Unit 10 - 1

Unit 10 The d-Block Elements

Section 10.1 General features of the d-block elements from Sc to Zn （1）Electronic configurations

d-block elements lie in between the s-block and p-block elements in the fourth, fifth and sixth

periods of the Periodic Table. The fourth period begins with two s-block elements : 19K ([Ar]4s1) and 20Ca ([Ar]4s2). The next ten d-block elements from 21Sc to 30Zn are remarkably similar to one another in their properties and are all metals. The reason for the similarity is that, considering from Sc to Zn., while the nuclear charge increases by 1 unit, each additional electron is entering the inner 3d orbitals, it helps to shield the 4s electrons from the increased nuclear charge, with the result that the effective nuclear charge remains fairly constant across the ten d-block elements. The sizes of the atoms ,the magnitudes of the first ionization enthalpies and electronegativities are therefore very similar.

The d-block elements are frequently referred to as transition elements because they are in

transition from the electropositive elements of the s-block to the electronegative elements of the p-block. The electronic configurations of the d-block elements from Sc to Zn are given below :

Element Sc Ti V Cr Mn Fe Co Ni Cu Zn At. No. Z 21 22 23 24 25 26 27 28 29 30

Electronic Configuration

[Ar] 3d14s2

[Ar] 3d24s2

[Ar] 3d34s2

[Ar] 3d54s1

[Ar] 3d54s2

[Ar] 3d64s2

[Ar] 3d74s2

[Ar] 3d84s2

[Ar] 3d104s1

[Ar] 3d104s2

[Ar] = 1s22s22p63s23p6 （a）Write down the electron configurations from Sc to Zn by using ‘electron-in-boxes’ notation for the

3d and 4s orbitals. 3d 4s Scandium [Ar] Titanium [Ar] Vanadium [Ar] Chromium [Ar] Manganese [Ar] Iron [Ar] Cabalt [Ar] Nickel [Ar] Copper [Ar] Zinc [Ar]

（b）Suggest reasons why atoms of chromium and copper have only one 4s electron while the other

d-elements have two 4s electrons. For chromium, the 3d54s1 configuration is at a lower energy level than 3d44s2 because the

former has more unpaired electrons. The extra repulsion between paired electrons, as compared to unpaired electrons, outweighs the small energy difference between the 3d and 4s level. For copper, the 3d104s1 is more stable than 3d94s2 because there appears to be extra stability associated with a full 3d subshell.

Unit 10 - 2 （2）d-block elements as metals

Most of the d-block elements have a close-packed structure in which each atom has twelve nearest

neighbours. Furthermore, they have a relatively low atomic radius because the electrons being added to the inner 3d subshell are nearer the nucleus than the electrons in the outermost 4s orbital. Consequently, the double effect of close packing and small atomic size results in strong metallic bonds between atoms. Hence, the d-block elements are typical metals, being good conductors of heat and electricity, hard, strong, malleable, ductile, lustrous, and silver-white in colour, and generally they have much higher melting and boiling points than the s-block elements. It is also possible to make alloys containing transition elements in a wide range of composition, as a result of their similar atomic radii. These general physical properties of d-block elements, together with their fairly low chemical reactivity, make transition metals extremely useful as structural metals.

Iron is the most important structural metal. Its great advantage of being much cheaper to

produce outweighs its great disadvantage of suffering from corrosion. Another advantage is that it can be converted into steel, which is harder and has superior resistance to corrosion.

Titanium has a larger atomic radius than iron, and is therefore less dense. It does not corrode as

iron does. Its combination of mechanical strength and low density makes it attractive for use in aircraft components. It is used in the construction of space shuttles, being better able than steel to withstand the high temperatures that are experienced when a space shuttle re-enters the earth’s atmosphere. The high cost of titanium has however limited its use.

The high thermal conductivity of copper leads to its use for cooking ware. The high electrical

conductivity makes copper wire admirably suitable for electrical circuits and cables. The resistance to corrosion makes copper useful for water pipes. Alloys of copper are coinage metal (Cu, Ni), brass (Cu, Zn) and bronze (Cu, Sn).

Zinc is used in galvanizing steel and in the production of brass (Cu, Zn). Transition elements have been defined as elements which form some compounds in which there is

an incomplete subshell of d electrons. Scandium (3d0 in compounds) and Zinc (3d10 in compounds) are excluded by this definition. However, it is convenient to include these metals in a treatment of transition elements, on account of the chemical resemblance of their compounds to transition metal compounds.

