the first transition series

45.1 Introduction
45.2 General Features of the d-Block Elements from Sc to Zn
45.3 Characteristic Properties of the d-Block Elements and their Compounds
by : Sudhir Kumar PGT (Chem)
KV 1 Pathankot
d & f -block
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
Introduction
Scandium
Titanium
Vanadium
Chromium
Manganese
Zinc
Copper
Nickel
Cobalt
Iron
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Strictly speaking, scandium (Sc) and zinc (Zn) are not transitions elements
Sc forms Sc3+ ion which has an empty d sub-shell (3d0)
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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
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General Features of the d-Block Elements from Sc to Zn
Electronic Configurations
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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
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Electronic configurations of the first series of d-block elements
Element
Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc
21 22 23 24 25 26 27 28 29 30
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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
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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
45.2 General Features of the d-Block Elements from Sc to Zn (SB p.167)
d-Block Elements as Metals
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Transition elements have similar atomic radii which make 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 to its alloy (duralumin)
Cr is for conferring inertness on stainless steel
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Iron is used to make ships
Tsing Ma Bridge is constructed of steel
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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
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 Radii
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Variations in atomic and ionic radii of the first series of d-block elements
N
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-shell
the screening and repulsive effects of the electrons in the 3d sub-shell increase
the effective nuclear charge increases slowly
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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
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Comparison of Some Physical and Chemical Properties between d-Block and s-Block Metals
Density
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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
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Ionization Enthalpy
1st
2nd
3rd
4th
Sc Ti V Cr Mn Fe Co Ni Cu Zn
632 661 648 653 716 762 757 736 745 908
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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
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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
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Successive ionization enthalpies exhibit a similar gradual increase across the first transition series
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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
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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.
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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
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Electronegativity
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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
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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
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Melting Point and Hardness
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The melting points of the d-block metals are much higher than those of the s-block metals
Reasons:
1. d-block metal atoms are small in size and closely packed in the metallic lattice. All Group I metals and some Group II metals 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 thus the metallic bond strength is very strong
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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
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Check Point 45-2
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 delocalizing 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.
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General Features of the d-Block Elements from Sc to Zn (
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 oxides and H2
Zn(s) + H2O(g) ZnO(s) + H2(g)
3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)
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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 States
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Oxidation states of the elements of the first transition series in their oxides and chlorides
Oxidation state
+2
TiO VO CrO MnO FeO CoO NiO CuO ZnO TiCl2 VCl2 CrCl2 MnCl2 FeCl2 CoCl2 NiCl2 CuCl2 ZnCl2
+3
Sc2O3 Ti2O3 V2O3 Cr2O3 Mn2O3 Fe2O3 Ni2O3·xH2O ScCl3 TiCl3 VCl3 CrCl3 MnCl3 FeCl3
+4
+5
V2O5
+6
CrO3
+7
Mn2O7
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Oxidation states of the elements of the first transition series in their compounds
Element
Sc Ti V Cr Mn Fe Co Ni Cu Zn
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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.
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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]3d0 configuration)
Mn2+ is more stable than Mn3+ ( [Ar]3d5 configuration)
Zn2+ is more stable than Zn+ ( [Ar]3d10 configuration)
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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
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In acidic medium, vanadium(V) state occurs as VO2+(aq); vanadium(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)
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Vanadium(V) ions can be reduced sequentially to vanadium(II) ions by the action of Zn powder and acid
The sequence of color changes forms a characteristic test for vanadium
VO2+(aq) VO2+(aq) V3+(aq) V2+(aq)
Zn
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The feasibility of the changes in oxidation number of vanadium can be predicted by using electrode potentials easily
–0.76
+1.00
+0.34
–0.26
V3+(aq) + e– V2+(aq)
E (V)
Half reaction
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Under standard conditions, Zn can reduce vanadium(V) to vanadium(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 = –0.76 V
2VO2+(aq) + Zn(s) + 4H+(aq)
2VO2+(aq) + Zn2+(aq) + 2H2O(l) Ecell = +1.76 V
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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 = –0.76 V
2VO2+(aq) + Zn(s) + 4H+(aq)
2V3+(aq) + Zn2+(aq)+ 2H2O(l) Ecell = +1.10 V
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Conclusion:
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
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 V
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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
Very pale pink Dark brown Black Green Purple
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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 reducing agents and is reduced to Mn2+
MnO2(s) + 4H+(aq) + 2e–
MnO2 is used in the laboratory production of chlorine
MnO2(s) + 4HCl(aq) MnCl2(aq) + 2H2O(l) + Cl2(g)
+2
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The most common Mn compound in +7 oxidation state is KMnO4 which is an extremely powerful oxidizing agent. Its oxidizing 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
MnO4–(aq) + 2H2O(l) + 3e– MnO2(s) + 4OH–(aq)
E = +0.59 V
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Mn(II)
Mn(III)
Mn(IV)
Mn(VI)
Mn(VII)
Check Point 45-3
(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
(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.
