INORGANIC CHEMISTRY - AS1. Reactions of Group 1 (the alkali metals) with oxygen.

  burn   4Li (s) + O2 (g) à 2Li2O (s)

    Lithium oxideBurns with a carmine-red flame; the oxide is white.

        burn  

2Na(s) + O2 (g) à Na2O2 (s)    Sodium peroxide

Burns with a yellow flame. The peroxide is pale yellow; some oxide is formed as well.        burn  

K(s) + O2 (g) à KO2 (s)    Potassium superoxide

Burns with a lilac flame; the superoxide is yellow.     

Rubidium and caesium react similarly to potassium to give superoxides; RbO2 is orange, CsO2 is red.

2. Reactions of the alkali metals with water.

2Li (s) + 2H2O (l) à 2LiOH (aq) + H2(g)

  All the metals react in a similar way. Lithium reacts slowly, melting and rushing about the surface of the water. Sodium reacts vigorously, and may catch fire; potassium reacts violently and always inflames. Rubidium and caesium react extremely violently, and would sink but for the fact they don’t survive for long enough.

3. Reactions of Group 2 (the alkaline earth metals) with oxygen.

  burn   2Mg (s) + O2(g) à 2MgO(s)

      The metal burns vigorously with a brilliant white flame.

  All the group 2 metals react in a similar way, though barium forms substantial amounts of barium peroxide Ba2O2. Calcium burns with a brick-red flame, strontium with a crimson flame, barium with a pale apple green flame. All of the oxides are white, ionic compounds.

4. Reactions of the alkaline earth metals with chlorine.

  heat   Mg(s) + Cl2(g) à MgCl2(s)

  All of the metals react similarly to give white, ionic chlorides.

5. Reactions of the alkaline earth metals with water.

  heat   Mg(s) + H2O(g) à MgO(s) + H2(g)

      Magnesium reacts very slowly with cold water, but rapidly with steam.

           

Ca(s) + 2H2O(l) à Ca(OH)2(aq) + H2(g)

Calcium reacts quite quickly with cold water to give a milky suspension of calcium hydroxide, some of which dissolves. Strontium and barium react similarly, the reaction of barium being vigorous and giving a colourless solution of barium hydroxide – the most soluble of the group 2 hydroxides.

6. Reactions of group 1 and group 2 oxides with cold water.

Group 1:    

Li2O(s) + H2O(l) à 2LiOH(aq)     

Na2O2(s) + 2H2O(l) à 2NaOH(aq) + H2O2(aq)     

2KO2(s) + 2H2O(l) à 2KOH(aq) + H2O2(aq) + O2(g)

      The superoxides of rubidium and caesium react similarly.

     

Group 2:    

MgO(s) + H2O(l) à Mg(OH)2(aq)

  All the group 2 oxides react similarly; the hydroxides are sparingly soluble and give white suspensions except for barium hydroxide. Calcium oxide reacts very exothermically.

7. Reactions of group 1 and group 2 oxides with dilute acids.

  All the reactions are shown as ionic equations with H+(aq) ions.

Group 1:    

Li2O(s) + 2H+(aq) à 2Li+(aq) + H2O(l)     

Na2O2(s) + 2H+(aq) à 2Na+(aq) + H2O2(aq)     

2KO2(s) + 2H+(aq) à 2K+(aq) + H2O2(aq) + O2(g)     

The superoxides of rubidium and caesium react similarly.     

Group 2:    

MgO(s) + 2H+(aq) à Mg2+(aq) + H2O(l)

  All the group 2 oxides react similarly except with sulphuric acid. In this case magnesium oxide reacts as given, but the sulphates of Ca – Ba are insoluble. The oxide therefore reacts superficially but reaction then ceases.

8. The thermal stability of group 1 and 2 carbonates and nitrates

Group 1 carbonates:    

Li2CO3 (s) à 2 Li2O (s) + CO2 (g)     

The rest are stable     

Group 2 carbonates:    

CaCO3 (s) à CaO (s) + CO2 (g)

The group 2 carbonates become more stable down the group. Explained in terms of polarising power of the cation – small cations polarise the carbonate ion and produce a more stable oxide lattice

Group 1 nitrates:    

4 LiNO3(s) à 2 Li2O (s) + 4 NO2(g) + O2 (g)     

2 NaNO3 (s) à 2 NaNO2 (s) + O2 (g)     

Group 2 nitrates:    

2 Mg(NO3)2 (s) à 2 MgO (s) + 4 NO2(g) + O2 (g)

9. The solubility of group 2 hydroxides and sulfates

Hydroxides increase in solubility down group; sulfates decrease in solubility down groupMg2+ (aq) + 2 OH- (s) à Mg(OH)2 (s); Ba2+ (aq) + SO4

2- (aq) à BaSO4 (s)

1. The qualitative analysis of halide ions.

The test solution is made acidic with nitric acid, which decomposes carbonates or sulphites that would interfere with the test. Silver nitrate solution is added:

Ag+(aq) + Cl- (aq) à AgCl(s)    white precipitate     

Ag+(aq) + Br- (aq) à AgBr(s)

    cream precipitate     

Ag+(aq) + I- (aq) à AgI(s)    yellow precipitate

The precipitates are then treated with ammonia solution:

dilute ammoniaAgCl(s) + 2NH3(aq) à [Ag(NH3)2]+ (aq) + Cl- (aq)

    colourless solution      concentrated ammonia

AgBr(s) + 2NH3(aq) à [Ag(NH3)2]+ (aq) + Br - (aq)    colourless solution

  Silver iodide does not react with ammonia – it is too insoluble.

 

2. The reaction of halide salts with concentrated sulphuric acid.

In the reaction with chlorides, H2SO4 acts as an acid displacing the more volatile HCl; with bromides and iodides, the halide ions are more easily reducible and the halogen is produced as well as other products. The sodium salts are typical.

Chlorides:     NaCl(s) + H2SO4(l) à NaHSO4(s) + HCl(g)

      Bromides:    

NaBr(s) + H2SO4(l) à NaHSO4(s) + HBr(g)2HBr(g) + H2SO4(l) à Br2(g) + SO2(g) + 2H2O(l)

     

Iodides:     NaI(s) + H2SO4(l) à NaHSO4(s) + HI(g)2HI(g) + H2SO4(l) à I2(s) + SO2(g) + 2H2O(l)6HI(g) + H2SO4(l) à 3I2(s) + S(s) + 4H2O(l)8HI(g) + H2SO4(l) à 4I2(s) + H2S(s) + 4H2O(l)

 

3. The disproportionation reactions of chlorine and chlorate(I).

  Chlorine is both oxidised and reduced when it reacts with water or with sodium hydroxide solution – disproportionation reactions.

Chlorine with water:          

Cl2(aq) + H2O(l) à HOCl(aq) + HCl (aq)     

Chlorine with cold dilute NaOH solution:      

Cl2(aq) + 2NaOH(aq) à NaCl (aq) + NaOCl (aq) + H2O(l)     

Chlorine and hot concentrated NaOH solution:      

3Cl2(aq) + 6NaOH(aq) à 5NaCl(aq) + NaClO3(aq) + 3H2O(l)     

Chlorate(I) ions on heating in solution:      

3 OCl- (aq) à ClO3 - (aq) + 2Cl – (aq)

 

  4. The displacement reactions of halide ions by halogens.

  Chlorine will displace both bromide and iodide ions; bromine will displace iodide ions.

Cl2(aq) + 2Br – (aq) à 2Cl – (aq) + Br2(aq)This reaction is used to manufacture bromine from seawater.

      Cl2(aq) + 2I – (aq) à 2Cl – (aq) + I2(aq)

      Br2(aq) + 2I – (aq) à 2Br – (aq) + I2(aq)

INDUSTRIAL PROCESSES  1. The Haber process for the production of ammonia.

  Nitrogen and hydrogen are reacted over a catalyst of iron with potassium oxide promoter at about 400oC and 110 atm pressure. Improvements in the process have led to a considerable reduction in the pressures used over the more familiar 350 – 1000 atm.

N2 (g) + 3H2 (g) 2NH3 (g) H = – 92 kJ mol-1

A high yield of ammonia is favoured by a low temperature (exothermic l à r) and a high pressure (reduction in the number of gas molecules l à r).

  A sensible rate is favoured by a high temperature and the presence of the catalyst. The temperature used is therefore a compromise between yield and rate.

  Objections to high pressure from exam candidates sometimes verge on the apocalyptic. Industry is perfectly well used to dealing with high pressures; the principal objection to them is the fuel costs in driving the compressors.

2. The Ostwald process for the production of nitric acid.

  Ammonia is oxidised by excess air (1:9 ratio by volume) over a 90% Pt/10% Rh catalyst at a pressure of 4 – 10 atm and a temperature of 900oC. The NO gas produced is then absorbed in water.

  Ammonia can be oxidised by oxygen in two ways, only the second of which is useful:

4NH3(g) + 3O2 (g) à 2N2(g) + 6H2O (g) H = – 1636 kJ mol – 1

      4NH3(g) + 5O2 (g) à 4NO (g) + 6H2O (g) H = – 900 kJ mol – 1

  The second reaction is favoured by

        moderate pressure        excess oxygen        a high temperature consistent with sensible rate, catalyst effectiveness, and operating costs.

  The gas mixture is cooled to around 40oC, and air added at a pressure of 7 – 12 atm. This favours the conversion of nitrogen monoxide to nitrogen dioxide. This is reacted with water in the series of reactions below. The NO produced in the last two reactions is recycled.

2NO (g) + O2 (g) 2NO2 (g)2 NO2 (g) à N2O4 (g)

      N2O4 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)

4HNO2 (aq) 2 NO (g) + N2O4 (aq) + 2H2O (l)3N2O4 (g) + 2H2O (l) 4HNO3 (aq) + 2NO (g)

3. The manufacture of sulphuric acid in the Contact Process.

  Sulphur dioxide is oxidised by oxygen over a vanadium(V) oxide catalyst. The equilibrium is exothermic, so the temperature of around 400oC is a compromise between high yield (favoured by low temperatures) and a sensible rate (favoured by high temperatures). A pressure around 2atm is used to drive the gases through the apparatus.

2SO2 (g) + O2 (g) à 2SO3 (g) H = – 192 kJ mol-1

      The sulphur trioxide is absorbed in 98% sulphuric acid, the 98.5% sulphuric acid produced then being diluted with water back to 98%. Direct reaction of sulphur trioxide with water is violent.

     

SO3 (g) + H2O (in H2SO4) à H2SO4 (l)

  4. The Hall-Heroult method for extraction of aluminium.

  Bauxite consists of aluminium oxide with iron(III) oxide (3 – 25%), silica (1 – 7%) and titanium dioxide (2 – 3%). The amphoteric aluminium oxide is dissolved by treating the finely-ground rock with 10% aqueous sodium hydroxide solution at about 4 atm and 150oC for 1 – 2 hours. Very little acidic SiO2 reacts under these conditions.

