the first transition series

1New Way Chemistry for Hong Kong A-Level Book 4

The First Transition Series

45.145.1 IntroductionIntroduction

45.245.2 General Features of the General Features of the dd-Block Elements -Block Elements from Sc to Znfrom Sc to Zn

45.345.3 Characteristic Properties of the Characteristic Properties of the dd-Block -Block Elements and their CompoundsElements and their Compounds

Chapter 45Chapter 45

45.1 Introduction (SB p.164)

The first transition series

d-Block elements (transition elements):

• Lie between s-block and p-block elements

• Occur in the fourth and subsequent periods

• All contains incomplete d sub-shell (i.e. 1 – 9 electrons) in at

least one of their oxidation state

Scandium

Titanium

Vanadium

ChromiumManganese

ZincCopper

NickelCobaltIron

• Strictly speaking, scandium (Sc) and zinc (Zn) are not

transitions elements

∵ Sc forms Sc3+ ion which has an empty d sub-shell

Zn forms Zn2+ ion which has a completely filled d

sub- shell (3d10)

• Cu+ is not a transition metal ion as it has a completely filled d sub-shell

• Cu2+ is a transition metal ion as it has an incompletely filled d sub-shell

• Cu shows some intermediate behaviour between transition

and non-transition elements because of two oxidation states,

Cu(I) & Cu(II)

45.2 General Features of the d-Block Elements from Sc to Zn (SB p.165)

Electronic ConfigurationsElectronic Configurations

Relative energy levels of orbitals before and after filling with electrons

• Before filling electrons, the energy of 4s sub-shell is

lower than that of 3d sub-shell

4s sub-shell is filled before 3d sub-shell

• Once the 4s sub-shell is filled, the energy will increase

The lowest energy sub-shell becomes 3d sub-shell,

so the next electron is put into 3d sub-shell

Element Atomic number

Electronic configuration

Scandium

Titanium

Vanadium

Chromium

Manganese

Cobalt

Nickel

Copper

[Ar]3d14s2

[Ar]3d24s2

[Ar]3d34s2

[Ar]3d54s1

[Ar]3d54s2

[Ar]3d64s2

[Ar]3d74s2

[Ar]3d84s2

[Ar]3d104s1

[Ar]3d104s2

Electronic configurations of the first series of d-block elements

• Cr is expected to be [Ar] 3d44s2 but the actual configuration is [Ar] 3d54s1

• Cu has the electronic configuration of [Ar] 3d104s1 instead of [Ar] 3d94s2

• This can be explained by the fact that a half-filled or

fully-filled d sub-shell provides extra stability

• d-block elements are typical metals

(1) good conductors of heat and electricity, hard, strong, malleable, ductile and lustrous

(2) high melting and boiling points except Hg is a liquid at room temperture

• These properties make d-block elements as good construction materials

e.g. Fe is used for construction and making machinery

Ti is used to make aircraft and space shuttles

d-Block Elements as Metalsd-Block Elements as Metals

• Transition elements have similar atomic radii which ma

ke them possible for the atom of one element to replace

those of another element in the formation of alloy

e.g. Mn is for conferring hardness and wearing resistance t

o its alloy (duralumin)

Cr is for conferring inertness on stainless steel

Iron is used to make ships Tsing Ma Bridge is constructed of steel

Tungsten in a light bulb

The statue is made of alloy of copper and zinc

Titanium is used in making aircraft

Jewellery made of gold

Observations:

• d-block metals have smaller atomic radii than s-block metals

• The atomic radii of the d-block metals do not show much variation across the series

• The atomic radii decrease initially, remain almost constant in the middle and then increase at the end of series

Atomic Radii and Ionic RadiiAtomic Radii and Ionic Radii

Variations in atomic and ionic radii of the first series of d-block elements

• The atomic size reduces at the beginning of the series

∵ increase in effective nuclear charge with atomic

numbers

the electron clouds are pulled closer to the nucleus

causing a reduction in atomic size

• The atomic size decreases slowly in the middle of the series

∵ when more and more electrons enter the inner 3d sub-

the screening and repulsive effects of the electrons in

the 3d sub-shell increase

the effective nuclear charge increases slowly

• The atomic size increases at the end of the series

∵ the screening and repulsive effects of the 3d

electrons reach a maximum

• The reasons for the trend of the ionic radii of the d-block

elements are similar to those for the atomic radii.

• Remember that the electrons have to be removed from the

4s orbital first

Comparison of Some Physical and Chemical Properties between d-Block and s-Block Metals

Density

Densities (in g cm-3) of the s-block metals and the first series of d-block metals

• d-block metals are generally denser than the s-block

because most of the d-block metals have close-packed

structures while most of the s-block metals do not.

• The densities increase generally across the first series of

d-block metals. This is in agreement with the general

decrease in atomic radius across the series

Ionization Enthalpy

ElementIonization enthalpy (kJ mol–1)

1st 2nd 3rd 4th

• 1st I.E. of d-block metals are greater than those of s-block

elements in the same row of the Periodic Table.

