Group 17

The halogens: colour, physical state and the trend in volatility

Group 17 contains fluorine, chlorine, bromine, iodine and astatine; the AS course runs from fluorine to iodine. Every halogen atom has the outer configuration ns2 np5, so it is one electron short of the noble gas configuration of the element after it. That one fact drives the whole topic: a halogen atom either gains one electron to become a halide ion, X-, or shares one electron in a covalent bond, and it is the most electronegative element in its period. In the elemental state the halogens exist as diatomic molecules - F2, Cl2, Br2 and I2 - each held together by a single covalent bond.

The colours are examinable facts. Fluorine is a very pale yellow gas, chlorine a pale yellow-green gas, bromine a dark red-brown liquid that gives off an orange-brown vapour, and iodine a shiny grey-black solid. Warm iodine gently and it sublimes, turning straight into a striking violet vapour without first melting, and the vapour re-deposits as dark crystals on a cold surface. Sublimation is used in the laboratory to purify iodine.

The trend you must describe is the trend in volatility, and it runs down the group. Volatility is the ease with which a substance turns into a gas, so a volatile substance has a low boiling point and a high vapour pressure. Fluorine and chlorine are gases at room temperature, bromine is a liquid and iodine is a solid, so the boiling points rise steadily from fluorine to iodine and the volatility decreases from fluorine to iodine.

There is a second, separate trend in this group that is much easier to confuse with the first: the strength of the X-X covalent bond also falls from chlorine to iodine. The two trends run in the same direction but for completely different reasons, and the physical state at room temperature is decided by the intermolecular forces, not by the covalent bond. Keep that distinction in mind for every question about melting and boiling.

Why volatility falls: instantaneous dipole-induced dipole forces

Halogen molecules are simple molecular substances. A molecule of Cl2 is symmetrical and its two atoms have the same electronegativity, so the covalent bond is non-polar and the molecule has no permanent dipole. Chlorine still condenses to a liquid at -34 °C, so there must be an attraction between the molecules. That attraction is the instantaneous dipole-induced dipole force, also called a van der Waals or dispersion force. It arises because the electrons in a molecule are not stationary: at any instant the electron cloud may be unevenly distributed, giving the molecule a temporary, instantaneous dipole. That dipole distorts the electron cloud of a neighbouring molecule and induces a matching dipole in it, and the two dipoles then attract each other.

The strength of this force depends on how polarisable the molecule is - how easily its electron cloud can be distorted - and that depends on how many electrons it has and how spread out they are. Going down Group 17 the molecules contain more electrons and the atoms get bigger, so Cl2, Br2 and I2 become progressively more polarisable and the instantaneous dipole-induced dipole forces between the molecules become stronger. F2 is the smallest and least polarisable, so its forces are the weakest.

Stronger forces between the molecules means more energy is needed to separate them, so the boiling point rises and the volatility falls. That is the whole explanation of the volatility trend, and it is the same argument that explains the rising boiling points of the alkanes or of the noble gases down a group.

Be precise about which bonds are broken. Melting and boiling a halogen break the intermolecular forces between the molecules; the X-X covalent bond inside the molecule stays intact. Bromine boils at 59 °C, but breaking the Br-Br bond needs 193 kJ mol-1 - far more energy than is available at that temperature. Writing that the covalent bonds break when a halogen boils is one of the most common ways to lose a mark in this topic.

Bond strength: the X-X bond becomes weaker down the group

The bond enthalpy of the X-X bond is the energy needed to break one mole of that bond in the gaseous molecule, X2(g) -> 2X(g). It falls steadily from chlorine to iodine, and fluorine breaks the pattern by being weaker than chlorine even though its atoms are much smaller.

From chlorine to iodine the explanation is straightforward. Going down the group the atoms get larger, so the two nuclei are further apart and the p orbitals that overlap to form the bond overlap less effectively, and the shared pair of electrons sits further from each nucleus. The attraction between the nuclei and the bonding pair is therefore weaker and less energy is needed to break the bond. This is why the I-I bond is the weakest of the three and iodine the most easily dissociated of the three molecules.