Al2O3 (s) + 6NaOH (aq) + 3H2O (l) à 2Na3Al(OH)6 (aq)     

The mixture is filtered, the solution cooled, and s3eed crystals of Al(OH)3 added. This causes precipitation of aluminium hydroxide, leaving sodium hydroxide in solution which is recovered and re-used:

  Al(OH)3   2Na3Al(OH)6 (aq) à Al2O3 (s) + 6NaOH (aq) + 3H2O (l)

      The mixture is filtered; some of the aluminium hydroxide is retained for use as seed, the remainder is heated to give pure alumina:

  heat   2Al(OH)3 (s) à Al2O3(s) + 3H2O (g)

  The alumina (5%) is dissolved in molten Na3AlF6 (95%) with calcium fluoride (5%) to lower the melting temperature of the mixture to around 950oC. The melt is electrolysed with carbon anodes in a carbon-lined steel box or ‘pot’, which is the Hall-Heroult cell. The layer of molten aluminium on the bottom of the cell is the effective cathode.

INORGANIC CHEMISTRY – A21. Reaction of the elements with oxygen.

Sodium: burn  

4Na(s) + O2(g) à 2Na2O(s)Burns with a yellow flame.

      Magnesium: burn  

2Mg (s) + O2 (g) à 2MgO (s)Burns with a brilliant white flame.

      Aluminium: burn  

4Al(s) + 3O2(g) à 2Al2O3 (s)The aluminium needs to be finely divided.

      Silicon: burn  

Si(s) + O2(g) à SiO2 (s)     

Phosphorus: burn   4P(s) + 5O2(g) à P4O10 (s)

Burns vigorously with a brilliant pinkish-white flame. White P inflames spontaneously in air; red P needs heating.

      Sulphur: burn  

S(s) + O2(g) à SO2(g)Burns with a brilliant blue flame.

      Chlorine does not react directly with oxygen.

2. Reaction of the elements with chlorine.

Sodium: burn   2Na(s) + Cl2(g) à 2NaCl(s)

      Magnesium: heat  

Mg(s) + Cl2(g) à MgCl2(s)     

Aluminium: heat   2Al(s) + 3Cl2(g) à 2AlCl3(s)

In the above two cases the dry chlorine is passed over the heated metal.     

Silicon: heat   Si(s) + 2Cl2(g) à SiCl4(l)

Dry chlorine is passed over heated silicon.     

Phosphorus: heat   2P(s) + 3Cl2(g) à 2PCl3(l)

Dry chlorine is passed over heated phosphorus; heating PCl3 with excess chlorine gives PCl5, a pale yellow solid:

PCl3(l) + Cl2(g) à PCl5(s)     

Sulphur: heat   2S(l) + Cl2(g) à S2Cl2(l)

3. Reaction of the elements with water.

Sodium:     2Na(s) + 2H2O(l) à 2NaOH(aq) + H2(g)

The reaction is vigorous and the sodium melts and floats on the water.     

Magnesium: heat   Mg(s) + H2O(g) à MgO(s) + H2(g)

Magnesium reacts slowly with cold water, but very exothermically with steam.     

Aluminium: heat   2Al(s) + 3H2O(g) à Al2O3(s) + 3H2(g)

Steam is passed over the finely-divided heated metal.

  Silicon, phosphorus and sulphur do not react with water.

  Chlorine:     Cl2(aq) + H2O(l) HOCl (aq) + HCl(aq)

The reaction is an equilibrium.

1. Basic oxides and hydroxides Na2O, NaOH, MgO and Mg(OH)2

Na2O (s) + H2O (l) à 2NaOH (aq)Na+ (aq) + OH - (aq) + H+ (aq) à Na+ (aq) + H2O (l)

      MgO (s) + 2H+ (aq) à Mg2+ (aq) + H2O (l)

Mg(OH)2 (s) + 2H+ (aq) à Mg2+ (aq) + 2H2O (l)

2. Amphoteric oxide and hydroxide Al2O3, Al(OH)3.

Basic properties:     Al2O3 (s) + 6H+ (aq) à 2Al3+ (aq) + 3H2O (l)

Aluminium oxide is very resistant to attack and this reaction is slow.Al(OH)3 (s) + 3H+ (aq) à Al3+ (aq) + 3H2O (l)

      Acidic properties:     Al2O3(s) + 6OH – (aq) + 3H2O(l) à 2 [Al(OH)6]3 – (aq)

    aluminate ionAl(OH)3 (s) + 3OH – (aq) à [Al(OH)6]3 – (aq)

3. Acidic oxides (acid anhydrides) SiO2, P4O10, SO2, SO3, Cl2O.

SiO2 (s) + 2 OH – (aq) à SiO3 2 – (aq) + H2O(l)

Silica’s giant covalent lattice means its is very resistant to attack; the alkali solution must be concentrated and needs heating.

      P4O10 (s) + 2H2O (l) à 4HPO3(aq)

This reaction is violent with cold water. On heating the solution:HPO3(aq) + H2O(l) à H3PO4(aq)

      SO2(g) + H2O (l) à H2SO3 (aq)SO3(g) + H2O (l) à H2SO4 (aq)

This latter reaction is violent, and is not used directly to make sulphuric acid.     

Cl2O (aq) + H2O(l) 2 HClO (aq)    chloric(I) acid

hypochlorous acid

In some cases the symbol ‘aq’ is used on the left-hand side of an equation to represent a large volume of solvent water.

 1. Chlorides that dissolve in water without hydrolysis.

NaCl(s) + aq à Na+ (aq) + Cl – (aq)MgCl2 (aq) + aq à Mg2+ (aq) + 2Cl – (aq)

 

2. Chlorides that hydrolyse in water.

Aluminium chloride is different from the other chlorides of period 3; it dissolves as ions, the acidity coming not from the presence of HCl but from the interaction of the hexaqua-aluminium ion with water in a deprotonation reaction similar to that which makes solutions of transition metal aqua ions acidic:

AlCl3 (s) + 6H2O (l) + aq à [Al(H2O)6]3+ (aq) + 3Cl – (aq)

[Al(H2O)6]3+ (aq) + H2O (l) à [Al(H2O)5 (OH)]2+ (aq) + H3O+ (aq)

The remaining chlorides of period 3 hydrolyse to give HCl as one of the products. All react with cold water; the phosphorus chlorides react violently. The HCl is shown in solution in the equations below, but some of it will invariably be evolved as misty acidic fumes, especially if only a small amount of water is used.

SiCl4 (l) + 2H2O (l) à SiO2 (s) + 4H+ (aq) + 4Cl – (aq)     

PCl3 (l) + 3H2O (l) à H3PO3 (aq) + 3H+ (aq) + 3Cl – (aq)     

Phosphorus pentachloride (phosphorus(V) chloride) reacts with water in two stages:PCl5(s) + H2O (l) à POCl3 (aq) + 2H+ (aq) + 2Cl – (aq)

POCl3(aq) + 3H2O (l) à H3PO4 (aq) + 3H+ (aq) + 3Cl – (aq)

  The hydrolysis of disulphur dichloride, S2Cl2, is complex. It cannot be represented by a single equation. The acidic mixture from the hydrolysis includes S, SO3

2 –, S2O3 2 – , H2S, and a range of thionic acids H2SnO6 where n is from 2 to 6 or even more.

  The dioxides of carbon and silicon are acidic; the dioxides and oxides of the remainder of group 4 are amphoteric.

  1. The oxides of carbon and of silicon; acidic oxides.

CO is often regarded as a neutral oxide. It does not react with water to give the predicted HCO2H or HCOOH, methanoic acid, though it will react with hot concentrated sodium hydroxide solution under pressure to give a solution of methanoate ions. It is really very weakly acidic.

      CO (g) + OH – (aq) à HCOO – (aq)

      Carbon dioxide is acidic, though the molecule H2CO3 is only partially formed in aqueous solution. The reaction between water and carbon dioxide is very slow, and a significant amount of CO2(aq) is present:

      CO2(aq) + H2O (l) à H+ (aq) + HCO3

– (aq)     

Silicon dioxide is acidic, but is so water-insoluble (owing to its giant covalent structure) that it reacts only with hot concentrated sodium hydroxide solution:

      SiO2 (s) + 2OH – (aq) à SiO3

2 – (aq) + H2O (l)     

The reaction of silica with calcium oxide in the blast furnace is an acid-base reaction of the Lewis type:

  500oC   SiO2 (s) + CaO (s) à CaSiO3 (l)

    slag

2. The oxides of germanium, tin and lead; amphoteric oxides.

  The Edexcel specification seems to include the oxides of germanium, wholly unimportant, and exclude those of tin which are perhaps more important. Fortunately they behave more or less similarly for all three elements, though the balance between acidic and basic properties changes with each element.

An important note.

The solutions of substances described as ‘germanates’, ‘stannates’ or ‘plumbates’, that is the products from the reaction of the oxides with alkali, are often not very well-characterised. Formulae for them were suggested many decades ago, but little work has been done on them and in any case their composition may well be variable depending on how they are made. Thus plumbates(II), produced from the reaction of PbO with alkali, may be represented [Pb(OH)6]4 – , HPbO2

– , PbO2 2 – , or PbO3

4 – . What teachers say they ‘are’ is usually related to what they were taught or to what their favourite textbook says. Any recognised representation is acceptable in an examination.

(a) The dioxides MO2. The acidic character of the dioxides decreases Ge < Sn < Pb.

  heat   GeO2 (s) + 2OH – (aq) + 2H2O (l) à [Ge(OH)6] 2 – (aq)

      The corresponding reaction for tin(IV) oxide requires concentrated alkali solution:

        heat  

SnO2 (s) + 2OH – (aq) + 2H2O (l) à [Sn(OH)6] 2 – (aq)     

The reaction with lead(IV) oxide requires molten alkali and gives a somewhat different product:

      PbO2 (s) + 2NaOH(l) à Na2PbO3 (s) + H2O (g)

  The basic character of the dioxides is illustrated by their reaction with concentrated HCl. The use of concentrated acid suppresses the hydrolysis of the chloride produced.

GeO2 (s) + 4HCl(aq) à GeCl4 (aq) + 2H2O(l)     

SnO2 (s) + 4HCl(aq) à SnCl4 (aq) + 2H2O(l)        <0oC  

PbO2 (s) + 4HCl(aq) à PbCl4 (l) + 2H2O(l)

(b) The oxides MO. Their acidic properties are shown by the reactions with aqueous alkali:

GeO (s) + 2OH – (aq) à GeO2 2 – (aq) + H2O (l)

      SnO (s) + 2OH – (aq) à SnO2

2 – (aq) + H2O (l)     

PbO (s) + 2OH – (aq) à PbO2 2 – (aq) + H2O (l)

  Their basic properties are shown in the reaction with concentrated HCl: apart from the case of PbO, this is shown simply as H+ (aq).

GeO (s) + 2H+(aq) à GeCl2(aq) + H2O (l)     

SnO (s) + 2H+(aq) à SnCl2(aq) + H2O (l)     

Lead(II) chloride is insoluble in water; in the presence of concentrated HCl it forms soluble complexes:

      PbO (s) + 2H+(aq) à PbCl2(s) + H2O (l)

      PbCl2(s) + Cl – (aq) à PbCl3

– (aq)PbCl2(s) + 2Cl – (aq) à PbCl4

2 – (aq)

(c) Trilead tetroxide, Pb3O4. This beautiful scarlet powder is a mixed oxide, containing both lead(II) and lead(IV) ions. It behaves as PbO2.2PbO.

  With dilute nitric acid, the PbO part reacts as a base to give lead(II) nitrate; PbO2 remains, since it is not basic enough to react with nitric acid under these conditions:

PbO2.2PbO (s) + 4HNO3 (aq) à PbO2 (s) + 2Pb(NO3)2 (aq) + 2H2O(l)

Its reaction with alkali is unimportant.

HYDROLYSIS OF CCl4 & SiCl4

The nature of the attack of OH- on CH3Cl is a well-established SN2 reaction. The hydrolysis of poly-halogenoalkanes does not involve a simple extension of this reaction. It is worth looking at the other halo-substituted methane derivatives before comparing CCl4 and SiCl4 hydrolysis, a popular question illuminating the differences between carbon and silicon.

 Dichloromethane.

The attack of hydroxide ion on this compound is initially SN2; the reaction is considerably slower than with chloromethane. The overall reaction is

CH2Cl2 + 2OH- à HCHO + 2Cl- + H2O.

The initial SN2 attack on dichloromethane is slow and gives chloromethanol:

CH2Cl2 + OH- à HOCH2Cl + Cl-.

Chloromethanol then reacts with hydroxide ion in a fast E2 step to give methanal, water and chloride ion.

 Trichloromethane (chloroform).

Trichloromethane would be expected to be even slower than dichloromethane in its reaction with hydroxide ions. It is not. It is faster, and does not involve hydroxide ion in an SN2 mechanism. Instead a highly reactive electron-deficient intermediate CCl2 is formed after abstraction of a proton from trichloromethane, hydroxide acting as a base:

HO- + HCCl3 à HOH + CCl3- (fast)

CCl3- à CCl2 + Cl- (slow).

CCl2 is electrophilic and reacts with water to give either carbon monoxide and chloride ions, or methanoate ion and chloride ions. The

net reaction is

3 CCl2 + 5H2O à CO + 2HCOO- + 8H+ + 6Cl- .

The overall reaction of trichloromethane with hydroxide ions is therefore

3CHCl3 + 3 OH- + 2H2O à CO + 2HCOO- + 8H+ + 9Cl- .

 Tetrachloromethane and silicon tetrachloride.

Tetrachloromethane reacts with great difficulty with hydroxide ions, whereas silicon tetrachloride reacts vigorously with water alone. The differences between the molecules are:

o The C-Cl bond length is 175 pm, Si-Cl is 206 pm (1). o The C-Cl bond dissociation enthalpy is 327 kJ mol-1 in CCl4, although the average is given as 397 ± 29 kJ mol-1 (1); Si-Cl

bond dissociation enthalpy in SiCl4 is 406 kJ mol-1. o The C-Cl electronegativity difference is 0.5, Si-Cl is 1.2. This corresponds to an approximate ionic character of 6% in

C-Cl, 30% in Si-Cl (2). o Chlorine’s covalent radius is 99 pm, carbon’s 77 pm, silicon’s 118 pm. o The silicon atom has accessible ‘empty’ d-orbitals, whereas carbon does not.

The combined effect of these is:

o The large size of the chlorine atom means that there is considerable steric hindrance on the incoming hydroxide ion with tetrachloromethane and even more in the 5-co-ordinate transition state which SN2 produces. The energy barrier to reaction is thus much larger than in the case of chloromethane with its small hydrogen atoms. Tetrachloromethane is therefore difficult to hydrolyse.

o The C-Cl bond is weaker than the Si-Cl bond despite the greater length of the latter. This is due firstly to the higher degree of ionic character in the case of silicon tetrachloride, and probably also to non-bonded repulsions between the chlorine atoms in tetrachloromethane. This latter view is supported by the weakness of C-Cl in tetrachloromethane compared with the average value for this bond.

o The silicon atom being larger than carbon means that there is less of a problem with steric hindrance in SiCl4, so the necessary 5-coordinate transition state can be formed. It may be a true 5-covalent intermediate, an sp3d hybrid.

o The d-orbitals of silicon are available to accept the lone electron pair from the attacking hydroxide ion, so the Si-Cl bond does not have to break first; in any case this is easier for Si-Cl compared with C-Cl given the much higher polarity.

The net result is the much faster hydrolysis of the silicon halide.

THE EFFECT OF AQUEOUS ALKALI ON AQUA IONSYou might think that, since the effects of aqueous sodium hydroxide and of aqueous ammonia on solutions of the common aqua ions of the first transition series are required by most A level syllabuses, the reactions and their products are well-understood. Not a bit of it.

There are several problems in getting solid information:

different books often don’t agree – usually on the formulae of the product. This might be because one book was written from another and there were transcription errors here, but more likely because:

there is genuine uncertainty about the nature of the compounds, and perhaps one author made an arbitrary choice (for the sake of simplicity) from several options and this has become perpetuated. Not many authors write books by doing all the experiments themselves or by going to the ‘original literature’. In any case much of this work is from the late 19th or early 20th century and is confused because:

techniques of analysis left something to be desired, or perhaps there really are several possibilities from a given aqua ion depending on the conditions under which the reactions are performed. The situation is unlikely to become much clearer because:

research is expensive since someone has to do it, and the facilities and the people have to be paid for, and: the problem is pretty unimportant in terms of the progress of science.

But it is quite important for you if you have to answer a question on it! I have tried to assemble here all of the answers which are likely to be acceptable, but please do not think that I have the answers – I am not psychic. Firstly though, a word on the nature of the ‘hydroxide’ precipitates generally and how you should/might write equations.

 

Transition metal hydroxides and the equations for their formation.

The simplest way for the Universe to have been constructed would be for all aqua ions to give the corresponding hydroxides when sodium hydroxide is added:

Mn+(aq) + nOH- (aq) à M(OH)n (s)

Alas, it is not that simple.

The transition metal ions are the hexaquo species [M(H2O)6]n+, octahedrally co-ordinated. The following three questions arise:

1 Does one start from this hexaqua species in any equation?

Yes.

2 Do the water molecules remain in the hydroxide that is precipitated?

Not for long.

3 Should these be shown in the equation?

Probably better not, but it wouldn’t be marked wrong.

The precipitation of the hydroxide is a sequential process of removing protons from the ligand water molecules in an acid-base reaction. The precipitation of copper(II) hydroxide from aqueous copper(II) solutions is as good an example as any, and the reactions which have been suggested in recent London mark schemes are:

[Cu(H2O)6]2+ + OH- à [CuOH(H2O)5] + + H2O

[CuOH(H2O)5] + + OH- à [Cu(OH)2(H2O)4] + H2O

Future mark schemes will change their mind, since this is not the answer though it is an acceptable answer.

The nature of the precipitate depends on how the reaction is done. If sodium hydroxide solution is added to copper(II) sulphate solution, the precipitate is greenish-blue and is probably CuSO4.3Cu(OH)2, that is the basic sulphate rather than the true hydroxide Cu(OH)2. Precipitates from solutions of copper(II) nitrate or copper(II) chloride give Cu(NO3)2.3Cu(OH)2 or CuCl2.3Cu(OH)2. This is NOT EXPECTED as an answer to an A level question – but it is what is formed. Basic salts are not in the London scheme, or in any other at A level. It explains why precipitates which are often given the same formula do not actually look the same.

If copper(II) sulphate solution is added to sodium hydroxide solution, the precipitate is blue and is gelatinous, and is probably Cu(OH)2

. Addition of ammonia to a boiling solution of copper(II) sulphate until the green precipitate becomes blue, followed by washing, and then warming with sodium hydroxide solution will give crystalline Cu(OH)2.

The composition of the precipitate is in any case variable; many oxides and hydroxides are not exactly stoichiometric, that is their formulae are approximate. In addition there is the probability that the solid will occlude (i.e. contain in pockets in the solid lattice) some of the precipitating reagent, sodium hydroxide or ammonia; this is very difficult or impossible to eliminate (Partington 1926).

Why, then, do we not tell the truth? I haven’t, in the appropriate section of my module 1 book. It is because I am limited by the syllabus – it does not include basic salts (nor should it, since they aren’t very important) and because lots of people think that it is easy for students to become overburdened with information. I think that, although you should not expect to have to reproduce it in an exam, you should know something of the background to a very complex problem. As landscape. Any other approach is patronising. Indeed, too little information makes a problem harder. Some will strongly disagree – but you don’t have to read on! But please do….

The simple lesson? That the equations that you use to represent the precipitation of metal hydroxides from alkaline solution are inevitably approximate, because the composition of the precipitate is also approximate. So what to do in your answer?

A simple solution is to write, for copper(II),

[Cu(H2O)6]2+ + 2 OH- à Cu(OH)2 + 6 H2O,

a representation which will get you the credit. If, that is, the question is about copper…..Similar versions for other aquo ions apply.

REACTIONS OF VANADIUMVanadium is not a common element – there are no concentrated ores – and it is chiefly used as an alloying metal for cutting and other tool steels, e.g. spanners. Its oxidation states of +5, +4, +3 and +2 are easily produced sequentially by reduction of solutions of vanadium(V) salts since the reduction potentials for the different states are more or less evenly and quite widely spaced.

Sulphur dioxide will reduce V(+5) to V(+4) only; zinc in hydrochloric acid will reduce V(+5) step by step to V(+2). The half equations and the overall equation is given for each of these reactions.

The aqueous chemistry of vanadium(V) is complex and depends on pH. Here we take strongly acidic solutions in which the yellow ion VO2

+ is the species present. The colour changes for the complete reduction sequence are:

+5 +5 and +4 +4 +3 +2 yellow green blue green lavenderVO2

+ VO2+ & VO2+ VO2+ [V(H2O)6]3+ [V(H2O)6]2+

The water ligands are not shown in the equations below, but remember that they are there for V(+3) and V(+2).

 The reduction equations.

+5 à +4:     with aqueous SO2:    

      2VO2

+(aq) + 4H+(aq) + 2e – à 2VO2+ (aq) + 2H2O (l)SO3

2 – (aq) + H2O (l) à SO42 – (aq) + 2H+(aq) + 2e – (aq)

2VO2+(aq) + SO3

2 – (aq) + 2H+(aq) à 2VO2+ (aq) + SO42 – (aq) + H2O (l)

      with zinc in HCl:          

2VO2+(aq) + 4H+(aq) + 2e – à 2VO2+ (aq) + 2H2O (l)

Zn(s) à Zn2+ (aq) + 2e –

2VO2+(aq) + 4H+(aq) + Zn(s) à 2VO2+ (aq) + Zn2+ (aq) + 2H2O (l)

           

+4 à +3 with zinc in HCl          

2VO2+ + 2e – + 4H+ (aq) à 2V3+ (aq) + 2H2O (l)Zn(s) à Zn2+ (aq) + 2e –

2VO2+ + Zn(s) + 4H+ (aq) à 2V3+ (aq) + Zn2+ (aq) + 2H2O (l)            

+3 à +2 with zinc in HCl          

2V3+(aq) + 2e – à 2V2+ (aq)Zn(s) à Zn2+ (aq) + 2e –

2V3+(aq) + Zn(s) à 2V2+ (aq) + Zn2+ (aq)

INORGANIC TESTS AND IDENTIFICATIONSColour

Unless the anion is a complex ion containing a transition metal, main group salts are colourless (white). Significant exceptions are AgBr, cream, and PbI2 and AgI, both yellow; all three are insoluble in water. Oxides are often brightly coloured; they are insoluble in water.

Salts of transition elements are usually coloured; however freshly prepared anhydrous CuSO4 is nearly white, CuI is pale cream – and insoluble in water.

     vanadium:o       (V) is yellowo       (IV) is blue (not quite the same hue as copper(II) ions)o       (III) is greeno       (II) is lavender

     chromium:o       (III) is green or purplish green; some crystalline salts are deep purple,  e.g. chromium(III) potassium sulphate-24 water (chrome alum)o       (VI) is yellow in CrO4

2-, orange in Cr2O72-.

     manganese:o       (II) is very pale pink o       (VI) is deep green in MnO4

2- (stable only in alkaline solutions)o       (VII) is purple in MnO4

- .

     iron:o       (II) is pale greeno       (III) is yellow in solution; some solid salts, e.g. iron(III) ammonium      sulphate, are amethyst

Commentary

     The colour found in oxides and in some other compounds such as AgBr and PbI2 is due to interactions in the crystal lattice. Lead(II) iodide when dissolved in hot water is colourless, forming shining yellow plates as it crystallises.

      Colour in transition metal ions is usually due to electron transitions within the d-shell. Intensely coloured ions with the metal in its highest oxidation state (e.g. Mn(VII), Cr(VI), Fe(VI)) derive the colour from electron transitions between the metal and the oxygen atoms.

 

     cobalt:o       (II) (hydrated) is red; anhydrous is deep blue. This is the basis of the use of CoCl2 paper to show the presence of water.o       (III) is red, indistinguishable from Co(II) by colour alone.

     nickel:o       (II) is green – a more intense colour than that of iron(II)

     copper:o       (II) is blue or occasionally bright green in solution (CuCl4

2-); solids may be eithero       (I) is nearly white (as CuI) and insoluble in water.

Solubility The following are insoluble (or nearly so) in water:

     halides: Pb2+, Ag+

     sulphates: Ba2+, Sr2+, Ca2+, Pb2+

     carbonates: all except those of the alkali metals. Aluminium carbonate does not exist.

The action of heat on the solid A little of the compound is heated in a dry ignition tube.

      The very pale colour of manganese(II) derives from the fact that the electron transitions are quantum-mechanically forbidden, so that the probability of their occurrence is low.

     sublimes: ammonium salt

     evolution of carbon dioxide: carbonate (except group 1 and 2)

     evolution of carbon monoxide and carbon dioxide: ethanedioate C2O42-

     evolution of ammonia: ammonium salt

     evolution of O2: group 1 nitrate or, if coloured, higher oxide e.g. PbO2

      The ammonium salts do not sublime, really; they thermally dissociate into substances that recombine to the ammonium compound on cooling.

     evolution of O2 and brown NO2: nitrate (not group 1 [except LiNO3])

     evolution of SO2: sulphite or sulphate

     evolution of water vapour accompanied by a colour change, suggests a hydrated salt of:

o       blue à   white à    black: copper(II)

o       green à    yellow à    red or black: iron(II)

o       yellow à    red or black: iron(III)

o       violet à    green: chromium(III)

The colour change may reverse on cooling:

     white à yellow: zinc(II)

     yellow à red: lead(II)

     red à black: iron(III).

The flame test Although used here qualitatively, the use of a flame photometer enables the quantitative measurement of the

concentrations of these ions, e.g. in biological fluids, using their flame emissions.

Ammonium nitrate can explode on heating.

Clean a flame wire (platinum is best, but is also around £60 or $100 a go – so you’ll likely get nichrome) by dipping it repeatedly in concentrated hydrochloric acid and holding it in a roaring Bunsen flame until there is no trace of colour in the flame.

To test a substance, moisten it with a few drops of concentrated hydrochloric acid, pick up the paste on the wire loop, and hold it to the edge of a roaring Bunsen flame.

      Once you have cleaned the wire, don't touch the end of it; you'll put enough salty sweat on it to give an intense sodium flame.

     bright yellow: sodium. This test is very sensitive, and it may be a good idea to do a blank test (i.e. no salt) on the hydrochloric acid that you use.

     lilac: potassium. Sodium is nearly always present, but can be filtered out by blue glass, when the potassium flame appears red.

     deep red: strontium or lithium.

     apple green: barium.

     brick red or orange-red: calcium.

     green with blue centre: copper

 

      Lithium and strontium cannot be told apart by a flame test alone.

The action of sodium hydroxide solution DEPROTONATIONTo a solution of the test substance, sodium hydroxide solution is added drop by drop until it is in excess (test with

red litmus paper). Some solutions may be acidic to begin with (test the original solution with blue litmus paper); in such cases nothing will happen until the acid has all been neutralised.

     white precipitate insoluble in excess NaOH: Mg2+, Ca2+.

     white precipitate soluble in excess NaOH: Al3+, Zn2+, Pb2+.

     blue precipitate insoluble in excess NaOH turning black on warming: Cu2+

     green precipitate insoluble in excess NaOH: Ni2+

     grey-green precipitate, turning brown on standing: Fe2+

     green precipitate soluble in excess NaOH: Cr3+.

     rust coloured (foxy red) precipitate: Fe3+.

     beige precipitate turning brown on standing: Mn2+

     blue precipitate (turning grey) from a red solution: Co2+

      The chemistry of the tests with sodium hydroxide and with ammona is covered in more detail in the cation analysis page.

The action of aqueous ammonia solution DEPROTONATION followed by LIGAND EXCHANGETo a solution of the test substance, ammonia solution is added drop by drop until it is in excess (test with red litmus

paper). Some solutions may be acidic to begin with (test the original solution with blue litmus paper); in such cases nothing will happen until the acid has all been neutralised.

     white precipitate insoluble in excess ammonia: Mg2+, Ca2+, Al3+, Pb2+.

     white precipitate soluble in excess ammonia:  Zn2+,

     blue precipitate turning to a deep blue solution with excess ammonia: Cu2+

     green precipitate soluble in excess ammonia to give pinkish solution: Ni2+

     grey-green precipitate, turning brown on standing: Fe2+

     green precipitate in soluble in excess ammonia: Cr3+.

     rust coloured (foxy red) precipitate: Fe3+.

     beige precipitate turning brown on standing: Mn2+

     blue precipitate from a red solution giving a blue solution with excess ammonia: Co2+

 

The action of dilute hydrochloric acid A little of the substance is added to about 5cm3 of dilute HCl. If there is no reaction warm gently. Test any gases

evolved.

     CO2  evolved with vigorous effervescence: carbonate or bicarbonate

     NO2 (brown) evolved: nitrite. The solution will be pale blue.

     SO2 evolved: sulphite.

     ethanoic acid evolved (smell of vinegar): ethanoate.

 

The action of concentrated sulphuric acid A little of the solid substance is added to about 2cm3 of concentrated sulphuric acid (CARE! Corrosive). If there is

no reaction the mixture is warmed cautiously.

     HCl evolved as steamy acidic fumes: chloride

     brown fumeso       HBr + Br2 + SO2 from a bromide – these will turn a drop of silver nitrate on a glass rod cloudy.o       NO2 from a nitrate or nitrite.

     purple fumes, brown mess, smell of bad eggs – HI + I2 + H2S from an iodide.

     CO and CO2 evolved: ethanedioate.

     charring: ethanedioate or ethanoate.

 

Tests for gases

These require careful technique. If the gas is to be collected it can be done by sucking the gas into a teat pipette.

     hydrogen: ignites with a squeaky pop. Very unlikely product since it is only obtained from reactive metals or from hydrides.

     oxygen: relights a glowing splint. This sometimes pops but not in the same squeaky way that hydrogen does.

     carbon monoxide: burns with a blue flame but does not explode.

     carbon dioxide: turns limewater milky.

     sulphur dioxide: when passed into acidified potassium dichromate solution turns it green.

     halogen hydrides: turn a drop of silver nitrate on a glass rod cloudy: HCl white, HBr cream, HI yellow.

     chlorine:o       turns blue litmus red, then bleaches it

o       turns moist starch-iodide paper blue-black

     bromine (brown fumes)o       reddens and then slowly bleaches blue litmus papero       turns fluorescein paper scarlet

     iodine (violet fumes) turns starch-iodide paper blue-black.

     nitrogen dioxide (brown fumes)o       holding a copper turning in the fumes intensifies the brown colourationo       turns moist starch iodide paper blue but does not affect silver nitrate solution

     ammonia:o       turns red litmus paper blueo gives white fumes with concentrated hydrochloric acid vapours.

CATIONS

1 Sodium Na+: 1.1 Flame test: covered earlier. There are no simple reagents that will precipitate sodium compounds.

2 Potassium K+: 2.1 Flame test: covered earlier. There are no simple reagents that will precipitate potassium compounds.

3 Magnesium Mg2+: 3.1 Sodium hydroxide solution precipitates white magnesium hydroxide, insoluble in excess NaOH:

Mg2+(aq) + 2OH-(aq) à Mg(OH)2(s)

3.2 Ammonia only partially precipitates magnesium hydroxide, and not at all in the presence of ammonium ions. This is because magnesium hydroxide is fairly soluble, and the small concentration of OH- ions in ammonia becomes even smaller in the presence of ammonium ions and is not sufficient to produce the precipitate.