Unit 10 - 3 （3）Comparison of physical properties and reactions with water between d-block and s-block metals

Atomic radii

Element K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Atomic radius /nm

0.24

0.20

0.16

0.15

0.14

0.13

0.14

0.13

0.13

0.13

0.13

0.13

There is a general decrease in atomic radii from K to Zn. In traversing the series of d-block

metals from scandium to zinc, the nuclear charge is increasing, but electrons are being added to an inner 3d subshell. These inner d electrons shield the outer 4s electrons from the increasing nuclear charge much more effectively than outer shell electrons and consequently the decrease in radius is much less rapidly than s-block metals. For example, atomic radius varies from 0.16 nm (Sc) to 0.15nm (Ti), a difference of only 6.3%. In s-block metals, atomic radius varies from 0.24 nm (K) to 0.20 nm (Ca), a difference of 16.7%.

Ionization enthalpies

Element K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn First ionization enthalpy kJmol-1

420 590 630 660 650 650 720 760 760 740 750 910

The trends in ionization enthalpies and electronegativity depend largely on atomic radius. The

closer an electron shell is to the nucleus, the more energy is required to remove an electron from it and the higher the ionization enthalpy. For the d-block metals, as the effective nuclear charge increases only slightly from Sc to Zn and the decrease in radius is less marked than in s-block metals, their first ionization enthalpies increase only slightly.

The first, second and third ionization enthalpies of d-block metals are given below : Element Sc Ti V Cr Mn Fe Co Ni Cu Zn First ionization enthalpy kJmol-1

630 660 650 650 720 760 760 740 750 910

Second I.E. kJmol-1 1240 1310 1410 1590 1510 1560 1640 1750 1960 1700 Third I.E. kJmol-1 2390 2650 2870 2990 3260 2960 3230 3390 3560 3800

Discuss and explain any irregularities in the values of ionization enthalpies.

（a）The first ionization enthalpy of zinc is exceptionally greater than that of copper. The electronic configuration of zinc is probably more stable in having full 3d and 4s subshells.

（b）The second ionization enthalpy of Chromium is slightly greater than that of its preceding and

succeeding neighbours. The second ionization enthalpy involves the removal of an electron from a half-filled subshell (3d5), which has extra stability :

The case is similar for copper which possesses a full 3d subshell : （c）The third ionization enthalpy of manganese is greater than its preceding and subsequent neighbours.

The third ionization enthalpy involves the removal of an electron from a half-filled subshell (3d5), which has extra stability :

Unit 10 - 4 Melting points and hardness Element K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Melting point ℃ 64 850 1540 1680 1900 1890 1240 1540 1500 1450 1080 420 Hardness 0.5 1.5 3.0 4.5 6.1 9.0 5.0 4.5 - - 2.8 2.5 Melting points

Most of the d-block metals have a close packed metallic lattice structure, in which each atom has 12 nearest neighbours. The atomic radius of these elements is relatively small. The d-block metals have therefore much higher melting points than s-block metals, since these small sized atoms are closely-packed in the metal lattice, with 3d and 4s electrons participating in metallic bonding by delocalizing into the electron sea. The strength of metallic bond in d-block metals is thus very strong, as reflected in high melting points.

Chromium, manganese and zinc have lower melting points, corresponding to weaker metallic

bonding. Chromium has half-filled 3d and 4s subshells, whereas manganese has a half-filled 3d subshell and a full 4s subshell. Moreover, zinc has completely filled 3d and 4s subshells. These electronic configurations have extra stability and make the outer electrons less available for delocalization. As a result, metallic bonding is weakened, and these elements have lower melting points.

In the s-block metals metallic bonding is weaker because the atomic radius is larger. The

difference is most marked when comparing d-block metals with Group I metals, which have the largest atomic radii and do not have close-packed structures. In addition, they have only one valency electron per atom to contribute to the electron sea. Hardness

Hardness measures the ability of the substances to scratch, abrade or indent one another. The hardness of the metals is again a reflection of the strength of the metallic bonds. Therefore, a comparison of the hardness between d-block and s-block metals parallels what has been described in the comparison of melting points. On the Mohs scale, hardness of diamond is 10 while that of finger-nail is 2.5. Reactions with water s-block metals usually react vigorously with water. e.g. K(s) + H2O(l) → KOH(aq) + 1 H2(g)

d-block metals react only very slowly with cold water. The rusting of iron is a slow process requiring both water and air. Other elements (e.g. Ti and Cr) often have an inherent oxide layer on its surface that resists corrosion by water and air. Iron, chromium and zinc react with steam at elevated temperatures to yield hydrogen.

e.g. 3 Fe(s) + 4 H2O(g) →

2 Cr(s) + 3 H2O(g) → Cr2O3(s) + 3H2(g) Scandium is similar to calcium in its reactivity with water :

Sc(s) + 3H2O(l) →

Unit 10 - 5

Section 10.2 Characteristic properties of the d-block elements and their compounds Section 10.2 A Variable oxidation states （1）Interpretation of variable oxidation states in terms of electronic structures and successive ionization enthalpies

d-block elements have electrons of similar energy in both 3d and 4s subshells. Thus one particular element can form ions of roughly the same stability by losing different number of electrons. Transition elements from titanium to copper exhibit two or more oxidation states in their compounds. For example, the common oxidation states of iron are iron(II) and iron(III) :

3d 4s Fe [Ar]3d64s2 Fe2+ [Ar]3d6 Fe3+ [Ar]3d5 Note that once the 3d orbitals are occupied by electrons, they repel the 4s electrons further from

the nucleus, to an energy level higher than the 3d orbitals now occupied. Thus d-block elements lose electrons from the 4s subshell first, before the 3d subshells, during the formation of ions.

