What this chapter covers11
Group 17 is the topic where one idea does almost all the work. A halogen atom needs a single electron to complete its outer shell, and everything in this chapter — which halogen displaces which, why iodine cannot make iron(III) iodide, why concentrated sulfuric acid gives hydrogen chloride but not hydrogen bromide, why silver iodide will not dissolve in ammonia — comes back to how hard the atom pulls on that one electron, and to how tightly the ion that results holds on to it. Down the group the atom gets bigger, the pull gets weaker, and every trend follows.
What topic 11 asks you to do
11.1 Physical properties of the Group 17 elements
11.1.1 describe the colours and the trend in volatility of chlorine,
bromine and iodine
11.1.2 describe and explain the trend in the bond strength of the halogen molecules
11.1.3 interpret the volatility of the elements in terms of instantaneous dipole–induced
dipole forces
11.2 The chemical properties of the halogen elements and the hydrogen halides
11.2.1 describe the relative reactivity of the elements as oxidising
agents
11.2.2 describe the reactions of the elements with hydrogen and explain their relative
reactivity
11.2.3 describe the relative thermal stabilities of the hydrogen halides and explain these in
terms of bond strengths
11.3 Some reactions of the halide ions
11.3.1 describe the relative reactivity of halide ions as reducing
agents
11.3.2 describe and explain the reactions of halide ions with: (a) aqueous silver ions
followed by aqueous ammonia (the formation and formula of the [Ag(NH3)2]+
complex is not required) (b) concentrated sulfuric acid, to include balanced chemical equations
11.4 The reactions of chlorine
11.4.1 describe and interpret, in terms of changes in oxidation
number, the reaction of chlorine with cold and with hot aqueous sodium hydroxide and recognise these
as disproportionation reactions
11.4.2 explain, including by use of an equation, the use of chlorine in water purification to
include the production of the active species HOCl and ClO− which kill bacteria
Read the verbs. 11.1.1 says describe the colours of chlorine, bromine and iodine — three elements, not five, and fluorine is not among them. 11.1.2 and 11.1.3 say explain and interpret, which means an argument is wanted, not a list. 11.3.2 says balanced chemical equations in so many words, so the equations for concentrated sulfuric acid have to be produced rather than described. And 11.4.1 says in terms of changes in oxidation number: an answer to that statement that never mentions an oxidation number has missed the point of the question.
What this page covers, and what it leaves out
This page is topic 11 and nothing else. Fluorine appears throughout, because two of the arguments here only make sense if you can see the element that breaks them — the F–F bond and the behaviour of fluoride — but it is flagged as out of statement wherever it is used, and you will never be asked for the colour or volatility of fluorine.
One piece of A Level machinery is borrowed and labelled as such. Standard electrode potentials, E°, are topic 24, not topic 11. They are used on this page because they turn "chlorine is a stronger oxidising agent than bromine" into a number you can subtract, and because every prediction the models make is then made honestly rather than looked up. Every AS argument is given in full alongside, and no AS question will require you to use an E° value.
The halogens, atom by atom11.1
The chemistry follows from the electronic structure, so start there. Every halogen has the outer configuration ns2np5 — seven outer electrons, one short of a full shell. The difference between one halogen and the next is not the number of outer electrons, which never changes; it is how far out they are and how well they are shielded from the nucleus.
| Element | Proton number | Electronic configuration | Outer electrons | Ion formed | Covalent radius / pm | Electronegativity |
|---|---|---|---|---|---|---|
| F fluorine (out of statement) | 9 | 1s22s22p5 | 7 | F− | 71 | 3.98 |
| Cl chlorine | 17 | 1s22s22p63s23p5 | 7 | Cl− | 99 | 3.16 |
| Br bromine | 35 | [Ar]3d104s24p5 | 7 | Br− | 114 | 2.96 |
| I iodine | 53 | [Kr]4d105s25p5 | 7 | I− | 133 | 2.66 |
The one sentence this chapter rests on
Every halogen atom needs one electron. What changes down the group is the distance from the nucleus to the place that electron has to go, and the number of full inner shells shielding it on the way. Both changes weaken the attraction, so the halogen becomes a weaker oxidising agent down the group — and, for exactly the same reason, the ion that results holds its electrons more loosely, so the halide becomes a better reducing agent down the group. Those two sentences, running in opposite directions from the same cause, are most of topic 11.
The model below plots any of the measured properties of the four elements, and measures the shape of what it has drawn rather than asserting it. Two of these trends are not smooth, and it is worth seeing which.
"Down the group the nuclear charge increases, so the attraction is stronger"
The nuclear charge really does rise — from +9 at fluorine to +53 at iodine. But 44 extra inner electrons come with it, and they shield the outer shell almost completely. The net pull felt by an electron arriving in the outer shell is roughly the same in all four; what differs is how far away that shell is. Quoting nuclear charge on its own reverses every trend in this chapter.
