The d-Block Elements.

The d-Block Elements

Introduction d-block elements  locate between the s-block and p-block
 known as transition elements  occur in the fourth and subsequent periods of the Periodic Table

d-block elements period 4 period 5 period 6 period 7

Introduction Transition elements are elements that contain an incomplete d sub-shell (i.e. d1 to d9) in at least one of the oxidation states of their compounds. 3d0 3d10

Introduction Sc and Zn are not transition elements because
They form compounds with only one oxidation state in which the d sub-shell are NOT imcomplete. Sc  Sc3+ 3d0 Zn  Zn2+ 3d10

Introduction Cu+ 3d not transitional Cu Cu d9 transitional

The first transition series
the first horizontal row of the d-block elements

Characteristics of transition elements
(d-block metals vs s-block metals) Physical properties vary slightly with atomic number across the series (cf. s-block and p-block elements) Higher m.p./b.p./density/hardness than s-block elements of the same periods. Variable oxidation states (cf. fixed oxidation states of s-block metals)

Characteristics of transition elements
4. Formation of coloured compounds/ions (cf. colourless ions of s-block elements) 5. Formation of complexes 6. Catalytic properties

Electronic Configurations
The building up of electronic configurations of elements follow:  Aufbau principle  Pauli exclusion principle  Hund’s rule

3d and 4s sub-shells are very close to each other in energy. Relative energy of electrons in sub-shells depends on the effective nuclear charge they experience. Electrons enter 4s sub-shell first Electrons leave 4s sub-shell first

Cu Cu2+ Relative energy levels of orbitals in atom and in ion

Valence electrons in the inner 3d orbitals Examples:  The electronic configuration of scandium: 1s22s22p63s23p63d14s2  The electronic configuration of zinc: 1s22s22p63s23p63d104s2

Electronic configuration
Electronic configurations of the first series of the d-block elements Element Atomic number Electronic configuration Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc 21 22 23 24 25 26 27 28 29 30 [Ar] 3d 14s2 [Ar] 3d 24s2 [Ar] 3d 34s2 [Ar] 3d 54s1 [Ar] 3d 54s2 [Ar] 3d 64s2 [Ar] 3d 74s2 [Ar] 3d 84s2 [Ar] 3d 104s1 [Ar] 3d 104s2

A half-filled or fully-filled d sub-shell
has extra stability

d -Block Elements as Metals
d-Block elements are typical metals Physical properties of d-Block elements :  good conductors of heat and electricity  hard and strong  malleable and ductile

Physical properties of d-Block elements:  lustrous  high melting points and boiling points Exceptions : Mercury  low melting point  liquid at room temperature and pressure

 extremely useful as construction materials  strong and unreactive

Iron  used for construction and making machinery nowadays  abundant  easy to extract cheap

Iron  corrodes easily  often combined with other elements to form steel  harder and more resistant to corrosion

Titanium Corrosion resistant, light, strong and withstand large temperature changes  used to make aircraft and space shuttles  expensive

The similar atomic radii of the transition metals facilitate the formation of substitutional alloys  the atoms of one element to replace those of another element  modify their solid structures and physical properties

Chromium  confers inertness to stainless steel Manganese confers hardness & wearing resistance to its alloys e.g. duralumin : alloy of Al with Mn/Mg/Cu

Atomic Radii and Ionic Radii
Two features can be observed: 1. The d-block elements have smaller atomic radii than the s-block elements 2. The atomic radii of the d-block elements do not show much variation across the series

Variation in atomic radius of the first 36 elements

On moving across the Period,
(i) Nuclear charge  (ii) Shielding effect (repulsion between e-)  (i)  (ii) (i) > (ii) (ii) > (i)

At the beginning of the series  atomic number   effective nuclear charge   the electron clouds are pulled closer to the nucleus  atomic size 

In the middle of the series
 more electrons enter the inner 3d sub-shell  The inner 3d electrons shield the outer 4s electrons effectively  the effective nuclear charge experienced by 4s electrons increases very slowly  only a slow decrease in atomic radius in this region

At the end of the series  the screening and repulsive effects of the electrons in the 3d sub- shell become even stronger  Atomic size 

Comparison of Some Physical and Chemical Properties between the d-Block and s-Block Elements
Many of the differences in physical and chemical properties between the d-block and s-block elements  explained in terms of their differences in electronic configurations and atomic radii

1. Density Densities (in g cm–3) of the s-block elements and the first series of the d-block elements at 20C

1. Density d-block > s-block
 the atoms of the d-block elements are generally smaller in size 2. are more closely packed (fcc/hcp vs bcc in group 1) 3. have higher relative atomic masses

1. Density The densities  generally increase across the first series of the d-block elements  1. general decrease in atomic radius across the series 2. general increase in atomic mass across the series

Ionization enthalpy (kJ mol–1)
K  Ca (sharp ) ; Ca  Sc (slight ) Element Ionization enthalpy (kJ mol–1) 1st 2nd 3rd 4th K Ca 418 590 3 070 1 150 4 600 4 940 5 860 6 480 Sc Ti V Cr 632 661 648 653 1 240 1 310 1 370 1 590 2 390 2 720 2 870 2 990 7 110 4 170 4 770

Ionization enthalpy (kJ mol–1)
Sc  Cu (slight ) ; Cu  Zn (sharp ) Element Ionization enthalpy (kJ mol–1) 1st 2nd 3rd 4th Cr Mn Fe Co Ni Cu Zn 653 716 762 757 736 745 908 1 590 1 510 1 560 1 640 1 750 1 960 1 730 2 990 3 250 2 960 3 230 3 390 3 550 3 828 4 770 5 190 5 400 5 100 5 690 5 980

2. Ionization Enthalpy The first ionization enthalpies of the d-block elements  greater than those of the s-block elements in the same period of the Periodic Table  1. The atoms of the d-block elements are smaller in size 2. greater effective nuclear charges

Sharp  across periods 1, 2 and 3
Slight  across the transition series

2. Ionization Enthalpy Going across the first transition series
 the nuclear charge of the elements increases  additional electrons are added to the ‘inner’ 3d sub-shell

2. Ionization Enthalpy The screening effect of the additional 3d electrons is significant The effective nuclear charge experienced by the 4s electrons increases very slightly across the series For 2nd, 3rd, 4th… ionization enthalpies, slight and gradual  across the series are observed.