The oxidation states occurring in compounds of d-block elements from Sc to Zn are given : Sc Ti V Cr Mn Fe Co Ni Cu Zn Oxidation states 7 that occur 6 6 6 in compounds 5 5 5 5 5 (common 4 4 4 4 4 4 4 oxidation states 3 3 3 3 3 3 3 3 3 are in bold print) 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1

Generalizations : （a）The common oxidation states for each element include +2 or +3 or both. +3 states are relatively

more common at the beginning of the series, whereas +2 states are more common towards the end. （b）The highest oxidation states up to manganese correspond to the involvement of all the electron

outside the argon core : 4 for Ti, 5 for V, 6 for Cr and 7 for Mn. After this, the increasing nuclear charge binds the d electrons more strongly and so one of the common oxidation states is that which involves the weakly held in the outer 4s shell only : 2 for Fe, 2 for Co, 2 for Ni and 1 for Cu.

（c）Ti, V, Cr and Mn never form simple ions in their highest oxidation state since this would result in

ions of extremely high charge density. Hence the compounds of these elements in which they exhibit their highest oxidation state are either covalently bonded or contain complex ions, e.g. VO3

-, CrO3, Cr2O7

2-, Mn2O7, MnO4-.

（d）The stability of Mn(II) and Fe(III) can be explained by the fact that both of their electronic structure

is 3d5. This half-filled subshell in these ions has a particular stability. The Cu(I) ion with its full 3d10 electronic structure, is less stable than the Cu(II) ion (3d9) in aqueous solution because the following disproportionation reaction takes place readily :

Unit 10 - 6 （2）Common oxidation states of Vanadium and their interconversions

The common oxidation states of vanadium are +5, +4, +3 and +2. Vanadium(V)

The higher oxidation of vanadium, +5 and +4, do not exist as hydrated ions of the type Vn+(aq)

because the small, highly charge V5+ and V4+ ions polarize the attracted water molecules so strongly that the O-H bonds are broken and polyatomic ions with covalent bonded oxygen atoms are formed.

The principal oxidation state of vanadium is +5, found in oxides and ions:

VO2+ VO3

- V2O5 Name dioxovanadium(V) ion trioxovanadate(V) ion vanadium(V) oxideColour

/

Vanadium(V) oxide is used mainly as a catalyst in the industrial conversion of SO2(g) to SO3(g) (the

Contact Process) for the manufacture of sulphuric(VI) acid. The activity of V2O5 as a heterogeneous catalyst may be linked to its reversible loss of oxygen that occurs at high temperatures.

In alkaline medium the stable species of vanadium(V) is VO3

-(aq). The solution of vanadium(V)

can be made by dissolving about 3g of ammonium trioxovanadate(V) in 40 cm3 of 2M NaOH and then adding 80 cm3 of 1M H2SO4. The white solid (NH4VO3) turned red and dissolved to form a yellow solution owing to the presence of dioxovanadium(V) ions, VO2

+, in acid solution : VO3

-(aq) + 2H+

(aq) →

VO2+

(aq) is an oxidizing agent and can be reduced to oxovanadium(IV) ion in acidic medium : VO2

+(aq) + 2H+

(aq) + e- → VO2+(aq) + H2O(l) EO = +1.00 V

Vanadium(IV)

The stable oxidation state of vanadium(IV) is oxovanadium(IV) ion, VO2+(aq), which is blue in

aqueous solution. VO2+

(aq) can be reduced to vanadium(III) ion in acidic medium : VO2+

(aq) + 2H+(aq) + e- → V3+

(aq) + H2O(l) EO = +0.337 V

Vanadium(III) and vanadium(II) In aqueous solution, vanadium(III) and vanadium(II) exist as hydrated ions, V3+

(aq) and V2+(aq).

Oxidation state +3 +2 Ion V3+ V2+ Electronic structure [Ar] 3d2 [Ar] 3d1 Colour in aqueous solution The half-equations for the conversion of vanadium(III) to vanadium(II) and vanadium(II) to

vanadium are as follows : V3+

(aq) + e- → V2+(aq) EO = -0.255 V

V2+(aq) + e- → V(s) EO = -1.18 V

In its lowest oxidation states, vanadium (as V2+) is a good reducing agent.

