* Cr O–
O–
O O
Unit 5 Transition metals, arenes and organic nitrogen chemistry
Chapter 11 Transition metals
and the d-block elements
eIt is a good idea to first draw the structural formula of the ion or molecule showing all the bonds. Chromium has six valence electrons (two 4s and four 3d) and can form six covalent bonds. A single bond is a pair of electrons shared and a double bond is two pairs shared. If an atom gains a single electron, it has a single negative charge. If it loses one electron, it has a single positive charge. In this dot-and-cross diagram, the extra electrons on the two negative oxygen atoms are shown as *.
b The structural formula of the VO2+ion is:
Its dot-and-cross diagram is:
eVanadium has five valence electrons, but one is lost in the formation of the positive ion, so it can form only four covalent bonds (two double bonds).
6 Titanium, calcium and zinc atoms have the following electron configurations:
Ti: [Ar] 3d24s2 Ca: [Ar] 4s2 Zn: [Ar] 3d104s2
Titanium can lose both 4s-electrons to form a Ti2+ion or both 4s- and one 3d-electron to form a Ti3+ion. The extra energy required to remove the third electron, which is of almost identical energy to the 4s-electrons, is compensated for by the extra hydration energy produced when the aqua ion is formed. Calcium has no extra electron of similar energy and so does not form a Ca3+ion. If zinc lost a 3d-electron, it would lose the consid-erable stability resulting from a full d-shell. This means that not enough energy is regained by hydration to make the formation of a Zn3+ion energetically feasible.
eThe extra energy required to remove another electron must be regained, either from the hydration enthalpy of the cation or from the lattice energy, depending on whether an aqueous ion or an ionic solid is formed. Remember that the values of both the hydration enthalpy and the lattice energy are greater if the ion is more highly charged.
7 a I Iron(III) oxide is a basic oxide and so reacts with acids to form a salt and water:
Fe2O3+ 6HCl → 2FeCl3+ 3H2O
I Iron(III) oxide is not amphoteric, so there is no reaction with sodium hydroxide.
b I Iron(VI) oxide is acidic. It is not amphoteric, so does not react with acids.
I As it is acidic, iron(VI) oxide reacts with alkalis to form a salt with the iron in the anion:
FeO3+ 2NaOH → Na2FeO4+ H2O
eEach of these reactions can be written as an ionic equation:
Fe2O3+ 6H+→ 2Fe3++ 3H2O FeO3+ 2OH–→ FeO42–+ H2O
These equations show clearly that the basic oxide forms the simple cation, Fe3+, and the acidic oxide forms an oxo-anion, FeO42–, which has one more oxygen atom than the parent oxide. The latter reaction is typical of acidic oxides. For example, carbon dioxide reacts with bases to form the carbonate anion, CO32–; sulfur trioxide, SO3, forms the sulfate ion, SO42–.
Remember that the oxides of transition metals in their lower valencies are basic and in their higher valencies are acidic.
8 The hexaaquachromium(III) ion is octahedral. Each bond angle between the central chromium ion and the oxygen atoms of the water molecules is 90°. The oxygen atoms are bonded to the chromium ion by dative bonds and to the hydrogen atoms by covalent bonds.
V O O
+V
O + O
Chapter 11 Transition metals and the d-block elements
eMake sure that the bonds to the chromium ion come from the oxygen atoms in the water. The complex ion has a charge of 3+, which must be included in the diagram that shows the shape. It is also a good idea to name the shape, in case your drawing is not clear.
9 a Chromium(II) chloride, CrCl2, and chromium(II) sulfate, CrSO4, are examples.
b Chromium(III) chloride, CrCl3, and chromium(III) sulfate, Cr2(SO4)3, are examples.
c Potassium dichromate(VI), K2Cr2O7, and potassium chromate(VI), K2CrO4, are examples.
10 a Copper(I) oxide, Cu2O, and copper(I) chloride, CuCl, are examples.
b Copper(II) sulfate, CuSO4, is an obvious example.
eCopper(I)oxide is the red solid formed when an aldehyde reduces Fehling’s solution. Copper(I)chloride is a colourless, insoluble chloride.
