7.2.1 Physical properties of the halogens

Colours and states of the halogens at room temperature

1. Chlorine is a pale green gas.
2. Bromine is a red-brown liquid that gives off an orange vapour.
3. Iodine is a dark grey solid that forms a purple vapour when warmed.
4. All three exist as molecules of two atoms, written ${Cl}_2$, ${Br}_2$ and $I_2$.
5. The colours deepen down the group, from pale green through red-brown to almost black.

Melting and boiling points rise down the group

1. Chlorine boils at about $-34{}^{\circ }C$, well below room temperature.
2. Bromine boils at about $59{}^{\circ }C$, so it is a liquid at room temperature.
3. Iodine melts at about $114{}^{\circ }C$, so it is a solid at room temperature.
4. Both melting point and boiling point increase steadily down the group.
5. The change of state from gas to liquid to solid is a direct consequence of that rise.

Larger molecules are held to one another more strongly

1. A halogen is a simple molecular substance, so melting separates whole molecules.
2. The intermolecular forces between molecules are what has to be overcome.
3. Molecules get larger down the group, from ${Cl}_2$ to ${Br}_2$ to $I_2$.
4. Larger molecules attract one another more strongly, so more energy is needed to separate them.
5. The covalent bonds inside each molecule are not broken when a halogen melts or boils.

Predicting fluorine and astatine

1. Fluorine sits above chlorine, so its melting and boiling points are lower still.
2. Fluorine is therefore a gas at room temperature, and its colour is paler than chlorine's.
3. Astatine sits below iodine, so its melting and boiling points are higher.
4. Astatine is therefore a solid at room temperature, and darker than iodine.
5. Each prediction extends the trend rather than recalling a measured value.

Reading a table of halogen data

1. Data are usually laid out in group order, from fluorine at the top downwards.
2. Checking that a quantity rises or falls steadily confirms which trend is present.
3. A state at room temperature is deduced by comparing $20{}^{\circ }C$ with the melting and boiling points.
4. A value above the boiling point means a gas, and below the melting point means a solid.
5. A missing entry is filled by placing the element in the order and reading the neighbouring values.

7.2.2 Reactions of the halogens

The chemical test for chlorine

1. Damp blue litmus paper is held in the gas being tested.
2. Chlorine turns the paper red and then bleaches it white.
3. The bleaching is the part that identifies chlorine, since other acidic gases only turn it red.
4. The paper must be damp, because the gas has to dissolve before it can act.
5. The test is carried out in a fume cupboard, as chlorine is toxic.

Halogens react with metals to form metal halides

1. A halogen reacts with a metal to give an ionic compound.
2. The metal atoms lose electrons and the halogen atoms gain them.
3. Sodium burns in chlorine with a bright yellow flame: $$2Na+{Cl}_2\to 2NaCl$$
4. Hot iron wool reacts with chlorine to give a brown solid: $$2Fe+3{Cl}_2\to 2{FeCl}_3$$
5. The product is named from the halogen, giving a chloride, bromide or iodide.

Predicting the reaction of other halogens with metals

1. Every halogen reacts with a metal in the same way, differing only in vigour.
2. The reaction gets less vigorous down the group, so iodine reacts more slowly than chlorine.
3. The general pattern lets any metal halide be predicted: $$metal+halogen\to metal\; halide$$
4. The charge on the metal ion decides the formula, so magnesium gives ${MgCl}_2$.
5. A halide ion always carries a charge of $1-$, whichever halogen it came from.

Halogens form hydrogen halides

1. A halogen reacts with hydrogen to give a compound of the two.
2. Hydrogen and chlorine give hydrogen chloride: $$H_2+{Cl}_2\to 2HCl$$
3. The bonding here is covalent, because both elements are non-metals.
4. Bromine gives hydrogen bromide and iodine gives hydrogen iodide, by the same pattern.
5. All the hydrogen halides are gases at room temperature.

Hydrogen halides dissolve to give acidic solutions

1. A hydrogen halide is very soluble in water.
2. On dissolving it splits into ions, releasing hydrogen ions into the solution.
3. Hydrogen ions are what make a solution acidic, so the solution has a low pH.
4. Hydrogen chloride dissolved in water is hydrochloric acid.
5. Hydrogen bromide and hydrogen iodide give hydrobromic and hydroiodic acid in the same way.

7.2.3 Relative reactivity and displacement reactions of the halogens

A more reactive halogen displaces a less reactive one

1. A halogen is added to a solution containing the halide ion of another halogen.
2. If the added halogen is more reactive, it takes the place of the one in the compound.
3. The displaced halogen is set free and colours the solution.
4. If the added halogen is less reactive, nothing happens at all.
5. Testing every pair in turn puts the halogens in order of reactivity.

What the displacement tests show

1. Chlorine displaces bromine, turning the solution orange: $${Cl}_2+2KBr\to 2KCl+{Br}_2$$
2. Chlorine displaces iodine, turning the solution brown: $${Cl}_2+2KI\to 2KCl+I_2$$
3. Bromine displaces iodine, again turning the solution brown: $${Br}_2+2KI\to 2KBr+I_2$$
4. Iodine displaces neither chlorine nor bromine, so those mixtures show no change.
5. Bromine does not displace chlorine, which completes the picture.

Reactivity decreases down the group

1. Chlorine displaces both of the others, so it is the most reactive of the three.
2. Bromine displaces iodine but not chlorine, so it sits between them.
3. Iodine displaces neither, so it is the least reactive of the three.
4. The order chlorine, bromine, iodine is therefore one of decreasing reactivity.
5. This runs opposite to Group 1, where reactivity increases down the group.

Predicting astatine

1. Astatine lies below iodine, so it is less reactive than iodine.
2. It would therefore displace none of chlorine, bromine or iodine from their compounds.
3. Any of the three would displace astatine from a solution of an astatide.
4. Chlorine and potassium astatide would react: $${Cl}_2+2KAt\to 2KCl+{At}_2$$
5. Fluorine, above chlorine, would displace every one of the others.

Electronic configuration explains the trend

1. Every halogen atom has seven outer electrons, and reacting means gaining one more.
2. Down the group each atom has more shells, so the outer shell sits further from the nucleus.
3. The inner shells also shield the outer shell from the nucleus's positive charge.
4. Distance and shielding together mean an incoming electron is attracted less strongly.
5. An electron that is gained less easily makes the halogen less reactive, which is the pattern observed.

7.2.4 Halogen displacement as a redox reaction

Electrons move from the halide ion to the halogen molecule

1. A halide ion carries an extra electron, which is what gives it its $1-$ charge.
2. In a displacement the more reactive halogen takes that electron.
3. Loss of electrons is oxidation and gain of electrons is reduction.
4. Both happen in the same reaction, which makes the displacement a redox change.
5. The electrons pass directly between the two species, with no circuit involved.

Half equations show each change separately

1. Bromide ions lose electrons, which is oxidation: $$2{Br}^{-}\to {Br}_2+2e^{-}$$
2. Chlorine molecules gain electrons, which is reduction: $${Cl}_2+2e^{-}\to 2{Cl}^{-}$$
3. Two halide ions are needed in each half equation, because a halogen molecule holds two atoms.
4. The electrons lost in one half equation equal the electrons gained in the other.
5. Adding the two together cancels the electrons and rebuilds the overall change.

Naming what is oxidised and what is reduced

1. The species that loses electrons is the one oxidised, which is the halide ion.
2. The species that gains electrons is the one reduced, which is the free halogen.
3. Charge is the giveaway: ${Br}^{-}$ to ${Br}_2$ is a loss, since the charge rises to zero.
4. ${Cl}_2$ to ${Cl}^{-}$ is a gain, since the charge falls from zero to $1-$.
5. Naming the substance rather than the element is what makes the answer unambiguous.

The ionic equation leaves out the spectators

1. Potassium bromide in solution is present as separate $K^{+}$ and ${Br}^{-}$ ions.
2. The potassium ions are unchanged on both sides, so they take no part.
3. Leaving them out gives the ionic equation: $${Cl}_2+2{Br}^{-}\to 2{Cl}^{-}+{Br}_2$$
4. Only the species whose charge changes remain in the equation.
5. The same ionic equation describes the reaction whichever soluble bromide was used.

A worked example: chlorine and potassium iodide

1. Chlorine is above iodine, so the iodide ions are the species that lose electrons.
2. The oxidation half equation is: $$2I^{-}\to I_2+2e^{-}$$
3. The reduction half equation is: $${Cl}_2+2e^{-}\to 2{Cl}^{-}$$
4. Adding them and cancelling the electrons gives: $${Cl}_2+2I^{-}\to 2{Cl}^{-}+I_2$$
5. The solution turns brown as iodine is released, which is the visible evidence.