CHAPTER 13

Practice Exercises

13.1 The d electron configurations are determined by the position of the metal ion in the periodic table: Fe3+, d5; Ni2+, d8; Pt2+, d8; Ir+, d8; Re+, d6.

13.3 The oxidation states are determined by summing the positive and negative charges within the neutral complex to give a net charge of zero. [Co(NH3)6]Cl3, Co3+; Na2[MnCl5], Mn3+; [Rh(en)2Cl2]Cl, Rh3+; K3[Cr(ox)3], Cr3+; [Co(en)(NH3)2(NCS)Cl]NO3, Co3+.

13.5 The compounds are named according to the rules in section 13.4.

(a) pentaamminecarbonatocobalt(III) nitrate(b) triamminetricarbonylmolybdenum(0)(c) pentaaquahydroxochromium(III) chloride(d) potassium hexacyanoferrate(III)(e) dichlorobisethylenediaminechromium(III) sulfate

13.7 Equilibrium [Ag+] =

3

27

1.9 10

1.6 10 0.246

= 2.0 10–9

M

Equilibrium [NH3] = 0.246 M

Equilibrium [Ag(NH3) 2 ] = 1.9 10–3

M

13.9 The number of unpaired electrons is given by filling the d orbitals according to the rules on page 533. The magnetic moments are obtained using the equation:

B( 2)n n

[NiCl6]2–, Ni4+, therefore d6 and will have either 0 unpaired electrons (low spin,

= 0 B) or 4 unpaired electrons (high spin, = 4.90 B)[CuCl6]

4–, Cu2+, therefore d9 and will have 1 unpaired electron. = 1.73 B

[ZnCl6]4–, Zn2+, therefore d10 and will have 0 unpaired electrons. = 0 B

Review Questions

13.1 (a) 23CO has a charge of –2, so Mn must be +2 to give overall neutrality.

(b) Cl (more electronegative) is –1, so Mo must be +5 to give overall neutrality.(c) Na is +1, so VO4 has overall charge –3; each O is –2, so V is +5.(d) O is –2, so Au must be +3 to give overall neutrality.

(e) H2O has zero charge, 2

4SO has a charge of –2, so Fe must be +3 to give

overall neutrality.

(f) 24SO has a charge of –2, so Ni must be +2 to give overall neutrality.

(g) K is +1, so MnO4 has charge –1; each O is –2, so Mn is +7 to give net charge of –1. (h) NH4 is +1, so WO4 has charge –2; each O is –2, so W is +6 to give net charge of –3.

(i) 24CrO has a charge of –2, so Pb must be +2 to give overall neutrality; each O is –2, so

Cr is +6 to give the net ionic charge of –2.(j) H

2O has no charge, each Cl is –1 and O is –2, so Zr must be +4 to give overall neutrality.

13.3 (a) 3d5

(b) 5d6

(c) 3d8

(d) 4d4

(e) 3d2

(f) 3d5

(g) 5d8

(h) 4d2

13.5 water, ammonia, carbon monoxide, as given in table 13.2

13.7 A bidentate ligand must have a lone pair of electrons on two atoms that can each form a coordinate covalent bond with a metal to give a ring structure which includes the metal.

13.9 EDTA4– has 10 potential donor atoms: 2 N atoms and 8 O atoms.

13.11 Because these are formed by coordinate bonds.

13.13 EDTA forms water soluble complexes with Ca2+ and Mg2+ that are responsible for hardness.

13.15 [Cr(en)3]3+ is expected to have the larger n value as it contains bidentate chelating ligands.

13.17 Octahedral

13.19 Isomers are compounds that have the same chemical formula, but different arrangements of the constituent atoms.

13.21 In a cis isomer, like groups are located adjacent to each other, whereas in a trans isomer, like groups are located opposite to each other.

13.23 Refer to Figure 13.26

13.25 An electron in the 2 2x yd

or 2z

d orbital has significant repulsive interactions with ligand electrons

as these orbitals point directly towards the ligands. The dxy, dxz, and dyz orbitals point between the ligands, so electrons in these orbitals do not experience such large repulsive interactions.

