Advanced A level inorganic chemistry exam revision notes for

Part 9: The chemistry of the Group 17 HALOGENS

9.2 Halogen displacement reactions, reactivity trend explained

9.3 Reaction of halogens with other elements including hydrogen and selected metals

 9.2 The halogen displacement reaction and Group 7 reactivity trend

The halogen displacement reactions are fully described and explained including the oxidation state changes and the group reactivity trend.

The reaction of halogens with other elements including both metals and non–metals is also described.

A few drops of chlorine water, bromine water and iodine water are added in turn to aqueous solutions of the salts  potassium chloride (KCl), potassium bromide (KBr) and potassium iodide (KI).

Three combinations produce a reaction (and three don't!). You can get 'simple' observations from the diagrams! A darkening effect compared to a water blank confirms a displacement reaction has happened. Chlorine displaces bromine from potassium bromide and iodine from potassium iodide.  Bromine only displaces iodine from potassium iodide and the least reactive iodine cannot displace chlorine or bromine from their salts.

On the basis that the most reactive element displaces a least reactive element the reactivity order must be:

fluorine > chlorine > bromine > iodine

The word, 'molecular' symbol and ionic equations for the 1 – 3 DISPLACEMENT REACTIONS on the diagram are given below.

This involves the interchange of the halogen oxidation states of 0 and -1

1. chlorine + potassium bromide ==> potassium chloride + bromine

Cl₂(aq) + 2KBr(aq) ==> 2KCl(aq) + Br₂(aq)

Cl₂(aq) + 2Br^–(aq) ==> 2Cl^–(aq) + Br₂(aq)

2. chlorine + potassium iodide ==> potassium chloride + iodine

Cl₂(aq) + 2KI(aq) ==> 2KCl(aq) + I₂(aq)

Cl₂(aq) + 2I^–(aq) ==> 2Cl^–(aq) + I₂(aq)

3. bromine + potassium iodide ==> potassium bromide + iodine

Br₂(aq) + 2KI(aq) ==> 2KBr(aq) + I₂(aq)

Br₂(aq) + 2I^–(aq) ==> 2Br^–(aq) + I₂(aq)

The halogen molecule is the electron acceptor (the oxidising agent) and is reduced by electron gain to form a halide ion.

The oxidation state of the halogen in the halogen molecule changes (reduces) from 0 to –1, electron gain, reduction

The halide ion is the electron donor (the reducing agent) and is oxidised by electron loss to form a halogen molecule

The oxidation state of the halogen in the halide ion changes (increases) from –1 to 0, electron loss, oxidation

chlorine molecule + bromide ion ==> chloride ion + bromine molecule

ionically the redox equations are written ...

1.  Cl₂(aq) + 2Br^–(aq) ==> 2Cl^–(aq) + Br₂(aq)

because the potassium ion, K+, is a spectator ion, that is, it does not take part in the reaction. The other two possible reaction equations involving (ii) chlorine + iodide and (iii) bromine + iodide, are similar to the example above.

2.  Cl₂(aq) + 2I^–(aq) ==> 2Cl^–(aq) + I₂(aq)

3.  Br₂(aq) + 2I^–(aq) ==> 2Br^–(aq) + I₂(aq)

Explaining the Reactivity Trend of the Group 7 Halogen

Period 2 halogen:   F [2.7] + e^– ==> F^– [2.8]^–

Period 3 halogen: Cl [2.8.7] + e^– ==> Cl^– [2.8.8]^–

Period 4 halogen: Br [2.8.18.7] + e^– ==> Br^– [2.8.18.8]^–

Period 5 halogen: I [2.8.18.18.7] + e^– ==> I^– [2.8.18.18.8]^–

• When a halogen atom reacts, it gains an electron to form a singly negative charged ion e.g. Cl + e^–  ==> Cl^– which has a stable noble gas electron structure like argon. (2.8.7 ==> 2.8.8)

• As you go down the group from one Group 7 halogen element down to the next .. F => Cl => Br => I ...

  • the atomic radius gets bigger due to an extra filled electron shell,

  • the outer electrons are further and further from the nucleus and are also shielded by the extra full electron shell of negative electron charge,

  • therefore the outer electrons are less and less strongly attracted by the positive nucleus as would be any 'incoming' electrons to form a halide ion (or shared to form a covalent bond).

• SO, this combination of factors means to attract an 8th outer electron is more and more difficult as you go down the group, so the element is less reactive as you go down the group, i.e. less 'energetically' able to form the X^– halide ion with increase in atomic number.

9.3 Reactions of halogens with other elements

9.3(a) Reaction of halogens with hydrogen H₂

• Halogens readily combine with hydrogen to form the hydrogen halides which are colourless gaseous covalent molecules.

  • (Halogen compounds – covalent bonding details revision notes)

• e.g. hydrogen + chlorine ==> hydrogen chloride

• H₂(g) + Cl₂(g) ==> 2HCl(g)

• The hydrogen halides dissolve in water to form very strong acids with solutions of pH1 e.g. hydrogen chloride forms hydrochloric acid in water HCl(aq) or H⁺Cl^–(aq) because they are fully ionised in aqueous solution even though the original hydrogen halides were covalent! An acid is a substance that forms H⁺ ions in water.