3.3 Sodium or potassium carbonate solution gives a white gelatinous precipitate of the basic carbonate Mg(OH)2.4MgCO3.5H2O.

4 Calcium Ca2+:

Commentary

 

 

 

 

 

 

4.1 Dilute sulphuric acid gives a white precipitate of calcium sulphate if the original solution is fairly concentrated:

Ca2+(aq) + SO42-(aq) à CaSO4(s)

4.2 Sodium or potassium carbonate solution precipitates white calcium carbonate:

Ca2+(aq) + CO32-(aq) à CaCO3(s)

      Calcium sulphate solution is permanently hard water; its solubility is about 0.05 mol dm-3 at 25oC.

4.3 Ammonium ethanedioate solution precipitates white calcium ethanedioate from neutral or alkaline solutions.      Calcium ethanedioate (calcium oxalate) is

The precipitate dissolves readily in dilute acid:

Ca2+(aq) + C2O42-(aq) à CaC2O4(s)

5 Chromium(III), [Cr(H2O)6]3+:

found in rhubarb leaves - it is what makes them poisonous.

5.1 Sodium hydroxide solution precipitates grey-green chromium(III) hydroxide, which reacts with excess NaOH to give a deep green solution of the chromite ion:

[Cr(H2O)6]3+(aq) + 3OH-(aq) à Cr(OH)3(s) + 6H2O(l)

Cr(OH)3(s) + 3OH-(aq) à [Cr(OH)6]3-(aq)

      The nature of transition metal hydroxides when precipitated from aqueous solution is covered in more detail on another page.

5.2 Ammonia precipitates grey-green chromium(III) hydroxide, which with excess ammonia forms pinkish solutions of ammines. The initial equation is as 5.1 top; then

Cr(OH)3(s) + 4NH3(aq) + 2H2O(l)            à [Cr(NH3)4(H2O)2]3+(aq) + 3OH-(aq)

The exact composition of the ammine will depend on the amount of ammonia used.

5.3 Oxidising agents convert chromium(III) to yellow chromate(VI) in the presence of alkali. Hydrogen peroxide is favoured:

2[Cr(H2O)6]3+(aq) + 3H2O2(aq) + 10OH-(aq) à 2CrO42-(aq) + 14H2O(l)

     Ammines are transition metal complexes with ammonia, NH3. Do not confuse them with amines RNH2.

The test solution is made alkaline with NaOH solution, then an equal volume of ‘20 volume’ hydrogen peroxide solution is added and the mixture boiled.

6 Manganese(II) [Mn(H2O)6]2+: 6.1 Sodium hydroxide solution precipitates beige manganese(II) hydroxide. This is insoluble in excess reagent, and

rapidly darkens owing to oxidation to hydrated manganese(IV) oxide:

[Mn(H2O)6]2+(aq) + 2OH-(aq) à Mn(OH)2(s) + 6H2O(l)

     20-volume hydrogen peroxide solution gives 20cm3 of oxygen per cm3 of solution that is decomposed to oxygen and water.

4Mn(OH)2(s) + 2H2O(l) + O2(aq) à 4MnO2.H2O(s)

6.2 Sodium or potassium carbonate solution gives a white precipitate of manganese(II) carbonate:

[Mn(H2O)6]2+(aq) + CO32-(aq) à MnCO3(s) + 6H2O(l)

            6.3 Lead(IV) oxide in the presence of nitric acid converts manganese(II) into purple manganate(VII); an alternative oxidising agent is sodium bismuthate(V), NaBiO3:

5PbO2(s) + 2Mn2+(aq) + 6H+(aq) à 2MnO4

-(aq) + 5Pb2+(aq) + 2H2O(l)

5BiO3-(aq) + 14H+(aq) + 2Mn2+(aq) à

                                                  5Bi3+(aq) + 2MnO4-(aq) + 7H2O(l)

A small amount of lead(IV) oxide or of sodium bismuthate(V) is added to the test solution, 6 or so drops of concentrated nitric acid added, and the mixture boiled. Filtration will give a purple filtrate if Mn2+ was present.

7 Iron(III), [Fe(H2O)6]3+: 7.1 Sodium hydroxide solution precipitates foxy-red iron(III) hydroxide, insoluble in excess alkali:

[Fe(H2O)6]3+(aq) + 3OH-(aq) à Fe(OH)3(s) + 6H2O(l)

7.2 Ammonia reacts in the same way as sodium hydroxide – iron does not form complexes with ammonia.

7.3 Potassium hexacyanoferrate(II) precipitates dark blue potassium iron(III) hexacyanoferrate(II), Prussian Blue:

K+(aq) + Fe3+(aq) + [Fe(CN)6]4-(aq) à KFeIII[FeII(CN)6](s)

8 Iron(II), [Fe(H2O)6]2+: 8.1 Sodium hydroxide precipitates dirty green iron(II) hydroxide; on standing the surface of the precipitate turns

foxy-red owing to air oxidation to iron(III) hydroxide:

[Fe(H2O)6]2+(aq) + 2OH-(aq) à Fe(OH)2(s) + 6H2O(l)

     Prussian Blue is the pigment that is used to print 'blueprints'. It was also used for Prussian army uniforms, and for the locomotives and rolling stock of the Somerset & Dorset Joint Railway, pre-1923.

8.2 Ammonia behaves similarly to sodium hydroxide – again there are no ammine complexes.

 

8.3 Potassium hexacyanoferrate(III) precipitates dark blue potassium iron(II) hexacyanoferrate(III), Turnbull’s Blue:

K+(aq) + Fe2+(aq) + [Fe(CN)6]3-(aq) à KFeII[FeIII(CN)6](s)

8.4 Potassium manganate(VII) oxidises iron(II) to iron(III) in acidic solution. The purple manganate colour is lost and the resulting solution of iron(III) is yellow:

MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) à

                                                   Mn2+(aq) + 5Fe3+(aq) + 4H2O(aq)

8.5 Potassium dichromate(VI) oxidises iron(II) to iron(III) in acidic solution. The orange dichromate colour is lost and the resulting solution is green owing to presence of chromium(III) which masks the iron(III) colour:

Cr2O72-(aq) + 6Fe2+(aq) + 14H+(aq) à

2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

9 Copper(II), [Cu(H2O)6]2+: 9.1 Sodium hydroxide solution gives a pale blue precipitate usually described as copper(II) hydroxide:

[Cu(H2O)6]2+(aq) + 2OH-(aq) à Cu(OH)2(s) + 6H2O(l)

 

      Although Turnbull's Blue and Prussian Blue don't have quite the same colour, they are in fact the same substance.

In fact this precipitate is the basic salt – thus from copper(II) sulphate solution the product is basic copper sulphate:

2[Cu(H2O)6]2+(aq) + 2OH-(aq) + SO42-(aq) à

Cu(OH)2.CuSO4(s) + 12H2O(l)

To get real copper(II) hydroxide, the solution of the copper(II) salt must be added to excess sodium hydroxide; the precipitate is then blue rather than turquoise

     Basic copper sulphate or basic copper carbonate is responsible for the green patina - verdigris - found on weathered copper roofs. The sulphate is present in

industrial surroundings.

In both cases CAUTIOUS warming of the mixture causes the precipitate to lose water and turn black as CuO(s) is formed.

      The loss of water by heating in an aqueous medium is unusual.

9.2 Ammonia solution initially gives a blue precipitate as for sodium hydroxide. Further addition of ammonia gives deep blue soluble cuprammine complexes whose composition depends on the amount of ammonia present. The reaction is usually represented

Cu(OH)2(s) + 4NH3(aq) + 2H2O(l) à [Cu(NH3)4(H2O)2]2+(aq) +2OH-(aq)

9.3 Concentrated HCl or a saturated solution of sodium chloride gives a bright green solution containing CuCl42-

ions; dilution with water causes the solution to turn pale blue as the [Cu(H2O)6]2+ ion re-forms.

[Cu(H2O)6]2+(aq) + 4Cl-(aq) à CuCl42-(aq) + 6H2O(l)

9.4 Sodium carbonate solution precipitates greenish-blue basic carbonates of indefinite composition.

9.5 Potassium iodide solution precipitates iodine and copper(I) iodide; this reaction is used volumetrically for the estimation of copper, the liberated iodine being titrated with sodium thiosulphate solution:

2[Cu(H2O)6]2+(aq) + 4I-(aq) à 2CuI(s) + I2(aq) + 6H2O(l)

The mixture turns a sludgy brown colour; addition of sodium thiosulphate solution leaves a creamy-pink precipitate of copper(I) iodide.

10 Zinc(II), [Zn(H2O)6]2+: 10.1 Sodium hydroxide solution precipitates white zinc(II) hydroxide, easily soluble in excess sodium hydroxide to

give sodium zincate(II), because the hydroxide is amphoteric:

[Zn(H2O)6]2+(aq) + 2OH-(aq) à Zn(OH)2(s) + 6H2O(l)

Zn(OH)2(s) + 2OH-(aq) à Zn(OH)42-(aq)

10.2 Ammonia solution initially precipitates zinc(II) hydroxide, as with sodium hydroxide. Excess ammonia causes

    The cuprammins solution has the bizarre property of being able to dissolve cellulose - bits of filter paper, for example. The resulting viscous substance can be extruded into a bath of dilute sulphuric acid to give viscose fibre, which is a form of artificial silk.

the precipitate to disappear owing to the formation of ammine complexes – a different reason from that in 10.1:

Zn(OH)2(s) + 4NH3(aq) à [Zn(NH3)4]2+(aq) + 2OH-(aq)

10.3 Sodium carbonate solution precipitates a white basic carbonate:

5Zn2+(aq) + 2CO32-(aq) + 6OH-(aq) à 2ZnCO3.3Zn(OH)2(s)

This latter substance is what you get when you buy zinc carbonate – which is why on heating it gives off water vapour as well as carbon dioxide.

             10.4 Sodium or potassium chromate(VI) solution gives a precipitate of yellow zinc chromate(VI); this is readily soluble in acid.

[Zn(H2O)6]2+(aq) + CrO42-(aq) à ZnCrO4(s) + 6H2O(l)

 

11 Aluminium, Al[H2O] 3+: 11.1 Sodium hydroxide solution precipitates white gelatinous aluminium hydroxide. This reacts with excess NaOH

to give a colourless solution of sodium aluminate:

[Al(H2O)6] 3+(aq) + 3OH-(aq) à Al(OH)3(s) + 6H2O(l)

 Al(OH)3(s) + 3OH-(aq) à [Al(OH)6]3-(aq)

11.2 Ammonia solution precipitates white aluminium hydroxide, as with NaOH; however it does not react further with excess ammonia.

 

12 Lead(II), Pb2+: 12.1 Dilute hydrochloric acid or other soluble chlorides precipitate white lead(II) chloride from moderately

concentrated solutions of lead(II) salts.