11 a There are many examples of chromium(III) complexes, such as the hexaaqua species, [Cr(H2O)6]3+, and the soluble [Cr(OH)6]3–, which is produced when excess sodium hydroxide is added to a precipitate of chromium(III) hydroxide.
b The three copper(II) complexes that you should know are the hydrated copper(II) ion, [Cu(H2O)6]2+, the deep blue ammine [Cu(NH3)4(H2O)2]2+, produced when excess ammonia is added to a precipitate of copper(II) hydroxide, and yellow [CuCl4]2–, which is produced when concentrated hydrochloric acid is added to a copper(II) chloride solution.
eBe careful when working out the charge on a complex ion. Add together the charge on the metal ion (3+ for chromium(III)) and the total charge of the ligands (6– for the chromium–hydroxide complex). If a complex has six ligands (a coordination number of six), its shape is octahedral. Complex ions with four ligands will be either tetrahedral, for example [CuCl4]2–, or square planar, for example [Pt(NH3)2Cl2].
12 a A substance that is green absorbs the complementary colour red.
b A substance that is red absorbs the complementary colour green.
eThe ligands split the energy levels of the d-orbitals. When white light is shone into the solution, some of the light is absorbed and the energy is used to promote an electron from one of the lower split d-orbitals to a higher one. This is known as a d–d transition.
13 In hydrated ions, the water molecules cause the d-orbitals to split into a lower-energy set of three and an upper set of two levels. The electron configurations of the two ions are:
Ti4+: [Ar] 3d04s0
Chapter 11 Transition metals and the d-block elements
The hydrated Ti4+ion has no electrons in d-orbitals and so promotion of an electron from a lower to a higher level is impossible. The Ti3+ion has one d-electron and can absorb visible light. This causes the d-electron to be promoted to one of the higher split 3d energy levels. The removal of some of the frequencies of visible light means that the ion is coloured.
eThe colour of complex ions is due to the absorption of light energy as the electron is pushed up into a higher energy level. This is different from the process that gives rise to emission spectra in flame tests, in which heat pushes the electron up to an unstable higher level. Light is then given off as the electron falls back to its ground state.
14 When excess ammonia is added to hydrated copper(II) sulfate solution, ligand exchange takes place and four of the water molecules are replaced by ammonia ligands. The different ligands produce a new colour. Ammonia is a stronger ligand than water and the absorption moves from orange (complementary colour: blue) towards yellow (complementary colour: violet).
eIn this question, the metal and its oxidation state stay the same. However, the ligands are different, hence the change in colour. A stronger ligand will move the absorption to a higher energy, which is in the general direction red → violet.
15 The colour of a complex ion depends on three factors:
I the particular transition metal
I its oxidation state
I the nature of the ligands
eThis assumes that the transition metal ion has between one and nine d-electrons, so that d–d transitions can take place.
If the ion has zero or ten electrons, its complex will be colourless.
16 There are no ligands in anhydrous copper(II) sulfate, so all five d-orbitals have the same energy and are not split.
Copper(I) ions have the electronic configuration [Ar] 3d10and so all the d-orbitals are full. The ligands split the energy levels of the five d-orbitals, but there is no empty or half-filled orbital into which an electron can be promoted. Thus the ion, even when complexed, cannot absorb radiation in the visible region.
eLigands are needed to provide the electric field that splits the d-orbitals into three levels of lower energy and two of higher energy. Light energy can then promote an electron from a lower to a higher d-orbital as long as there is an empty or half-filled d-orbital. This causes the ion to be coloured.