Electron has to be removed from completely filled 3p subshell
Cr+ Mn2+ Fe3+ 3d10 d10/s2

2. Ionization Enthalpy The first few successive ionization enthalpies for the d-block elements  do not show dramatic changes  4s and 3d energy levels are close to each other

3. Melting Points and Hardness
d-block >> s-block  1. both 4s and 3d e- are involved in the formation of metal bonds 2. d-block atoms are smaller

K has an exceptionally small m.p. because it has an more open b.c.c. structure.

Sc Ti V Cr Mn Fe Co Ni Cu Zn  Unpaired electrons are relatively more involved in the sea of electrons

Sc Ti V Cr Mn Fe Co Ni Cu Zn 3d 4s   Sc   Ti   V m.p.  from Sc to V due to the  of unpaired d-electrons (from d1 to d3)

Sc Ti V Cr Mn Fe Co Ni Cu Zn 3d 4s    Fe    Co    Ni 2. m.p.  from Fe to Zn due to the  of unpaired d-electrons (from 4 to 0)

Sc Ti V Cr Mn Fe Co Ni Cu Zn 3. Cr has the highest no. of unpaired electrons but its m.p. is lower than V. 3d 4s   Cr It is because the electrons in the half-filled d-subshell are relatively less involved in the sea of electrons.

Why is Hg a liquid at room conditions ?
Sc Ti V Cr Mn Fe Co Ni Cu Zn 4. Mn has an exceptionally low m.p. because it has the very open cubic structure. Why is Hg a liquid at room conditions ? All 5d and 6s electrons are paired up and the size of the atoms is much larger than that of Zn.

The hardness of a metal depends on  the strength of the metallic bonds The metallic bonds of the d-block elements are stronger than those of the s-block elements  much harder than the s-block elements

Mohs scale : - A measure of hardness
Talc Diamond 10 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

4. Reaction with Water In general, the s-block elements
 react vigorously with water to form metal hydroxides and hydrogen The d-block elements  react very slowly with cold water  react with steam to give metal oxides and hydrogen

4. Reaction with Water 2K(s) + 2H2O(l)  2KOH(aq) + H2(g)
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g) Ca(s) + 2H2O(l)  Ca(OH)2(aq) + H2(g) Zn(s) + H2O(g)  ZnO(s) + H2(g) 3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)

d-block compounds vs s-block compounds A Summary : -
Ions of d-block metals have higher charge density  more polarizing  1. more covalent in nature 2. less soluble in water 3. less basic (more acidic) Basicity : Fe(OH)3 < Fe(OH)2 << NaOH Charge density : Fe3+ > Fe2+ > Na+

d-block compounds vs s-block compounds A Summary : -
4. less thermally stable e.g. CuCO3 << Na2CO3 5. tend to exist as hydrated salts e.g. CuSO4.5H2O, CoCl2.2H2O 6. hydrated ions undergo hydrolysis more easily e.g. [Fe(H2O)6]3+(aq) + H2O  [Fe(OH)(H2O)5]2+(aq) + H3O+ acidic

Variable Oxidation States
One of the most striking properties  variable oxidation states The 3d and 4s electrons are  in similar energy levels  available for bonding

Variable Oxidation States
Elements of the first transition series  form ions of roughly the same stability by losing different numbers of the 3d and 4s electrons

Oxidation states of the elements of the first transition series in their oxides and chlorides
Oxides / Chloride +1 Cu2O Cu2Cl2 +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 TiO2 VO2 MnO2 TiCl4 VCl4 CrCl4 +5 V2O5 +6 CrO3 +7 Mn2O7

Possible oxidation state Possible oxidation state
Oxidation states of the elements of the first transition series in their compounds Element Possible oxidation state Sc Ti V Cr Mn Fe Co Ni Cu Zn Element Possible oxidation state Sc Ti V Cr Mn Fe Co Ni Cu Zn +3 +2

1. Scandium and zinc do not exhibit variable oxidation states
Scandium of the oxidation state +3  the stable electronic configuration of argon (i.e. 1s22s22p63s23p6) Zinc of the oxidation state +2  the stable electronic configuration of [Ar] 3d10

2. (a). All elements of the first transition
2. (a) All elements of the first transition series (except Sc) can show an oxidation state of +2 (b) All elements of the first transition series (except Zn) can show an oxidation state of +3

3. Manganese has the highest oxidation state +7
E.g. MnO4-, Mn2O7 Mn7+ ions do not exist.

The +7 state of Mn does not mean that all 3d and 4s electrons are removed from Mn to give Mn7+.
Instead, Mn forms covalent bonds with oxygen atoms by making use of its half filled orbitals

Draw the structure of Mn2O7

3. Manganese has the highest oxidation state +7
The highest possible oxidation state = the total no. of the 3d and 4s electrons  inner electrons (3s, 3p…) are not involved in covalent bond formation

4. For elements after manganese, there is a reduction in the number of possible oxidation states
The 3d electrons are held more firmly  the decrease in the number of unpaired electrons  the increase in nuclear charge

5. The relative stability of various oxidation states is correlated with the stability of electronic configurations Stability : - Mn2+(aq) > Mn3+(aq) [Ar] 3d [Ar] 3d4 Major factor Fe3+(aq) > Fe2+(aq) [Ar] 3d [Ar] 3d6 : Fe3+ > Fe2+ Major factor