Fluorine is the exception. Its atoms are the smallest in the group, so you would expect the strongest bond, and the F-F bond is indeed very short - yet its bond enthalpy is lower than that of Cl-Cl. The reason is the lone pairs. Each fluorine atom carries three lone pairs in a very small atom, so the non-bonding pairs on the two atoms sit close together and repel each other strongly, and that repulsion partly cancels the gain from the short bond. Bond strength does not follow size alone.

Do not read too much into the X-X bond enthalpy when you are judging reactivity. Fluorine has one of the weakest X-X bonds in the group and is by far the most reactive halogen, but the vigour of a halogen's reactions depends on the balance between the energy needed to break the X-X bond and the energy released when the new bonds form. Fluorine always wins on the second term, because every bond it forms is very strong.

The halogens as oxidising agents, and displacement reactions

A halogen acts as an oxidising agent by taking electrons from another species: X2 + 2e- -> 2X-. Every halogen atom needs just one electron to complete its outer shell, so all four are oxidising agents, but their oxidising power decreases down the group: fluorine > chlorine > bromine > iodine. Fluorine is the strongest oxidising agent of all the elements.

The reason is the same increase in size and shielding that runs through the whole group. To accept an electron, a halogen atom has to attract it from a distance. Going down the group the atomic radius increases and the extra shells shield the incoming electron from the nucleus more effectively, so the attraction is weaker and the atom is less ready to gain an electron. Electronegativity falls down the group in exactly the same way - the two trends are two views of the same fact.

You can see the order directly in a displacement reaction, in which a more reactive halogen takes the place of a less reactive one in its compound. Chlorine water added to colourless potassium bromide solution turns it orange-brown, because bromine is being formed: Cl2(aq) + 2KBr(aq) -> 2KCl(aq) + Br2(aq), or as an ionic equation, Cl2(aq) + 2Br-(aq) -> 2Cl-(aq) + Br2(aq). Chlorine water added to colourless potassium iodide solution turns it brown, and if starch is present the mixture goes blue-black, which confirms iodine: Cl2(aq) + 2KI(aq) -> 2KCl(aq) + I2(aq). Bromine water displaces iodine from potassium iodide in the same way, but iodine water cannot displace either chloride or bromide.

The rule is worth learning as a single sentence: a halogen can displace the halide ion of any halogen below it in the group, and never one above it. No reaction is seen when chlorine water is added to potassium fluoride, or when iodine water is added to potassium bromide. This is what makes the displacement reactions a test of relative oxidising power, and it is also the chemistry used to extract bromine from seawater, where chlorine is bubbled through the water to displace bromide ions.

• Oxidising power: F2 > Cl2 > Br2 > I2, because the atoms become larger and more shielded down the group.
• A halogen displaces the halide ion of any halogen below it: chlorine displaces bromide and iodide, bromine displaces iodide only, iodine displaces neither.
• Chlorine water + potassium bromide solution -> orange-brown solution of bromine.
• Chlorine water + potassium iodide solution -> brown solution, or blue-black if starch is present.

Reactions with hydrogen, and the thermal stability of the hydrogen halides

Every halogen reacts with hydrogen to give a hydrogen halide: X2 + H2 -> 2HX. The reactivity falls down the group, and the conditions needed become harsher to match. Fluorine reacts explosively with hydrogen even in the dark and at low temperature, which is why the reaction is never carried out in a school laboratory. Chlorine reacts explosively in sunlight, because the reaction is photochemical and light starts a chain of free-radical steps, and it also reacts steadily when hydrogen is burned in a jet of chlorine. Bromine needs heating before it will react and a platinum catalyst is sometimes used. Iodine reacts very slowly, needs strong heating, and the reaction is reversible and never goes to completion: H2(g) + I2(g) <-> 2HI(g).