17 a The green solution containing nickel(II) ions forms a green precipitate that does not dissolve on addition of excess sodium hydroxide. The equation is:
[Ni(H2O)6]2+(aq) + 2OH–(aq) → [Ni(H2O)4(OH)2](s) + 2H2O(l)
With ammonia solution, the green solution forms a green precipitate that dissolves in excess ammonia to form a pale blue solution:
[Ni(H2O)6]2+(aq) + 2NH3(aq) → [ Ni(H2O)4(OH)2](s) + 2NH4+
[Ni(H2O)4(OH)2](s) + 4NH3(aq) → [Ni(NH3)4(H2O)2]2+(aq) + 2H2O(l) + 2OH–(aq)
b The blue solution containing copper(II) ions forms a blue precipitate that remains on addition of excess sodium hydroxide:
[Cu(H2O)6]2+(aq) + 2OH–(aq) → [Cu(H2O)4(OH)2](s) + 2H2O(l)
With ammonia, the pale blue solution forms a blue precipitate that dissolves, in a ligand-exchange reaction, to form the dark blue copper–ammonia complex:
[Cu(H2O)6]2+(aq) + 2NH3(aq) → [Cu(H2O)4(OH)2](s) + 2NH4+
[Cu(H2O)4(OH)2](s) + 4NH3(aq) → [Cu(NH3)4(H2O)2]2+(aq) + 2H2O(l) + 2OH–(aq)
Chapter 11 Transition metals and the d-block elements
eThe formulae for the hydroxide precipitates can be written as Ni(OH)2and Cu(OH)2. The reaction of transition metal ions with hydroxide ions and with ammonia to form the precipitate of metal hydroxide is called ‘deprotonation’.
18 The green precipitate obtained when alkali is added to a solution of iron(II) ions is iron(II) hydroxide — a deprotonation reaction. Addition of hydrogen peroxide to this precipitate oxidises it to iron(III) hydroxide, which is red-brown.
eThere will also be some fizzing when the hydrogen peroxide is added, because it decomposes to water and oxygen gas.
The iron(III)hydroxide acts as a catalyst for this decomposition.
19 Hydrated chromium(III) ions are deprotonated by water and by hydroxide ions.
a In aqueous solution, the solvent water molecules partially deprotonate the hydrated ion:
[Cr(H2O)6]3+(aq) + H2O(l) [Cr(H2O)5(OH)]2+(aq) + H3O+(aq)
eThe production of H3O+ions makes the solution acidic.
b When some sodium hydroxide is added, ligand water molecules are successively deprotonated and a green precipitate of chromium(III) hydroxide is obtained:
[Cr(H2O)6]3+(aq) + 3OH–(aq) → [Cr(H2O)3(OH)3](s) + 3H2O(l)
With excess sodium hydroxide, further deprotonation takes place and a clear, green solution is obtained:
[Cr(H2O)3(OH)3](s) + 3OH–(aq) → [Cr(OH)6]3–(aq) + 3H2O(l) 20 a The two reduction half-equations are:
Fe3++ e− Fe2+ E° = +0.77 V
FeO4−+ 8H++ 3e− Fe3++ 4H2O E° = +2.20 V
The value of E°cellis found by reversing the second equation (therefore changing the sign of the E° value) and then adding the two E° values. This is done in order to get the reactant, Fe3+ions, on the left.
E°cell= −(+2.20) + (+0.77) = −1.43V
E°cellis negative, so the disproportionation reaction will not take place (under standard conditions).
b The two reduction half-equations are:
[Fe(CN)6]3−+ e− [Fe(CN)6]4− E° = +0.36 V [Co(NH3)6]3++ e− [Co(NH3)6]2+ E° = +0.10 V
The reactants are [Fe(CN)6]3−and [Co(NH3)6]2+, so the second equation has to be reversed. The E° values are then added.
E°cell= +0.36 + (−0.10) = +0.26 V
E°cellis positive, so this redox reaction is thermodynamically feasible and will take place unless the activation energy is too high.