5. The relative stability of various oxidation states is correlated with the stability of electronic configurations Stability : - Zn2+(aq) > Zn+(aq) [Ar] 3d [Ar] 3d104s1 : Zn2+ > Zn+ Major factor

1. Variable Oxidation States of Vanadium and their Interconversions
The compounds of vanadium, vanadium  oxidation states of +2, +3, +4 and +5  forms ions of different oxidation states  show distinctive colours in aqueous solutions

Oxidation state of vanadium in the ion Colour in aqueous solution
Colours of aqueous ions of vanadium of different oxidation states Ion Oxidation state of vanadium in the ion Colour in aqueous solution V2+(aq) V3+(aq) VO2+(aq) +2 +3 +4 +5 Violet Green Blue Yellow

In an acidic medium  the vanadium(V) state usually occurs in the form of VO2+(aq) dioxovanadium(V) ion  the vanadium(IV) state occurs in the form of VO2+(aq) oxovanadium(IV) ion

In an alkaline medium  the stable form of the vanadium(V) state is VO3–(aq), metavanadate(V) or VO43–(aq), orthovanadate(V), in strongly alkaline medium

Compounds with vanadium in its highest oxidation state (i.e. +5)  strong oxidizing agents

Vanadium of its lowest oxidation state (i.e. +2)  in the form of V2+(aq)  strong reducing agent  easily oxidized when exposed to air

Interconversions of the common oxidation states of vanadium can be carried out readily in the laboratory The most convenient starting material  ammonium metavanadate(V) (NH4VO3)  a white solid  the oxidation state of vanadium is +5

1. Interconversions of Vanadium(V) species VO2+(aq) V2O5(s) VO3(aq) VO43(aq) OH H+ Yellow orange yellow colourless Vanadium(V) can exist as cation as well as anion

1. Interconversions of Vanadium(V) species VO2+(aq) V2O5(s) VO3(aq) VO43(aq) OH H+ Yellow orange yellow colourless Amphoteric In acidic medium In alkaline medium

1. Interconversions of Vanadium(V) species VO2+(aq) V2O5(s) VO3(aq) VO43(aq) OH H+ Yellow orange yellow colourless Amphoteric In acidic medium In alkaline medium Give the equation for the conversion : V2O5  VO2+ V2O5(s) + 2H+(aq)  2VO2+(aq) + H2O(l)

1. Interconversions of Vanadium(V) species VO2+(aq) V2O5(s) VO3(aq) VO43(aq) OH H+ Yellow orange yellow colourless Amphoteric In acidic medium In alkaline medium Give the equation for the conversion : V2O5  VO3 V2O5(s) + 2OH(aq)  2VO3(aq) + H2O(l)

1. Interconversions of Vanadium(V) species VO2+(aq) V2O5(s) VO3(aq) VO43(aq) OH H+ Yellow orange yellow colourless Amphoteric In acidic medium In alkaline medium Give the equation for the conversion : VO3  VO2+ VO3(aq) + 2H+(aq)  VO2+(aq) + H2O(l)

VO43(aq) H3O+ 8H2O orthovanadate(V) ion V5+ ions does not exist in water since it undergoes vigorous hydrolysis to give VO43 The reaction is favoured in highly alkaline solution

V  VO43(aq) orthovanadate(V) ion
Cr  CrO42(aq) chromate(VI) ion Mn  MnO4(aq) manganate(VII) ion Draw the structures of VO43, CrO42 and MnO4

6H2O VO3(aq) + 6H3O+ The reaction is favoured in alkaline solution
Metavanadate(V) ion The reaction is favoured in alkaline solution VO3 is a polymeric anion like SiO32

Metavanadate(V) ion, (VO3)nn

VO2+(aq) H3O+ 4H2O The reaction is favoured in acidic solution

The action of zinc powder and concentrated hydrochloric acid  vanadium(V) ions can be reduced sequentially to vanadium(II) ions

VO2+(aq)  yellow Zn conc. HCl VO2+(aq)  blue Zn conc. HCl V3+(aq)  green Zn conc. HCl V2+(aq) violet

(a) (b) (c) (d) VO2+(aq) VO2+(aq) V3+(aq) V2+(aq) Colours of aqueous solutions of compounds containing vanadium in four different oxidation states: (a) +5; (b) +4; (c) +3; (d) +2

The feasibility of the changes in oxidation state of vanadium
 can be predicted using standard electrode potentials Half reaction (V) 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) –0.76 +1.00 +0.34 –0.26

Under standard conditions  zinc can reduce 1. VO2+(aq) to VO2+(aq) > 0 2. VO2+(aq) to V3+(aq) > 0 3. V3+(aq) to V2+(aq) > 0

2 × (VO2+(aq) + 2H+(aq) + e– VO2+(aq) + H2O(l)) = V –) Zn2+(aq) + 2e– Zn(s) = –0.76 V 2VO2+(aq) + Zn(s) + 4H+(aq) 2VO2+(aq) + Zn2+(aq) + 2H2O(l) = V

2 × (VO2+(aq) + 2H+(aq) + e– V3+(aq) + H2O(l)) = V –) Zn2+(aq) + 2e– Zn(s) = –0.76 V 2VO2+(aq) + Zn(s) + 4H+(aq) 2V3+(aq) + Zn2+(aq) + 2H2O(l) = V

2 × (V3+(aq) + e– V2+(aq)) = –0.26 V –) Zn2+(aq) + 2e– Zn(s) = –0.76 V 2V3+(aq) + Zn(s) 2V2+(aq) + Zn2+(aq) = V

2. Variable Oxidation States of Manganese and their Interconversions
 show oxidation states of +2, +3, +4, +5, +6 and +7 in its compounds The most common oxidation states  +2, +4 and +7

Oxidation state of manganese in the ion
Colours of compounds or ions of manganese in different oxidation states Ion Oxidation state of manganese in the ion Colour Mn2+ Mn(OH)3 Mn3+ MnO2 MnO43 MnO42– MnO4– +2 +3 +4 +5 +6 +7 Very pale pink Dark brown Red Black Bright blue Green Purple

(a) (b) (c) Mn2+(aq) Mn(OH)3(aq) MnO2(s) Colours of compounds or ions of manganese in differernt oxidation states: (a) +2; (b) +3; (c) +4

(d) (e) MnO42–(aq) MnO4–(aq) Colours of compounds or ions of manganese in differernt oxidation states: (d) +6; (e) +7

2. Variable Oxidation States of Manganese and their Interconversions
Manganese of the oxidation state +2  the most stable at pH 0 Mn2+ Mn3+ +1.50V Mn 1.18V MnO4 +1.51V MnO2 +1.23V

Explosive on heating and extremely oxidizing
Mn(VII) Explosive on heating and extremely oxidizing +7 +6 +4 2 2KMnO K2MnO4 + MnO2 + O2 heat  in ON = 2(+2) = +4  in ON = (1) + (3) = 4

Mn(VII) 4MnO4 + 4H+ 4MnO2 + 2H2O + 3O2
+4 +7 2 4MnO4 + 4H MnO H2O + 3O2 light  in ON = 6(+2) = +12  in ON = 4(3) = 12 The reaction is catalyzed by light Acidified KMnO4(aq) is stored in amber bottle

Oxidizing power of Mn(VII) depends on pH of the solution
In an acidic medium (pH 0) MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l) = V In a neutral or alkaline medium (up to pH 14) MnO4–(aq) + 2H2O(l) + 3e– MnO2(s) + 4OH (aq) = V

Why is the Eo of MnO4 MnO42 Eo = +0.56V
not affected by pH ? MnO4(aq) + e MnO42 Eo = +0.56V The reaction does not involve H+(aq) nor OH(aq)

In an acidic medium (pH 0)
MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l) = V In a neutral or alkaline medium (up to pH 14) MnO4–(aq) + 2H2O(l) + 3e– MnO2(s) + 4OH (aq) = V Under what conditions is the following conversion favoured? MnO4(aq) + e MnO42 Eo = +0.56V When [OH(aq)] > 1M

Predict if Mn(VI) Mn(VII) + Mn(IV) is feasible at (i) pH 0 and (ii) pH 14
MnO42(aq) + 4H+(aq) + 2e MnO2(s) + 2H2O(l) Eo = +2.26V MnO42(aq) + 2H2O(l) + 2e MnO2(s) + 4OH(aq) Eo = +0.60V MnO4 + e MnO42 Eo = +0.56V (1) (2) (3) At pH 0 (1) 2(3) 3MnO42(aq) + 4H+(aq) MnO4(aq) + MnO2(s) + 2H2O(l) Eocell = +1.70V (feasible) Mn(VI) is unstable in acidic medium At pH 14 (2) 2(3) 3MnO42(aq) + 2H2O(l) MnO4(aq) + MnO2(s) + 4OH(aq) Eocell = +0.04V (much less feasible)

Mn(IV) Oxidizing in acidic medium
MnO2(s) + 4H+(aq) + 2e– Mn2+(aq) + 2H2O(l) = 1.23 V Used in the laboratory production of chlorine MnO2(s) + 4HCl(aq)  MnCl2(aq) + 2H2O(l) + Cl2(g)

Mn(IV) Reducing in alkaline medium
Oxidized to Mn(VI) in alkaline medium 2MnO OH + O2  2MnO42 + 2H2O

7B MnO2 is oxidized to MnO42 in alkaline medium
2MnO OH + O2  2MnO42 + 2H2O Suggest a scheme to prepare MnO4 from MnO2 1. 2MnO OH + O2  2MnO42 + 2H2O 2. 3MnO42 + 4H+  2MnO4 + MnO2 + 2H2O 3. Filter the resulting mixture to remove MnO2 7B

Cu+(aq) + e  Cu(s) Eo = +0.52V
Cu2+(aq) is more stable than Cu+(aq) The only copper(I) compounds which can be stable in water are those which are (i) insoluble (e.g. Cu2O, CuI, CuCl) (ii) complexed with ligands other than water e.g. [Cu(NH3)4]+ Cu+(aq) + e  Cu(s) Under these conditions, [Cu+(aq)]   Equil. Position shifts to left

2Cu2+(aq) + 4I(aq)  2CuI(s) + I2(aq)
Estimation of Cu2+ ions 2Cu2+(aq) + 4I(aq)  2CuI(s) + I2(aq) unknown excess white fixed I2(aq) + 2S2O32(aq)  2I(aq) + S4O62(aq) standard solution

Formation of Complexes
Another striking feature of the d-block elements is the formation of complexes

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. The molecules or ions that donate lone pairs of electrons to form the dative covalent bonds are called ligands.

A ligand  can be an ion or a molecule having at least one lone pair of electrons that can be donated to the central metal atom or ion to form a dative covalent bond

Complexes can be electrically neutral Ni(CO)4 [Co(H2O)6]3+ positively charged [Fe(CN)6]3 negatively charged

A co-ordination compound is either
a neutral complex e.g. Ni(CO)4 or made of a complex ion and another ion e.g. [Co(H2O)6]Cl3  [Co(H2O)6] Cl K3[Fe(CN)6]  3K+ + [Fe(CN)6]3

Criteria for complex formation
1. Presence of vacant and low-energy 3d, 4s, 4p and 4d orbitals in the metal atoms or ions to accept lone pairs from ligands. 2. High charge density of the central metal ions.

Diagrammatic representation of the formation of a complex

3d 4s 4p 4d 3d 4s 4p 4d sp3d2 hybridisation [Co(H2O)6]2+ Co :  
The six sp3d2 orbitals accept six lone pairs from six H2O. Arranged octahedrally to minimize repulsion between dative bonds. sp3d2 hybridisation 

1. Complexes with Monodentate Ligands
A ligand that forms one dative covalent bond only is called a monodentate ligand. Examples: neutral  CO, H2O, NH3 anionic  Cl–, CN–, OH–

In the formation of complexes, classify the transition metal ion and the ligand as a Lewis acid or base. Explain your answer briefly. The transition metal ion is the Lewis acid since it accepts lone pairs of electrons from the ligands in forming dative covalent bonds. The ligand is the Lewis base since it donates a lone pair of electrons to the transition metal ion in forming dative covalent bonds.

What is the oxidation state of the central metal ?
Cr3+ Zn2+

Co3+

Fe3+ Co2+

2. Complexes with Bidentate Ligands
A ligand that can form two dative covalent bonds with a metal atom or ion is called a bidentate ligand. A ligand that can form more than one dative covalent bond with a central metal atom or ion is called a chelating ligand.

Ethylenediamine (H2NCH2CH2NH2)
Oxalate (C2O42–) ethylenediamine oxalate ion The term chelate is derived from Greek, meaning ‘claw’. The ligand binds with the metal like the great claw of the lobster.

ethylenediamine oxalate ion

3. Complexes formed by Multidentate Ligands
Ligands that can form more than two dative covalent bonds to a metal atom or ion are called multidentate ligands. Some ligands can form as many as six bonds to a metal atom or ion. Example:  ethylenediaminetetraacetic acid (abbreviated as EDTA)

ethylenediaminetetraacetate ion
EDTA forms six dative covalent bonds with the metal ion through six atoms giving a very stable complex.  hexadentate ligand ethylenediaminetetraacetate ion

Structure of the complex ion formed by iron(II) ions and EDTA
? 2 EDTA4 Fe2+ [FeEDTA]2 Structure of the complex ion formed by iron(II) ions and EDTA

Uses of EDTA 1. Determining concentrations of metal ions by complexometric titrations e.g. determination of water hardness 2. In chelation therapy for mercury poisoning and lead poisoning Poisonous Hg2+ and Pb2+ ions are removed by forming stable complexes with EDTA. 3. Preparing buffer solutions ( ) 4. As preservative to prevent catalytic oxidation of food by metal ions.

The central metal atom or ion in the complex
The coordination number of the central metal atom or ion in a complex is the number of dative covalent bonds formed by the central metal atom or ion in a complex. Complex The central metal atom or ion in the complex Coordination number [Ag(NH3)2]+ Ag+ 2 [Cu(NH3)4]2+ Cu2+ 4 [Fe(CN)6]3– Fe3+ 6

4. Nomenclature of Transition Metal Complexes with Monodentate Ligands
IUPAC conventions 1. (a) For any ionic compound  the cation is named before the anion (b) If the complex is neutral  the name of the complex is the name of the compound

1. (c) In naming a complex (which may be neutral, a cation or an anion)
 the ligands are named before the central metal atom or ion  the liqands are named in alphabetical order (prefixes not counted) (d) The number of each type of ligands are specified by the Greek prefixes 1  mono- 2  di 3  tri 4  tetra- 5  penta- 6  hexa-

1. (e) The oxidation number of the metal ion in the complex is indicated immediately after the name of the metal using Roman numerals K3[Fe(CN)6] potassium hexacyanoferrate(III) [CrCl2(H2O)4]Cl tetraaquadichlorochromium(III) chloride [CoCl3(NH3)3] triamminetrichlorocobalt(III)

2. (a) The root names of anionic ligands
always end in “-o” CN– cyano Cl– chloro Br bromo I iodo OH hydroxo NO2 nitro SO42 sulphato H hydrido (b) The names of neutral ligands are the names of the molecules  except NH3, H2O, CO and NO

Nitrogen monoxide (NO)
Neutral ligand Name of ligand Ammonia (NH3) Water (H2O) Carbon monoxide (CO) Nitrogen monoxide (NO) Ammine Aqua Carbonyl Nitrosyl

3. (a) If the complex is anionic
 the suffix “-ate” is added to the end of the name of the metal,  followed by the oxidation number of that metal Formula Name of the complex [CoCl4]2 tetrachlorocobaltate(II) ion [Fe(CN)6]3 hexacyanoferrate(III) ion [CuCl4]2– tetrachlorocuprate(II) ion

Name in anionic complex
Names of some common metals in anionic complexes Metal Name in anionic complex Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Platinum Titanate Vanadate Chromate Manganate Ferrate Cobaltate Nickelate Cuprate Zincate Platinate

3. (b) If the complex is cationic or neutral
 the name of the metal is unchanged  followed by the oxidation number of that metal Formula Name of the complex [CrCl2(H2O)4]+ tetraaquadichlorochromium(III) ion [CoCl3(NH3)3] triamminetrichlorocobalt(III)

(a) Write the names of the following compounds.
(i) [Fe(H2O)6]Cl2 (ii) [Cu(NH3)4]Cl2 (iii) [PtCl4(NH3)2] (iv) K2[CoCl4] (v) [Cr(NH3)4SO4]NO3 (vi) [Co(H2O)2(NH3)3Cl]Cl (vii) K3[AlF6]

(i) [Fe(H2O)6]Cl2 (ii) [Cu(NH3)4]Cl2 (iii) [PtCl4(NH3)2] (iv) K2[CoCl4] (v) [Cr(NH3)4SO4]NO3 Hexaaquairon(II) chloride Tetraamminecopper(II) chloride Diamminetetrachloroplatinum(IV) Potassium tetrachlorocobaltate(II) Tetraamminesulphatochromium(III) nitrate

(a) (vi) [Co(H2O)2(NH3)3Cl]Cl
triamminediaquachlorocobalt(II) chloride (vii) K3[AlF6] potassium hexafluoroaluminate Al has a fixed oxidation state (+3) no need to indicate the oxidation state

[Co(NH3)5Cl]Cl2 (NH4)2[TiCl6] [Fe(H2O)4(OH)2]
(b) Write the formulae of the following compounds. (i) pentaamminechlorocobalt(III) chloride (ii) Ammonium hexachlorotitanate(IV) (iii) Tetraaquadihydroxoiron(II) [Co(NH3)5Cl]Cl2 (NH4)2[TiCl6] [Fe(H2O)4(OH)2]

Coordination number of the central metal atom or ion
Stereo-structures of complexes Coordination number of the central metal atom or ion Shape of complex Example 2 linear [Ag(NH3)2]+ [Ag(CN)2]– sp hybridized

Stereo-structures of complexes Coordination number of the central metal atom or ion Shape of complex Example 4 Tetrahedral [Zn(NH3)4]2+ [CoCl4]2+ sp3 Square planar [Cu(NH3)4]2+ [CuCl4]2– dsp2

Tetra-coordinated Complexes
Tetrahedral complexes  tetrahedral shape [Co(H2O)6]2+ Octahedral, pink blue

(b) Square planar complexes  have a square planar structure

Example:

Stereo-structures of complexes Coordination number of the central metal atom or ion Shape of complex Example 6 Octahedral [Cr(NH3)6]3+ [Fe(CN)6]3– sp3d2

Hexa-coordinated Complexes
Example:

6. Displacement of Ligands and Relative Stability of Complex Ions
Different ligands have different tendencies to bind with the metal atom/ion  ligands compete with one another for the metal atom/ion. A stronger ligand can displace a weaker ligand from a complex.

6. Displacement of Ligands and Relative Stability of Complex Ions
Stronger ligand Weaker ligand [Fe(H2O)6]2+(aq) + 6CN–(aq) Hexaaquairon(II) ion [Fe(CN)6]4–(aq) + 6H2O(l) Hexacyanoferrate(II) ion Reversible reaction Equilibrium position lies to the right Kst  mol6 dm18

[Ni(H2O)6]2+(aq) + 6NH3(aq) Hexaaquanickel(II) ion
Stronger ligand Weaker ligand [Ni(H2O)6]2+(aq) + 6NH3(aq) Hexaaquanickel(II) ion [Ni(NH3)6]2+(aq) + 6H2O(l) Hexaamminenickel(II) ion The greater the equilibrium constant, the stronger is the ligand on the LHS and the more stable is the complex on the RHS The equilibrium constant is called the stability constant, Kst

Consider the general equilibrium system below,
[M(H2O)x]m+ + xLn [M(L)x](m-xn)+ + xH2O Units = (mol dm3)-x Kst measures the stability of the complex, [M(L)x](m-xn)+, relative to the aqua complex, [M(H2O)x]m+

Relative strength of some ligands bonding with copper(II) ions
monodentate bidentate multidentate TAS Expt 6 Relative strength of some ligands bonding with copper(II) ions

What is the Kst of the formation of [Cu(H2O)4]2+(aq) ?
Equilibrium Kst ((mol dm–3)–n) [Cu(H2O)4]2+(aq) + 4Cl–(aq) [CuCl4]2–(aq) + 4H2O(l) [Cu(H2O)4]2+(aq) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 4H2O(l) [Cu(H2O)4]2+(aq) + 2H2NCH2CH2NH2(aq) [Cu(H2NCH2CH2NH2)2]2+(aq) + 4H2O(l) [Cu(H2O)4]2+(aq) + EDTA4–(aq) [CuEDTA]2–(aq) + 4H2O(l) 4.2 × 105 1.1 × 1013 1.0 × 1.0 × What is the Kst of the formation of [Cu(H2O)4]2+(aq) ?

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

Factors affecting the stability of complexes
The charge density of the central ion 7.7 × 104 4.5 × 1033 [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) Kst (mol6 dm18) Equilibrium ≈ 1024 ≈ 1031 [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)

2. The nature of ligands Ability to form complex : - CN > NH3 > Cl > H2O [Zn(CN)4]2 Kst = 5  1016 mol4 dm12 [Zn(NH3)4]2+ Kst = 3.8  109 mol4 dm12 [Cu(NH3)4]2+ Kst = 1.1  1013 mol4 dm12 [CuCl4]2+ Kst = 4.2  105 mol4 dm12

3. The pH of the solution In acidic solution, the ligands are protonated  lone pairs are not available  the complex decomposes Equilibrium position shifts to the right [Cu(NH3)4]2+(aq) + 4H2O(l) [Cu(H2O)4]2+(aq) + 4NH3(aq) NH4+(aq) H+(aq)

Consider the stability constants of the following 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 What will be formed when CN–(aq) is added to a solution of [Ag(NH3)2]+? [Ag(CN)2](aq) and NH3

Consider the stability constants of the following 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 What will be formed when NH3(aq) is added to a solution of [Ag(CN)2]–? No apparent reaction

Only free CN is poisonous Kst  1  1024 mol6 dm18
FeSO4(aq) is used as the antidote for cyanide poisoning Very stable Only free CN is poisonous [Fe(H2O)6]2+(aq) + 6CN(aq) [Fe(CN)6]4 + 6H2O(l) Kst  1  1024 mol6 dm18 Why is Fe2(SO4)3(aq) not used as the antidote ? Fe3+(aq) is too acidic. [Fe(H2O)6]3+(aq) + H2O(l) [Fe(H2O)5OH]2+(aq) + H3O+(aq)