13.27 Each complex has a characteristic value of ∆. As the value of ∆ changes, so does the frequency of light that a complex absorbs.

13.29 The spectrochemical series is a list of ligands arranged in order of increasing ability to produce a large value of ∆. This series is determined by the frequencies of light absorbed by a series of complexes containing various ligands.

13.31 Both high-spin and low-spin complexes are possible for d4, d5, d6 and d7 electron configurations.

13.33 This is a Co(III) complex and therefore has a d6 electron configuration. As it is diamagnetic, it has no unpaired electrons and must therefore be low spin, as shown below:

13.35 The main differences are that haemoglobin transports O2 whereas myoglobin stores it, and myoglobin contains a single haem group whereas haemoglobin contains four such groups.

13.37 Metallurgy is the science and technology of metals and is concerned with the procedures and chemical processes that are used to separate metals from their ores and the preparation of metals for practical use.

13.39 Gold can be separated from rock and sand because gold is denser than the rock and sand, and remains behind in the pan when the rock and sand are washed away.

13.41 Cu2S + 3O2 → 2Cu2O + 2SO2

2PbS + 3O2 → 2PbO + 2SO2

SO2 + CaCO3 → CaSO3 + CO2

The CaSO3 produced in the final reaction above is converted to sulfuric acid.

13.43 Coke is coal that has been heated strongly in the absence of air. It is composed almost entirely of carbon.

13.45 Refining is the purification of a metal after it has been reduced to the metallic state.

13.47 The coinage metals are those that have been used since antiquity for coin production: copper, silver, and gold. All are in group 11 of the periodic table. They are characterised by high electrical conductivity, good ductility and low chemical reactivity, in particular resistance to oxidation. Hence they are used for money (a vanishing use in technologically advanced countries), electrical wire, jewellery and other decorative objects.

13.49 Titanium is used as an engineering metal because of its relatively low density, high strength, resistance to corrosion, and ability to withstand high temperatures, all of which make it a favoured structural material.

Review Problems

13.51 The net charge is −3 and the formula is [Fe(CN)6]3–.

13.53 [Cr(NH3)2(NO2)4]–

13.55 (a) hexaamminenickel(II) ion(b) triamminetrichlorochromium(III) (c) hexanitrocobaltate(III) ion(d) diamminetetracyanomanganate(II) ion(e) trioxalatoferrate(III) ion or trisoxalatoferrate(III) ion(f) diiodoargentate(I) ion (g) diaquabis(ethylenediamine)cobalt(III) sulfate (h) pentaamminechlorochromium(III) sulfate (i) potassium trioxalatocobaltate(III) or potassium trisoxalatocobaltate(III)

13.57 The coordination number is six and the oxidation number of the iron atom is +2.

13.59 (a) The three N atoms are the donor atoms.(b) The coordination number of Co is 6.(c) There are two possible isomers of the complex, which arise from the different ways in

which the two dien ligands can be coordinated relative to each other:

CoH2N

H2N NH2

NH2

HN

NH

CoH2N

HN NH2

NH

H2N

NH2

3+ 3+

(d) [Co(dien)2]3+ has the larger n, as it contains two polydentate ligands.

(e)

H2N

HN

NH

NH2

13.61

Pt

Cl

Br

NH3

NH3 Pt

Cl

H3N

Br

NH3

cis trans

13.63 (a) [Cr(H2O)6]3+, as the oxidation state of the metal ion is higher. Cr(III) is smaller than Cr(II)

and the ligands can approach more closely, thereby giving a larger Δo

(b) [Cr(en)3]3+, as en is a stronger field ligand than Cl–

13.65 [Cr(CN)6]3–, as this has the strongest field ligands and so will have the greatest value of Δo

13.67 (a) The value of increases down a group. Therefore, [RuCl(NH3)5]2+ has the greater value of

Δ.(b) The value of increases with oxidation state of the metal. Therefore, [Ru(NH3)6]

3+ has the greater value of Δ.