21 a amount of KMnO4in titre = 0.0500 mol dm–3× 0.0200 dm3= 0.00100 mol
b The oxidation number of manganese changes from +7 to +2, so each manganese gains five electrons.
moles of electrons gained = 5 × 0.00100 = 0.00500 mol
c amount of V3+ions = 0.100 mol dm–3× 0.0250 dm3= 0.00250 mol
d 0.00250 mol of V3+ions provides (to the MnO4–ions) 0.00500 mol of electrons, so each V3+ion loses two electrons. As two electrons are lost, the oxidation number of vanadium increases by two from +3. Hence its oxidation state after reaction is +5.
Chapter 11 Transition metals and the d-block elements
emoles = concentration (in mol dm–3) ×volume (in dm3), so the titre in cm3must be divided by 1000 to change it into a volume in dm3.
In acid solution, manganate(VII) ions are reduced to Mn2+ions.
Reduction is gain of electrons. The number gained by each manganese equals the change in its oxidation number (from +7 to +2 = 5).
Oxidation is loss of electrons. The change in oxidation number equals the number of electrons lost by each vanadium ion.
22 a The reduction half-equations with their standard electrode potentials are:
VO2++ 2H++ e–→ VO2++ H2O E° = +1.00 V VO2++ 2H++ e–→ V3++ H2O E° = +0.34 V S + 2H++ 2e–→ H2S E° = +0.14 V
To check whether H2S will reduce VO2+to VO2+, the sign of E° for the third equation must be reversed and added to that of the first:
E°cellfor the reduction of VO2+to VO2+= +1.00 + (–0.14) = +0.86 V This is positive, so the reduction is possible.
To check whether the VO2+ions can be reduced further, the sign of E° for the third equation has to be reversed and added to that of the second equation:
E°cellfor the reduction of VO2+to V3+= +0.34 + (–0.14) = +0.20 V, which is also positive, so this reduction is feasible.
As both reductions are feasible, the VO2+ions will be reduced to V3+ions.
b Adding the first two equations to the reversed third equation gives the overall equation:
VO2++ 2H++ H2S → V3++ 2H2O + S
eThese standard electrode potentials can be found in a data book or on pages 187 and 228 in Chapters 10 and 11 of the textbook.
Two electrons have to be added to the VO2+ion to reduce it from the +5 to the +3 state. H2S has to lose two electrons for the sulfur to become oxidised from the –2 to the zero oxidation state, so there will be one of each species on the left-hand side of the equation.
23 a The standard reduction potentials are:
V3++ e–→ V2+ E° = –0.26 V
VO2++ 2H++ e–→ V3++ H2O E° = +0.34 V VO2++ 2H++ e–→ VO2++ H2O E° = +1.00 V
1–
2O2+ 2H++ 2e–→ H2O E° = +1.23 V
As the standard reduction potential of oxygen is more positive than all three potentials for vanadium going from +2 to +3, +3 to +4 and +4 to +5, oxygen should oxidise V2+ions to VO2+ions in acid solution.
b The change in oxidation number of the vanadium is from +2 to +5, a change of 3. Each oxygen atom in O2
changes by two, so there must be 2V2+and 3 × 1–2O2in the overall equation, so that both change by a total of six. The overall reaction is obtained by adding the four equations:
2V2+→ 2V3++ 2e–
Chapter 11 Transition metals and the d-block elements
2V3++ 2H2O → 2VO2++ 4H++ 2e– 2VO2++ 2H2O → 2VO2++ 4H++ 2e–
eNote that the 6H+on the left cancel with six of the 8H+on the right; three of the 4H2O on the left cancel with the 3H2O on the right; the 6e–on the left cancel with the 6e–on the right.
E°cellfor the oxidation of V2+to V3+is +1.23 + (+0.26) = +1.49 V E°cellfor the oxidation of V3+to VO2+is +1.23 + (–0.34) = +0.89 V E°cellfor the oxidation of VO2+to VO2+is +1.23 + (–1.00) = +0.23 V
All the E°cellvalues are positive, so all the reactions will take place. This means that oxygen under standard conditions should oxidise V2+to vanadium in the +5 state.
