Suggested languages for you:

Americas

Europe

|
|

# Reactions of Halides

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?

A halide ion is a negative anion formed from a halogen atom. They have a charge of -1. The term halide is also used to describe a compound made up of a halogen atom bonded to a less electronegative species.

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:

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

Did you know that some halide ions are essential to our health? For example, scientists believe that an iodine deficiency can cause intellectual disability. That’s slightly concerning when you find out that almost two billion people worldwide are deficient in iodine! Researchers in China estimate that such a deficiency reduces the average citizen’s IQ score by 12 points. Iodine is found naturally in soil but levels can vary across the globe, so plants are an unreliable source - your best bet is seafood. Vegetarians can get their iodine from iodised salt, which is standard sodium chloride mixed with trace amounts of iodide salts.

On the other hand, chloride ions are important for plant growth too. Cereal grains typically contain between 10-20 ppm chlorine, and growth suffers severely if chloride levels in the soil fall below 2 ppm.

## 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.

Remember, acids are proton donors. Check out Brønsted-Lowry Acids and Bases for a full run-down of acids.

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:

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

Note that this isn’t a redox reaction - no electrons are transferred in the process.

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.

${\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)+\mathrm{HF}\left(\mathrm{aq}\right)⇌{\mathrm{H}}_{3}{\mathrm{O}}^{+}+{\mathrm{F}}^{-}\left(\mathrm{aq}\right)$

You might have learnt the term dissociate when it comes to strong and weak acids. Weak acids only partially dissociate in solution, whereas strong acids fully dissociate. Dissociate simply means to split into separate parts. When it comes to acids, they split apart into ions. In this case, dissociation is just another way of saying ionisation. Therefore, hydrofluoric acid is a weak acid because it is only partially ionised.

### 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:

${\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{OH}+\mathrm{HBr}\to {\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{Br}+ {\mathrm{H}}_{2}\mathrm{O}$

Here's the mechanism for the reaction between ethanol and hydrogen bromide.

1. One of the oxygen atom's lone pairs of electrons attacks the partially positive hydrogen atom in hydrogen bromide, adding the hydrogen to the alcohol.
2. Water is eliminated from the alcohol, leaving behind a carbocation.
3. The negative bromide ion adds to the carbocation, forming a halogenoalkane.

### Reaction with ammonia

Hydrogen halides react with ammonia to produce ammonium halides.

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

$\mathrm{HCl}\left(\mathrm{g}\right)+ {\mathrm{NH}}_{3}\left(\mathrm{g}\right)\to {\mathrm{NH}}_{4}\mathrm{Cl}\left(\mathrm{s}\right)$

## 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.

You must use nitric acid to acidify the solution. Using hydrochloric acid or sulfuric acid, say, would also lead to a white carbonate or sulfate precipitate forming - another example of a false-positive 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.

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

Why do we add ammonia? Well, it’s all to do with something called the solubility product value. Let’s investigate it further.

When ionic compounds dissolve in water, they separate into ions. If the concentration of ions reaches a certain value, the compound will form a precipitate. In other words, it won’t dissolve anymore. This value varies for each compound and is known as the solubility product value. You find it by multiplying the concentrations of the respective ions together. For example, the solubility product value of silver halides is shown below:

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\right]\left[{\mathrm{X}}^{-}\left(\mathrm{aq}\right)\right]$

So, if the concentration of silver ions multiplied by the concentration of halide ions is less than or equal to the solubility product value, your silver halide will dissolve and no precipitate will form. But as soon as the product of the two concentrations exceeds the solubility value - boom! A precipitate forms.

The higher the solubility product value, the more soluble the compound is, as you need more ions in solution before they form a precipitate. For example, silver chloride has a higher solubility product value than silver iodide - it is more soluble.

Let’s go back to our reaction. Adding silver nitrate to a solution containing halide ions forms a silver halide, $\mathrm{AgX}$. When silver halides dissolve in ammonia solution, they then form complex ions. Silver is a transition metal and ammonia is an example of a ligand - a species with a lone pair of electrons that can bond to transition metals with a dative covalent bond, also called a coordinate bond. In this case, each positive silver ion bonds to two neutral ammonia molecules. The resulting complex has a positive charge and is attracted to the negative halide ions in solution. This forms a salt: a diamine silver halide.

Diamine silver halide

This uses up some of the silver ions in solution. The concentration of silver ions in solution has decreased. If we now multiply this concentration by the concentration of halide ions, we should get a lower value. There is a greater chance that this value will now be lower than the solubility product value - if it is, the compound will dissolve.

Simply put, adding ammonia reduces the ion concentration in solution, meaning the product of ion concentrations is more likely to be lower than the solubility product value. The compound is therefore more likely to dissolve.

The following table should help you put all this new information together.

Compound solubility table

## Reactions of halides as reducing agents

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

An oxidising agent oxidises other species and is itself reduced in the process.

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

A reducing agent reduces other species and is itself oxidised in the process.

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.

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.

Halocarbons, also known as organohalides, are molecules containing one or more halogen atoms bonded to a carbon atom in an organic compound.

Let's explore how they react.