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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 Complexes
The molecules or ions that form the dative covalent bonds are called ligands
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Examples of ligands:
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Depending on the overall charge of the complex formed, complexes are classified into 3 main types: cationic, neutral and anionic complex
Cationic complex ions
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Neutral complex
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The coordination number of the central metal atom or ion in 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 the central Cu2+ ion, so the coordination number is 4
The most common coordination numbers are 4 and 6
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For the first series of d-block metals, complexes are formed using 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 ligands
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 covalent character
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Diagrammatic representation of the formation of a complex
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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
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(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,
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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
Name of ligand
Bromide (Br–) Chloride (Cl–) Cyanide (CN–) Fluoride (F–) Hydroxide (OH–) Sulphate(VI) (SO42–) Amide (NH2–)
Bromo Chloro Cyano Fluoro Hydroxo Sulphato Amido
Ammonia (NH3) Water (H2O) Carbon monoxide (CO) Nitric oxide (NO)
Ammine Aqua Cabonyl Nitrosyl
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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)
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(b) If the complex is cationic or neutral, then the name of the metal is unchanged.
e.g. [CrCl2(H2O)4]+ dichlorotetraaquachromium(III) ion
Titanate Chromate Manganate Ferrate Cobaltate Nickelate Cuprate Zincate Platinate
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Examples:
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2. Neutral complex
(i) [Fe(H2O)6]Cl2
(ii) [Cu(NH3)4]Cl2
Answer
(a) (i) Hexaaquairon(II) chloride
(b) Write the formulae of the following compounds.
(i) chloropentaamminecobalt(III) chloride
(ii) ammonium hexachlorotitanate(IV)
Answer
(b) (i) [CoCl(NH3)5]Cl2
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Displacement of Ligands and Relative Stability of Complex Ions
The tendency to donate unshared electrons to form dative covalent bonds varies with different ligands
Different ligands form dative covalent bonds of different strength with the metal atom or ion
The ligand within a complex can be replaced by another ligand if the incoming ligand can form a stronger bond with the metal atom or ion
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A stronger ligand (e.g. CN–, Cl–) can displace a weaker ligand (e.g. H2O) from a complex, and a new complex is formed
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
e.g. [Fe(H2O)6]2+(aq) + 6CN–(aq) [Fe(CN)6]4–(aq) + 6H2O(l)
hexaaquairon(II) ion hexacyanoferrate(II) ion
hexaaquanickel(II) ion hexaamminenickel(II) ion
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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 by Cl– ligands
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[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
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Overall equation:
Overall stability constant of [CuCl4]2–(aq) is:
which is given by:
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The larger the overall stability constant, the more stable is the 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 stable complex with Cu2+ ion
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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– ligand to replace a remaining bonded H2O decreases
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NH3 forms a more stable complex with Cu2+ ion than Cl– and H2O ligands do
NH3 can displace both H2O ligands from [Cu(H2O)4]2+(aq) and 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) + 4H2O(l)
[Cu(H2O)4]2+(aq) + NH3(aq) [Cu (NH3)(H2O)3]2+(aq) + H2O(l)
K1 = 1.9 104 dm3 mol–1
[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
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The overall stability constant for [Cu(NH3)4]2+(aq) is larger than that for [CuCl4]2–(aq)
NH3 is a stronger ligand compared with Cl– or H2O
[Cu(NH3)4]2+(aq) is more stable than [CuCl4]2–(aq)
Overall stability constant of [Cu(NH3)4]2+(aq) is:
which is given by Kst = K1 K2 K3 K4 = 1.1 1013 dm12 mol–4
By adding the above 4 equations, overall equation is obtained.