Colour, state and volatility11.1.1
This is a "describe" statement, so it is marked on precision. Vague colours lose marks; so does giving a colour without the state it belongs to, or describing iodine vapour when the question asked about the solid.
| Element | State at room temperature | Colour | Melting point / K | Boiling point / K | Volatility |
|---|---|---|---|---|---|
| Cl2 chlorine | gas | greenish-yellow | 171.6 | 239.1 | most volatile |
| Br2 bromine | liquid | dark red-brown, giving a reddish-brown vapour | 265.9 | 332.0 | |
| I2 iodine | solid | shiny grey-black, giving a purple vapour | 386.9 | 457.6 | least volatile |
| F2 fluorine (out of statement) | gas | pale yellow | 53.5 | 85.0 |
Volatility
How readily a substance turns into a vapour. A more volatile substance has a lower boiling point. Volatility decreases down group 17: chlorine is already a gas, bromine has to be warmed slightly, iodine has to be warmed a good deal — and when it is, it sublimes straight to a purple vapour rather than melting first.
Say the colour and the state together
"Iodine is purple" is half a mark at best. Solid iodine is a shiny grey-black; the purple belongs to its vapour, and to its solution in a non-polar solvent such as cyclohexane. In water, iodine gives a yellow-brown solution, and in potassium iodide solution a deeper brown. Three different colours for one element, so the question's wording decides which one is wanted.
Two further colours are worth knowing because they turn up constantly in the rest of this chapter, and neither is the colour of the element itself.
| Halogen | In water | In a non-polar solvent (cyclohexane) |
|---|---|---|
| chlorine | very pale green, often looks colourless | very pale green |
| bromine | orange to yellow-brown | orange |
| iodine | yellow-brown (brown and deeper in potassium iodide solution) | violet |
Bromine and iodine are easy to confuse in water — both can be described as brown. Shaken with an organic solvent they separate cleanly into orange and violet, which is why that step appears in so many practical questions.
Why volatility falls down the group11.1.3
All four halogens exist as diatomic molecules: two atoms sharing one pair of electrons, with three lone pairs left on each atom. The molecule is symmetrical, so it has no permanent dipole. There is no hydrogen bonding and no dipole–dipole attraction. The only thing holding one molecule to the next is the weakest of all intermolecular forces.
Instantaneous dipole – induced dipole forces
The electrons in a molecule are in constant motion, so at any instant they are unevenly distributed and the molecule has a small, momentary dipole. That dipole repels the electrons of a neighbouring molecule, inducing a dipole in it that points the right way to be attracted. The effect lasts only an instant and then re-forms somewhere else, but averaged over a whole liquid it is a real, continuous attraction.
CIE uses the name instantaneous dipole – induced dipole forces. You will also see London forces, dispersion forces and van der Waals forces for the same thing; use the syllabus wording in an exam.
The strength of the attraction depends on how easily the electron cloud can be distorted, and that depends on how many electrons there are and how loosely the outermost of them are held. Chlorine has 34 electrons per molecule, bromine 70, iodine 106. Bigger cloud, larger momentary dipoles, stronger attraction between neighbours, more energy needed to separate them — so higher melting and boiling points, and lower volatility.
The three-mark answer
1. The halogens are simple molecules held together by instantaneous dipole – induced dipole forces. 2. Down the group the molecules have more electrons, so the electron cloud is more easily distorted and these forces are stronger. 3. More energy is therefore needed to separate the molecules, so melting and boiling points rise and volatility falls.
Notice what is not in that answer: the covalent bond. It is never broken when a halogen boils.
"The bonds get stronger down the group, so iodine boils at a higher temperature"
This is the single commonest error in 11.1.3, and it is wrong twice over. Boiling separates molecules, so the covalent bond is irrelevant to it; and the covalent bond actually gets weaker down the group, not stronger, as the next section shows. An answer that explains a boiling point with a bond energy is describing a different experiment.
The X–X bond, and the fluorine exception11.1.2
Now the bond inside the molecule. Breaking it is a different process from boiling, and it runs the opposite way.
| Molecule | Covalent radius of the atom / pm | X–X bond energy / kJ mol−1 |
|---|---|---|
| F2 | 71 | 158 |
| Cl2 | 99 | 242 |
| Br2 | 114 | 193 |
| I2 | 133 | 151 |
From chlorine to iodine the pattern is straightforward: the atoms get larger, the bond gets longer, and the shared pair sits further from both nuclei, so less strongly attracted. Bond energy falls. Fluorine is the exception — on that argument it should have the strongest bond of all, and instead it has almost the weakest.
Why F–F is so weak
Each fluorine atom carries three lone pairs, and the F–F bond is very short. That forces six lone pairs into a small volume, where they repel one another strongly. The repulsion works against the bonding and cancels a large part of it. Chlorine, bromine and iodine are all big enough for their lone pairs to keep out of one another's way, so the simple bond-length argument works for them.
This matters beyond a table of numbers. A weak F–F bond is cheap to break, and the bonds fluorine then makes are exceptionally strong — that combination is why fluorine reacts with almost everything, often violently.
How much of this does the statement want?
11.1.2 says "describe and explain the trend in the bond strength of the halogen molecules". The safe answer describes the fall from chlorine to iodine and explains it by increasing bond length and the shared pair being further from the nuclei. If you also know why fluorine breaks the pattern, say so — it is the kind of detail that separates a good answer from a complete one — but the trend and its explanation are what the statement asks for.
The halogens as oxidising agents11.2.1
An oxidising agent takes electrons from something else. A halogen molecule does this in one step, gaining two electrons in total — one for each atom:
X2 + 2e− → 2X−
How readily that happens is the whole of 11.2.1, and it depends on how strongly the atom attracts the electron it is taking. Down the group the atom is larger and the incoming electron ends up further from the nucleus, with more full inner shells in between. The attraction is weaker, so the halogen is a weaker oxidising agent.
The order to remember
Oxidising power decreases down group 17: F2 > Cl2 > Br2 > I2. Fluorine is the strongest oxidising agent of any element; iodine is a fairly mild one. Everything in this part of the chapter — displacement, the reaction with iron, the reaction with hydrogen — is that single order showing itself in a different experiment.
The number behind the order (A Level, topic 24)
Standard electrode potentials put a figure on it: E°(F2/2F−) = +2.87 V, E°(Cl2/2Cl−) = +1.36 V, E°(Br2/2Br−) = +1.07 V, E°(I2/2I−) = +0.54 V. A more positive value means a stronger oxidising agent, and the difference between two of them says whether one halogen will take electrons from another's halide. The models on this page use these numbers so that their predictions are computed rather than looked up. You are not expected to use them at AS; the argument from atomic size is the one that earns the marks.
Displacement reactions of the halide ions11.2.1
This is the standard test-tube demonstration of the order. Add a halogen to a solution containing a different halide ion. If the halogen added is the stronger oxidising agent, it takes the electrons and the other halogen is set free.
Cl2(aq) + 2Br−(aq) → 2Cl−(aq) + Br2(aq)
Cl2(aq) + 2I−(aq) → 2Cl−(aq) + I2(aq)
Br2(aq) + 2I−(aq) → 2Br−(aq) + I2(aq)
The three reactions that do not happen are just as important: bromine cannot displace chloride, and iodine cannot displace either chloride or bromide. A halogen only ever displaces one from below it in the group.
| added ↓ to → | potassium chloride | potassium bromide | potassium iodide |
|---|---|---|---|
| chlorine | — | orange solution: bromine displaced | brown solution, black solid if excess: iodine displaced |
| bromine | no reaction | — | brown solution: iodine displaced |
| iodine | no reaction | no reaction | — |
Writing the two half-equations
Take chlorine and potassium bromide. The bromide ions lose electrons:
2Br− → Br2 + 2e−
and the chlorine gains them:
Cl2 + 2e− → 2Cl−
Two electrons on each side, so they add directly with nothing to multiply. The potassium ions never appear: they start and finish as K+(aq) and are spectators.
Use the organic layer
Practical questions often ask you to add a little cyclohexane and shake. Bromine and iodine can both look brown in water; in the organic layer bromine is orange and iodine is violet, which settles the identification. The organic layer floats on top, because cyclohexane is less dense than water.
Heated iron wool11.2.1
Displacement compares halogens with each other. Passing a halogen over heated iron compares them against a fixed opponent, and produces a result that is easy to remember and easy to get wrong.
2Fe(s) + 3Cl2(g) → 2FeCl3(s)
2Fe(s) + 3Br2(g) → 2FeBr3(s)
Fe(s) + I2(g) → FeI2(s)
"The halogens react with iron to form iron(III) halides"
True for chlorine and bromine, false for iodine. Iodine is not a strong enough oxidising agent to take iron past +2, so the product is iron(II) iodide. Iron(III) iodide is not a stable compound: if you could make it, the iodide ions would immediately reduce the iron(III) back to iron(II). Writing FeI3 in an answer is a straightforward loss of marks, and it is exactly the error that a question about iron wool is set to catch.
Why this is a good test of the order
Iron can be oxidised to +2 easily and to +3 only with more effort. So the experiment sorts the halogens into "strong enough to reach +3" and "not strong enough", and the boundary falls between bromine and iodine. Chlorine does it vigorously with a bright glow; bromine does it less vigorously; iodine manages only a slow reaction with continued heating, and stops one electron short.
Putting an oxidation number on it11.2.1
Every reaction in this chapter is a redox reaction, and every one of them can be read by assigning oxidation numbers and seeing which way they move. It is worth being fluent at this now, because 11.4 asks for it explicitly.
The four labels, and what they mean
Oxidised — oxidation number rises; the species has lost electrons.
Reduced — oxidation number falls; the species has gained electrons.
Oxidising agent — the species that does the oxidising, which means it is itself
reduced.
Reducing agent — the species that does the reducing, which means it is itself
oxidised.
The last two are the ones that get crossed over under exam pressure. The agent is always the opposite of what it does.
Worked example: chlorine and potassium iodide
Cl2 + 2I− → 2Cl− + I2
Chlorine: 0 in Cl2, −1 in Cl−. Down by one per atom, so chlorine is reduced, and Cl2 is therefore the oxidising agent.
Iodine: −1 in I−, 0 in I2. Up by one per atom, so iodide is oxidised, and I− is the reducing agent.
Two electrons move in total, which matches the 2 in front of each ion.
The halogens with hydrogen11.2.2
Every halogen reacts with hydrogen to give a hydrogen halide, and the equation is the same in each case:
H2(g) + X2(g) → 2HX(g)
What changes is the conditions — and the conditions are what 11.2.2 is about. The statement asks you to describe the reactions and explain their relative reactivity, so both halves have to be there.
| Halogen | Conditions needed | Product |
|---|---|---|
| fluorine (out of statement) | explodes on contact, even in the cold and in the dark | HF |
| chlorine | explodes in sunlight or ultraviolet light; burns quietly with a pale flame if hydrogen is lit in chlorine | HCl |
| bromine | reacts slowly on heating, faster over a hot platinum catalyst | HBr |
| iodine | reacts only partially even on continued heating — an equilibrium is set up | HI, mixed with unreacted H2 and I2 |
The reactivity falls steeply down the group, in the same order as everything else in this topic. Iodine is the interesting case: the reaction does not stop because it has run out of reactant, but because hydrogen iodide decomposes almost as fast as it forms.