∵ the d-block metals are smaller in size than the s-block

metals, thus they have greater effective nuclear

charges

• For K, the 2nd I.E. is exceptionally higher than its 1st I.E

• For Ca, the 3rd I.E. is exceptionally higher than its 2nd I.E

∵ the electrons are come form the inner fully-filled

electron shells

• The first few successive I.E. for d-block elements do not

show dramatic changes

∵ removal of electrons does not involve the

disruption of inner electron shells

• The 1st I.E. of the d-block metals increase slightly and irregularly across the series

∵ Going across the first transition series, the nuclear charge of the elements increases, and additional electrons are found in the inner 3d sub-shell

The additional screening effect of the additional 3d electrons is so significant that the effective nuclear charge of the elements increases only very slowly across the series

• Successive ionization enthalpies exhibit a similar

gradual increase across the first transition series

• The increases in the 3rd and 4th ionization enthalpies

across the series are progressively more rapid

• Some abnormal high ionization enthalpy, e.g. 1st I.E. of Zn, 2nd I.E. of Cr & Cu and the 3rd I.E. of Mn

∵The removal of an electron from a fully-filled or half-filled sub-shell requires a relatively large amount of energy

Variation of successive ionization enthalpies of the first series of the d-block

elements

Check Point 45-1 Check Point 45-1

Explain the following variation in terms of electronic configurations.

(a) The second ionization enthalpies of both Cr and Cu are higher than those of their next elements respectively.

Answer(a) The second ionization enthalpies of both Cr and Cu are higher than those of their next elements respectively. In the case of Cr, the

second ionization enthalpy involves the removal of an electron from a half-filled 3d sub-shell, which has extra stability. Therefore, this second ionization enthalpy is relatively high. The case is similar for copper where its second ionization enthalpy involves the removal of an electron from a fully-filled 3d sub-shell which also has extra stability. Thus, its second ionization enthalpy is also relatively high.

Check Point 45-1 (cont’d) Check Point 45-1 (cont’d)

Explain the following variation in terms of electronic configurations.

(b) The third ionization enthalpy of Mn is higher than that of its next element.

Answer

(b) The third ionization enthalpy of Mn is higher than that of its next element. It is because its third ionization enthalpy involves the removal of an electron from a half-filled 3d sub-shell which has extra stability. Therefore, its third ionization enthalpy is relatively high.

Electronegativity

Electronegativity values of the s-block metals and the first series of the d-block

metals

• The electronegativity of d-block metals are generally

higher than those of the s-block metals

∵ Generally, d-block metals have smaller atomic

radii than s-block metals

the nuclei of d-block metals can attract the

electrons in a bond more tightly towards

themselves

• The electronegativity shows a slight increase generally

with increasing atomic numbers across the series

∵ Gradual increase in effective nuclear charge and

decrease in atomic radius across the series

The closer the electron shell to the nucleus, the

more strongly the additional electron in a bond is

attracted

Higher electronegativity

Melting Point and Hardness

Melting points (C) of the s-block metals and the first series of the d-block metals

• The melting points of the d-block metals are much higher tha

n those of the s-block metals

Reasons:

1. d-block metal atoms are small in size and closely packed in t

he metallic lattice. All Group I metals and some Group II meta

ls do not have close-packed structures

2. Both 3d and 4s electrons of d-block metals participate in

metallic bonding by delocalizing into the electron sea, and th

us the metallic bond strength is very strong

s-Block metals have only 1 to 2 valence electrons per atom d

elocalizing into the electron sea

• The hardness of a metal depends on the strength of the

metallic bonds

∵ The metallic bond of d-block metals is

stronger than that of s-block metals

d-block metals are much harder than the s-

block metals

What are the differences between the structures and bonding of the d-block and s-block metals? How do these differences affect their melting points?

Answer

The d-block metals are comparatively small, and the metallic atoms are closely packed in the metallic lattice. Besides, both the 3d and 4s electrons of the d-block metals participate in metallic bonding by del

ocalizing into the electron sea. The strength of metallic bond in these metals is thus very strong. In the case of s-block metals, the metallic radius is larger and most of them do not have close-packed structures. Also , as they have only one or two valence electrons per atom delocalizing into the electron sea, the metallic bond formed is weaker. Therefore, the d-block metals have a much higher melting point than the s-block metals.

Reaction with Water

• Generally, s-block metals (e.g. K, Na & Ca) react with

H2O vigorously to form metal hydroxides and H2

• d-block metals react only very slowly with cold water.

Zn and Fe are relatively more reactive

Zn and Fe react with steam to give metal oxid

es and H2

Zn(s) + H2O(g) ZnO(s) + H2(g)

3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)

45.3 Characteristic Properties of the d-Block Elements and their Compounds (SB p.175)

• d-block elements has ability to show variable oxidation states

∵ 3d & 4s electrons are of similar energy levels, the electrons in both of them are available for bonding

When the first transition elements react to form compounds, they can form ions of roughly the same stability by losing different numbers of electrons

Form compounds with a wide variety of oxidation states

Variable Oxidation StatesVariable Oxidation States

Oxidation state

Oxide/Chloride

+1 Cu2O

Cu2Cl2

+2 TiO VO CrO MnO FeO CoO NiO CuO ZnO

TiCl2 VCl2 CrCl2 MnCl2 FeCl2 CoCl2 NiCl2 CuCl2 ZnCl2

+3Sc2O3 Ti2O3 V2O3 Cr2O3 Mn2O3 Fe2O3 Ni2O3·xH2O

ScCl3 TiCl3 VCl3 CrCl3 MnCl3 FeCl3

+4 TiO2 VO2 MnO2

TiCl4 VCl4 CrCl4

+5 V2O5

+6 CrO3

+7 Mn2O7

Oxidation states of the elements of the first transition series in their oxides and chlorides

Element Possible oxidation state

+1 +2 +3 +4

+1 +2 +3 +4 +5

+1 +2 +3 +4 +5 +6

+1 +2 +3 +4 +5 +6 +7

+1 +2 +3 +4 +5 +6

+1 +2 +3 +4 +5

+1 +2 +3

Oxidation states of the elements of the first transition series in their compounds

Observations:

1. Sc and Zn do not exhibit variable oxidation states. Sc3+ has

electronic configuration of argon (i.e. 1s22s22p63s23p6). Zn2+

has the electronic configuration of [Ar] 3d10. Other oxidation

states are not possible.