Unit 10 - 7 Summary : Oxidation state +5 +4 +3 +2 Ion Colour in aq. solution VO2

+(aq) VO2+

(aq) V3+(aq) V2+

(aq) V(s) +1.00V +0.337V -0.255V -1.18V

The electrode potentials of some half-reactions are given : Half-reaction EO

Zn2+(aq) + 2e- → Zn(s) -0.76V

Cu2+(aq) + 2e- → Cu(s) +0.34V

I2(aq) + 2e- → 2I- (aq) +0.54V

MnO4-(aq) + 8H+

(aq) + 5e- → Mn2+(aq) + 4H2O(l) +1.51V

SO42-

(aq) + 2H+(aq) + 2e- → SO3

2_(aq) + H2O(l) +0.17V

Fe3+(aq) + e- → Fe2+

(aq) +0.77V Br2(aq) + 2e- → 2Br-

(aq) +1.09V

Complete the following table :

Test Observations Summary of reaction 1. NH4VO3(S) + acid The white solid turns red and dissolved to

form a yellow solution.

2. Vanadium(V) from 1. + zinc powder

The zinc effervesced and the yellow solution became green, blue, green again and eventually violet

3. Vanadium(V) + copper powder

Copper powder dissolved slowly to give a greenish blue solution

4. Vanadium(V) + iodide + thiosulphate

The yellow solution became a muddy brown. Addition of thiosulphate gave a clear solution.

5. Vanadium(II) + manganate(VII)

The violet solution became green, blue, green-yellow and finally pink.

6. Vanadium(V) + sulphite Add vanadium(II)

The yellow solution became blue. On adding V(II) the mixture became green

7. Vanadium(V) + V3+(aq) The mixture became blue.

8. VO2+(aq) + Fe3+

(aq) No reaction is observed. 9. VO2+

(aq) + Br-(aq) No reaction is observed.

10. V2+(aq) + Cu2+

(aq) A red precipitate of copper is formed. 11. V3+

(aq) + Fe3+(aq) The mixture became blue.

Write the redox equations involved for the test 2, 4, 6, 7 and calculate their EO

cell values. 2. 2VO2

+ + Zn + 4H+ → 2VO2+ + Zn2+ + 2H2O EOcell = +1.76 V

2VO2+ + Zn + 4H+ → 2V3+ + Zn2+ + 2H2O EOcell = +1.097 V

2V3+ + Zn → 2V2+ + Zn2+ EOcell = +0.505 V

Unit 10 - 8 （3）Common oxidation states of Manganese and their interconversions

The common oxidation states of manganese are +7, +6, +4, +3 and +2. Manganese(VII)

The most important compound containing manganese(VII) is potassium manganate(VII), KMnO4. This solid is highly soluble in water, forming a deep purple solution. The ion manganate(VII), MnO4

- , has a tetrahedral structure :

Solutions of potassium manganate(VII) are kept in brown bottles because in the presence of light

they slowly oxidizes water to oxygen : 4MnO4

-(aq) + 4H+

(aq) → Potassium manganate(VII) is a powerful oxidizing agent, its oxidizing power in solution depends

on the pH of the solution. Under acidic condition it is reduced to pink manganese(II) ions : MnO4

-(aq) + 8H+

(aq) + 5e- → Mn2+(aq) + 4H2O(l) EO = +1.51V

purple pink In neutral and alkaline solution the manganate(VII) is reduced to brown manganese(IV) oxide : MnO4

-(aq) + 2H2O(l) + 3e- → MnO2(s) + 4OH-

(aq) EO = +1.67V purple brown In more strongly alkaline solution, manganate(VII) is reduced to green manganate(VI) ions : MnO4

-(aq) + e- → MnO4

2-(aq) EO = +0.56V

purple green Potassium manganate(VII) is used widely as an oxidizing agent in volumetric analysis. It is used,

for example, to estimate iron(II) and ethanedioates : MnO4

-(aq) + 5Fe2+

(aq) + 8H+(aq) → Mn2+

(aq) + 5Fe3+(aq) + 4H2O(l)

2MnO4

-(aq) + 5C2O4

2-(aq) + 16H+

(aq) → The colour change is from purple to colourless, because the Mn2+

(aq) concentration is too low to colour the solution pink. Titration with potassium manganate(VII) requires no indicator, as the end point is shown by the appearance of its purple colour. Dilute sulphuric(VI) acid is usually used to acidify manganate(VII) because hydrochloric acid is oxidized to chlorine and nitric(V) acid is an oxidizing agent.

Manganese(VI)

Potassium manganate(VI), K2MnO4 , is a dark green solid. It is only stable in alkaline solution. In acidic solution it disproportionates to manganate(VII) and manganese(IV) oxide.

MnO4

2-(aq) + 4H+

(aq) + 2e- → MnO2(s) + 2H2O(l) EO =+2.26V MnO4

2-(aq) → MnO4

-(aq) + e- EO =-0.56V

overall equation:

Unit 10 - 9 Manganese(IV)

The most important manganese(IV) compound is manganese(IV) oxide, MnO2 . This is a dark brown compound which is insoluble in water. MnO2 is a powerful oxidizing agent in acidic medium :

MnO2(s) + 4H+(aq) + 2e- → Mn2+

(aq) + 2H2O(l) EO =+1.23V This is sometimes used as a laboratory preparation chlorine from conc. hydrochloric acid : MnO2(s) + 4HCl(aq) →

Manganese(III)

Manganese(III) only occurs in complexes. It is unstable. Under acid conditions manganese(III) disproportionates to manganese(II) and manganese(IV) oxide.

Mn3+(aq) + e- → Mn2+

(aq) EO =+1.51V Mn3+

(aq) + 2H2O(l) → MnO2(s) + 4H+(aq) + e- EO =-0.95V

overall equation:

Managanese(II)

This is the most stable form of manganese. It has the outer electron configuration 3d5. Each of the five 3d orbitals contains one electron. Manganese(II) ions are hydrated in aqueous solution forming the pale pink hexaaquamanganese(II) complex ion, [Mn(H2O)6]2+

(aq) , which is usually simplified as Mn2+

(aq). Manganese(II) ions are formed by the reaction between manganese metal and dilute hydrochloric acid :

Mn(s) + 2H+(aq) → Mn2+

(aq) + H2(g) Addition of alkali to manganese(II) solution gives a white precipitate of manganese(II) hydroxide : Mn2+

(aq) + 2OH-(aq) → Mn(OH)2(s)

pale pink white Manganese(II) can be oxidized to manganate(VII) by the very powerful oxidizing agent such as

bismuthate(V), BiO3-, in dilute nitric acid :

2Mn2+

(aq) + 5BiO3-(aq) + 14H+

(aq) → pale pink purple

Summary : Oxidation state +7 +6 +4 +3 +2 Ion/oxide Colour deep purple green dark brown red pale pink

Predict the feasibility of making Mn(III) from Mn(II) and Mn(VII) in acid conditions. The relevant electrode potentials are :

Mn3+(aq) + e- → Mn2+

(aq) EO =+1.51V MnO4

-(aq) + 8H+

(aq) + 5e- → Mn2+(aq) + 4H2O(l) EO = +1.51V

Combining the two half-equations gives :

EOcell =0.00V means that the reaction is in equilibrium under standard condition. It should be

possible to make the EOcell value positive by increasing the hydrogen ion concentration. This will shift

the equilibrium to the right.

Unit 10 - 10

Section 10.2 B Complex formation （1）Formation of transition metal complexes

Most transition metal chemistry takes places in aqueous solution. However, when Fe2+(aq) is

written, it refer not to simple Fe2+ ions surrounded by more or less mobile water molecules, but to the definite identifiable species Fe(H2O)6

2+. This is an example of a complex ion; the six water molecules are ligands attached by dative covalent bonds, as shown below :

Complexes are formed by the dative covalent bonds from the lone pairs of electrons on the ligands

to the vacant, low energy orbitals on the central d-block metal ion or atom. The central metal ion or atom uses empty hybrid orbitals formed by its vacant s, p and d orbitals

to accept lone pairs of electrons. The ligand is an anion or molecule containing lone pair of electrons that can be donated to the

central metal ion or atom with the formation of a dative covalent bond. A ligand which can only form one bond to a central metal ion or atom is called a monodentate ligand, e.g. H2O, NH3, Cl- and CN-.

A polydentate ligand can form more than one bond. Examples are the bidentate ligands

ethanedioate ion, 1,2-diaminoethane (en) and the multidentate ligand edta. For instance, in the hexacyanoferrate(II) ion, [Fe(CN)6]4-, cyanide ions form dative covalent bonds

with Fe2+ ion. The electronic configurations of Fe2+ and the complex ion are as follows : 3d 4s 3p

Fe2+ Hybridized Fe2+

d2sp3

[Fe(CN)6]4- d2sp3

The unpaired d electrons pair up to vacate orbitals in Fe2+ which can be occupied by lone pairs of

electrons. Six hybrid d2sp3 orbitals are formed. They are directed to the apices of an octahedron. Lone pairs of electrons from the cyanide ions coordinate into the empty hybrid orbitals.

Unit 10 - 11

The coordination number is the number of dative covalent bonds forming between the ligands and the central metal ion or atom. 2, 4 and 6 are the common coordination number. 2-coordinated complexes

Complexes with a coordination number of 2 have a linear arrangement of ligands. Example : [Ag(NH3)2]+

4-coordinated complexes Complexes with a coordination number of 4 usually have a tetrahedral arrangement of ligands. Example : [CoCl4]2-

A few complexes with a coordination number of 4 have a square planar arrangement of ligands. Example : [Cu(NH3)4]2+

6-coordinated complexes

Complexes with a coordination number of 6 have an octahedral arrangement of ligands. Example : [Fe(CN)6]3-

Example : [Fe(C2O4)3]3-

Unit 10 - 12 （2）Nomenclature of complexes

In the formula of a complex, the symbol for the central ion or atom appears first and is followed by the anionic ligands and then by neutral ligands. The formula for the complex may be enclosed in square brackets.

Example : [CoCl2(NH3)4]+ central ion : ligands : The complex can be cationic, neutral or anionic.

1. If the complex is neutral or a cation, its name gives the name and oxidation state of the central metal ion or atom preceded by the number and name of ligands attached to it. The name of ligands are as follows :

Ligand H2O NH3 Cl- CN- OH- CO edta Name aqua ammine chloro cyano hydroxo carbonyl edta

The prefixes : di, tri, tetra, penta and hexa are used to show the number of ligands. If several different ligands are present, they are listed a. in order of anionic ligand before neutral ligand, and b. in alphabetical order, and the prefixes, di, tri, etc., are not allowed to change this order.

Example : [Co(NH3)6]2+ Example : [CoCl2(NH3)4]+

Example : [FeCl2(H2O)4]+ Example : Ni(CO)4

2. If the complex is an anion, the suffix -ate follows the name of the metal, e.g. zincate and chromate. If the metal has a Latin name, then in the complex anion the Latin name of the metal is used, followed by the suffix -ate.

Central metal Fe Co Cu Ag Pt Pb Name

Examples are given in table : Formula of complex Name of complex [Fe(CN)6]4- [Fe(CN)6]3- [CuCl4]2- [CoCl3(H2O)3]- [Fe(OH)2(H2O)4]+ [Fe(OH)2(H2O)4] [Cu(NH3)4]2+ [Zn(OH)4]2- [Zn(edta)]2-

Give that names of the following complex compounds : （a）[CoCl2(NH3)4]Cl

Unit 10 - 13 （b）K2[CoCl4] （3）Structural isomers of transition metal complexes Isomerism is common in complex compounds. Structural isomers are compounds of the same molecular formula but different structures, i.e. different arrangement of ligands in complex compounds. Example 1

Cr(H2O)5ClBr has two structural isomers of different complex ions : Example 2

When hydrogen chloride is passed into cold chromium(III) nitrate solution, violet crystals A, CrCl3.6H2O, are obtained. These lose no water when stood over concentrated sulphuric acid in a desiccator, and 1 mol of A gives 3 mol of AgCl immediately, on addition of silver nitrate to the solution.

On cooling a hot solution of chromium(III) chloride, green crystals B, CrCl3.6H2O, are formed. B

loses water over concentrated sulphuric acid to give CrCl3.4H2O. On addition of silver nitrate, 1 mol of B gives 1 mol of AgCl at once.

By treatment of a solution of B successively with sulphuric and hydrochloric acids, the green crystals C, also with the formula CrCl3.6H2O, can be made. 1 mol of C loses 1 mol of water over concentrated sulphuric acid, gives 2 mol of AgCl at once. （a） Suggest structures for compounds A, B and C. （b） Suggest a structure for yet another compound with the formula CrCl3.6H2O. What would be its

behaviour with silver nitrate and with concentrated sulphuric acid ? A solution of this compound would not react with silver nitrate. However, 1 mol of this

compound would readily lose 3 mol of water over conc. sulphuric acid. Example 3

Complete the following table which gives the range of cobalt(III) chloride/ammonia complex compounds :

Molecular formula

Colour Structural formula

Complex ion No. of moles of AgCl precipitated per mol ofcomplex compound

CoCl3.6NH3 yellow CoCl3.5NH3 purple CoCl3.4NH3 green CoCl3.5NH3.H2O red

Unit 10 - 14 （4）Displacement of ligands and relative stability of complex ions Stability constants

In aqueous solution, the commonest ligand is water and transition metal ions contain complex ions with formulae such as :

[Cu(H2O)6]2+(aq) [Fe(H2O)6]3+

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

The H2O ligands in an aqueous transition metal complex can be displaced by another ligand L, if

that ligand can form stronger bond with the transition metal ion. [M(H2O)m](aq) + mL(aq) [MLm](aq) + mH2O(l) The equilibrium constant for this reaction, Kst = unit : is known as the stability constant of the complex MLm at that temperature. Water is usually

present in such large quantity that its concentration is relatively unaffected. The term [H2O(l)] is therefore incorporated into the stability constant. Such a stability constant measures quantitatively the extent of formation of the complex in the above reaction, so that if Kst has a large magnitude, the complex is formed in large amount, i.e., it is relatively more stable.

For example, Copper(II) sulphate dissolved in water forms the hexaaquacopper(II) ion,

[Cu(H2O)6]2+(aq), which is responsible for the pale blue colour of the solution. Addition of chloride ions

in high concentration (e.g. concentrated hydrochloric acid) leads to the stepwise replacement of H2O ligands by Cl- ligands :

K at 25℃(mol-1dm3) pale blue [Cu(H2O)6]2+

(aq) + Cl-(aq) [Cu(H2O)5Cl]+

(aq) + H2O(l) K1 = 6.3 x 102 [Cu(H2O)5Cl]+

(aq) + Cl-(aq) [Cu(H2O)4Cl2](aq) + H2O(l) K2 = 4.0 x 101

green [Cu(H2O)4Cl2](aq) + Cl-

(aq) [Cu(H2O)3Cl3]-(aq) + H2O(l) K3 = 5.4

yellow [Cu(H2O)3Cl3]-

(aq) + Cl-(aq) [CuCl4]2-

(aq) + 3H2O(l) K4 = 3.1 Adding the above four equations together gives the over reaction : [Cu(H2O)6]2+

(aq) + 4Cl-(aq) [CuCl4]2-

(aq) + 6 H2O(l) It can be shown that the overall stability constant of [CuCl4]2-

(aq) , Kst = Note that the [CuCl4]2-

(aq) complex is yellow in colour but when mixed with the blue [Cu(H2O)6]2+

(aq) ion a yellow-green solution is often formed. Kst reflects the stability of the complex formed. The larger the stability constant, the more stable

is the complex.

Unit 10 - 15 Displacement of ligands Ammonia is an even stronger ligand than Cl-. It will displace both water from [Cu(H2O)6]2+

(aq) and Cl- from [CuCl4]2-

(aq) , forming the deep blue tetraamminecopper(II) ion, [Cu(NH3)4]2+(aq) :

[Cu(H2O)6]2+(aq) + 4NH3(aq) [Cu(NH3)4]2+

(aq) + 6 H2O(l) Kst = The stability constant for [Cu(NH3)4]2+

(aq) is larger than that for [CuCl4]2-(aq) , showing that NH3 is a

stronger ligand than either Cl- or H2O. When ammonia solution is added to a solution of [CuCl4]2-(aq) ,

NH3 molecules displace Cl- from [CuCl4]2- forming [Cu(NH3)4]2+ and the colour change from yellow to deep blue.

[CuCl4]2-(aq) + 4NH3(aq) [Cu(NH3)4]2+

(aq) + 4Cl-(aq)

Kc = The equilibrium constant for the displacement is large. Thus, complexing reactions usually

involve competitions between different ligands for metal ions. In this case, ammonia acts as a stronger ligand than Cl-.

Relative stability of complex ions Factors affecting the stability of complexes are : （a）Ligand strength If a ligand can donate its lone pair of electrons more readily, the greater will be the stability of

complexes formed by it. In general, （b）Charge density of central metal ion The higher the charge density of the central metal ion, the greater is the electrostatic attraction for

the lone pairs of electrons from the ligands and thus the more stable a complex tends to be formed.

Example : [Fe(CN)6]3- Kst = 1.0 x 1031 mol-6dm18 [Fe(CN)6]4- Kst = 2.0 x 108 mol-6dm18

The extraordinary difference between two stability constants demonstrates the fact that Fe3+ has a higher charge and smaller ionic radius than that of Fe2+. （c）Chelate effect A polydentate ligand tends to form more stable complexes than monodentate ones. It is because

the coordination of one end of a polydentate ligand will greatly enhance the chance for the coordination of the other ends.

Example : [Co(NH3)6]2+ Kst = 2.3 x 104 mol-6dm18

[Co(edta)]2- Kst = 1.26 x 1016 mol-1dm3

Unit 10 - 16

One of the most fascinating and intriguing demonstration of the relative stability of complexes is outlined in the tests shown in table :

Ligand H2O Cl- NH3 edta4-

Complex ion Colour Blue yellow deep blue light blue

Stability constant

- 3.9 x 105 mol-4dm12

1.1 x 1013 mol-4dm12

6.3 x 1018 mol-1dm3

Test √ = displacement (change of colour) = none Add H2O Add Cl-

Add NH3 Add edta4-

One of the ligand exchange reactions appeared to be readily reversible. Which one was this ? H2O and Cl- ligands appear to be readily interchangeable when concentrations are changed. Adding

water to [CuCl4]2-(aq) will reduce the concentration of Cl- ions sufficiently to convert a significant

proportion of [CuCl4]2- ions to [Cu(H2O)6]2+ ions and so change of the colour. [CuCl4]2-

(aq) + 6 H2O(l) [Cu(H2O)6]2+(aq) + 4Cl-

(aq)

Use of edta in titrations Edta forms such stable complexes with many metal ions that the reactions can be used in titrations to

measure their concentrations. Another complex, less stable but more highly coloured, is used as an indicator. This is simply an application of the displacement of ligands.