13.69 [CoA6]3+. This is the complex with the largest Δo value, so we expect the eg orbitals in this

complex to be of higher energy than those in [CoB6]3+. Co(II) is a d7 ion and the electron to be

removed is in the eg orbital set. Those in the highest energy orbitals are the easiest to remove.

13.71 This is a weak field complex of Co2+ and it is a high–spin d7 case. It cannot be diamagnetic. If it were low spin, it would still have one unpaired electron because of the odd number of d electrons.

13.73 (a) Each Cl is –1, and NH3 is neutral, so Ru has oxidation state +2. Ru is in group 8 (8 valence electrons), giving d6.

(b) Each I is –1, and en is neutral, so Cr has oxidation state +3. Cr is in group 6 (6 valence electrons), giving d3.

(c) Each Cl is –1, and trimethylphosphine is neutral, so Pd has oxidation state +2. Pd is in group 10 (10 valence electrons), giving d8.

(d) Each Cl is –1, and NH3 is neutral, so Ir has oxidation state +3. Ir is in group 9 (9 valence electrons), giving d6.

(e) CO is a neutral molecule, so Ni has oxidation state 0. Ni is in group 10 (10 valence electrons), s2d8.

(f) Each Cl is –1, and en is neutral, so Rh has oxidation state +3. Rh is in group 9 (9 valence electrons), giving d6.

(g) Each Br is –1, and CO is neutral, so Mo has oxidation state +2. Mo is in group 6 (6 valence electrons), giving d4.

(h) Each Cl is –1, and Na is +1, so Ir has oxidation state +3. Ir is in group 9 (9 valence electrons), giving d6.

(i) Each Cl is –1, and NH3 is neutral, so Ir has oxidation state +3. Ir is in group 9 (9 valence electrons), giving d6.

(j) Cl is –1, and CO is neutral, so Mn has oxidation state +1. Mn is in group 7 (7 valence electrons), giving six valence electrons. Recall that in transition metal cations, the (n-1)d orbitals are always of lower energy than the n sorbital. Thus, there are 6 d electrons.

13.75 The structure of a metal complex usually is octahedral (6 donor atoms), tetrahedral (4 donor atoms), or square planar (4 donor atoms):

(a) 6 NH3

ligands in an octahedron around the central Ru:

RuH3N

H3N NH3

NH3

NH3

NH3

2+

(b) 2 I– ligands at opposite ends of 1 axis, 2 en ligands in a square plane around the central Cr:

CrH2N

H2N NH2

NH2

I

I

I

(c) Square planar arrangement about Pd, with 2 Cl– ligands adjacent each to other:

Pd

Cl

Cl

P(CH3)3

P(CH3)3

(d) Octahedral arrangement around Ir, with 3 Cl– ligands in a triangular face:

IrCl

Cl NH3

NH3

NH3

Cl

(e) Tetrahedral arrangement of CO ligands about a central Ni:

CO

Ni

OCCO

CO

13.77 (a) The cis-tetraamminechloronitrocobalt(III) ion has six ligands and an octahedral geometry. The cis indicates that the chlorine and nitro ligands are adjacent:

CoO2N

Cl NH3

NH3

NH3

NH3

+

(b) The amminetrichloroplatinate(II) ion has a square planar geometry:

Pt

Cl

Cl

Cl

NH3

-

(c) In trans-diaquabis(ethylenediamine) copper(II), ethylenediamine is a bidentate ligand, giving a coordination number of 6 and an octahedral geometry. The trans means that the water ligands are opposite each other in the complex:

CuH2N

H2N NH2

NH2

OH2

OH2

2+

(d) The tetrachloroferrate(III) ion has a tetrahedral geometry:

Cl

Fe

ClCl

Cl

-

(e) The tetrachloroplatinate(II) ion in potassium tetrachloroplatinate has a square planar geometry:

Pt

Cl

Cl

Cl

ClK2

(f) The pentaammineaquachromium(III) ion has 6 ligands and an octahedral geometry:

CrH3N

H2O NH3

NH3

NH3

NH3

I3

(g) In the tris(ethylenediamine)manganese(II) ion, ethylenediamine is a bidentate ligand, so the coordination number is 6 and the complex has an octahedral geometry:

MnH2N

H2N NH2

NH2

NH2

NH2

Cl2

(h) Pentaammineiodocobalt(III) has 6 ligands and an octahedral geometry:

CoH3N

I NH3

NH3

NH3

NH3

(NO3)2

13.79 (a) Ir3+ is d6, and NH3 generates a relatively large splitting. All the electrons are paired and therefore the complex is diamagnetic: = 0 B

(b) Cr2+ is d4, and water generates a relatively small splitting. The complex is paramagnetic with 4 unpaired electrons. = 4.90 B

(c) Pt2+ is d8, and the complex is square planar. Inspection of the splitting diagram for square planar complexes (Figure 13.27) shows that electrons fill the dxz, dyz, 2z

d and dxy orbitals,

so this complex is diamagnetic. = 0 B

(d) Pd(0) has a d10 configuration, so all orbitals are filled and this complex is diamagnetic. = 0 B

(e) Ru2+ has a d6 configuration, and CN– generates a relatively large splitting, so all electrons are paired, generating a diamagnetic species. = 0 B

(f) Co3+ is d6, and NH3 generates a relatively large splitting, so all the electrons are paired and the complex is diamagnetic. = 0 B

(g) Co2+ is d7; tetrahedral complexes have a 2–3 splitting pattern (Figure 13.28), so the lower levels are filled and there is one electron in each of the three upper levels; the complex is paramagnetic with 3 unpaired electrons. = 3.87 B

(h) Pt2+ is d8, and the complex is square-planar (remember that en is a bidentate ligand). Inspection of the splitting diagram for square planar complexes (Figure 13.27) shows that electrons fill the dxz, dyz, 2z

d and dxy orbitals, so this complex is diamagnetic. = 0 B

13.81 The colours of transition metal complexes are generally determined by d-d transitions between levels whose energy separation is determined by the spectrochemical series. NH3 is a stronger field ligand than H2O, so the energy gap (Δo) in [Cr(NH3)6]

3+ is greater than that in [Cr(OH2)6]3+.

According to figures 13.30 and 13.31, a violet colour results when light of 560 nm is absorbed and an orange colour results when light of 480 nm is absorbed. Photon energy is inversely

proportional to wavelength, so the orange complex has the larger energy gap and is [Cr(NH3)6]3+

.

13.83 (a) The coordination number is the number of coordinate bonds formed between metal and ligands. There are six aqua monodentate ligands, so CN = 6.

(b) The oxidation number is determined by examining the net charge and correcting for charges on all species other than the transition metal. Here, en is neutral. The sulfate anion contributes a –2 charge, so the oxidation number of Mn is +2. Mn is in group 7 of the periodic table, so it has (7 – 2) = 5 d electrons and its valence configuration is 3d5.

(c) A coordination number of 6 implies an octahedral geometry.

(d) With an odd number of electrons, it is not possible for all of them to be paired, so this complex is paramagnetic.

(e) The problem states that the complex is high spin, so each electron occupies a different d orbital and all have the same spin, giving five unpaired electrons.

(f) With five unpaired electrons we predict = 5.92 B from the equation:

B( 2)n n .

13.85 The wavelength of light that is absorbed provides a measure of the crystal field splitting energy:

o = E 34 8 1 23 1

9

(6.626 10 J s)(2.998 10 m s )(6.022 10 mol )

465 10 mAhcN

= 2.57 105 J mol–1

= 257 kJ mol–1

The complex absorbs visible light at 465 nm, in the blue. The colour of the complex will be the complementary colour to blue. The data in table 13.4 show that this colour is orange.