24 a The standard reduction potentials are:
VO2++ 2H++ e–→ V3++ H2O E° = +0.34 V VO2++ 2H++ e–→ VO2++ H2O E° = +1.00 V
1–
2 I2+ e–→ I– E° = +0.54 V
The E° value of iodine is more positive than that for the first equation but is less positive than that for the second equation. Thus, iodine will oxidise V3+to VO2+, but no further.
b The equation is:
V3++ H2O + 1–2I2→ VO2++ 2H++ I–
eAnother way of answering this question is to work out the values of E°cellfor the two possible oxidations.
G Oxidation of V3+to VO2+: E°cell= +0.54 + (–0.34) = +0.20 V, which is positive and so the reaction is feasible.
G Oxidation of VO2+to VO2+: E°cell= +0.54 + (–1.00) = –0.46 V, which is negative and so the reaction will not take place.
25 The standard reduction potentials are:
MnO2+ 4H++ 2e–→ Mn2++ 2H2O E° = +1.23 V
Cl2+ 2e–→ 2Cl– E° = +1.36 V
Br2+ 2e–→ 2Br– E° = +1.07 V
E°cellfor the oxidation of chloride ions by MnO2= +1.23 + (–1.36) = –0.13 V, which is negative and so the reaction will not take place under standard conditions.
E°cellfor the oxidation of bromide ions by MnO2= +1.23 + (–1.07) = +0.16 V, which is positive and so the reaction is feasible.
eDo not state that a reaction will take place if the E°cellvalue is positive. The activation energy might be too high for the reaction to happen rapidly enough to be observed. The correct terminology is that the reaction is feasible or thermo-dynamically spontaneous.
26 The disproportionation reaction would be:
3Cr2+→ 2Cr3++ Cr E°cellfor this is given by:
Cr2++ 2e–→ Cr E° = –0.91 V
2Cr2+→ 2Cr3++ 2e– E° = –(–0.41) = +0.41 V
E°cell= –0.91 +0.41 = –0.50 V, which is negative, so disproportionation will not occur.
11–2O2+ 6H++ 6e–→ 3H2O 2V2++ 11–2O2+ H2O → 2VO2++ 2H+
Chapter 11 Transition metals and the d-block elements
27 a There are several possible answers to this question:
I Add a solution of the disodium salt of EDTA, followed by a few drops of alkali to form the EDTA4−ion.
Then add to a solution of chromium(III) sulfate. Ligand exchange takes place:
[Cr(H2O)6]3++ EDTA4−→ [Cr(EDTA)]−+ 6H2O
I Add a solution containing excess 1,2-diaminoethane (often written as ‘en’) to a solution of chromium(III) sulfate. Ligand exchange takes place:
[Cr(H2O)6]3++ 3en → [Cr(en)3]3++ 6H2O
I Add excess sodium hydroxide to a solution of chromium(III) sulfate. Complete deprotonation takes place and the complex ion chromium(III) [Cr(OH)6]3−is formed.
[Cr(H2O)6]3++ 6OH−→ [Cr(OH)6]3−+ 6H2O
b Chromate(III) ions are oxidised in alkaline solution by hydrogen peroxide. Excess aqueous sodium hydroxide is added to a solution of chromium(III) sulfate, followed by hydrogen peroxide solution. The mixture is then warmed on a water bath. The result is a yellow solution of sodium chromate(VI), containing CrO42−ions.
c Powdered zinc and concentrated hydrochloric acid are added to a solution of chromium(III) sulfate in a conical flask fitted with a Bunsen valve. The colour of the solution changes as chromium(II) ions are formed.
The function of the Bunsen valve is to exclude air, which would oxidise the chromium(II) back to chromium(III).
28 a Ligand exchange is a reaction in which one type of ligand in a d-block metal complex ion is replaced by another, either totally or partially. An example is the addition of concentrated hydrochloric acid to a solution containing hydrated copper(II) ions:
[Cu(H2O)6]2++ 4Cl−→ [CuCl4]2−+ 6H2O
b A catalyst is a substance that speeds up a chemical reaction without itself being used up in the reaction. A heterogeneous catalyst is in a different phase from the reactants. An example is the solid iron catalyst used in the Haber process involving the reaction between nitrogen gas and hydrogen gas.