H2(g) + I2(g) ⇌ 2HI(g)
Why the reactivity falls
Two bond energies decide it. The halogen molecule has to be split, which costs energy, and the H–X bond has to be made, which releases it. Down the group the H–X bond made gets steadily weaker — 562, 431, 366, 299 kJ mol−1 — so less and less energy is released, while the X–X bond broken changes far less. The reaction becomes less exothermic at every step, and by iodine it is barely worth the molecule's while.
The model below builds the enthalpy of formation of each hydrogen halide out of nothing but bond energies, and then checks the answer against the measured value.
The step that is nearly always dropped
An enthalpy of formation starts from the elements in their standard states. Hydrogen and chlorine are gases, so there is nothing to do. But bromine is a liquid and iodine a solid, and before you can use a bond energy you have to get them into the gas phase — +15.4 kJ mol−1 for half a mole of bromine, +31.2 for half a mole of iodine. Leave those out and the calculated value for HBr comes out about 15 kJ mol−1 too negative and for HI about 31 too negative, which is exactly the size of the error you would expect.
Thermal stability of the hydrogen halides11.2.3
Heat a hydrogen halide strongly enough and it splits back into its elements:
2HX(g) → H2(g) + X2(g)
How hot you have to make it is the trend 11.2.3 wants, and the explanation the statement asks for is a single quantity: the strength of the H–X bond.
Thermal stability decreases down group 17
HF and HCl are thermally stable — they do not decompose at any temperature you can reach with a Bunsen burner. HBr is slightly decomposed by a red-hot wire. HI decomposes readily on heating, filling the tube with purple iodine vapour. The reason is that the H–X bond gets weaker down the group: the halogen atom is larger, so the bonding pair sits further from its nucleus and is held less strongly.
The numbers, if you want them
Taking 2HX apart costs 2 × D(H–X), and gives back D(H–H) + D(X–X). For hydrogen chloride that is 2(431) − 436 − 242 = +184 kJ mol−1. For hydrogen iodide it is 2(299) − 436 − 151 = +11 kJ mol−1. Eleven kilojoules is almost nothing — ordinary heating supplies it easily, which is why the decomposition of HI is something you can watch and the decomposition of HCl is not.
"HF is the least stable because hydrogen bonding makes it different"
Hydrogen bonding does make HF different — it is why HF has a much higher boiling point than HCl, out of line with the rest of the series. But boiling point and thermal stability are different properties. Thermal stability is about breaking the H–F bond itself, which at 562 kJ mol−1 is the strongest of the four, so HF is the most thermally stable, not the least. Do not let one anomaly get imported into the wrong argument.
The reactivity argument in your own words11.2
Everything in part two rests on one chain of reasoning: bigger atom, weaker pull on the electron being taken, weaker oxidising agent. The exercise below is a check on whether you can produce that chain rather than recognise it. Every blank is marked against the data in the last section of this page, not against a stored answer key.
The halide ions as reducing agents11.3.1
Turn the chapter over. So far the halogen has been taking an electron; now the halide ion is the one giving one away.
2X− → X2 + 2e−
The two trends are the same fact seen from opposite ends. Fluorine grabs an electron hard, so fluoride hangs on to it hard and is a poor reducing agent. Iodine takes an electron only reluctantly, so iodide gives one up readily and is a good reducing agent.
Reducing power increases down group 17
I− > Br− > Cl− > F−. The larger the ion, the further its outer electrons are from the nucleus and the more inner shells shield them, so the more easily one is lost. This is the reverse of the order for the halogens as oxidising agents, and it has to be, because they are two descriptions of the same electron.
Two orders, one cause — and the marks are in saying so
A common exam question gives you one trend and asks you to deduce the other. The link is worth one clean sentence: the halogen's attraction for an added electron weakens down the group, so the halide's hold on that electron weakens too. A candidate who learns the two orders separately will sooner or later write them the same way round.
| Ion | Ionic radius / pm | As a reducing agent | Oxidised by |
|---|---|---|---|
| F− | 133 | essentially useless | nothing in this topic |
| Cl− | 181 | very weak | nothing in this topic |
| Br− | 196 | moderate | concentrated H2SO4, chlorine |
| I− | 220 | good | concentrated H2SO4, chlorine, bromine |
Silver nitrate, then ammonia11.3.2(a)
This is the standard test for a halide ion in solution, and it has two steps. The first step gives a precipitate; the second step is what makes the test actually work.