[Cu(H2O)4]2+(aq) + Cl(aq) [Cu(H2O)3Cl]+(aq) + H2O(l)
K1 = 6.3102 mol1 dm3 [Cu(H2O)3Cl]+(aq) + Cl(aq) [Cu(H2O)2Cl2](aq) + H2O(l) K2 = 40 mol1 dm3 [Cu(H2O)2Cl2](aq) + Cl(aq) [Cu(H2O)Cl3](aq) + H2O(l) K3 = 5.4 mol1 dm3 [Cu(H2O)Cl3](aq) + Cl(aq) [CuCl4]2(aq) + H2O(l) K1 = 3.1 mol1 dm3 [Cu(H2O)4]2+(aq) + 4Cl(aq) [CuCl4]2(aq) + 4H2O(l) Kst = K1  K2  K3  K4 = 4.2  105 mol4 dm12

K1 > K2 > K3 > K4 Reasons : Statistical effect On successive displacement, less water ligands are available to be displaced.

K1 > K2 > K3 > K4 Reasons : 2. Charge effect
On successive displacement, the Cl experiences more repulsion from the complex [Cu(H2O)4] Cl attraction [Cu(H2O)Cl3] Cl repulsion

Formula of copper(II) complex
Colours of some copper(II) complexes Formula of copper(II) complex Colour of the complex [Cu(H2O)4]2+ [CuCl4]2– [Cu(NH3)4]2+ [Cu(H2NCH2CH2NH2)]2+ [Cu(EDTA)]2– Pale blue Yellow Deep blue Violet Sky blue The displacement of ligands are usually accompanied with easily observable colour changes

Coloured Ions The colours of many gemstones are due to the presence of small quantities of d-block metal ions

Coloured Ions 3d10 : Zn2+, Cu+; 3d0 : Sc3+, Ti4+
Most of the d-block metals  form coloured compounds  due to the presence of the incompletely filled d orbitals in the d-block metal ions Which aqueous transition metal ion(s) is/are not coloured ? 3d10 : Zn2+, Cu+; 3d0 : Sc3+, Ti4+

Number of unpaired electrons in 3d orbitals Colour in aqueous solution
Colours of some d-block metal ions in aqueous solutions Number of unpaired electrons in 3d orbitals d-Block metal ion Colour in aqueous solution Sc3+ Ti4+ Zn2+ Cu+ Colourless 1 Ti3+ V4+ Cu2+ Purple Blue

Colours of some d-block metal ions in aqueous solutions Number of unpaired electrons in 3d orbitals d-Block metal ion Colour in aqueous solution 2 V3+ Ni2+ Green 3 V2+ Cr3+ Co2+ Violet Pink

Colours of some d-block metal ions in aqueous solutions Number of unpaired electrons in 3d orbitals d-Block metal ion Colour in aqueous solution 4 Cr2+ Mn3+ Fe2+ Blue Violet Green 5 Mn2+ Fe3+ Very pale pink Yellow

Colours of some d-block metal ions in aqueous solutions
Co2+(aq) Zn2+(aq) Fe3+(aq) Colours of some d-block metal ions in aqueous solutions

Colours of some d-block metal ions in aqueous solutions
Mn2+(aq) Fe2+(aq) Cu2+(aq) Colours of some d-block metal ions in aqueous solutions

Light reflected or transmitted
A substance absorbs visible light of a certain wavelength  reflects or transmits visible light of other wavelengths (complimentary colour)  appears coloured Coloured ion Light absorbed Light reflected or transmitted [Cu(H2O)4]2+(aq) Yellow Blue [CuCl4]2(aq)

Complimentary colour chart
Violet Blue light absorbed Appears yellow Magenta Green Blue Yellow Red Cyan Yellow light absorbed Appears blue Greenish yellow

Coloured Ions The absorption of visible light is due to the d-d electronic transition 3d  3d i.e. an electron jumping from a lower 3d orbital to a higher 3d orbital

In gaseous state, the five 3d orbitals are degenerate i.e. they are of the same energy level In the presence of ligands, The five 3d orbitals interact with the orbitals of ligands and split into two groups of orbitals with slightly different energy levels

distributes along x and y axes
distributes along z axis Interact more strongly with the orbitals of ligands The splitting of the degenerate 3d orbitals of a d-block metal ion in an octahedral complex

Higher energy eg

Criterion for d-d transition : -
presence of unpaired d electrons in the d-block metal atoms or ions Or presence of incompletely filled d-subshell d-d transition is possible for 3d1 to 3d9 arrangements d-d transition is NOT possible for 3d0 & 3d10 arrangements

H2O as ligand  Cu2+  3d9 : d-d transition is possible

*Cu2+ 3d9 : d-d transition is possible H2O as ligand
Yellow light absorbed, appears blue  *Cu2+  3d9 : d-d transition is possible

Fe2+  3d6 : d-d transition is possible

*Fe2+ 3d6 : d-d transition is possible
Magenta light absorbed, appears green  *Fe2+  3d6 : d-d transition is possible

 Zn2+  3d10 : d-d transition NOT possible

Sc3+ 3d0 : d-d transition NOT possible

E E depends on the nature and charge of metal ion
[Fe(H2O)6]2+ green, [Fe(H2O)6]3+ yellow E depends on the nature and charge of metal ion [Cu(H2O)4]2+ blue, [CuCl4]2 yellow 2. the nature of ligand

Coloured Ions Why does Na+(aq) appear colourless ?
3d0 : d-d transition is NOT possible 2p  3s transition involves absorption of radiation in the UV region.