[Cu(H2O)4]2+(aq) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 4H2O(l)
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The displacement of the H2O ligands in [M(H2O)m] by another ligand L can be represented as:
[M(H2O)m] + mL [MLm] + mH2O
The stability constant for the complex [MLm] at a given temp.:
1024
1031
[Fe(H2O)4]3+(aq) + 4Cl–(aq) [FeCl4]–(aq) + 4H2O(l)
1 10–2
Equilibrium
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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
4.8 107
[Cu(H2O)4]2+(aq) + 4Cl– [CuCl4]2–(aq) + 4H2O(l) [Cu(H2O)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)
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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
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 10
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Check Point 45-5
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
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Check Point 45-5 (cont’d)
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
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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;
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Shape
Examples:
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(b) Square planar complexes
Examples:
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2. Hexa-coordinated complexes
For complexes with coordination no. of 6, the
ligands occupy octahedral position to minimize the repulsion from six electron pairs around the central metal ion
Examples:
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Shapes of tetra- and hexa-coordinated complexes
[Cu(NH3)4]2+
[CuCl4]2–
Square planar
[Zn(NH3)4]2+
[CoCl4]2–
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Shapes of tetra- and hexa-coordinated complexes (cont’d)
[Cr(NH3)6]3+
[Fe(CN)6]3–
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Isomer
1. Structural isomers
2. Geometrical isomers
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1. Structural isomers
Structural isomers are isomers that have different ligands 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)3Cl3] • 3H2O brown
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2. Geometrical isomers
Geometrical isomers are isomers that have different arrangement of ligands in space
Only square planar and octahedral complexes have geometrical isomers
(a) Square planar complexes
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Example:
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(b) Octahedral complexes
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Example:
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Example:
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Shape of complex
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Shape of complex
Check Point 45-6
(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]
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(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(H2O)5Cl]Cl2 • H2O, [Cr(H2O)4Cl2]Cl • 2H2O and
[Cr(H2O)3Cl3] • 3H2O) have different numbers of free Cl– ions. One way 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.
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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 Ions
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When a substance absorbs visible light of a certain wavelength, light of wavelengths of other regions of the visible light spectrum will be reflected or transmitted.
the substance will appear coloured
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If the energy involved in electronic transition does not fall into visible light region, the substance will not appear coloured
s-block and p-block elements are usually colourless because an electronic transition is from one principle energy level to a higher one
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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
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When a sufficient amount of energy is absorbed,
electrons will be promoted from 3d orbitals at lower energy level to those at the higher energy level
The energy required for the d-d transition falls within the visible light spectrum.
This leads to light absorption, and reflects the remainder of the visible light
d-block metal ions have specific colours
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The colours of some hydrated d-block metal ions
Number of unpaired d electrons
Hydrated ion
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The colours of some hydrated d-block metal ions (cont’d)
Number of unpaired d electrons
Hydrated ion
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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
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The colors of hydrated metal ions are determined by the oxidation 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 radiation absorbed during electronic transition
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Catalytic Properties of Transition Metals and their Compounds
The use of some d-block metals and their compounds as catalysts in industry
Catalytic oxidation of ammonia
(Manufacture of nitric(V) acid)
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l)
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d-block metals and their compounds exert their catalytic actions in either heterogeneous catalysis or homogeneous catalysis
The function of a catalyst is to provide an alternative pathway of lower activation energy
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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 reaction surface for the reactants to come close together and react
Heterogeneous Catalysis
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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)
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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
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Summary:
In heterogeneous catalysis, the d-block metals or compounds provide a suitable reaction surface for the reaction to take place
the presence of partly-filled d-orbitals
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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
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The standard e.m.f. calculated for the reaction is a highly positive value
there is high tendency for the forward reaction to occur
However the reaction is very slow due to kinetic factors
e.g. reaction between peroxodisulphate(VI) ions and iodide ions
S2O82–(aq) + 2I–(aq) 2SO42–(aq) + I2(aq)
Ecell = +1.47 V
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The Fe2+(aq) are subsequently oxidized by S2O82–(aq) and Fe3+(aq) ions are regenerated
2Fe2+(aq) + S2O82–(aq) 2Fe3+(aq) + 2SO42–(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 V
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Fe(III) ions catalyze the reaction by acting as an intermediate for the transfer of electrons between peroxodisulphate(VI) and iodide ions
The overall reaction:
+) 2Fe2+(aq) + S2O82–(aq) 2Fe3+(aq) + 2SO42–(aq)
2I–(aq) + S2O82–(aq) I2(aq) + 2SO42–(aq)
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Answer
Check Point 45-7
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
(Given: S2O82–(aq) + 2e– 2SO42–(aq) E = +2.01V
I2(aq) + 2e– 2I–(aq) E = +0.54V)
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)
MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l)
eqm
4
eqm
2
4
2
eqm
2
4
st

the first transition series

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