• Bromine forms hydrogen bromide gas HBr(g), which dissolved in water forms hydrobromic acid HBr(aq). Iodine forms hydrogen iodide gas HI(g), which dissolved in water forms hydriodic acid HI(aq). Note the group formula pattern.

The mechanism of the direct combination of chlorine and bromine (X₂) with hydrogen is a classic case of a free radical chain reaction.

(i) initiation step: X₂ ==> 2X^.

Homolytic bond fission by heat or light to give two halogen free radicals.

(ii) propagation steps: X^. + H₂ ==> HX + H^.  

followed by    H^. + X₂ ==> HX + X^.

Two steps giving the product and a free radical to continue chain reaction

(iii) termination steps: H^. + X^. ==> HX   or   2H^. ==> H₂   or   2X^. ==> X₂

The three possible ways of ending a chain sequence.

9.3 (b) Reaction of halogens with Group 1 Alkali Metals Li Na K etc.

• Alkali metals burn very exothermically and vigorously when heated in chlorine to form colourless crystalline ionic salts e.g. NaCl or Na⁺Cl^–. This is a very expensive way to make salt! Its much cheaper to produce it by evaporating sea water!

• e.g. sodium + chlorine ===> sodium chloride

• 2Na(s) + Cl₂(g) ===> 2NaCl(s)

• The sodium chloride is soluble in water to give a neutral solution pH 7, universal indicator is green. The salt is a typical ionic compound i.e. a brittle solid with a high melting point. Similarly potassium and bromine form potassium bromide KBr, or lithium and iodine form lithium iodide LiI.  Again note the group formula pattern.

•  (Halogen compounds – ionic bonding details revision notes)

9.3(c) Reaction of halogens with other metals (iron and aluminium)

• Reaction of element with chlorine: 

  • 

  • Burns when heated strongly in chlorine gas to form the white^* solid aluminium chloride on heating in chlorine gas.

  • aluminium + chlorine ==> aluminium chloride

  • 2Al(s) + 3Cl₂(g) ==> 2AlCl₃(s)

  • ^* It is often a faint yellow in colour, due to traces of iron forming iron(III) chloride.

  

• The structure of Aluminium chloride, a curious substance in its behaviour!

  • The solid, AlCl₃, consists of a layered ionic lattice of Al³⁺ ions, each surrounded by six Cl^– ions, though you would expect some covalent character via the strong polarising power of the Al³⁺ ion and its effect on the outer electron clouds of the chloride ions.

  • However, rather curiously, on heating solid aluminium chloride, as you approach the melting point the electrical conductivity increases rapidly (ions getting more mobile) but falls to almost zero at the melting point. There is also a large 45% decrease in density at the same time!

  • What appears to happen on heating, at just below ~180^oC, the thermal kinetic energy of vibration of the ions in the lattice is sufficient to cause some freedom of movement - rise in electrical conduction.

  • BUT, at the melting point the AlCl₃ 'ionic lattice' breaks down and sublimation takes place (s ==> g), NOT a melt of ions! The reason being a volatile covalent molecule is formed.

  • In the melting process, the nature of the aluminium–chlorine bond changes from 'ionic' to covalent and the co–ordination number of the aluminium ion changes from an 'ionic' six to a 'covalent' four, forming the readily vapourised covalent dimer molecule, Al₂Cl₆, shown above.

  • So the overall change is from an ionic type lattice to a molecular lattice - when the vapour solidifies on a cold surface.

    • Note on other aluminium halides.

    • Aluminium fluoride is a 'classic' high melting ionic lattice compound with melt to give an electrically conducting fluid of mobile ions.

    • Aluminium bromide solid consists of a lattice of covalent dimer molecules with no apparent ionic character.

    • So in order of ionic character: AlF₃ > AlCl₃ > AlBr₃ (> AlI₃), which fits in with the bond character expected from a decreasing difference in the electronegativity between aluminium and the halogen.

• You can also prepare iron(III) chloride in the same way.

• iron + chlorine ===> iron(III) chloride_{(brown solid)}

• 2Fe(s) + 3Cl₂(g) ===> 2FeCl₃(s)

• Molecular iron(III) chloride is formed and consists of dimer molecules, Fe₂Cl₆

• so the equation should really be written as: 2Fe(s) + 3Cl₂(g) ===> Fe₂Cl₆(s)

• If the iron - halogen experiment is repeated with bromine the reaction is less vigorous and iron(III) bromide is formed.

  • The exothermic nature of the reaction may or may not be seen?

  • The dimer molecules are present in the brown solid.

  • 2Fe(s) + 3Br₂(g) ===> Fe₂Br₆(s)

  • The reaction is easily demonstrated by warming a little bromine with iron wool in a fume cupboard!

• When iron wool is heated with iodine there is little reaction, a small amount of iron(II) iodide is formed.

  • Fe(s) + I₂(s) ===> FeI₂(s)

  • Fe³⁺ is sufficient in oxidising power to oxidise an iodide ion to iodine,

  • so FeI₂ is formed, not FeI₃.

• Note that these reactions with iron also illustrate the halogen reactivity series.