      Chromates are used as the pigments to produce the popular yellow lines on roads.

Pb2+(aq) + 2Cl-(aq) à PbCl2(s)

The precipitate dissolves in an excess of concentrated hydrochloric acid to give yellow solutions containing various species including PbCl3

- and PbCl42-.

12.2 Sodium hydroxide solution gives a white precipitate of lead(II) hydroxide which is soluble in excess NaOH to give colourless sodium plumbate(II)

Pb2+(aq) + 2OH-(aq) à Pb(OH)2(s)

Pb(OH)2(s) + 2OH-(aq) à [Pb(OH)4]2-(aq)

12.3 Ammonia solution precipitates white lead(II) hydroxide; this does not dissolve in excess ammonia, because no complexes are formed and ammonia is not a strong enough base to bring out the amphoteric properties of lead(II) hydroxide.

12.4 Dilute sulphuric acid gives a white precipitate of lead(II) sulphate:

                        Pb2+(aq) + SO42-(aq) à PbSO4(s)

12.5 Potassium chromate(VI) solution gives a bright yellow precipitate of lead(II) chromate(VI):

Pb2+(aq) + CrO42-(aq) à PbCrO4(s)

12.6 Potassium iodide solution gives a bright yellow precipitate of lead(II) iodide. This will dissolve in boiling water to give a colourless solution – the colour comes from interactions in the crystal lattice rather than from coloured ions:

Pb2+(aq) + 2I-(aq) à PbI2(s) 

13 Ammonium ion NH4+:

13.1 Heat: all ammonium salts decompose on heating; ammonium nitrate may explode. Ammonium chloride and sulphate give products that recombine on cooling, so that the salts apparently sublime.

NH4Cl(s) l NH3(g) + HCl(g)

NH4NO3(s) à N2O(g) + 2H2O(g)

2(NH4)2SO4(s) l 2NH3(g) + SO3(g) + H2O(g)

Ammonium dichromate(VI) decomposes spectacularly on ignition in a reaction that is oxidation of the cation by the anion; the initially orange solid gives a fluffy green product of much larger volume:

(NH4)2Cr2O7(s) à N2(g) + Cr2O3(s) + 4H2O(g)

13.2 Alkalis (sodium hydroxide, calcium hydroxide) liberate ammonia from ammonium salts on warming with the solution, or even from a mixture of the solids. This is because OH- is a stronger base than ammonia, so removes a hydrogen ion from the ammonium ion:

NH4+(aq) + OH-(aq) à NH3(g) + H2O(l)

The test solution is warmed with sodium hydroxide solution and the vapours tested with moist red litmus paper. It is important to test the vapours immediately heating begins, since the ammonia is lost very quickly and by the time the solution boils it may well have all gone.

 

 

14. Cobalt (II), [Co(H2O)6]2+:            The (+2) state is the most stable for simple cobalt salts; they are coloured pink or blue. The pink colour of the hydrated ion [Co(H2O)6]2+ changes to blue on heating, on dehydration, or in the presence of concentrated acids.

             14.1 Sodium hydroxide solution precipitates a blue basic salt which on warming with excess sodium hydroxide forms solid pink cobalt(II) hydroxide. With a solution of cobalt(II) nitrate the equations are:

[Co(H2O)6]2+ (aq) +  NO3 – (aq) + OH – (aq)  → Co(OH)NO3 (s) + 6H2O(l)

        blue

   

        Basic salts, usually hydroxides or carbonates, contain not only the expected hydroxide or carbonate ion but also the anion from the original salt. The precipitate from cobalt(II) nitrate solution is therefore different from

Co(OH)NO3 (s)  +  OH – (aq)  → Co(OH)2 (s)  +  NO3 –  (aq)

     pink

If the pink precipitate is boiled for some time (or is warmed with hydrogen peroxide solution) it is converted to brownish-black cobalt(III) hydroxide:

4Co(OH)2 (s)  +  2H2O (l)  +  O2 (aq) →  4Co(OH)3 (s)

             14.2 Ammonia solution gives a blue basic precipitate as with sodium hydroxide; excess ammonia converts this to a brown solution of a cobaltammine which turns red on exposure to air giving an ammine of cobalt(III). The overall reaction is:

Co(OH)NO3 (s) + 28NH3(aq) + 6H2O(l) + O2(aq)   → 4[Co(NH3)6](OH)3(aq) + 4NH4NO3(aq)

 

that which would be given from cobalt(II) chloride. The latter would be Co(OH)Cl.

15. Nickel(II), [Ni(H2O)6]2+:              15.1 Sodium hydroxide solution gives a green gelatinous precipitate of nickel(II) hydroxide, insoluble in excess of the reagent:

[Ni(H2O)6]2+ (aq)  +  2OH – (aq)  →  Ni(OH)2 (s)  +  6H2O(l)

               15.2 Ammonia solution gives a green precipitate of a basic salt which dissolves readily in excess ammonia solution to form a blue-violet solution of complex nickel ammines. With nickel(II) chloride the reactions are:

[Ni(H2O)6]2+(aq) + Cl – (aq) + NH3(aq) →  Ni(OH)Cl(s) + NH4+(aq) + 5H2O(l)

Ni(OH)Cl(s)  + 5NH3(aq) + H2O(l) →  [Ni(NH3)4](OH)2(aq)  +  NH4+(aq) + Cl – (aq)

Ni(OH)Cl(s)  + 7NH3(aq) + H2O(l) →  [Ni(NH3)6](OH)2(aq)  +  NH4+(aq) + Cl – (aq) 

 

  

ANIONS14 Chloride Cl-:

AgCl, PbCl2, Hg2Cl2 and CuCl are insoluble in water.Commentary

14.1 Concentrated sulphuric acid liberates steamy acidic fumes of HCl from solid chlorides:

NaCl(s) + H2SO4(l) à NaHSO4(s) + HCl(g)

     Sulphuric acid does not form sodium sulphate in this reaction even on strong heating - 'at no temperature attainable in glass', to use the phrase of a fine classic textbook.

14.2 Addition of silver nitrate solution to a solution of a chloride that has been acidified (test with blue litmus paper) with dilute nitric acid gives a white precipitate of silver chloride. The precipitate is readily soluble in dilute ammonia or in sodium thiosulphate solution:

Ag+(aq) + Cl-(aq) à AgCl(s)

AgCl(s) + 2NH3(aq) à [Ag(NH3)2]+(aq) + Cl-(aq)

AgCl(s) + 2S2O32-(aq) à [Ag(S2O3)2]3-(aq) + Cl-(aq)

Acidification with nitric acid is necessary to eliminate carbonate or sulphite, both of which interfere with the test by giving spurious precipitates.

Concentrated solutions of sulphates can give a precipitate of silver sulphate in this test; its appearance is wholly different

from AgCl. The latter is truly white; the sulphate is a pearly white, rather like pearlescent nail varnish.

 

15 Bromide Br-:

AgBr, PbBr2, Hg2Br2 and CuBr are insoluble in water.

15.1 Concentrated sulphuric acid gives a mixture of hydrogen bromide, bromine and sulphur dioxide with solid bromides; the HBr produced is oxidised by sulphuric acid. The mixture evolves steamy brownish acidic fumes:

NaBr(s) + H2SO4(l) à NaHSO4(s) + HBr(g)

 

2HBr + H2SO4 à Br2 + SO2 + 2H2O

 

15.2 Addition of silver nitrate solution to a solution of a bromide that has been acidified (test with blue litmus paper) with dilute nitric acid gives a cream precipitate of silver bromide. The precipitate is readily soluble in concentrated ammonia:

Ag+(aq) + Br-(aq) à AgBr(s)

AgBr(s) + 2NH3(aq) à [Ag(NH3)2]+(aq) + Br-(aq)

Acidification with nitric acid is necessary to eliminate carbonate or sulphite, both of which interfere with the test by

giving spurious precipitates.

15.3 Oxidising agents oxidise bromide to bromine, which is yellow or orange in aqueous solution. Bromine can be extracted from the solution by shaking with an immiscible organic solvent, for example hexane, the organic layer then turning orange.

A suitable oxidising agent is sodium chlorate(I); this is added to the test solution, followed by a little dilute hydrochloric acid and a few cm3 of hexane:

OCl-(aq) + 2H+(aq) + 2Br-(aq) à Br2(aq) + Cl-(aq) + H2O(l)

16 Iodide I-:

AgI, PbI2, Hg2I2 and CuI are insoluble in water.

16.1 Concentrated sulphuric acid gives a mixture of hydrogen iodide, iodine, hydrogen sulphide, sulphur and sulphur dioxide when added to solid iodides; the HI produced is oxidised by sulphuric acid. The mixture evolves purple acidic fumes, turns to a brown slurry, and is a mess:

NaI(s) + H2SO4(l) à NaHSO4(s) + HI(g)2HI + H2SO4 à I2 + SO2 + 2H2O

6HI + H2SO4 à 3I2 + S + 4H2O

8HI + H2SO4 à 4I2 + H2S + 4H2O 

     There are no state symbols in these equations because the mixture is such a mess, and is more sulphuric acid than

16.2 Addition of silver nitrate solution to a solution of an iodide that has been acidified (test with blue litmus paper) with dilute nitric acid gives a yellow precipitate of silver iodide. The precipitate is insoluble even in concentrated ammonia:

Ag+(aq) + I-(aq) à AgI(s)

Acidification with nitric acid is necessary to eliminate carbonate or sulphite, both of which interfere with the test by giving spurious precipitates.

16.3 Oxidising agents oxidise iodide to iodine, which is yellow or orange in aqueous solution. Iodine can be extracted from the solution by shaking with an immiscible organic solvent, for example hexane, the organic layer then turning purple.

A suitable oxidising agent is sodium chlorate(I); this is added to the test solution, followed by a little dilute hydrochloric acid and a few cm3 of hexane:

OCl-(aq) + 2H+(aq) + 2Ir-(aq) à I2(aq) + Cl-(aq) + H2O(l)

water.

16.4 Lead ethanoate or lead nitrate solutions give a bright yellow precipitate of lead(II) iodide with iodides:

Pb2+(aq) + 2I-(aq) à PbI2(s)

      The colour of lead(II) iodide comes from interactions in the lattice; dissolving the salt in boiling water gives a colourless solution

which deposits glittering yellow plates on cooling.

16.5 Solutions of copper(II) salts give a brown mixture containing iodine and copper(I) iodide when added to solutions of iodides. Addition of sodium thiosulphate solution decolourises the iodine and leaves pinkish-cream copper(I) iodide as a precipitate.

2Cu2+(aq) + 4I-(aq) à 2CuI(s) + I2(aq)

2S2O32-(aq) + I2(aq) à 2I-(aq) + S4O6

2-(aq)

17 Sulphite SO32-

Sulphurous acid is considerably stronger than carbonic acid, so sulphites do not give the effervescence that is characteristic of carbonates when dilute acid is added.