Step one. Acidify with dilute nitric acid — this removes carbonate and hydroxide ions, which would otherwise give precipitates of their own and ruin the test — then add aqueous silver nitrate:
Ag+(aq) + X−(aq) → AgX(s)
Step two. Add aqueous ammonia, dilute first and then concentrated, and see whether the precipitate dissolves.
| Halide | With silver nitrate | With dilute ammonia | With concentrated ammonia |
|---|---|---|---|
| fluoride | no precipitate — AgF is soluble | — | — |
| chloride | white precipitate | dissolves | dissolves |
| bromide | cream precipitate | does not dissolve | dissolves |
| iodide | pale yellow precipitate | does not dissolve | does not dissolve |
Why the ammonia step is the one that matters
White, cream and pale yellow are not easy to tell apart in a test tube, especially in poor light and especially when the precipitate is fine. The solubility in ammonia is unambiguous: either the cloudiness clears or it does not. The three silver halides become steadily less soluble down the group — their solubility products fall by seven powers of ten from silver chloride to silver iodide — so it takes more and more ammonia to pull the silver back into solution, and for silver iodide no concentration is enough.
What is doing the dissolving (beyond the statement)
Ammonia binds to a silver ion to form a soluble complex, which lowers the concentration of free Ag+ until the solid can no longer stay in equilibrium with it. The statement says in so many words that the formation and formula of the [Ag(NH3)2]+ complex is not required, so you will not be asked for it — but the model above uses it, because it is the only way to make the dilute-versus-concentrated boundary fall out of the numbers instead of being asserted. Answer the exam question with the observations.
"Silver halides are white, cream and yellow — that is the test"
Only half of it. An answer that stops at the colours cannot distinguish silver bromide from silver iodide reliably, and examiners know it. Quote the ammonia step as part of the test, and be precise about which ammonia: silver bromide needs the concentrated solution.
Naming an unknown halide11.3.2(a)
Practical questions usually run the test backwards: you are given the observations and asked which halide was present. The exercise below generates its observations from the same calculation the test model uses, so the reasoning is identical in both directions.
A worked identification
A solution is acidified with dilute nitric acid and silver nitrate is added. A cream precipitate forms. It does not dissolve in dilute aqueous ammonia, but dissolves in concentrated aqueous ammonia. Identify the anion and write an ionic equation.
Cream rules out chloride (white) and iodide (pale yellow); a precipitate at all rules out fluoride. The ammonia behaviour confirms it — insoluble in dilute, soluble in concentrated is the middle case. The anion is the bromide ion:
Ag+(aq) + Br−(aq) → AgBr(s)
Note the state symbols; a precipitation equation without them is usually marked down.
Concentrated sulfuric acid: the acid–base half11.3.2(b)
Add concentrated sulfuric acid to a solid sodium halide and two quite different things can happen. The first happens with every halide, and it is not a redox reaction at all.
Sulfuric acid behaving as an acid
Concentrated sulfuric acid is a stronger acid than any of the hydrogen halides, so it protonates the halide ion and drives out the hydrogen halide as a gas:
NaX(s) + H2SO4(l) → NaHSO4(s) + HX(g)
No oxidation number changes anywhere in that equation. Sodium stays +1, sulfur stays +6, the halogen stays −1.
Written out for each halide:
NaF + H2SO4 → NaHSO4 + HF
NaCl + H2SO4 → NaHSO4 + HCl
NaBr + H2SO4 → NaHSO4 + HBr
NaI + H2SO4 → NaHSO4 + HI
In every case you see steamy, acidic fumes of the hydrogen halide as it meets the moisture in the air. For sodium chloride, that is the whole story — which is exactly why this reaction is the standard laboratory preparation of hydrogen chloride.
Why it cannot be used to prepare HBr or HI
Because for those two the story does not end here. Some of the hydrogen halide that forms is immediately attacked by the acid that made it, and you end up with a mixture rather than a pure gas. If a question asks why concentrated sulfuric acid is unsuitable for preparing hydrogen bromide, that is the answer: the bromide reduces the acid.
Concentrated sulfuric acid: the redox half11.3.2(b)
Concentrated sulfuric acid is also an oxidising agent — its sulfur is at +6, the top of its range, with nowhere to go but down. Whether it actually oxidises the halide depends on how good a reducing agent that halide is, and 11.3.1 has already told you the order.
| Sodium halide | Is the halide oxidised? | Sulfur ends at | What you see |
|---|---|---|---|
| NaF | no | +6, unchanged | steamy fumes of HF (which attacks glass) |
| NaCl | no | +6, unchanged | steamy fumes of HCl only |
| NaBr | yes, partly | +4, in SO2 | steamy fumes, red-brown bromine, choking sulfur dioxide |
| NaI | yes, extensively | +4, 0 and −2 | steamy fumes, purple iodine vapour and black solid, sulfur dioxide, a yellow deposit of sulfur, and the rotten-egg smell of hydrogen sulfide |
The bromide can push sulfur down one step, from +6 to +4. The iodide, a much better reducing agent, pushes it all the way to −2, and because it does so in stages all three sulfur products are found together.
2HBr + H2SO4 → Br2 + SO2 + 2H2O
2HI + H2SO4 → I2 + SO2 + 2H2O
6HI + H2SO4 → 3I2 + S + 4H2O
8HI + H2SO4 → 4I2 + H2S + 4H2O
"Sulfuric acid oxidises the halide, so there is no HX"
Both reactions happen, and the acid–base one happens first with every halide including the iodide. Steamy fumes of hydrogen iodide are produced from sodium iodide; some of them are then oxidised. An answer listing the observations for sodium iodide that omits the steamy fumes is incomplete.
Building those equations from oxidation numbers11.3.2(b)
11.3.2(b) asks for balanced chemical equations, and four of them is more than anyone should try to memorise. They do not have to be memorised: each one can be rebuilt in about thirty seconds from the change in the oxidation number of sulfur.