13.87 The number of possible isomers of a complex is determined by its geometry and the number of ligands of each type:

(a)

IrCl

Cl NH3

NH3

NH3

Cl

IrCl

H3N NH3

NH3

Cl

Cl

fac mer

(b)

Pd

Cl

Cl

P(CH3)3

P(CH3)3Pd

P(CH3)3

Cl

P(CH3)3

Cl

trans cis

(c)

CrBr

Br CO

CO

CO

CO

CrBr

OC Br

CO

CO

CO

cis trans

(d) The possible cis/trans isomers are shown below. Note that the mirror image of each of these will give enantiomers.

CoH2N

H2N NH3

I

I

NH3

CoH2N

H2N NH3

NH3

I

I

CoH2N

H2N I

I

NH3

NH3

13.89 In the bidentate mode, two oxygen atoms of a carbonate anion bond to the metal. In the monodentate mode, just one oxygen atom bonds to the metal as shown:

CO

O

O

M O C

O

O

bidentate monodentate

13.91 Coordination complexes are named following the rules stated on pp 528–9: (a) cis-tetraaquadichlorochromium(III) chloride(b) bromopentacarbonylmanganese(I) (c) cis-diamminedichloroplatinum(II)(d) tetraamminezinc(II) sulfate(e) potassium hexacyanoferrate(III)(f) trans-tetraamminedichlorocobalt(III) chloride

13.93 Cu(II) in water forms an aqua complex. Addition of cyanide produces a colourless, insoluble salt. As this is colourless, it is no longer a Cu(II) complex, but a complex of the d10 Cu(I) ion: CuCN.

This dissolves in excess CN– to give the colourless tetracyanocopper(I) complex, [Cu(CN)4]–:

2Cu2+(aq)+ 4CN–(aq) 2CuCN (s) + (CN)2(g)

CuCN(aq) + 3CN–(aq) [Cu(CN)4]3–(aq)

13.95 The fac isomer of an octahedral complex has three like ligands arranged mutually cis:

13.97 When there are three or fewer valence electrons, they occupy the three t2g orbitals with unpaired

spins, so the d1

– d3

configurations always have the maximum number of spins. When there are eight or more valence electrons, they completely fill the three t2g orbitals and partially fill the two eg, regardless of the value of Δo. Only when there are between 4 and 7 valence electrons are there two distinct electron configurations; high-spin and low-spin values.

13.99 Orbital sketches show that the two orbitals experience quite different electron–electron repulsion from ligands in the xy plane in an octahedral environment. Because the lobes of the dxy orbital point between the ligands, there are fewer destabilising repulsive interactions than for the 2 2x y

d

orbital:

13.101 Superoxide dismutase catalyses the conversion of superoxide into molecular oxygen and hydrogen peroxide. This reaction occurs at a metal site that contains one Zn2+ ion and one Cu2+

ion, linked by a histidine ligand that bonds to both metal ions. An O atom binds to the Zn2+ ion:

2O2– + 2H+ SOD O2 + H2O2

N N

N

Cu2+

NN

N

Zn2+

O N

13.103 In order for electron transfer to occur from a cysteine ligand to a Cu ion, there must be a low energy, vacant orbital available to accommodate the transferredelectron. Because Cu is in group

11 of the periodic table, Cu+

has a d10 configuration, with all d orbitals filled, whereas Cu2+ has a d9 configuration and can accept one more electron:

d10 d9

electron

13.105 Visible spectroscopy is used to study coordination compounds that have an energy gap between the highest occupied and lowest unoccupied orbitals that matches the energy of visible light. Because Zn2+ has a d10 configuration, its d orbitals are completely filled and there is no possible

d-d transition in the visible region of the spectrum. In contrast, Co2+ has a d7

configuration, which has unfilled d orbitals. Consequently, metalloproteins that contain Co2+ absorb visible light, making it possible to study them with visible spectroscopy.