Chapter 11 Transition metals and the d-block elements
Summary worksheet (www.hodderplus.co.uk/philipallan)
1 B Ethene does not have a lone pair of electrons and so cannot be a ligand. Therefore, option B is the correct answer to this negative question. Options A, C and D can all act as ligands: F−has four lone pairs, the N of the –NH2group and the O−of the –COO−group both have lone pairs that can form bonds with transition metal ions.
2 D Here, five ions are formed from one ion and six molecules. Since this is a reduction in particles from seven to five, the entropy will decrease. In options A and C more particles are produced, so there will be an increase in entropy. In option B a gas is produced, so this too will have a positive ΔSsystem.
3 B Mn2+gives an off-white precipitate with alkali. This darkens on standing as it is oxidised slowly to MnO2. The cations in options A, C and D all give green precipitates with alkali. The addition of barium chloride showed that the substance is either a sulfate, a carbonate or a sulfite and the addition of hydrochloric acid proved it to be a sulfate.
4 D [CrCl4]−is tetrahedral, so option A is incorrect. The copper in [CuCl4]2−is in the +2 state, so option B is incorrect and the copper in [CuCl2]−is in the +1 state, so option C is also wrong. The complex ion in D has six ligands and so is octahedral, and the copper is in the +2 state. Therefore, option D is the correct answer.
5 A Option A is the electron configuration of chromium, which like copper breaks the 3dx4s2rule. Option B is zinc, which does not form an ion with an incomplete d-shell. Option C is scandium, which is always +3 in compounds and so also does not form an ion with an incomplete d-shell. The substance represented by the electron configuration in option D does not exist (it should be [Kr]4d75s2).
6 C Iron(II) sulfate gives a green precipitate of iron(II) hydroxide, which does not form a complex with ammonia, and so does not redissolve to form a solution. Silver(I) ions (option A) and copper(I) ions (option B) form colourless ammonia complexes. Nickel(II) (D) forms a blue complex. Therefore, options A, B and D all give hydroxide precipitates that redissolve in excess ammonia.
7 A Chromium(III), iron(III) and copper(II) compounds in solution give precipitates of hydroxides, but only chromium(III) hydroxide is amphoteric. Thus the precipitates remain with the iron(III) and copper(II) compounds, so B and C are incorrect. Copper(I) chloride, CuCl, is insoluble, so does not form a precipitate.
Therefore, D is also incorrect.
8 B All transition metals, i.e. those in the series Ti to Cu have more than one oxidation state in their compounds.
Not all transition metal ions are coloured — Ti4+and Cu+ are colourless, so option A is incorrect. Copper metal has the configuration [Ar] 3d104s1. It is a transition metal and it has no unpaired d-electrons, so option C is incorrect. Option D is false because copper(III) does not exist.
Chapter 11 Transition metals and the d-block elements
1 a The Hess’s law cycle is:
ΔH1= 6 × ΔHa(carbon) = 6 × +715 = +4290 kJ ΔH2= 6 × ΔHa(hydrogen) = 6 × +218 = +1308 kJ
ΔH3= heat change on making 6 × C–H bonds, 3 × C–C bonds and 3 × C=C bonds
= 6 × (–412) + 3 × (–348) + 3 × (–612) = –5352 kJ
ΔHf(‘cyclohexatriene’(g)) = ΔH1+ ΔH2+ ΔH3= 4290 + 1308 + (–5352)
= +246 kJ mol–1
The enthalpy of formation of the theoretical gaseous ‘cyclohexatriene’ is +246 kJ mol–1.
eRemember that bond making is always exothermic (negative ΔH) and that the enthalpy of atomisation of hydrogen is per mole of atoms of hydrogen produced.
b The Hess’s law cycle is:
b The Hess’s law cycle is: