Oxidising ability of the halogens
The halogens in their chemical reactions are generally
good oxidising agents. That
is they are electron acceptors, they oxidise
other substances and the halogens are themselves
reduced when they gain electrons.
Displacement reactions using halogens are a good way to
demonstrate the trend in the oxidising ability of the
halogens. Fluorine is the most
powerful oxidising agent in group 7 and the trend in ability of the
halogens to
accept electrons decreases as we descend group 7.
To explain why the oxidising ability of the halogens
decrease as we
descend group 7 we need to consider a range of factors. The oxidiation
reaction is simple enough, we can show it as:
½X2 + e → X-
where X is any halogen.
The factors we need to think about which will influence how good an oxidising agent
a halogen is are:
- The bond strength of the halogen molecules. From the
reduction equation above we can see that the covalent
bond between the halogen atoms in the X2 molecule must be
broken when the
halogen accepts an electron
to form a halide ion. The weaker the bond, the more likely this reaction is to occur. The table below lists the
bond
energies for the halogens.
| Halogen |
F2 |
Cl2 |
Br2 |
I2 |
| Bond enthalpy/kJmol-1 |
158 |
243 |
192 |
151 |
The bond enthalpies for the halogens
drop as we descend the group; however this pattern does not hold for fluorine,
its
bond is much weaker than might be expected. This is a consequence
of its small size; being so small its non-bonding electrons
in the p-orbitals on each
fluorine atom are relatively close to each other and this causes some repulsion between them which weakens the bond.
The ability of an atom to attract an electron from another substance
will depend upon the size (atomic radius) of the halogen atom,
the size of the nuclear charge and the amount of shielding
taking place. The electron affinity will give us a measure of an atoms
ability to attract an electron. The stronger
the attraction between any incoming electron and the nucleus
the more energy will be
released. The electron affinity is the amount of energy released by the following process:
X(g) + e → X-(g)
The table below lists the energy changes when 1 mole of electrons are gained by
each halogen to form a halide ion.
| Halogen |
F |
Cl |
Br |
I |
| electron affinity/kJmol-1 |
-328 |
-349 |
-324 |
-295 |
We can see that there is not a particularly large difference in the electron affinities
when we go from one halogen to the next.
The increasing nuclear charge as we decend group 7 is offset
by increased screening of the nucleus
and an increase in the atomic radius. Ultimately
the effective nuclear charge that an electron
will feel will probably be about the same for each halogen atom due to these
competing factors; for example fluorine (1s22s22p5) has 9
electrons; 4 of which are in the lower 1s and 2s sub-levels,
if we assume that the 5 electrons in the outer sub-level shield
the nucleus poorly then any incoming electron
will feel a nuclear charge of +5 (9 protons - 4 shielding electrons= +5). This effective nuclear charge is the
same for all the halogens, but of course as we descend the group
the atomic radius increases so any incoming
electron will be much further away and feel a much smaller attraction
to the effective nuclear charge.
However
fluorine does not fit the pattern
here, its electron affinity is lower than might be expected by
looking at the pattern for the other halogens.
This is simply because
the fluorine atom is so small
and any added electron will have to go into the lower 2p orbitals, which are
small when compared to the
3p, 4p and 5p orbitals of the other halogens. Being small
means that there will be extra repulsion between the electrons
and
this accounts for a lower electron affinity for fluorine.
- Finally you should consider the energy released when the halide ion
formed when the halogen gains an electron
goes into
solution or an ionic lattice. Since all the halide ions have a charge
of -1 the factor here that
will influence the amount of energy released will be the size of the ion.
The smaller the ion the more energy
will be released when it goes into solution or an ionic lattice, so fluorine wins here!
So the trend is oxidising power of the halogens
is a balance between all the factors listed above. Ultimately the
weak F-F bond and the fact that more energy is released
when it enters solution or an ionic lattice means that fluorine
is
the strongest oxidising of the halogens
and the ability of the halogens to act as oxidising agents decreases
as we descend group 7.
Halogen displacement reactions
Displacement reactions are a good way to show the ability of
halogens to act as oxidising agents. A typical displacement reaction
is shown opposite. In the first test-tube we have a solution of sodium iodide dissolved
in water. On top of this is added a few centimetres of an organic solvent
such as hexane or cyclohexane.
Cyclohexane is a very good
solvent
for halogens and given the choice between dissolving in water and dissolving
in cyclohexane a halogen
will always dissolve in cyclohexane before water. Cyclohexane does not dissolve in water but like oil simply floats on top of it which is why there are two layers shown in the first test-tube in the image opposite.
Sodium iodide being an ionic compound will dissolve
in water. This solution contains sodium ions (Na+) and iodide ions (I - ).