2. Except Sc, all elements have +2 oxidation state. Except Zn,

all elements have +3 oxidation state

3. The highest oxidation state is +7 at Mn. This corresponds to

removal of all 3d & 4s electrons. (Note: max.oxidation state is

NEVER greater than the total number of 3d & 4s electrons)

4. There is a reduction in the number of oxidation states after Mn.

∵ decrease in the number of unpaired electrons and

increase in nuclear charge which holds the 3d electrons

more firmly

5. The relative stability of various oxidation states can be

correlated -with the stability of empty, half-filled and fully-

filled configuration

e.g. Ti4+ is more stable than Ti3+ ( [Ar]3∵ d0 configuration)

Mn2+ is more stable than Mn3+ ( [Ar]3∵ d5 configuration)

Zn2+ is more stable than Zn+ ( [Ar]3∵ d10 configuration)

• Vanadium shows oxidation states from +2 to +5 in its compounds

• In these oxidation state, vanadium forms ions which have distinctive colours in aqueous solutions

Variable Oxidation States of Vanadium and their Interconversions

Ion Oxidation state Colour

Violet

Yellow

• In acidic medium, vanadium(V) state occurs as VO2+(aq); van

adium(IV) state occurs as VO2+(aq)

• In alkaline medium, vanadium(V) state occurs as VO3–(aq)

• Most compounds with vanadium(V) are good oxidizing agents while those with vanadium(II) are good reducing agents

• The starting material for the interconversions of common oxidation states of vanadium is ammonium vanadate(V) (NH4VO3)

• When NH4VO3 is acidified, vanadium exists in the form o

f VO2+(aq) which the oxidation state of +5

• Vanadium(V) ions can be reduced sequentially to vanadium(II) ions by the action of Zn powder and acid

• The sequence of colour changes forms a characteristic test for vanadium

VO2+(aq) VO2+(aq) V3+(aq) V2+(aq)

conc. HCl

conc. HClyellow violetgreenblue

• The feasibility of the changes in oxidation number of

vanadium can be predicted by using electrode potentials

easily

–0.76

–0.26

Zn2+(aq) + 2e– Zn(s)

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

VO2+(aq) + 2H+(aq) + e– V3+(aq) + H2O(l)

V3+(aq) + e– V2+(aq)

E (V)Half reaction

• Under standard conditions, Zn can reduce vanadium(V) to van

adium(IV) as the Ecell value is +ve

2 (VO2+(aq) + 2H+(aq) + e– VO2+(aq) + H2O(l)) E = +1.00 V

–) Zn2+(aq) + 2e– Zn(s) E = –0.7

2VO2+(aq) + Zn(s) + 4H+(aq)

2VO2+(aq) + Zn2+(aq) + 2H2O(l) Ecell = +1.7

• Further reduction of vanadium(IV) to vanadium(III) by Zn is

feasible as the Ecell value is +ve

2 (VO2+(aq) + 2H+(aq) + e– V3+(aq) + H2O(l)) E = +0.34 V

–) Zn2+(aq) + 2e– Zn(s) E = –0.7

2VO2+(aq) + Zn(s) + 4H+(aq)

2V3+(aq) + Zn2+(aq)+ 2H2O(l) Ecell = +1.1

• Further reduction of vanadium(III) to vanadium(II) by Zn is

also feasible

2 (V3+(aq) + e– V2+(aq)) E = +0.34 V

–) Zn2+(aq) + 2e– Zn(s) E = –0.76

2V3+(aq) + Zn(s) 2V2+(aq) + Zn2+(aq) Ecell = +0.50 VConclusion:

Zn acts as a strong reducing agent which reduces vanadium(V)

through vanadium(IV), vanadium(III) and finally to

vanadium(II) in an acidic medium

• Mn shows oxidation states from +2 to +7 in its compounds

• The most common oxidation states of Mn include +2, +4, +7

• Mn also forms coloured compounds or ions in these oxidation states

Variable Oxidation States of Manganese and their Interconversions

Ion/compound Oxidation state Colour

Mn(OH)3

MnO42–

MnO4–

Very pale pink

Dark brown

Purple

• Mn is most stable in +2 oxidation state

• The most common Mn compound in +4 oxidation state is MnO2 which is a strong oxidizing agent. It reacts with reducin

g agents and is reduced to Mn2+

MnO2(s) + 4H+(aq) + 2e–

Mn2+(aq) + 2H2O(l) E = +1.2

• MnO2 is used in the laboratory production of chlorine

MnO2(s) + 4HCl(aq) MnCl2(aq) + 2H2O(l) + Cl2(g)

very pale pink

• The most common Mn compound in +7 oxidation state is KM

nO4 which is an extremely powerful oxidizing agent. Its oxi

dizing power depends on pH

• In acidic medium, MnO4– ions are reduced to Mn2+ ions

MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l)

E = +1.23 V

purple

very pale pink

purple

• In alkaline medium, MnO4– ions are reduced to MnO2

MnO4–(aq) + 2H2O(l) + 3e– MnO2(s) + 4OH–(aq)

E = +0.59 V

Mn(II)

Mn(III)

Mn(IV)

Mn(VI)

Mn(VII)

(a) The oxidation numbers of copper in its compounds are +1 and +2.

