions are formed.

### Decreasing oxidizing ability from chlorine to bromine to iodine

The decreases in atomization energy between these three elements are small, and would tend to make the overall enthalpy change more negative down the group; this does not support the explanation. It is more instructive to consider the changes in electron affinity and hydration enthalpy down the group. The values below are calculated from the table above:

Change in electron affinity
(kJ mol-1)
Change in hydration enthalpy
(kJ mol-1)
Cl to Br +25 +29
Br to I +29 +42

Both of these effects contribute, but the change in the hydration enthalpy has a larger impact. Down the group, the ions experience a smaller attraction to water molecules as they increase in size. Although the ease with which an atom attracts an electron matters, it is not as important as the hydration enthalpy of the negative ion formed.

The faulty explanation misses the mark even if when restricted to chlorine, bromine and iodine.

### Contributors

• Jim Clark (ChemGuide)