When chlorine water is added to this test-tube and shaken for around 30 seconds you can see in the image opposite that the
cyclohexane layer has changed colour; it has turned a violet/purple colour.
It may be easier to understand what is happening here if we write an equation for the reaction:
sodium iodide(aq) + chlorine(aq) → sodium chloride(aq)
+ iodine(aq)
2NaI(aq) + Cl2(aq) → 2NaCl(aq)
+ I2(aq)
On the reactants side of the word equation we have two halogens present; iodine
in the form of iodide ions in the solution and chlorine from the chlorine water.
In a displacement reaction the more reactive
halogen will oxidise the less reactive halide ion present in a solution or a compound, in this case chlorine
will displace or oxidise the iodide ion from the sodium iodide solution. The chlorine effectively takes the place of the
iodide ion. The iodine is kicked out of the solution and now has a choice of places to go to; either into the
cyclohexane layer or into aqueous solution; it dissolves into the cyclohexane layer since halogens are more soluble in organic solvents than in water.
Another advantage
of cyclohexane is that the halogens show up as bright clear colours when dissoloved organic solvents, iodine is
purple, chlorine is green/yellow and bromine is red-brown. So the purple
colour of the cyclohexane
is due to the iodine dissolved in it!
Displacing bromine from a solution
In the example shown below we again have 2 halogens on the reactants side of the equation, these
are bromine in the form of a bromide ion from the sodium bromide solution and chlorine
from the chlorine water. Chlorine is more reactive than bromine so will displace or
oxidise it from the solution. The
bromide will be kicked out of solution and will dissolve in the cyclohexane solvent
turning it red/brown.
We can show this as:
sodium bromide(aq) + chlorine(aq) → sodium chloride(aq)
+ bromine(aq)
2NaBr(aq) + Cl2(aq) → 2NaCl(aq)
+ Br2(aq)
In this equation the chlorine has been reduced to chloride and the bromide ion has
been oxidised to bromine. Chlorine is the electron
acceptor or oxidising agent and the bromide is the electron donor or
reducing agent.
Displacement reactions????
However be careful as you can get caught out if you are not careful! Consider the reaction shown below.
sodium chloride(aq) + iodine(aq) → sodium chloride(aq)
+ iodine(aq)
2NaCl(aq) + I2(aq) → 2NaCl(aq)
+ I2(aq)
Here we have iodine and chlorine as the two halogens present on the reactant side of the equation.
Iodine is not able to oxidise the chloride ions present and so no reaction will happen.
Reaction of the halogens with iron(II) chloride solutions
Displacement reactions are one way to show the oxidising ability of the halogens, another method is the reactions of the halogens with solutions containing the iron ions Fe2+ and Fe3+ions. Iron (II) chloride solution is a pale green colour while iron (III) chloride solution is a pale yellow colour; as shown in the image opposite.
We can demonstrate the oxidising ability of the halogens by reacting them with a solution of Iron(II) chloride. Here chlorine and bromine are strong
enough oxidising agents to oxidise the Fe2+ ions to Fe3+ ions and in the process the chlorine and bromine are both reduced to
chloride and bromide ions. We can show this as:
2Fe2+(aq) + Cl2(aq) → 2Fe3+(aq) + 2Cl-(aq)
and also:
2Fe2+(aq) + Br2(aq) → 2Fe3+(aq) + 2Br-(aq)
The pale green solution containing the Fe2+ ions will change colour to a pale yellow solution containing Fe3+ when
chlorine and bromine are added to it however this is not what happens with iodine. Iodine being the poorest oxidising agent in group 7 is not
able to oxidise the Fe2+ ions; in fact a solution of Fe3+ will oxidise iodide ions (I-) to form
iodine (I2)
2I-(aq) + 2Fe3+(aq) → I2(s) + 2Fe2+(aq)
Here the soluble iodide ions (I-(aq)) are oxidised to form the almost insoluble iodine. A pale brown solution
of iodine in water will form along with a solid precipitate of insoluble iodine on the bottom of the beaker or test-tube.
Key Points
- The oxidising ability of the halogens
decreases as we descend group 7. The reasons for this are:
- The weakness of the F-F bond
- The fluoride ion will release the more energy when it dissolves or forms part of an ionic lattice making it more stable
due to its
small size while larger halide ions will release much less energy.
- A more reactive halogen will displace a less reactive halogen
from its solutions.
- Cyclohexane or hexane are good solvents for the halogens.
When halogens dissolve in cyclohexane, it gives clear and vivid colours, for example, a green solution for
chlorine, red-brown for bromine and lilac for
iodine.
Practice questions
Next