d10 d7

13.107 Consult section 13.6 for the chemical reactions of various metallurgical processes:(a) CuFeS2 + 3O2 CuO + FeO + 2SO2

(b) Si + O2 + CaO CaSiO3

(c) TiCl4 + 4Na 4NaCl + Ti

13.109 limestone

100162 g = 170 g

95.5m = 0.170 kg

13.111 Titanium (group 4 of the periodic table) occurs naturally as Ti(IV), a very stable oxidation state that cannot be reduced to free metal except by more active metals such as Na and Al. Because these metals must be produced electrolytically, metallurgists in antiquity were unable to reduce titanium ores to Ti metal. Copper (group 11 of the periodic table) occurs naturally as CuS, an ore that can be reduced to the free metal by roasting in air, a technology that was readily available inantiquity.

13.113 Copper tarnish is a mixture of copper(II) tetrahydroxysulfate and copper(II) dihydroxycarbonate, and is formed when trace amounts of sulfur dioxide, carbon dioxide and water vapour in the atmosphere react with the metal:

3Cu(s) + 2H2O(g) + SO2(g) + 2O2(g) Cu3(OH)4SO4(s)2Cu(s) + H2O(g) + CO2(g) + O2(g) Cu2(OH)2CO3(s)

Additional Exercises

13.115 (a) [Cr(OH2)5Cl]2+

(b) The correct formula of the compound is [Cr(OH2)5Cl]Cl2

(c)

CrCl

H2O OH2

OH2

OH2

OH2

2+

(d) There is only one possible isomer of the complex ion.

13.117 In a linear geometry with the ligands along the z axis, the dz2orbital experiences greater

electron–electron repulsion than any of the other four d orbitals, so this orbital lies highest in energy. The two orbitals that are in the xy plane experience the least repulsion and are lowest in energy:

13.119 The complex is octahedral and the CO molecule is a ligand that causes a large splitting, so the energy level splitting is expected to be large. Tungsten(0) is a d6 ion and all electrons fill the lower levels. The complex is colourless because the energy gap is sufficiently large that visible light has insufficient energy to be absorbed:

13.121 Determine whether each compound is high or low spin using the position of the ligand in the spectrochemical series.

(a) CN– generates a large splitting, so we expect a low-spin complex; Fe2+ is d6

and the complex is octahedral:

(b) Cl– generates a small splitting, so we expect a high-spin complex; Mn2+ is d5

and the complex is tetrahedral:

(c) Coordination complexes of 4d elements are almost always low-spin; Rh3+ is d6

and the complex is octahedral:

(d) H2O is medium-field, but lower oxidation states of 3d elements tend to be high-spin; Co2+

is d7

and the complex is octahedral:

13.123

Empirical formula Molecular formula Structure

PtCl4.2NH3 [Pt(NH3)2Cl4]

PtCl

Cl NH3

NH3

Cl

Cl

cis and transisomers possible

PtCl4.3NH3 [Pt(NH3)3Cl3]Cl

PtCl

Cl NH3

NH3

Cl

NH3

Cl

mer and facisomers possible

PtCl4.4NH3 [Pt(NH3)4Cl2]Cl2

PtCl

Cl NH3

NH3

NH3

NH3

Cl2

cis and transisomers possible

PtCl4.5NH3 [Pt(NH3)5Cl]Cl3

PtH3N

Cl NH3

NH3

NH3

NH3

Cl3

PtCl4.6NH3 [Pt(NH3)6]Cl4

PtH3N

H3N NH3

NH3

NH3

NH3

Cl4

The names of the complexes are as follows:[Pt(NH3)2Cl4] diamminetetrachloroplatinum(IV)[Pt(NH3)3Cl3]Cl triamminetrichloroplatinum(IV) chloride[Pt(NH3)4Cl2]Cl2 tetraamminedichloroplatinum(IV) chloride[Pt(NH3)5Cl]Cl3 pentaaminechloroplatinum(IV) chloride[Pt(NH3)6]Cl4 hexaammineplatinum(IV) chloride