Catalytic Properties of Transition Metals and their Compounds
The d-block metals and their compounds  important catalysts in industry and biological systems

V2O5 or vanadate(V) (VO3–)
The use of some d-block metals and their compounds as catalysts in industry d-Block metal Catalyst Reaction catalyzed V V2O5 or vanadate(V) (VO3–) Contact process 2SO2(g) + O2 (g) SO3(g) Fe Haber process N2(g) + 3H2(g) NH3(g)

The use of some d-block metals and their compounds as catalysts in industry
Reaction catalyzed Ni Hardening of vegetable oil (Manufacture of margarine) RCH = CH2 + H2  RCH2CH3 Pt Catalytic oxidation of ammonia (Manufacture of nitric(V) acid) 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)

The d-block metals and their compounds exert their catalytic actions in either  heterogeneous catalysis  homogeneous catalysis

Generally speaking, the function of a catalyst  provides an alternative reaction pathway of lower activation energy  enables the reaction to proceed faster than the uncatalyzed one

1. Heterogeneous Catalysis
The catalyst and reactants  exist in different states The most common heterogeneous catalysts  finely divided solids for gaseous reactions

A heterogeneous catalyst provides a suitable reaction surface for the reactants to come close together and react.

Example: The synthesis of gaseous ammonia from nitrogen and hydrogen (i.e. Haber process) N2(g) + 3H2(g) NH3(g)

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 (i.e. one nitrogen and three hydrogen molecules)  very small

The four reactant molecules  collide in proper orientation in order to form the product The bond enthalpy of the reactant (N  N),  very large  the reaction has a high activation energy

In the presence of iron as catalyst  the reaction proceeds much faster  provides an alternative reaction pathway of lower activation energy

Fe is a solid H2, N2 and NH3 are gases The catalytic action occurs at the interface between these two states The metal provides an active reaction surface for the reaction to occur

1. Gaseous nitrogen and hydrogen molecules  diffuse to the surface of the catalyst 2. The gaseous reactant molecules  adsorbed (i.e. adhered) on the surface of the catalyst

The iron metal  many 3d electrons and low-lying vacant 3d orbitals  form bonds with the reactant molecules  adsorb them on its surface  weakens the bonds present in the reactant molecules

2. The free nitrogen and hydrogen atoms  come into contact with each other  readily to react and form the product 3. The weak interaction between the product and the iron surface  gaseous ammonia molecules desorb easily

The catalytic mechanism of the formation of gaseous ammonia from nitrogen and hydrogen

43.3 Characteristic Properties of the d-Block Elements and their compound (SB p.162) 1. Heterogeneous Catalysis Sometimes, the reactants  in aqueous or liquid state Other example: The decomposition of hydrogen peroxide 2H2O2(aq)  2H2O(l) + O2(g) MnO2(s) as the catalyst

Energy profiles of the reaction of nitrogen and hydrogen to form gaseous ammonia in the presence and absence of a heterogeneous catalyst

2. Homogeneous Catalysis
A homogeneous catalyst  the same state as the reactants and products  the catalyst forms an intermediate with the reactants in the reaction  changes the reaction mechanism to an another one with a lower activation energy

In homogeneous catalysis, the ability of the d-block metals to exhibit variable oxidation states enables the formation of the reaction intermediates. Example: The reaction between peroxodisulphate(VI) ions (S2O82–) and iodide ions (I–)

Peroxodisulphate(VI) ions  oxidize iodide ions to iodine in an aqueous solution  themselves being reduced to sulphate(VI) ions S2O82–(aq) + 2I–(aq) SO42–(aq) + I2 (aq)

The reaction is very slow due to strong repulsion between like charges. Iron(III) ions  take part in the reaction by oxidizing iodide ions to iodine  themselves being reduced to iron(II) ions 2I–(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq) = V

Iron(II) ions  subsequently oxidized by peroxodisulphate(VI) ion  the original iron(III) ions are regenerated 2Fe2+(aq) + S2O82–(aq) 2Fe3+(aq) + 2SO42–(aq) = V

The overall reaction: 2I–(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq) = V 2Fe2+(aq) + S2O82–(aq) +) 2Fe3+(aq) + 2SO42–(aq) = V S2O82–(aq) + 2I–(aq) 2SO42–(aq) + I2(aq) = V Feasible reaction

43.3 Characteristic Properties of the d-Block Elements and their compound (SB p.164) 2. Homogeneous Catalysis Iron(III) ions  catalyze the reaction  acting as an intermediate for the transfer of electrons between peroxodisulphate(VI) ions and iodide ions

Peroxodisulphate(VI) ions  oxidize Fe2+ to Fe3+ Iodide ions  reduce Fe3+ to Fe2+

The End

Check Point 43-3E Energy profiles for the oxidation of iodide ions by peroxodisulphate (VI) ions in the presence and absence of a homogeneous catalyst

Let's Think 3 Besides iron(III) ions, iron(II) ions can also catalyze the reaction between peroxodisulphate(VI) ions and iodide ions. Why? Answer

Iron(II) ions catalyze the reaction by reacting with the peroxodisulphate(VI) ions first. 2Fe2+(aq) + S2O82–(aq) 2Fe3+(aq) + 2SO42–(aq) The iron(III) ions formed then oxidize the iodide ions. 2Fe3+(aq) + 2I–(aq) Fe2+(aq) + I2(aq) In this way, the reaction between peroxodisulphate(VI) ions and iodide ions is catalyzed. Back

Check Point 43-3E 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) = V MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l) = V Sn4+(aq) + 2e– Sn2+(aq) = V (Given: S2O82–(aq) + 2e– SO42–(aq) = V I2(aq) + 2e– 2I–(aq) = V) Answer

Those redox systems with greater than +0. 54 V and smaller than +2
Those redox systems with greater than V and smaller than V are able to catalyze the oxidation of iodide ions by peroxodisulphate(VI) ions in an aqueous solution. Therefore, the following two redox systems are able to catalyze the reaction. Cr2O72–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l) MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l) Back

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