(i) Give the names, formulae and colours of compounds formed between copper and oxygen.

(ii) Is copper more stable in the oxidation state of +1 or +2?

Answer(a) (i) Copper(I) oxide Cu2O – reddish brown

Copper(II) oxide CuO – black

(ii) +2

(b) Explain the following:

(i) When iron(II) sulphate(VI) (FeSO4) is required, it has to be freshly prepared.

(ii) When aluminium reacts with chlorine and hydrogen chloride respectively, aluminium chloride (AlCl3) is formed in both cases. However, two different products are produced when iron reacts with these two chemicals respectively.

Answer

(b) (i) Iron(II) sulphate(VI) solution cannot be stored for a long time. It will be oxidized by air to form iron(III) sulphate(VI).

(ii) Aluminium has only one oxidation state (+3) in its compounds, whereas iron has two (+2 & +3). Iron reacts with the oxidizing agent Cl2 to form FeCl3 but with the non-oxidizing agent HCl to give FeCl2.

A complex is formed when a central metal atom or ion is

surrounded by other molecules or ions which form dative

covalent bonds with the central metal atom or ion.

Formation of ComplexesFormation of Complexes

• The molecules or ions that form the dative covalent bonds ar

e called ligands

• In a ligand, there is at least one atom having a lone pair of e

lectrons which can be donated to the central metal atom or i

on to form a dative covalent bond

Examples of ligands:

• Depending on the overall charge of the complex formed,

complexes are classified into 3 main types: cationic, neutral

and anionic complex

Cationic complex ions

Neutral complex

Anionic complex ions

• The coordination number of the central metal atom or ion i

n a complex is the number of ligands bonded to this metal

atom or ion

e.g. in [Cu(NH3)4]2+(aq), there are 4 ligands are bonded to th

e central Cu2+ ion, so the coordination number is 4

• The most common coordination numbers are 4 and 6

• For the first series of d-block metals, complexes are formed u

sing the 3d, 4s, 4p and 4d orbitals present in the metal atoms

or ions

• Due to the presence of vacant, low energy orbitals, d-block

metals can interact with the orbitals of the surrounding ligand

• Due to the the relatively small sizes and high charge of

d-block metal ions, they introduce strong polarization on the

ligands. This favours the formation of bonds of high covalen

t character

Diagrammatic representation of the formation of a complex

• Complexes are named according to the rules recommended by IUPAC

Nomenclature of Complexes

The rules of naming a complex are as follow:

1. (a) For any ionic compound, the cation is named before the anion.

(b) If the compound is neutral, then the name of the complex is name of the compound

(c) In naming a complex, the ligands are named before the central metal atom or ion, negative ones first an

d then neutral ones

(d) The number of each type of ligands are specified by the

Greek prefixes: mono-, di-, tri-, tetra-, penta-, hexa-, etc.

(e) The oxidation number of the metal ion in the complex is

named immediately after it by Roman numerals

Therefore,

K3[Fe(CN)6] potassium hexacyanoferrate(III)

[CrCl2(H2O)4]Cl dichlorotetraaquachromium(III) chloride

[CoCl3(NH3)] trichlorotriamminecobalt(III)

Note: in the formulae, the complexes are always enclosed in [ ]

2. (a) The root names of anionic ligands always end in -o.e.g. CN– cyano

Cl– chloro

(b) The names of neutral ligands are the names of the molecules, except NH3, H2O, CO and NO

e.g. NH3 ammine

H2O aqua

Anionic ligandName of ligan

dNeutral ligand

Name of ligand

Bromide (Br–)Chloride (Cl–)Cyanide (CN–)Fluoride (F–)Hydroxide (OH–)Sulphate(VI) (SO4

–)Amide (NH2

BromoChloroCyanoFluoroHydroxoSulphatoAmido

Ammonia (NH3)Water (H2O)Carbon monoxide (CO)Nitric oxide (NO)

AmmineAquaCabonylNitrosyl

3. (a) If the complex is anionic, then the suffix -ate is

attached to the name of the metal, followed by the

oxidation state of that metal

e.g. K2CoCl4 potassium tetrachlorocobaltate(II)

K3Fe(CN)6 potassium hexacyanoferrate(III)

[CuCl4]2– tetrachlorocuprate(II) ion

(b) If the complex is cationic or neutral, then the name of the

metal is unchanged.

e.g. [CrCl2(H2O)4]+ dichlorotetraaquachromium(III) ion

[CoCl3(NH3)3] trichlorotriamminecobalt(III)

Metal Name in anionic complex

TitaniumChromiumManganeseIronCobaltNickelCopperZincPlatinum

TitanateChromateManganateFerrateCobaltateNickelateCuprateZincatePlatinate

Examples:

1. Ionic complexes

2. Neutral complex

(a) Name the following compounds.

