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Reactions of Halides

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Chemistry

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.

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

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.

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

Remember, acids are proton donors. Check out Brønsted-Lowry Acids and Bases for a reminder.

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

We often simplify the hydronium ion to just a proton:

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.

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 or .

For example, reacting ethanol with hydrogen bromide gives chloroethane:

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 onto the alcohol.
  2. Water is eliminated from the alcohol, leaving behind a carbocation.
  3. The negative bromide ion adds onto the carbocation, forming a halogenoalkane.

Reaction with silver nitrate and ammonia

Another useful reaction to learn is the one between acidified silver nitrate solution, , and various halide ions. This is one way of identifying halide ions in solution. Adding ammonia solution afterwards will help confirm your results.

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.

Reactions of Halides results table testing for halides StudySmarterA table to record your results when testing for halide ions. Anna Brewer, StudySmarter Originals

The results

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

Reactions of Halides results table testing for halides StudySmarterThe results table, completed. Anna Brewer, StudySmarter Originals

Writing equations

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

We can simplify this into the following ionic equation.

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:

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, . 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.

Reactions of Halides diamine silver halide StudySmarterDiamine silver halide. Anna Brewer, StudySmarter Originals

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.

For more information on complex ions, head over to Transition Metals.

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.

Reactions of Halides RAD OAT StudySmarterThe acronyms OIL RIG and RAD OAT help you remember electron movement. Anna Brewer, StudySmarter Originals

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.

Reactions of Halides fluorine fluoride ion Study SmarterA fluorine atom, left, and a fluoride ion, right. Anna Brewer, StudySmarter Originals

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

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.

Reactions of Halides fluorine chlorine electron shells configuration StudySmarterThe electron shells of fluorine and chlorine. Anna Brewer, StudySmarter Originals, created using images from commons.wikimedia.org

Fluoride ions have the electron configuration . Chloride ions have the electron configuration . 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 electron - and losing electrons is exactly what reducing agents do.

Reaction with sulfuric acid

All the halide ions react with concentrated sulfuric acid, but the reactions produce a variety of different products depending on 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:

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.

Reactions of Halides sodium fluoride sulfuric acid StudySmarterThe oxidation states of sulfur in the reaction between sodium fluoride and sulfuric acid. Anna Brewer, StudySmarter Originals

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:

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:

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.

For more information on oxidation states, check out Redox.

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.

Reactions of Halides iodide sulfuric acid reaction StudySmarterThe reaction between sodium iodide and sulfuric acid. Anna Brewer, StudySmarter Originals

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.

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:

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.

Reactions of Halides

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?

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Answer

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

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Question

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

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Answer

Strong acid

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Question

Hydrogen fluoride dissolves in water to form a ______.


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Answer

Weak acid

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Question

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


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Answer

Acidified silver nitrate solution followed by ammonia solution.

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Question

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

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Answer

True

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Question

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

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Answer

False

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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? 

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Answer

True

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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?


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Answer

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

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Question

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

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Answer

Fluorine

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Question

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


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Answer

True

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Question

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


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Answer

False

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Question

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


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Answer

True

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Question

Chloride ions can reduce sulfuric acid. True or false? 

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Answer

False

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Question

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


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Answer

Bromine

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Question

Which statement is correct?


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Answer

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

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Question

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


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Answer

  • 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)

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Question

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


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Answer

No observable reaction.

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Question

Why must the reaction between sodium iodide and silver nitrate solution be carried out in a fume cupboard?

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Answer

The reaction produces sulfur dioxide, a toxic gas.

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