The four-step method, on 8HI + H2SO4
1. Find the change in sulfur. It starts at +6 in H2SO4 and finishes at −2 in H2S. That is a fall of 8, so one sulfur atom gains 8 electrons.
2. Count how many halide ions supply them. Each I− goes from −1 to 0 and gives up one electron, so you need 8 HI.
3. Write the halogen product. Eight iodine atoms make 4 I2.
4. Balance oxygen with water, then check hydrogen. The four oxygens in H2SO4 have to go somewhere, and none of them is in H2S, so they leave as 4 H2O. Hydrogen: 8 from the HI and 2 from the acid makes 10 on the left; 2 in H2S and 8 in the water makes 10 on the right. It balances.
8HI + H2SO4 → 4I2 + H2S + 4H2O
The same four steps give the other two iodide equations — sulfur to 0 needs 6 electrons and so 6 HI; sulfur to +4 needs 2 and so 2 HI — and the bromide equation, where sulfur only falls to +4.
The halide ions in your own words11.3
Part three has two threads — reducing power increasing down the group, and the silver-halide test — and they are connected by the same argument about ionic size. Check that you can produce both.
Chlorine in water11.4.2
Chlorine dissolves in water and, to a small extent, reacts with it. What it does there is the first of the two disproportionations in this topic.
Cl2(aq) + H2O(l) ⇌ HCl(aq) + HOCl(aq)
Follow the chlorine. It starts at 0 in Cl2. In hydrochloric acid it is −1; in chloric(I) acid, HOCl, it is +1. One chlorine atom has gone down and the other has gone up, in the same reaction, with no other element changing at all.
Disproportionation
A reaction in which the same element is simultaneously oxidised and reduced. Some atoms of it end up in a higher oxidation state and some in a lower one than they started in. Recognising a disproportionation is done by assigning oxidation numbers and finding one element that moves in both directions.
Naming and writing HOCl
HOCl is chloric(I) acid; the older name, hypochlorous acid, is still widely used. Its anion, ClO−, is the chlorate(I) ion, or hypochlorite. The Roman numeral is the oxidation number of chlorine, and it is what distinguishes chlorate(I) from chlorate(V), ClO3−, which appears later in this part. Writing the formula as HOCl rather than HClO is a small kindness to yourself: it puts the atoms in the order they are actually bonded, and it makes it obvious that the hydrogen is on the oxygen.
In sunlight, a solution of chlorine in water does something different. The chloric(I) acid decomposes and oxygen is given off, so a bottle of chlorine water left on a windowsill slowly loses its pale green colour:
2HOCl(aq) → 2HCl(aq) + O2(g)
How to spot a disproportionation in three seconds
Look for an element that appears once on the left and twice on the right — or the other way round for the reverse process. Then check its oxidation number in all three places. If one is higher and one is lower than the single starting value, it is disproportionation. In Cl2 + H2O → HCl + HOCl, chlorine appears once on the left and twice on the right, which is the clue before any arithmetic is done.
Chlorine in cold, dilute sodium hydroxide11.4.1
Replace the water with cold dilute sodium hydroxide and the same disproportionation happens, but it goes to completion instead of settling at an equilibrium — the alkali removes the acids as they form.
Cl2(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O(l)
Interpreting it in terms of oxidation number
Chlorine begins at 0. One atom is reduced to −1 in sodium chloride; the other is oxidised to +1 in sodium chlorate(I), NaClO. One electron is transferred, from the atom going up to the atom going down. That is the sentence 11.4.1 is asking for, and an answer without oxidation numbers in it is not answering the statement as written.
The resulting mixture of sodium chloride and sodium chlorate(I) solution is household bleach. The chlorate(I) ion is the active ingredient — a good oxidising agent, which is how it both kills bacteria and destroys the coloured compounds in a stain.
Chlorine in hot, concentrated sodium hydroxide11.4.1
Change the conditions and chlorine goes much further up. With hot, concentrated sodium hydroxide the oxidised product is not chlorate(I) but chlorate(V), with chlorine at +5:
3Cl2(aq) + 6NaOH(aq) → 5NaCl(aq) + NaClO3(aq) + 3H2O(l)
Where the 5 : 1 comes from
This equation is not worth memorising, because it builds itself. One chlorine atom rising from 0 to +5 must give up five electrons. Each chlorine atom falling from 0 to −1 accepts only one. So five atoms must be reduced for every one oxidised — which is the 5 in front of NaCl and the 1 in front of NaClO3.
Six chlorine atoms in total means 3 Cl2 on the left. Six chlorine-containing products means six sodium ions, so 6 NaOH. Those six hydroxides bring six oxygens; three of them end up in the chlorate(V) ion, so the other three leave as 3 H2O — which also takes care of the six hydrogens. Every coefficient has come out of one electron count.
The conditions are part of the answer
These two reactions differ only in their conditions, so a question that asks for one of them is really asking whether you noticed which. Cold and dilute gives chlorate(I); hot and concentrated gives chlorate(V). Quoting the right equation under the wrong conditions scores nothing.