(i) [Fe(H2O)6]Cl2

(ii) [Cu(NH3)4]Cl2

(iii) [PtCl4(NH3)2]

(iv) K2[CoCl4] Answer(a) (i) Hexaaquairon(II) chloride

(ii) Tetraamminecopper(II) chloride

(iii) Tetrachlorodiammineplatinum(IV)

(iv) Potassium tetrachlorocobaltate(II)

(b) Write the formulae of the following compounds.

(i) chloropentaamminecobalt(III) chloride

(ii) ammonium hexachlorotitanate(IV)

(iii) dihydroxotetraaquairon(II)Answer

(b) (i) [CoCl(NH3)5]Cl2

(ii) (NH4)2[TiCl6]

(iii) [Fe(H2O)4(OH)2]

Displacement of Ligands and Relative Stability of Complex Ions

• The tendency to donate unshared electrons to form dative covale

nt bonds varies with different ligands

• Different ligands form dative covalent bonds of different streng

th with the metal atom or ion

• The ligand within a complex can be replaced by another liga

nd if the incoming ligand can form a stronger bond with the

metal atom or ion

• When different ligands are present, they compete for a metal ion

• A stronger ligand (e.g. CN–, Cl–) can displace a weaker ligand (e.g. H2O) from a complex, and a new complex is form

e.g. [Fe(H2O)6]2+(aq) + 6CN–(aq) [Fe(CN)6]4–(aq) + 6H2O(l)hexaaquairon(II) ion hexacyanoferrate(II) ion

[Ni(H2O)6]2+(aq) + 6NH3(aq) [Ni(NH3)6]2+(aq) + 6H2O(l)hexaaquanickel(II) ion hexaamminenickel(II) ion

• Complex ions are usually coloured and the colours are related to the types of ligands present

Displacement of ligands usually associated with colour changes which can be followed during experiments easily

Example:

• 0.5 M CuSO4 solution is put into a test tube. The complex

ion present is [Cu(H2O)6]2+ which is pale blue

• Conc. HCl is added dropwise to the CuSO4 solution

• The solution turns from pale blue to green and finally to yellow

• This is due to the stepwise replacement of H2O ligands b

y Cl– ligands

• Each stage is charaterized by an equilibrium constant called the stepwise stability constant

[Cu(H2O)4]2+(aq) + Cl–(aq) [Cu(H2O)3Cl]+(aq) + H2O(l)

K1 = 6.3 102 dm3 mol–1

[Cu(H2O)3Cl]+(aq) + Cl–(aq) [Cu(H2O)2Cl2](aq) + H2O(l)

K2 = 4.0 101 dm3 mol–1

[Cu(H2O)Cl3]–(aq) + Cl–(aq) [CuCl4]2–(aq) + H2O(l)

K4 = 3.1 dm3 mol–1

[Cu(H2O)2Cl2](aq) + Cl–(aq) [Cu(H2O)Cl3]–(aq) + H2O(l)

K3 = 5.4 dm3 mol–1

Overall equation:

[Cu(H2O)4]2+(aq) + 4Cl–(aq) [CuCl4]2–(aq) + 4H2O(l)

Overall stability constant of [CuCl4]2–(aq) is:

which is given by:

Kst = K1 K2 K3 K4 = 4.2 105 dm12 mol–4

)](Cl[)](]O)[[Cu(H

)](][[CuClK

• The larger the overall stability constant, the more stable is t

he complex

• In this example, the overall equilibrium lies mainly on the right

and [CuCl4]2–(aq) is predominant over [Cu(H2O)4]2+(aq)

Cl– ligands can replace H2O ligands to form a more stab

le complex with Cu2+ ion

• The stepwise stability constant decreases from K1 to K4

Reasons:

1. When the central Cu2+ ion is surrounded by an increasing

number of Cl– ligands, the chance for an addition Cl– liga

nd to replace a remaining bonded H2O decreases

2. There is a progressive change from a cationic complex to

a neutral complex, and then anionic complex. Due to the

electrostatic repulsion between anionic complex and Cl–

ions, the approach of Cl– ligands becomes more difficult

• NH3 forms a more stable complex with Cu2+ ion than Cl– a

nd H2O ligands do

• NH3 can displace both H2O ligands from [Cu(H2O)4]2+(aq) an

d Cl– ligands from [CuCl4]2–(aq), forming the deep blue [Cu

(NH3)4]2+(aq) ion

[Cu(H2O)4]2+(aq) + 4NH3(aq)

[Cu(NH3)4]2+(aq) + 4H2

• The displacement also occurs in stepwise reaction

[Cu(H2O)4]2+(aq) + NH3(aq) [Cu (NH3)(H2O)3]2+(aq) + H2O(l)

K1 = 1.9 104 dm3 mo

[Cu(NH3)(H2O)3]2+(aq) + NH3(aq) [Cu (NH3)2(H2O)2]2+(aq) + H2O(l)

K2 = 3.9 103 dm3 mol–1

[Cu(NH3)2(H2O)2]2+(aq) + NH3(aq) [Cu (NH3)3(H2O)]2+(aq) + H2O(l)

K3 = 1.0 103 dm3 mol–1

[Cu(NH3)3(H2O)]2+(aq) + NH3(aq) [Cu (NH3)4]2+(aq) + H2O(l)

K4 = 1.5 102 dm3 mol–1

• Overall stability constant of [Cu(NH3)4]2+(aq) is:

which is given by Kst = K1 K2 K3 K4 = 1.1 1013 dm12 mol–

)](NH[)](]O)[[Cu(H

)](])[[Cu(NHK

• The overall stability constant for [Cu(NH3)4]2+(aq) is larg

er than that for [CuCl4]2–(aq)

NH3 is a stronger ligand compared with Cl– or H2

O [Cu(NH3)4]2+(aq) is more stable than [CuCl4]2–(a

• By adding the above 4 equations, overall equation is obtained.