Get the explanation right, because it is not the X-X bond. The bond broken in every one of these reactions is the same H-H bond, so it cannot account for the difference. The deciding quantity is the bond enthalpy of the H-X bond that is formed, which falls down the group: H-F 562, H-Cl 431, H-Br 366 and H-I 299 kJ mol-1. Fluorine forms the strongest bond to hydrogen, so the reaction releases the most energy and is the most violent; iodine forms the weakest, so its reaction is slow and reversible. The fall in H-X bond enthalpy comes from the increasing size of the halogen atom, which makes the overlap between its orbital and the hydrogen 1s orbital less effective.

The same quantity explains the second trend, the thermal stability of the hydrogen halides. When a hydrogen halide is strongly heated it decomposes back to the elements: 2HX(g) <-> H2(g) + X2(g). Thermal stability decreases from HF to HI. Hydrogen fluoride is so stable that it can be heated to high temperatures with very little decomposition; hydrogen chloride decomposes only at very high temperature; hydrogen bromide decomposes more readily; and hydrogen iodide decomposes measurably on gentle heating, so a sample of HI turns brown as iodine forms. The reason is the H-X bond enthalpy again: the weaker the bond, the less energy is needed to break it, so the less thermally stable the compound.

Look at the three trends together, because they all point the same way: the H-X bond enthalpy, the thermal stability of the hydrogen halide and the reactivity of the halogen with hydrogen all decrease from fluorine to iodine. Keep them apart from the X-X bond enthalpy trend, which falls from chlorine to iodine for the same underlying reason - the increasing size of the halogen atom.

Halide ions as reducing agents and the silver nitrate test

A halide ion can act as a reducing agent by giving away its extra electron and turning back into the halogen: 2X- -> X2 + 2e-. Reducing power increases down the group, so the order is I- > Br- > Cl- > F-. The explanation is the size-and-shielding argument in reverse: a halide ion holds its outer electrons in shells that become larger and more shielded down the group, so the nucleus holds them less tightly and the ion loses one more easily. Iodide is oxidised readily, which is why iodine is released so readily from iodide solutions, while fluoride gives up an electron so reluctantly that it is not a useful reducing agent at all.

The first test for a halide ion uses aqueous silver ions, added as silver nitrate solution, and it identifies the ion from the colour of the precipitate: Ag+(aq) + X-(aq) -> AgX(s). Silver chloride is a white precipitate, silver bromide is cream and silver iodide is yellow. Silver fluoride is soluble, so a fluoride solution gives no precipitate - the one halide ion that fails the test entirely.

White and cream are easy to confuse, so the test is confirmed by adding aqueous ammonia. Silver chloride dissolves in dilute aqueous ammonia; silver bromide dissolves in concentrated aqueous ammonia but not in the dilute solution; silver iodide is insoluble in concentrated aqueous ammonia. The precipitate dissolves because the ammonia molecules bind to the silver ion and form a soluble complex ion, pulling silver ions out of the solid and leaving the halide ions in solution. You are not required to know the formula of that complex.

So the full test is a sequence: acidify the unknown solution with dilute nitric acid to remove any carbonate ions that would also give a white precipitate, add silver nitrate solution and note the colour, then add ammonia and note whether the precipitate dissolves. A white precipitate that dissolves in dilute ammonia confirms chloride; a cream precipitate that dissolves only in concentrated ammonia confirms bromide; a yellow precipitate that does not dissolve in concentrated ammonia confirms iodide.

• AgCl: white precipitate, dissolves in dilute aqueous ammonia.
• AgBr: cream precipitate, dissolves in concentrated aqueous ammonia but not in the dilute solution.
• AgI: yellow precipitate, insoluble even in concentrated aqueous ammonia.
• AgF is soluble, so fluoride ions give no precipitate.
• Acidify with dilute nitric acid first, so that carbonate ions do not also give a white precipitate.

Halide ions and concentrated sulfuric acid

Concentrated sulfuric acid is both a strong acid and an oxidising agent, so it reacts with a solid halide in two stages. Stage one is an acid-base reaction and it happens with every halide: the acid displaces the hydrogen halide, NaX(s) + H2SO4(l) -> NaHSO4(s) + HX(g), which is why steamy white fumes appear above the mixture. Stage two happens only if the hydrogen halide is a strong enough reducing agent to reduce the sulfur of the sulfuric acid. Because reducing power increases down the group, the further down the group you go the more of the acid is reduced and the more dramatic the observations become.

With a chloride, no redox reaction takes place. Hydrogen chloride is not a strong enough reducing agent to reduce sulfuric acid under these conditions, so NaCl(s) + H2SO4(l) -> NaHSO4(s) + HCl(g) is the only reaction and the only observation is the steamy white fumes of hydrogen chloride. No bromine, no sulfur dioxide and no sulfur appear.

With a bromide, the acid-base step gives hydrogen bromide, and that hydrogen bromide then reduces sulfur from +6 to +4: 2HBr(g) + H2SO4(l) -> Br2(g) + SO2(g) + 2H2O(l). You see steamy fumes of HBr and orange-brown bromine vapour, together with sulfur dioxide - a choking gas that turns acidified potassium dichromate(VI) paper from orange to green. Sulfur is not reduced below +4, because bromide is only a moderate reducing agent.

With an iodide, the acid-base step gives hydrogen iodide, and hydrogen iodide is a strong reducing agent, so sulfur is reduced all the way from +6 to +4, then to 0, then to -2. Three equations matter. Reducing sulfur to +4 gives 2HI(g) + H2SO4(l) -> I2(g) + SO2(g) + 2H2O(l). Reducing it further to 0 gives 6HI(g) + H2SO4(l) -> 3I2(g) + S(s) + 4H2O(l), when yellow solid sulfur is seen. Reducing it all the way to -2 gives 8HI(g) + H2SO4(l) -> 4I2(g) + H2S(g) + 4H2O(l), when hydrogen sulfide is produced - the gas with the smell of rotten eggs, which turns lead ethanoate paper black. Alongside these you see the violet vapour of iodine and, on cooling, dark solid iodine.

Chlorine with sodium hydroxide, and the purification of water

Chlorine reacts with cold dilute aqueous sodium hydroxide to give a mixture of a chloride and a chlorate(I): Cl2(g) + 2NaOH(aq) -> NaCl(aq) + NaClO(aq) + H2O(l). Look at the oxidation number of chlorine on both sides. In Cl2 it is 0; in NaCl, as the chloride ion, it is -1; in NaClO, as the chlorate(I) ion, it is +1. The chlorine has been reduced in one product and oxidised in the other at the same time, and that is the definition of disproportionation. The solution made this way is a bleach, and the species that attacks stains and bacteria is the chlorate(I) ion, ClO-.

Chlorine reacts differently with hot concentrated aqueous sodium hydroxide: 3Cl2(g) + 6NaOH(aq) -> 5NaCl(aq) + NaClO3(aq) + 3H2O(l). This time chlorine goes from 0 to -1 in the chloride and from 0 to +5 in the chlorate(V) ion, so this is disproportionation as well. The only difference between the two reactions is the temperature and the concentration of the alkali, so read the question carefully and look for the words cold and dilute or hot and concentrated before you choose an equation.

Chlorine and water. Chlorine added to water does not simply dissolve; it reacts with the water in a reversible disproportionation: Cl2(aq) + H2O(l) <-> HCl(aq) + HOCl(aq). Chlorine is 0 on the left, -1 in the HCl and +1 in the HOCl. The HOCl formed is chloric(I) acid, also called hypochlorous acid, a weak acid that ionises only partly: HOCl(aq) <-> H+(aq) + ClO-(aq). Both HOCl and ClO- are the active species that kill bacteria, by oxidising the organic material of the bacterial cell; the uncharged HOCl is the more effective of the two because it can pass through the bacterial cell wall. This is why chlorine is added to drinking water and to the water in swimming pools.

Chlorine is used on this huge scale because it is cheap and it leaves a residual disinfecting power that keeps the water safe as it travels through the pipes. The disinfectant works for the same reason a bleach works: HOCl and ClO- are strong oxidising agents, so they take electrons from the cell material of the microorganism, exactly as they take electrons from the coloured material of a dye.