Disproportionation, and how to spot it11.4.1
The statement asks you to recognise these as disproportionation reactions, so it is worth being able to justify the label rather than just apply it.
| Reaction | Element that moves | From | Down to | Up to |
|---|---|---|---|---|
| Cl2 + H2O → HCl + HOCl | chlorine | 0 | −1 | +1 |
| Cl2 + 2NaOH → NaCl + NaClO + H2O | chlorine | 0 | −1 | +1 |
| 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O | chlorine | 0 | −1 | +5 |
| 3ClO− → 2Cl− + ClO3− | chlorine | +1 | −1 | +5 |
The fourth row is why heating matters. Chlorate(I) is not stable when warm: it disproportionates again, to chloride and chlorate(V). So the hot, concentrated conditions do not so much choose a different reaction as let the first product carry on reacting.
"Chlorine is oxidised because it gains electrons"
Two errors at once, and both are common enough to be worth naming. Gaining electrons is reduction, never oxidation. And in a disproportionation chlorine is doing both things, so neither word on its own describes it — the correct statement is that chlorine is simultaneously oxidised and reduced, with the oxidation numbers quoted to prove it.
Chlorine in water treatment11.4.2
11.4.2 asks for an explanation including by use of an equation, and it names the two species that do the work. Start with the equation:
Cl2(aq) + H2O(l) ⇌ HCl(aq) + HOCl(aq)
and then the second step, because chloric(I) acid is a weak acid and partly ionises:
HOCl(aq) ⇌ H+(aq) + ClO−(aq)
What kills the bacteria
HOCl and ClO− — the two species the statement names. Both are powerful oxidising agents, and they kill bacteria by oxidising the molecules those cells are built from. The chlorine also keeps algae out of reservoirs and treatment tanks, and because a little stays dissolved all the way along the pipes, it goes on protecting the water after it leaves the works.
Of the two, the uncharged molecule is much the more effective, because it crosses a bacterial cell membrane easily and a negative ion does not. Which of them you mostly have is decided by the pH.
The benefits, and the risks
For. Chlorination is one of the most effective public-health measures ever adopted. Waterborne cholera, typhoid and dysentery killed enormous numbers of people in cities before it, and very few after it. It is cheap, it works at very low concentrations, and the residual chlorine protects the water in the distribution system.
Against. Chlorine is itself toxic, so a leak at a treatment works is dangerous to the people nearby. It also reacts with the organic matter dissolved in natural water to form chlorinated hydrocarbons, some of which are suspected of being carcinogenic. And because it is added to a public supply, nobody drinking the water can easily opt out of the decision — which is the ethical objection, and a different kind of argument from the other two.
A question asking you to "discuss" this wants both sides and a conclusion. The conclusion most chemists and public-health authorities reach is that the risks are real but small, and are enormously outweighed by the deaths that chlorination prevents.
Chlorate ions as oxidising agents11.4
Chlorate(I) and chlorate(V) hold chlorine well above its usual oxidation state, so both are ready oxidising agents — which is what bleach is for. Working out the equations for their reactions is the same electron count used for the sodium hydroxide equations, run in the other direction.
Reading one of them
ClO− + H2O + 2Fe2+ → Cl− + 2OH− + 2Fe3+
Chlorine falls from +1 in ClO− to −1 in Cl−, so it gains two electrons. Iron rises from +2 to +3, one electron each, so it takes two iron(II) ions to supply them. Everything else is bookkeeping: the oxygen from the chlorate(I) and the water leaves as two hydroxide ions, and the charge comes out at +3 on both sides.
Ten things that lose marks in this topic11
Every one of these has appeared on the page above. Collected here, they make a short checklist to run through before a topic 11 question.
| # | The error | What to write instead |
|---|---|---|
| 1 | Explaining boiling points with the covalent bond | Boiling separates molecules; the forces overcome are instantaneous dipole – induced dipole forces, which strengthen as the number of electrons rises |
| 2 | Saying the X–X bond strengthens down the group | It weakens from chlorine to iodine, because the bond is longer and the shared pair is further from both nuclei |
| 3 | "Iodine is purple" | Solid iodine is grey-black; its vapour and its solution in a non-polar solvent are purple; in water it is yellow-brown |
| 4 | Writing FeI3 | Iodine oxidises iron only to +2: Fe + I2 → FeI2 |
| 5 | Getting oxidising and reducing agents the wrong way round | The oxidising agent is itself reduced; the reducing agent is itself oxidised |
| 6 | Saying HF is the least thermally stable because of hydrogen bonding | Hydrogen bonding raises HF's boiling point; thermal stability follows the H–X bond energy, and H–F is the strongest, so HF is the most stable |
| 7 | Stopping the halide test at the colours | Quote the ammonia step too, and say which ammonia: silver bromide needs concentrated |
| 8 | Omitting the steamy fumes from the observations with sodium iodide | The acid–base reaction happens with every halide; HI is produced and then partly oxidised |
| 9 | Answering 11.4.1 without an oxidation number | The statement says "in terms of changes in oxidation number" — quote 0 → −1 and 0 → +1 or +5 |
| 10 | Quoting the wrong sodium hydroxide equation | Cold and dilute gives chlorate(I); hot and concentrated gives chlorate(V) |
If you remember one thing
Down group 17 the atom gets bigger and holds an added electron less strongly. That makes the halogen a weaker oxidising agent down the group, and the halide a better reducing agent down the group. Displacement, iron wool, hydrogen, thermal stability, concentrated sulfuric acid and the silver halide test are six different experiments that all report the same fact.