[Cu(H2O)4]2+(aq) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 4H2O(l)

]L[]O)[M(H

• The displacement of the H2O ligands in [M(H2O)m] by anoth

er ligand L can be represented as:

[M(H2O)m] + mL [MLm] + mH2O

• The stability constant for the complex [MLm] at a given tem

8 10–2

[Fe(H2O)6]2+(aq) + 6CN–(aq) [Fe(CN)6]4–(aq) + 6H2O(l)

[Fe(H2O)6]3+(aq) + 6CN–(aq) [Fe(CN)6]3–(aq) + 6H2O(l)

[Fe(H2O)4]3+(aq) + 4Cl–(aq) [FeCl4]–(aq) + 4H2O(l)

1 10–2Cr(OH)3(aq) + OH–(aq) [Cr(OH)4]–(aq)

Kst ((mol dm–3)–n)Equilibrium

Equilibrium Kst ((mol dm–3)–n)

[Co(H2O)6]2+(aq) + 6NH3(aq) [Co(NH3)6]2+(aq) + 6H2O(l)

[Co(H2O)6]3+(aq) + 6NH3(aq) [Co(NH3)6]3+(aq) + 6H2O(l)

7.7 104

4.5 1033

[Ni(H2O)6]2+(aq) + 6NH3(aq) [Ni(NH3)6]2+(aq) + 6H2O(l) 4.8 107

[Cu(H2O)4]2+(aq) + 4Cl– [CuCl4]2–(aq) + 4H2O(l) [Cu(H2

O)4]2+(aq) + NH3(aq) [Cu(NH3)(H2O)3]2+(aq) + H2O(l)

[Cu(NH3)(H2O)3]2+(aq) + NH3(aq) [Cu(NH3)2(H2O)2]2+(aq) + H2O(l)

[Cu(NH3)2(H2O)2]2+(aq) + NH3(aq) [Cu(NH3)3(H2O)]2+(aq) + H2O(l)

[Cu(NH3)3(H2O)]2+(aq) + NH3(aq) [Cu(NH3)4]2+(aq) + H2O(l)

[Cu(H2O)4]2+(aq) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 4H2O(l)

4.8 105

1.9 104 (K1)

3.9 103 (K2)

1.0 103 (K3)

1.5 102 (K4)

1.1 1013

(Kst = K1K2K3K

Equilibrium Kst ((mol dm–3)–n)

[Zn(H2O)4]2+(aq) + 4CN–(aq) [Zn(CN)4]2– (aq) + 4H2O(l)

[Zn(H2O)4]2+(aq) + 4NH3(aq) [Zn(NH3)4]2+(aq) + 4H2O(l)

Zn(OH)2(s) + 2OH–(aq) [Zn(OH)4]2– (aq)

5 1016

3.8 109

• As shown in the table, the values of stability constants

are very large

The complex ions of the d-block metals are

generally very stable

Answer the following questions by considering the stability constants of the silver complexes.

Ag+(aq) + 2Cl–(aq) [AgCl2]–(aq) Kst = 1.1 105 mol–2 dm6

Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) Kst = 1.6 107 mol–2 dm6

Ag+(aq) + 2CN–(aq) [Ag(CN)2]–(aq) Kst = 1.0 1021 mol–2 dm6

(a) Give the most stable and the least stable complexes of silver.Answer

(a) The most stable complex of silver is [Ag(CN)2]–(aq), whereas the least stable one is [AgCl2]–(aq)

Answer the following questions by considering the stability constants of the silver complexes.

Ag+(aq) + 2Cl–(aq) [AgCl2]–(aq) Kst = 1.1 105 mol–2 dm6

Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) Kst = 1.6 107 mol–2 dm6

Ag+(aq) + 2CN–(aq) [Ag(CN)2]–(aq) Kst = 1.0 1021 mol–2 dm6

(b) (i) What will be formed when CN–(aq) is added to a solution of [Ag(NH3)2]+?

(ii) What will be formed when NH3(aq) is added to a solution of [Ag(CN)2]–? Answer

(b) (i) [Ag(CN)2]–(aq) and NH3(aq)

(ii) No reaction

Stereostructures of Tetra- and Hexa- Coordinated Complexes

• The spatial arrangement of ligands around the central metal atom or ion in a complex is referred to as the stereochemistry of the complex

• The coordination number of the central metal atom or ion is determined by:

1. The size of the central metal atom or ion;

2. The number and the nature of vacant orbitals of the d-block metal atoms or ions available for the formation of dative covalent bonds

Shape1. Tetra-coordinated complexes

(a) Tetrahedral complexes

Tetrahedral shape is a common geometry of tetra-coordinated complexes

Examples:

(b) Square planar complexes

Some tetra-coordinated complexes show a square planar structure

Examples:

2. Hexa-coordinated complexes

For complexes with coordination no. of 6, the ligands occupy octahedral position to mini

mize the repulsion from six electron pairs around the central metal ion

Examples:

[Cu(NH3)4]2+

[CuCl4]2–

Square planar

[Zn(NH3)4]2+

[CoCl4]2–

Tetrahedral

ExampleShape of complexCoordination number of the central metal atom or ion

Shapes of tetra- and hexa-coordinated complexes

[Cr(NH3)6]3+

[Fe(CN)6]3–

Octahedral

ExampleShape of complexCoordination number of the central metal atom or ion

Shapes of tetra- and hexa-coordinated complexes (cont’d)

Isomer

• Isomers of complexes are classified into:

1. Structural isomers

2. Geometrical isomers

Isomers are different compounds that have the same

molecular formula

1. Structural isomers

Structural isomers are isomers that have different l

igands bonded to the central metal atom or ion

Example: Cr(H2O)6Cl3 has four structural isomers

which have different colours:

[Cr(H2O)6]Cl3 violet

[Cr(H2O)5Cl]Cl2 • H2O light green

[Cr(H2O)4Cl2]Cl • 2H2O dark green

[Cr(H2O)3Cl3] • 3H2O brown

2. Geometrical isomers

Geometrical isomers are isomers that have different arr

angement of ligands in space

• Only square planar and octahedral complexes have geometrical isomers

(a) Square planar complexes

(i) Square planar complexes of the form [Ma2b2]

may exist in cis- or trans- form

Example:

Isomers in which two ligands of the same type occupy ad

jacent corners of the square are called cis-isomer

Isomers in which two ligands of the same type occupy op

posite corners of the square are called trans-isomer

(ii) Square planar complexes of the form [Ma2bc] may

also exist in cis- or trans- form

(b) Octahedral complexes

(i) Octahedral complexes of the form [Ma4b2] may

exist in cis- or trans- form

Example:

(ii) Octahedral complexes of the form [Ma3b3] may

exist in fac- or mer- form

Example:

Shape of complex

Chemical formula

Geometrical isomer

Square planar

[Ma2b2]

cis trans

[Ma2bc]

cis trans

Shape of complex

Chemical formula Geometrical isomer

Octahedral

[Ma4b2]

cis trans

[Ma3b3]

fac mer

(a) Are there any geometrical isomers for a complex of the form [Ma2b2]? Explain your answer with suitable drawings.

(M represents the central metal ion, a and b are two different kinds of ligands.)

Answer

(a) The square planar complex of the form [Ma2b2] may exist in cis and trans forms.

There is no geometrical isomer for a tetrahedral complex of the form [Ma2b2]

(b) Besides using colours, suggest two experimental methods to distinguish between the four isomers of Cr(H2O)6Cl3:[Cr(H2O)6]Cl3, [Cr(H2O)5Cl]Cl2 • H2O, [Cr(H2O)4Cl2]Cl • 2H2O, [Cr(H2O)3Cl3] • 3H2O.

Answer

(b) The four isomers of chromium(III) (i.e. [Cr(H2O)6]Cl3, [Cr(H2

O)5Cl]Cl2 • H2O, [Cr(H2O)4Cl2]Cl • 2H2O and [Cr(H2O)3Cl3] • 3H2O) have different numbers of free Cl– ions. One wa

y to distinguish them is by the use of acidified silver nitrate(V) solution. When excess AgNO3(aq) is added to one mole of each of the isomers, [Cr(H2O)6]Cl3 gives three moles of AgCl, [Cr(H2O)5Cl]Cl2 • H2O gives two moles of AgCl, [Cr(H2O)4Cl2]Cl • 2H2O gives one mole of AgCl, and [Cr(H2O)3Cl3] • 3H2O does not give AgCl.

Another way to distinguish them is by measuring their electrical conductivities. As the electrical conductivity depends on the number of ions formed when dissolved in water, [Cr(H2O)6]Cl3 has the highest electrical conductivity whereas [Cr(H2O)3Cl3] • 3H2O has the least.

• The natural colours of precious gemstones are due to the existence of small quantities of d-block metal ions

• Most of the d-block metals form coloured compounds and most of their complexes are coloured too

∵ the presence of incompletely filled d orbitals in the d-block metal ions

Coloured IonsColoured Ions

• When a substance absorbs visible light of a certain wavele

ngth, light of wavelengths of other regions of the visible lig

ht spectrum will be reflected or transmitted.

the substance will appear coloured

• The absorption of light energy is associated with electronic t

ransition (i.e. electron jumping from a lower energy level to

a higher one). The energy required for electronic transition i

s quantized

• If the energy involved in electronic transition does not fall

into visible light region, the substance will not appear col

• s-block and p-block elements are usually colourless beca

use an electronic transition is from one principle energy l

evel to a higher one

the energy involved is too high in energy and it fall

s into ultraviolet region

• For the d-block elements, the five 3d orbitals are degenerate in gaseous ions

• However, under the influence of a ligand, the 3d orbitals will split into 2 groups of orbitals with slightly different energy levels

due to the interaction of the 3d orbitals with the electron clouds of the ligands

• When a sufficient amount of energy is absorbed,

electrons will be promoted from 3d orbitals at lower en

ergy level to those at the higher energy level

• The energy required for the d-d transition falls within th

e visible light spectrum.