 

17.1 Dilute hydrochloric acid on warming with a sulphite evolves sulphur dioxide; this turns acidified potassium dichromate(VI) solution (or paper) green:

SO32-(aq) + 2H+(aq) à H2O(l) + SO2(g)

17.2 Barium chloride solution gives a white precipitate of barium sulphite; addition of dilute hydrochloric acid causes the precipitate to dissolve without effervescence:

SO32-(aq) + Ba2+(aq) à BaSO3(s)

    

This reaction is the basis for the volumetric estimation of copper; the  iodine liberated from a known amount of a copper(II) solution is titrated with standard sodium thiosulphate solution.

 

 

18 Sulphate SO42-:

BaSO4, SrSO4 and PbSO4 are insoluble; CaSO4 is sparingly soluble.18.1 Barium chloride solution added to the test solution acidified with dilute hydrochloric acid gives a white precipitate of barium sulphate:

Ba2+(aq) + SO42-(aq) à BaSO4(s)

HSO4- does the same thing with barium ions; however the original test solution would then be very acidic, so that should

be tested for.

18.2 Lead ethanoate solution gives a precipitate of white lead sulphate:

Pb2+(aq) + SO42-(aq) à PbSO4(s)

19 Nitrate NO3-

Since all nitrates are water soluble, there is no precipitation reaction for this ion.

 

19.1 Solid nitrates decompose on heating; those of group 1 (except Li) give the nitrite and oxygen;

2NaNO3(s) à 2NaNO2(s) + O2(g)

      The addition of HCl destroys any carbonate or sulphite ions present so prevents the spurious positive result due to the precipitation of these barium salts.

    Barium nitrate solution can also be used instead of barium chloride.

All others give the metal oxide, nitrogen dioxide, and oxygen. A brown gas is emitted that re-lights a glowing splint:

2Pb(NO3)2(s) à 2PbO(s) + O2(g) + 2NO2(g)

 

19.2 Nitrate ions are reduced to ammonia by boiling with aluminium or with Devarda’s Alloy in sodium hydroxide solution. Devarda’s Alloy contains aluminium, zinc and copper. Since ammonium ions also give ammonia with NaOH, the test solution must be boiled with NaOH and the vapour tested for ammonia; if present heating must continue until all the ammonia has gone. The mixture is then cooled, Devarda’s Alloy (or a piece of aluminium foil) added, and the mixture re-heated. A gas that turns moist red litmus paper blue indicates nitrate in the original solution:

3NO3-(aq) + 8Al(s) + 18H2O(l) + 21 OH-(aq) à

                                                                    8[Al(OH)6]3-(aq) + 3NH3(g)

Not an equation to be remembered!

20 Carbonate CO32- (see also 21)

Only the alkali metal and ammonium carbonates are water soluble. Some carbonates (e.g. zinc, copper(II)) are basic carbonates and contain a proportion of the hydroxide in their structure.

20.1 Heating decomposes all but the alkali and alkaline earth metal carbonates (at Bunsen temperatures) giving the oxide and carbon dioxide:

CuCO3(s) à CuO(s) + CO2(g)20.2 Dilute hydrochloric acid gives vigorous effervescence with carbonates, evolving carbon dioxide:

CO32-(aq or s) + 2H+(aq) à H2O(l) + CO2(g)

     Bicarbonates also give this effervescence. The reaction of carbonates

21 Hydrogen carbonate (bicarbonate) HCO3-:

Only the alkali metal and ammonium bicarbonates are obtainable as solids; group 2 bicarbonates exist only in solution.

21.1 Addition of calcium chloride solution to a bicarbonate solution gives no precipitate since calcium bicarbonate is soluble; this distinguishes it from carbonate, which does give a precipitate. On heating the calcium chloride/bicarbonate mixture a white precipitate appears since the bicarbonate decomposes to carbonate:

Ca2+(aq) + 2HCO3-(aq) à CaCO3(s) + CO2(g) + H2O(l)

22 Ethanoate CH3COO-:

22.1 Ethanoates on heating with dilute hydrochloric acid give ethanoic acid, recognisable by its vinegary smell.

22.2 Neutral iron(III) chloride solution added to neutral solutions of ethanoate ion give a deep red colouration owing to formation of iron(III) ethanoate.

23 Ethanedioate C2O42-:

23.1 Addition of concentrated sulphuric acid to a solid ethanedioate salt gives a mixture of carbon monoxide and carbon dioxide from dehydration of the ethanedioic acid formed:

HOOC-COOH à H2O + CO2 + CO

23.2 Addition of a solution of an ethanedioate to a little potassium manganate(VII) solution acidified with dilute sulphuric acid causes the purple colour to disappear:

with acid is exothermic; bicarbonates react endothermically.

2MnO4-(aq) + 5C2O4

2-(aq) + 16H+(aq) à 2Mn2+(aq) + 10CO2(g) + 8H2O(l)

23.3 Addition of a solution of calcium chloride to a solution of an ethanedioate salt gives a white precipitate of calcium ethanedioate:

Ca2+(aq) + C2O42-(aq) à CaC2O4(s)

The precipitate dissolves readily in dilute hydrochloric acid.

24 Chromate(VI) CrO42- and dichromate(VI) Cr2O7

2-.

These ions are related through the equilibrium

Cr2O72-(aq) + 2OH-(aq)          2CrO4

2-(aq) + H2O(l)

In alkaline solution the yellow chromate(VI) dominates, in acidic solution orange dichromate(VI). All dichromate(VI) salts are soluble; addition of dichromate(VI) ions to solutions of ions of metals which have insoluble chromate(VI) salts leads to the precipitation of chromates. This means that the only dichromates that can exist are those of group 1 metals, ammonium, magnesium, calcium and strontium.

25.1 Addition of barium chloride solution to a chromate(VI) or dichromate(VI) solution precipitates bright yellow barium chromate(VI):

Ba2+(aq) + CrO42-(aq) à BaCrO4(s)

25.2 A solution of dichromate(VI) ions is an oxidising agent; oxidation is shown by the solution turning from orange to green (Cr(III)). The following can be oxidised:

(a) iron(II) to iron(III):

     The addition of a heavy metal ion to potassium dichromate solution precipitates the chromate and therefore moves the equilibrium to the right hand side. If the chromate is sparingly soluble (e.g. strontium) the supernatant liquid will remain yellow. Very insoluble chromates,

Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) à 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

(b) iodide to iodine (the solution turns murky greenish-brown):

Cr2O72-(aq) + 14H+(aq) + 6I-(aq) à 2Cr3+(aq) + 7H2O(aq) + 3I2(aq)

(c) sulphite to sulphate:

Cr2O72-(aq) + 8H+(aq) + 3SO3

2-(aq) à 2Cr3+(aq) + 4H2O(l) + 3SO42-(aq)

(d) nitrite to nitrate:

Cr2O72-(aq) + 8H+(aq) + 3NO2

-(aq) à 2Cr3+(aq) + 4H2O(l) + 3NO3-(aq)

 

such as lead, remove all the colour from the supernatant liquid.

25.3 Hydrogen peroxide reacts with acidified dichromate(VI) solutions to give a blue compound that rapidly turns green and evolves oxygen. The blue compound can be extracted into an organic solvent such as butan-1-ol.

Cr2O72-(aq) + 8H+(aq) + 3H2O2 (aq) à 2Cr3+(aq) + 7H2O(l) + 3 O2(g)

 

25.4 Acidified potassium dichromate(VI) solutions oxidise primary alcohols to aldehydes and then acids, secondary alcohols to ketones. Ethanol can be used to test for dichromate(VI) ions, therefore, the solution turning green and the apple smell of ethanal being evident.

      The blue compound is CrO5, which contains a peroxy structure. It is covalent, and is stable in organic solvents though not in water.

INORGANIC PROBLEM 1A is a white compound of lead; on heating it gives yellow B and a colourles gas that turns limewater milky. If B is heated in air for several hours at 470oC it is converted into a scarlet powder C. C contains 90.66% lead and 9.34% oxygen by mass.

C on heating with dilute aqueous nitric acid gives a colourless solution D and a brown solid E. If sodium hydroxide solution is added to solution D, a white gelatinous precipitate is formed, which with excess sodium hydroxide solution gives a colourless solution G.

Compound E reacts with concentrated hydrochloric acid to give a white solid I and a green gas J. I is soluble in hot water but insoluble in cold, and forms soluble complex ions such as K with excess chloride ions. With potassium iodide solution J gives a brown solution of L, which on shaking with hexane gives a deep purple organic layer.

With potassium iodide solution D gives a bright yellow solid M; this is insoluble in cold water, but in hot water gives a colourless solution.

Identify all the substances A - M, and write equations for all the reactions involved.

 

INORGANIC PROBLEM 2Copper reacts with 50% nitric acid to give a blue solution A and a brown gas B. If the solution A is diluted and sodium hydroxide solution added cautiously, a gelatinous blue precipitate C is obtained, which if warmed forms a black solid G.

Addition of concentrated ammonia solution to C gives a deep blue solution that contains the ion D. Addition of concentrated hydrochloric acid to C gives a green solution of the ion E, which on dilution with water gives a solution of the ion F.

If the brown gas B is passed into water a mixture of acids H and I is formed in a disproportionation reaction.

Identify A - I, giving equations for and explaining each reaction as far as possible.

 

INORGANIC PROBLEM 3A white powder A on heating turns yellow and evolves a gas C that turns limewater milky, as well as water vapour. The yellow residue B turns white on cooling, but will turn yellow again when heated.

B reacts with dilute sulphuric acid to give a colourless solution D. If dilute sodium hydroxide solution is added to D, a white precipitate E is produced which with excess alkali gives a colourless solution F.

With dilute ammonia solution D gives a white precipitate G; adding excess ammonia causes the mixture to become clear, giving solution H.

Identify all the substances A – H, and account as thoroughly as you can for the observations.

 INORGANIC PROBLEM 4

A is a metal that reacts moderately quickly with dilute sulphuric acid to give a pale green solution B and a gas C. If solution B is allowed to crystallise, the pale green solid obtained gives on strong heating a solid H, and two gases I and J. These gases are compounds of non-metallic elements, one of which is in a different oxidation state in each compound. If J is passed into a solution of barium chloride there is a vigorous reaction and a white precipitate K is formed.

A reacts with steam in an equilibrium reaction to give a solid M, which contains A in two different oxidation states, and the gas C; this reaction was at one time used industrially to make C.

A reacts with gaseous HCl on heating to give K, a white solid, and C. Hydrated K is pale green. With chlorine, A gives a brown covalent solid L which sublimes on heating; aqueous solutions of the hydrate of L react with copper metal, and for that reason are used to etch printed-circuit boards in electronics.