Self-test11
Thirty questions across the whole of topic 11. Each one explains itself after you answer, including why the wrong options are wrong.
Definitions to learn11
Volatility
How readily a substance vaporises. A more volatile substance has a lower boiling point. Volatility decreases down group 17.
Instantaneous dipole – induced dipole forces
Attractions between molecules that arise because the electrons in a molecule are in constant motion, giving it a momentary dipole, which induces a dipole in a neighbouring molecule. They strengthen as the number of electrons in the molecule rises.
Oxidising agent
A species that takes electrons from another species, and is itself reduced in doing so. Oxidising power decreases down group 17.
Reducing agent
A species that gives electrons to another species, and is itself oxidised in doing so. The reducing power of the halide ions increases down group 17.
Disproportionation
A reaction in which the same element is simultaneously oxidised and reduced — some atoms of it end in a higher oxidation state and some in a lower one than they began in.
Thermal stability
How resistant a compound is to being decomposed by heat. For the hydrogen halides it decreases down the group, because the H–X bond gets weaker.
Chlorate(I) and chlorate(V)
ClO−, in which chlorine is at +1, and ClO3−, in which chlorine is at +5. The Roman numeral gives the oxidation number of the chlorine. HOCl is chloric(I) acid.
Data used on this page11
These are reference values, not the official data booklet
The numbers below are consistent with one another and are what every model on this page computes from, which is why they are printed rather than hidden. They are not taken from the Cambridge data booklet, and small differences are normal — bond energies in particular are averages, and published values for H–Cl range from about 428 to 432 kJ mol−1 depending on the source. In an examination, use the values supplied with the paper.
Physical properties of the elements
| Element | Melting point / K | Boiling point / K | Covalent radius / pm | Ionic radius of X− / pm | Electronegativity | Electrons per X2 |
|---|---|---|---|---|---|---|
| F2 | 53.5 | 85.0 | 71 | 133 | 3.98 | 18 |
| Cl2 | 171.6 | 239.1 | 99 | 181 | 3.16 | 34 |
| Br2 | 265.9 | 332.0 | 114 | 196 | 2.96 | 70 |
| I2 | 386.9 | 457.6 | 133 | 220 | 2.66 | 106 |
Energetics
| Element | First ionisation energy / kJ mol−1 | Electron affinity / kJ mol−1 | X–X bond energy / kJ mol−1 | H–X bond energy / kJ mol−1 | ΔHf° of HX(g) / kJ mol−1 | E°(X2/2X−) / V |
|---|---|---|---|---|---|---|
| F | 1681 | −328 | 158 | 562 | −273.3 | +2.87 |
| Cl | 1251 | −349 | 242 | 431 | −92.3 | +1.36 |
| Br | 1140 | −325 | 193 | 366 | −36.4 | +1.07 |
| I | 1008 | −295 | 151 | 299 | +26.5 | +0.54 |
H–H bond energy 436 kJ mol−1. The enthalpy needed to take half a mole of the element into the gas phase from its standard state is 0 for fluorine and chlorine, +15.4 kJ mol−1 for bromine and +31.2 kJ mol−1 for iodine. E°(Fe3+/Fe2+) = +0.77 V.
Silver halides
| Silver halide | Colour | Ksp / mol2 dm−6 | In dilute ammonia (2 mol dm−3) | In concentrated ammonia (15 mol dm−3) |
|---|---|---|---|---|
| AgF | no precipitate — soluble | — | — | — |
| AgCl | white | 1.77 × 10−10 | dissolves | dissolves |
| AgBr | cream | 5.35 × 10−13 | does not dissolve | dissolves |
| AgI | pale yellow | 8.52 × 10−17 | does not dissolve | does not dissolve |
The solubility of each silver halide in ammonia is computed on this page from its Ksp together with the overall formation constant of the diamminesilver(I) ion, taken as 1.1 × 107 dm6 mol−2. The syllabus states that the complex itself is not required; it is used here only so that the dilute-versus- concentrated boundary is calculated rather than asserted. A precipitate is counted as having dissolved when the calculated solubility exceeds 0.01 mol dm−3.
Chlorine in water
| Quantity | Value at 298 K | Used for |
|---|---|---|
| Ka of chloric(I) acid, HOCl | 2.9 × 10−8 mol dm−3 | the proportion of free chlorine present as HOCl rather than ClO− |
| pKa of chloric(I) acid | 7.54 | the pH at which the two are present in equal amounts |
| Hydrolysis constant of chlorine in water | 4.2 × 10−4 mol2 dm−6 | how far Cl2 + H2O ⇌ H+ + Cl− + HOCl proceeds |
Summary of the trends
| Property | Down group 17, fluorine to iodine | Because |
|---|---|---|
| atomic radius | increases | one more occupied shell at each step |
| electronegativity | decreases | the bonding pair is further out and better shielded |
| volatility | decreases | more electrons, so stronger instantaneous dipole – induced dipole forces |
| X–X bond energy | decreases from chlorine to iodine; fluorine is anomalously low | longer bond; and in F2, lone-pair repulsion |
| H–X bond energy | decreases | the bonding pair is further from the halogen's nucleus |
| thermal stability of HX | decreases | follows the H–X bond energy |
| oxidising power of X2 | decreases | weaker attraction for the electron being gained |
| reducing power of X− | increases | the same electron is held less tightly in the larger ion |
| solubility of AgX | decreases | — |