This leads to light absorption, and reflects the

remainder of the visible light

d-block metal ions have specific colours

Number of unpaired d electrons

Hydrated ion Colour

0 Sc3+

Colourless

1 Ti3+

Purple

Violet

The colours of some hydrated d-block metal ions

Number of unpaired d electrons

Hydrated ion Colour

4 Cr2+

Violet

5 Mn2+

Very pale pink

Yellow

Co2+(aq) Zn2+(aq) Fe3+(aq) Mn2+(aq) Fe2+(aq) Cu2+(aq)

The colours of some hydrated d-block metal ions (cont’d)

• For d-d electronic transition and absorption of visible light

to occur, there must be unpaired d electrons in the d-block

metal atoms or ions

Sc3+ and Zn2+ are colourless due to the empty 3d

sub-shell and the fully-filled 3d sub-shell respect

• The colours of hydrated metal ions are determined by the o

xidation states of the particular d-block elements

e.g. Fe2+(aq) is green while Fe3+(aq) is yellow

different oxidation states are caused by different

numbers of d electrons in the d-block metal ion

this has direct effects on the wavelength of the radiati

on absorbed during electronic transition

Catalytic Properties of Transition Metals and their CompoundsCatalytic Properties of Transition Metals and their Compounds

Catalytic oxidation of ammonia

(Manufacture of nitric(V) acid)

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l)

Hardening of vegetable oil (Manufacture of margarine)

RCH = CH2 + H2 RCH2CH3 NiNi

Haber process

N2(g) + 3H2(g) 2NH3(g)FeFe

Contact process

2SO2(g) + O2(g) 2SO3(g)

V2O5 or vanadate(V)(VO3

Reaction catalyzedCatalyst d-block element

The use of some d-block metals and their compounds as catalysts in industry

• d-block metals and their compounds exert their catalyt

ic actions in either heterogeneous catalysis or homog

eneous catalysis

• The function of a catalyst is to provide an alternative

pathway of lower activation energy

enabling the reaction to proceed faster than the

uncatalyzed one

• In heterogeneous catalysis, the catalyst and reactants

are in different phases

• The most common heterogeneous catalysts are finely

divided solids for gaseous reactions

• A heterogenous catalyst provides a suitable reactio

n surface for the reactants to come close together an

d react

Heterogeneous Catalysis

• In the absence of a catalyst, the formation of gaseous ammonia proceeds at an extremely low rate

∵ the probability of collision of four gaseous molecules is very small

the four reactant molecules have to collide in

a proper orientation in order to give products

the bond enthalpy of N N is very large

the reaction has a high activation energy

e.g.: Synthesis of gaseous ammonia from N2 and H2

N2(g) + 3H2(g) 2NH3(g)

• In the presence of iron catalyst, the reaction

proceeds faster as it provides an alternative

reaction pathway

• The catalyst exists in a different phase from that of

both reactant and products

• The catalytic action occurs at the interface between

two phases, and the metal provides an active

reaction surface for the reaction to occur

The catalytic mechanism of the formation of NH3(g) from N2(g) and H2(g)

Energy profiles of the reaction pathways in the presence and absence of a heterogeneous catalyst

Summary:

• In heterogeneous catalysis, the d-block metals or compoun

ds provide a suitable reaction surface for the reaction to t

ake place

∵ the presence of partly-filled d-orbitals

this enables the metals to accept electrons from

reactant particles on one hand and donate electro

ns to reactant particles on the other

• A homogenous catalyst is in the same phase as the reactants and products

• The catalyst forms an intermediate with the reactants

it changes the reaction mechanism to a new one with a lower activation energy

• The ability of d-block metals to exhibit variable oxidation states enables the formation of the reaction intermediates

Homogeneous Catalysis

• e.g. reaction between peroxodisulphate(VI) ions and iodide ions

S2O82–(aq) + 2I–(aq) 2SO4

2–(aq) + I2(aq)

Ecell = +1.47 V

• The standard e.m.f. calculated for the reaction is a highly

positive value

there is high tendency for the forward reaction to

• However the reaction is very slow due to kinetic factors

• The Fe2+(aq) are subsequently oxidized by S2O82–(aq) and

Fe3+(aq) ions are regenerated

2Fe2+(aq) + S2O82–(aq) 2Fe3+(aq) + 2SO4

2–(aq)

Ecell = +1.24 V

• In catalytic process, Fe3+(aq) ions oxidizes I–(aq) to I2(aq)

with themselves being reduced to Fe2+(aq)

2I–(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq)

Ecell = +0.23

• Fe(III) ions catalyze the reaction by acting as an intermed

iate for the transfer of electrons between peroxodisulph

ate(VI) and iodide ions

The overall reaction:

2I–(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq)

+) 2Fe2+(aq) + S2O82–(aq) 2Fe3+(aq) + 2SO4

2–(aq)

2I–(aq) + S2O82–(aq) I2(aq) + 2SO4

2–(aq)

Energy profiles for the oxidation of I–(aq) ions by S2

O82–(aq) ions in the presen

ce and absence of a homogeneous catalyst

Which of the following redox systems might catalyze the oxidation of iodide ions by peroxodisulphate(VI) ions in an aqueous solution?

Cr2O72–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l)

E = +1.33V

E = +1.52V

Sn4+(aq) + 2e– Sn2+(aq) E = +0.15V

(Given: S2O82–(aq) + 2e– 2SO4

2–(aq) E = +2.01VI2(aq) + 2e– 2I–(aq) E = +0.54V)

Answer

Systems with E greater than +0.54V and smaller than +2.01V are able to catalyze the oxidation of iodide ions by peroxodisulphate(VI) ions in an aqueous solution. Hence, the following two redox systems are able to catalyze the reactions.

Cr2O72–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l)

The END

the first transition series

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