Solution B gives a dirty-green precipitate D if sodium hydroxide is added; this precipitate remains in excess sodium hydroxide. The same reaction is seen with the addition of ammonia. If D is allowed to stand in air it forms a foxy-red compound E, which is also obtained if sodium hydroxide solution is added to solutions of L.

Acidified solutions of B decolourise potassium manganate(VII) solution giving the manganese-containing ion N as one of the products.

Identify all of the substances A – N, and deduce as much as you can about the processes occurring.

INORGANIC ANSWERS

1

The evolution of carbon dioxide on heating A suggests that it is a carbonate, lead(II) carbonate; the residue B is therefore an oxide, lead(II) oxide:

A à B:         PbCO3 à PbO + CO2

On prolonged heating at 470oC lead(II) oxide (which is also called litharge) partially oxidises to the mixed oxide Pb3O4 which contains both lead(II) and lead(IV) ions.

B à C:        6PbO + O2 à 2 Pb3O4

Pb3O4 is scarlet and is also known as red lead and as minium. It behaves as PbO2.2PbO. With nitric acid the basic lead(II) oxide portion reacts to give a solution of lead(II) nitrate, D, and a residue of lead(IV) oxide, E:

C à D + E:    Pb3O4 + 4HNO3 à 2Pb(NO3)2 + PbO2 + 2H2O.

With lead(II) nitrate solution sodium hydroxide gives an initial precipitate of lead(II) hydroxide F; this substance is amphoteric and reacts with more sodium hydroxide to give a solution of sodium plumbate(II), G, which is colourless:

D à F:        Pb(NO3)2 + 2NaOH à Pb(OH)2 + 2NaNO3

or, ionically, Pb2+ + 2OH- à Pb(OH)2.

F à G:        Pb(OH)2 + 2NaOH à Na2Pb(OH)4

Or, ionically, Pb(OH)2 + 2OH - à [Pb(OH)4]2-.

Lead(IV) oxide is a strong oxidising agent and oxidises concentrated HCl to chlorine, J, and lead(II) chloride I. Lead(II) chloride complexes with chloride ions to give various species such as PbCl3

- and PbCl42-, which are yellow and water-soluble:

E à I + J:   PbO2 + 4HCl à PbCl2 + 2H2O + Cl2.

J à K:         PbCl2 + Cl- à PbCl3- .

Chlorine oxidises potassium iodide solution to iodine, L, which in the presence of excess iodide ions gives a brown solution of the tri-iodide ion:

J à L: Cl2 + 2I- à I2 + 2Cl-

             I2 + I- à I3- .

Lead(II) nitrate solution gives a precipitate of lead(II) iodide M with potassium iodide solution. Lead(II) iodide is bright yellow when solid, but gives a colourless solution in hot water, from which shimmering yellow plates crystallise on cooling. The colour arises from interactions in the crystal lattice and not from the ions themselves, both of which are colourless:

D à M:             Pb2+ + 2I- à PbI2 .

2The reaction of copper with nitric acid is complicated and depends on the concentration of the acid and its temperature. With 50% acid the initial reaction, to give copper(II) nitrate A and nitrogen dioxide B is:

Cu + 4HNO3 à Cu(NO3)2 + 2NO2 + 2H2O.

If sodium hydroxide is added to a solution of copper(II) nitrate, a pale blue precipitate is given, usually represented as copper(II) hydroxide.

Cu(NO3)2 + 2NaOH à Cu(OH)2 + 2NaNO3

The reaction is actually with the hydrated copper ions themselves, which are deprotonated:

[Cu(H2O)6]2+ + 2OH- à Cu(OH)2 + 6H2O.

However, if as here the sodium hydroxide is added to the solution of the copper(II) salt, the precipitate is not the hydroxide, but the basic nitrate Cu(OH)2.Cu(NO3)2 (or sulphate, or chloride, depending on the copper(II) salt employed). To get the true hydroxide, the copper(II) salt has to be added to sodium hydroxide solution. The business of hydroxide precipitates is convoluted to say the least - take a look at my article on the subject. For the purposes of this answer, C is copper(II) hydroxide.

On heating (even in aqueous suspension) copper(II) hydroxide loses water to give G, copper(II) oxide, as a black precipitate:

C à G: Cu(OH)2 à CuO + H2O.

Copper(II) nitrate solution reacts with aqueous ammonia in a ligand replacement reaction to give the tetrammine complex, which is deep blue and is D:

A à D: [Cu(H2O)6]2+ + 4NH3 à [Cu(NH3)4(H2O)2]2+ + 4H2O

If a large excess of chloride ions is added to D, ligand replacement occurs to give the tetrahedral CuCl4

2- ion, E, which is bright green:

D à E: [Cu(NH3)4(H2O)2]2+ + 4H + + 4Cl- à [CuCl4]2- + 2H2O + 4NH4 +

Dilution of this solution E results in replacement of the chloride ligands by water, and solution F contains hexaquacopper(II), pale blue:

E à F: [CuCl4]2- + 6H2O à [Cu(H2O)6]2+ + 4Cl - .

 

Nitrogen dioxide reacts with water to give a mixture of nitrous and nitric acids; the nitrogen goes from oxidation state (+4) in NO2 to (+3) in nitrite and (+5) in nitrate. Nitrogen is simultaneously oxidised and reduced so the reaction is disproportionation:

B à H + I:           2NO2 + H2O à HNO2 + HNO3

 3The evolution of carbon dioxide C from A suggests that A is a carbonate; the yellow residue that turns white on cooling is zinc oxide, C. This colour change is unique, and depends on zinc ions moving a little in the crystal lattice as it is heated, returning to their original positions on cooling. You might therefore say that A is ZnCO3. This substance does not exist; zinc carbonate is a basic salt, ZnCO3.2Zn(OH)2, that is has zinc hydroxide incorporated in the lattice. Zinc hydroxide loses water on heating and also forms zinc oxide. Its presence accounts for the water vapour that was observed:

A à B + C: ZnCO3.2Zn(OH)2 à 3ZnO + 2H2O + CO2.

Zinc oxide gives a solution of zinc sulphate, D, with dilute sulphuric acid, but all the

subsequent reactions are of the zinc ion [Zn(H2O)6]2+, so I shall use ionic equations and leave the spectator sulphate ion out:

B à D: ZnCO3.2Zn(OH)2 + 6H+ + 15 H2O à 3 [Zn(H2O)6]2+ + CO2.

This is perhaps a bit daunting, and since basic salts are not in the Edexcel syllabus a satisfactory equation is

B à D: ZnCO3 + 2H+ + 6H2O à [Zn(H2O)6]2+ + H2O + CO2.

Solution D with sodium hydroxide gives a precipitate of zinc hydroxide, E, which being amphoteric reacts further with hydroxide ions to give the zincate ion F, which is soluble and colourless:

D à E: [Zn(H2O)6]2+ + 2OH- à Zn(OH)2 + 6H2O

E à F: Zn(OH)2 + 2OH- à [Zn(OH)4]2-

Ammonia also precipitates zinc hydroxide G from aqueous solutions of zinc ions. In excess ammonia this reacts further to give the colourless and soluble tetramminezinc(II) ion:

Zn(OH)2 + 4NH3 à [Zn(NH3)4]2+ + 2OH-

Basic zinc carbonate occurs naturally as the mineral calamine, which has long been used as part of a lotion for skin disorders, and which is used to promote healing. Zinc oxide, as an ointment or in plasters or dressings, is used for the same purpose, and although poisonous in large quantities zinc is a necessary trace element for health.

4The colours involved in this reaction scheme should suggest that A is iron. With dilute sulphuric acid iron gives a pale green solution  of iron(II) sulphate B, and hydrogen, C:

A à B + C: Fe + H2SO4 à FeSO4 + H2

The pale green ion is [Fe(H2O)6]2+.

If solution B is allowed to crystallise, the iron(II) sulphate (also called green vitriol) gives iron(III) oxide H on strong heating, and sulphur dioxide (sulphur(IV) oxide) I and sulphur trioxide (sulphur(VI) oxide) J:

B à H + I + J: 2FeSO4 à Fe2O3 + SO2 + SO3.

The iron(III) oxide formed is a very fine powder, and is used as a polishing powder in the jewellery trade. It is sometimes called jewellers’ rouge.

If J is passed into a solution of barium chloride there is a vigorous reaction since sulphuric acid is formed rather violently from sulphur trioxide and water. That is why, in the Contact Process, the sulphur trioxide is absorbed in concentrated sulphuric acid and the mixture then diluted. The white precipitate K is barium sulphate:

J à K:       SO3 + H2O à 2H+ + SO42- (but see my article on the pH of sulphuric acid);

followed by Ba2+ + SO42- à BaSO4.

Iron reacts with steam in an equilibrium reaction to give tri-iron tetroxide M, which contains iron in oxidation states (+2) and (+3); Fe2O3.FeO; and hydrogen, C. This reaction was at one time used industrially to make hydrogen. The iron oxide formed was returned to the blast furnace.

A    à      M + C:   3Fe  +   4H2O   àFe3O4  +  4H2

With gaseous HCl on heating iron gives white K, anhydrous iron(II) chloride, and hydrogen:

A à K:     Fe + 2HCl à FeCl2 + H2

Hydrated iron(II) chloride is pale green. With chlorine on heating iron gives brown covalent iron(III) chloride L which sublimes on heating:

A à L: 2Fe + 3Cl2 à 2FeCl3

Aqueous solutions of iron(III) chloride react with copper metal to give iron(II) ions and copper(II) ions, and for that reason are used to etch printed-circuit boards in electronics:

2 [Fe(H2O)6]3+ + Cu + 6H2O à 2 [Fe(H2O)6]2+ + [Cu(H2O)6]2+

The solution of iron(II) ions, B gives a dirty-green precipitate of iron(II) hydroxide D if sodium hydroxide is added; this precipitate remains in excess sodium hydroxide since it is not amphoteric. The same reaction is seen with the addition of ammonia, the latter acting simply as an alkali.

B à D: [Fe(H2O)6]2+ + 2OH- à Fe(OH)2 + 6H2O

If D is allowed to stand in air it forms a foxy-red compound E, which is iron(III) hydroxide produce by air oxidation of iron(II) hydroxide. This is also obtained if sodium hydroxide solution is added to solutions of L:

L à E: [Fe(H2O)6]3+ + 3OH- à Fe(OH)3 + 6H2O

Acidified solutions of iron(II) ions decolourise potassium manganate(VII) solution being oxidised to iron(III) and giving the hexaquamanganese(II) ion N also:

B    à      F  +  N    5[Fe(H2O)6]2+ + 8H+ + MnO4- à 5[Fe(H2O)6]3+ + Mn2+ + 4H2O