Did you know that some halide ions are essential for our health? For example, iodine is needed for thyroid function.

• Halides can react in multiple different ways. Here, we will look at how they act as reducing agents, and the trend in this reaction as you move down the group in the periodic table.
• We’ll also look at the reactions of hydrogen halides and organohalides.

What are halides?

As an example, chlorine forms halide ions that we call chloride ions, Cl^-. Chlorine also reacts with sodium to produce a sodium halide known as sodium chloride:

${\mathrm{Cl}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-}{\mathrm{Cl}}_2+2\mathrm{Na}\to 2\mathrm{NaCl}$

Reactions of hydrogen halides

Hydrogen halides consist of a hydrogen atom covalently bonded to a halogen atom. They're formed when hydrogen reacts with a halogen. Let's look at how hydrogen halides react with water, alcohols, and ammonia.

Reaction with water

Hydrogen chloride, hydrogen bromide and hydrogen iodide all react with water to form a strong acid.

For example, hydrogen chloride dissolves in water to produce hydrochloric acid, which consists of a hydronium ion and a chloride ion.

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

We often simplify the hydronium ion to just a proton:

$\mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

Hydrochloric acid, hydrobromic acid, and hydroiodic acid are all strong acids. This means that they ionise fully in solution. However, hydrofluoric acid is a weak acid, meaning it only partially ionises in solution. Although when you add hydrogen fluoride to water, it does ionise, the ions are attracted to each other so strongly that some of the ions form tightly-bound ion pairs. Because not all of the ions are free in the solution, we say that hydrofluoric acid is weak.

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+\mathrm{HF}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)$

Reaction with alcohols

Reacting hydrogen halides with an alcohol gives an alkyl halide, also known as a halogenoalkane. We'll look at these later on. You can also use phosphorus halides such as PCl₅ or PBr₃.

For example, reacting ethanol with hydrogen bromide gives chloroethane:

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}+\mathrm{HBr}\to {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+\hspace{0.17em}{\mathrm{H}}_2\mathrm{O}$

Reaction with ammonia

Hydrogen halides react with ammonia to produce ammonium halides.

For example, reacting hydrogen chloride with ammonia produces ammonium chloride:

$\mathrm{HCl}\left(\mathrm{g}\right)+\hspace{0.17em}{\mathrm{NH}}_3\left(\mathrm{g}\right)\to {\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)$

Reaction of halides with silver nitrate and ammonia

Another useful reaction of halides to learn is their reaction with acidified silver nitrate solution, AgNO₃. This is one way of identifying halide ions in solution. Adding ammonia solution afterward helps confirm your results.

Experiment

To carry out the test, add a few drops of nitric acid to an unknown halide in solution. The acid reacts with any soluble carbonate or hydroxide impurities that would give a false result.

Next, add a few drops of silver nitrate solution and make note of any observable changes in a lab notebook or another suitable place. Do any precipitates form? If so, what colour are they?

If chlorine, bromine, or iodine are present, they should form a precipitate. This is because they react with the silver nitrate solution to form insoluble silver halides - silver chloride, bromide, and iodide respectively. Fluorine won’t produce any observable results because silver fluoride is soluble in water.

You can then test the compounds further by adding ammonia solution. Do any of the precipitates dissolve in dilute ammonia solution? How about if the solution is concentrated? It can help to make a table to record your observations in, like the one shown here.

A table to record your results when testing for halide ions

With any luck, you’ll produce the following results.

The results table, completed

Equation

The general equation for the reaction between a sodium halide and silver nitrate solution is given below.

$\mathrm{NaX}\left(\mathrm{aq}\right)+{\mathrm{AgNO}}_3\left(\mathrm{aq}\right)\to \mathrm{AgX}\left(\mathrm{s}\right)+{\mathrm{NaNO}}_3\left(\mathrm{aq}\right)$

We can simplify this into the following ionic equation.

${\mathrm{X}}^{-}\left(\mathrm{aq}\right)+{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to \mathrm{AgX}\left(\mathrm{s}\right)$

Reactions of halides as reducing agents

We have explored how halogens can act as oxidising agents (see Reactions of Halogens).

Halide ions do quite the opposite - they act as reducing agents.

Do you remember the two acronyms, OIL RIG and RAD OAT? They help you remember the movement of electrons in redox reactions and in reactions involving oxidising or reducing agents.

The acronyms OIL RIG and RAD OAT help you remember electron movement

This means that a reducing agent donates electrons to another species. The other species gains these electrons and is reduced. The reducing agent loses electrons and so is oxidised.

How do halide ions act as reducing agents? You know that a halide is a negative anion. It contains an extra electron compared to the halogen in its elemental state. Halide ions can react by losing this extra electron to form a neutral halogen atom.

A fluorine atom (left) and a fluoride ion (right)

Let’s consider a general reaction between a halide ion, which we’ll call X^-, and another substance, which we’ll call Y:

$2{\mathrm{X}}^{-}+2\mathrm{Y}\to 2{\mathrm{Y}}^{-}+{\mathrm{X}}_2$

Note the following:

• The halide loses an electron. It is oxidised.
• The other species gains an electron. It is reduced.
• The halide reduces the other species. Therefore, the halide is a reducing agent.