### 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.

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

${\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{Cl}+{\mathrm{OH}}^{-}\to {\mathrm{CH}}_{2}{\mathrm{CH}}_{2}+{\mathrm{H}}_{2}\mathrm{O}$

Check out Nucleophilic Substitution Reactions and Elimination Reactions to find out more about these types of reaction, including mechanisms and examples.

### 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.

Why don't aryl halides take part in substitution reactions? It is because the C-X bond in an aryl halide is much stronger than the C-X bond in an alkyl halide. There are two reasons for this.

Firstly, the C-X bond in aryl halides is much shorter than the C-X bond in alkyl halides. This makes it stronger.

Secondly, aryl halides show resonance. This means that its electron bonding can't be described by a single structure. You might know from Reactions of Benzene that benzene contains delocalised pi electrons found in a plane above and below the carbon ring, which makes the molecule more stable. It also makes the C-C single bonds in benzene behave a little like C=C double bonds.

In chlorobenzene, one of the lone pairs of electrons on the chlorine atom also gets involved in resonance, meaning the C-X single bond takes on some of the character of a C=X double bond. Double bonds are much stronger than single bonds and so the overall strength of the C-X bond increases.

## Reactions of Halides - Key takeaways

• A halide is a negative ion with a charge of -1 formed from a halogen atom.
• Hydrogen halides react in water to form acids. Hydrogen chloride, bromide, and iodide all produce strong acids whereas hydrogen fluoride produces a weak acid.
• Hydrogen halides react with alcohols to form a halogenoalkane and water.
• You can use acidified silver nitrate solution followed by ammonia to test for halide ions in solution.
• Halide ions can act as reducing agents. A reducing agent reduces another species and is oxidised in the process.
• Halides become better reducing agents as you move down the group in the periodic table.
• All halide ions react with concentrated sulfuric acid, but only bromide and iodide ions are strong enough reducing agents to reduce it.
• Alkyl halides react in nucleophilic substitution reactions and elimination reactions.
• Aryl halides react in halogen-metal exchange reactions.

No. The reactions are irreversible.

Reacting potassium and iodine produces the metal halide potassium iodide.

A precipitation reaction occurs, forming an insoluble salt, AgX.

Halides become more reactive as you move down the group in the periodic table.

Halides are halogen atoms that have each gained an electron to form a negative anion with a charge of -1. An example is the chlorine anion.

## Final Reactions of Halides Quiz

Question

What is a halide?

A halogen atom that has gained an electron to form a negative anion with a charge of -1.

Show question

Question

Hydrogen chloride, bromide and iodide all dissolve in water to form a ______.

Strong acid

Show question

Question

Hydrogen fluoride dissolves in water to form a ______.

Weak acid

Show question

Question

Name a reactant or combination of reactants used to test for halide ions.

Acidified silver nitrate solution followed by ammonia solution.

Show question

Question

Bromide ions react with acidified silver nitrate solution to form a cream precipitate which dissolves in concentrated ammonia solution. True or false?

True

Show question

Question

Chloride ions react with acidified silver nitrate solution to form a yellow precipitate which dissolves in dilute ammonia solution. True or false?

False

Show question

Question

Iodide ions react with acidified silver nitrate solution to form a yellow precipitate which is insoluble in all concentrations of ammonia solution. True or false?

True

Show question

Question

To test for halide ions, you can use a combination of acidified silver nitrate solution and ammonia solution. Which acid is used to acidify the silver nitrate solution, and why?

Nitric acid. Sulfuric acid and hydrochloric acid would also produce a white precipitate, a false-positive result.

Show question

Question

Which of the following halide ions is the best reducing agent? Explain your answer.

Fluorine

Show question

Question

Reducing agents donate electrons in redox reactions. True or false?

True

Show question

Question

Reducing agents are reduced in redox reactions. True or false?

False

Show question

Question

All halide ions react with sulfuric acid to produce a hydrogen halide. True or false?

True

Show question

Question

Chloride ions can reduce sulfuric acid. True or false?

False

Show question

Question

Which of the following are produced when bromide ions react with sulfuric acid?

Bromine

Show question

Question

Which statement is correct?

The sulfur in sulfur dioxide has an oxidation state of +4.

Show question

Question

What can you expect to observe when iodide ions react with sulfuric acid?

• Steamy white fumes (hydrogen iodide)
• A yellow solid (sulfur)
• A black solid (iodine)
• You may also see the bubbling of a colourless gas (sulfur dioxide) and smell rotten eggs (hydrogen sulfide)

Show question

Question

The reaction between fluoride ions and acidified silver nitrate solution produces _____. Explain your answer.

No observable reaction.

Show question

60%

of the users don't pass the Reactions of Halides quiz! Will you pass the quiz?

Start Quiz

## Study Plan

Be perfectly prepared on time with an individual plan.

## Quizzes

Test your knowledge with gamified quizzes.

## Flashcards

Create and find flashcards in record time.

## Notes

Create beautiful notes faster than ever before.

## Study Sets

Have all your study materials in one place.

## Documents

Upload unlimited documents and save them online.

## Study Analytics

Identify your study strength and weaknesses.

## Weekly Goals

Set individual study goals and earn points reaching them.

## Smart Reminders

Stop procrastinating with our study reminders.

## Rewards

Earn points, unlock badges and level up while studying.

## Magic Marker

Create flashcards in notes completely automatically.

## Smart Formatting

Create the most beautiful study materials using our templates.