Trends in reducing ability

You might remember that halogens become better oxidising agents as you move up the group in the periodic table. However, this trend reverses when it comes to reducing ability. In general, halides become better reducing agents as you move down the group in the periodic table.

Why is this the case? Let’s look at the electronic structures of fluorine and chlorine, by way of an example.

The electron shells of fluorine and chlorine

Fluoride ions have the electron configuration 1s² 2s² 2p⁶. Chloride ions have the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶. Chloride is a larger ion than fluoride as it has more electron shells. This means that chloride’s outer shell electron is further from its nucleus than fluoride’s. The attraction between this outer shell electron and the nucleus is weaker and so it is easier to lose the outermost electron - and losing electrons is exactly what reducing agents do.

Reaction of halides with sulfuric acid

All the halide ions react with concentrated sulfuric acid, but the reactions produce a variety of different products. This dependson the halide used. Some halides are able to reduce the sulfur in sulfuric acid, while others are not.

We use sodium halide salts as a source of halide ions. Let’s explore each of the reactions in turn.

Fluoride ions and sulfuric acid

Sodium fluoride reacts with concentrated sulfuric acid to produce hydrogen fluoride and sodium hydrogensulfate:

$2\mathrm{NaF}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)+2\mathrm{HF}\left(\mathrm{g}\right)$

You’ll see a white solid - sodium hydrogensulfate - and the steamy fumes of hydrogen fluoride.

Notice that this isn’t a redox reaction - fluoride ions are not a strong enough reducing agent to reduce the sulfur in sulfuric acid. All of the oxidation states stay the same. Instead, this is an acid-base reaction.

The oxidation states of sulfur in the reaction between sodium fluoride and sulfuric acid

Chloride ions and sulfuric acid

Sodium chloride reacts in a similar way. Once again, chloride ions aren’t strong enough to reduce sulfur dioxide. The only reaction is an acid-base reaction, producing steamy white fumes of hydrogen chloride and the white solid sodium hydrogensulfate:

$\mathrm{NaCl}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to {\mathrm{NaHSO}}_4\left(\mathrm{s}\right)+\mathrm{HCl}\left(\mathrm{g}\right)$

Bromide ions and sulfuric acid

We now know that reducing ability increases as you move down the group on the periodic table. This means that bromide ions are a much better reducing agent than fluoride and chloride ions. In fact, bromide ions can reduce sulfuric acid. When sodium bromide reacts with sulfuric acid, we still get the same acid-base reaction that we saw earlier, but we also get an additional redox reaction producing bromine and sulfur dioxide:

$\mathrm{NaBr}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to {\mathrm{NaHSO}}_4\left(\mathrm{s}\right)+\mathrm{HBr}\left(\mathrm{g}\right)2{\mathrm{H}}^{+}+2{\mathrm{Br}}^{-}+{\mathrm{H}}_2{\mathrm{SO}}_4\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2+{\mathrm{Br}}_2$

Look at the oxidation states in this reaction:

• Bromine goes from -1 to +0.
• Sulfur goes from +6 to +4.
• Bromide ions lose electrons and are oxidised.
• Sulfur gains electrons and is reduced.

Therefore, bromide ions are a strong enough reducing agent to reduce sulfur.

Iodide ions and sulfuric acid

The trend continues down the group - iodide ions are even better at reducing other species than bromide ions! Four separate reactions take place.

• Firstly, an acid-base reaction produces hydrogen iodide.
• Next, iodide ions reduce sulfur from an oxidation state of +6 in sulfuric acid to +4 in sulfur dioxide.
• Iodide ions then reduce sulfur atoms further to elemental sulfur with an oxidation state of +0.
• They can also reduce sulfur further still into hydrogen sulfide. In this molecule, sulfur has an oxidation state of -2.

The next table provides an overview of the different reactions, oxidation states involved and what you should expect to see.

The reaction between sodium iodide and sulfuric acid

In summary:

• Fluoride and chloride ions don’t reduce sulfuric acid.
• Bromide ions reduce sulfur from an oxidation state of +6 to +4.
• Iodide ions, on the other hand, reduce sulfur from an oxidation state of +6 all the way down to -2!

Reactions of alkyl halides and aryl halides

Alkyl halides and aryl halides are types of halocarbons.

Reactions of alkyl halides

Alkyl halides are also known as halogenoalkanes and contain a halogen bonded to a carbon in an alkane. They are useful because they can be turned into molecules with a variety of other functional groups in the following reactions.

• Nucleophilic substitution of halogenoalkanes can produce alcohols, nitriles, and primary amines.
• Elimination of halogenoalkanes produces alkenes.

For example, eliminating chloroethane using ethanoic sodium hydroxide produces ethene and water:

Reactions of aryl halides

Aryl halides contain a halogen bonded to a carbon in an aromatic benzene ring. Unlike their alkyl halide cousins, they are relatively unreactive, and don't take part in elimination or substitution reactions. However, they can take part in metal-halogen exchange reactions. In these reactions, the halogen atom is swapped for a metal ion, resulting in a metal ion bonded to an aromatic benzene ring.