25 cm3 of an orange solution Fe3+ ions of unknown concentration was titrated with 0.10 M edta4-,

using a little 2-hydroxybenzoic acid as an indicator. When 20.0 cm3 of edta solution had been added, the violet colour of 2-hydroxybenzoate complex disappeared, leaving a clear yellow solution. Calculate the concentration of Fe3+ ions in the original solution.

Investigate the hardness of water using edta titration Hardness in water is caused by calcium and magnesium ions. Both these ions complex strongly with

edta. To a 200 cm3 sample of tap water were added an alkaline buffer and a few drops of Eriochrome Black T. The metal ion-indicator complex colour of red is seen at the beginning of the titration. As the edta solution is added, metal ions are removed from the indicator and complex with edta. At the end-point, the blue colour of the free indicator is seen. A volume of 3.50 cm3 of 0.100 mol dm-3 edta was used in titration. Find the concentration of calcium and magnesium ions in the water. The complexes have the formulae Ca(edta) and Mg(edta).

Metal ion-indicator (red) + edta → Metal-edta + Indicator (blue)

Unit 10 - 17

Section 10.2 C Catalytic properties of transition metals and their compounds （1）Introduction

Transition metals and their compounds are important catalysts in industry and in biological systems.

Some of the transition metals including copper, manganese, iron, cobalt, nickel and chromium are essential for the effective catalytic activity of various enzymes. One of the most important enzymes containing copper is cytochrome oxidase. This enzyme is involved in the process energy is obtained from the oxidation of food. In the absence of copper, cytochrome oxidase is completely inhibited and the animal or plant is unable to metabolize food effectively.

Numerous transition metals and their compounds are important industrial catalysts. A list of some

of the more important examples is shown in table :

Transition metal

Substance used as catalyst

Reaction catalysed

Ti TiCl3 Polymerization of ethene :

V V2O5 or vanadate(V)

Contact Process :

Fe Fe or Fe2O3

Haber Process :

Ni Ni Manufacture of margarine :

Cu Cu or CuO

Oxidation of ethanol to ethanal :

Pt Pt Contact Process :

Pt Pt Manufacture of nitric acid :

Transition metal and their compounds can catalyse reactions because they are able to introduce an entirely new reaction mechanism with a lower activation energy than the uncatalysed reaction. Since the activation energy of the catalysed reaction is lower, the reaction rate is faster. Chemist believe that the catalytic activity of transition metals and their compounds depends on their ability to exist in various oxidation states.

Unit 10 - 18 （2）Heterogeneous catalysis In the heterogeneous catalysis, the presence of partly filled d orbitals in the transition metals and their compounds enables it to accept electrons from reactant particles on one hand and donate electrons to the reactants particles on the other. Use of Fe in Haber Process In the absence of a catalyst, the formation of gaseous ammonia from nitrogen and hydrogen depends on the simultaneous collision of four gaseous molecules, the probability of which is extremely small. The synthesis of ammonia with iron as a catalyst in Haber Process provides an example of heterogeneous catalyst : N2(g) + 3H2(g) 2NH3(g) The iron metal is in the solid phase, whereas the hydrogen and nitrogen are in the gaseous phase. The catalytic process actually occurs at the interface between these two phases, with the iron metal providing an active reaction surface for the reactants. The catalytic reaction occurs in a sequence of steps : 1. N2(g) and H2(g) diffuse to the surface of the catalyst iron metal. 2. The gases are then adsorbed on the iron metal surface. The iron metal owes its catalytic effect by

having numerous 3d electrons and many vacant 3d orbitals. It is therefore capable of temporarily donating or accepting electrons from species adsorbed on its surface. This results in a sort of chemisorption, with weak bonds forming between the catalyst surface and the reactant particles.

3. Chemisorption weakens significantly the N≡N bond, thus providing a different reaction path with

lower activation energy for the reaction. Specifically, N2 dissociates into weakly adsorbed N atoms at a moderate temperature of around 500℃, whereas H2 dissociates at temperatures as low as -196℃ into weakly adsorbed H atoms. These atoms are highly reactive, and meet each other quite readily to result in the formation of NH3.

4. The NH3 molecules desorb easily as a result of the weak interaction with the iron surface. The energy profile for the synthesis of ammonia in the presence of iron and in the absence of a heterogeneous catalyst is shown as follows : Potential energy

Reaction Co-ordinate

Unit 10 - 19 Use of MnO2 in the decomposition of hydrogen peroxide The decomposition of hydrogen peroxide as catalysed by manganese(IV) oxide is another example of heterogeneous catalysis. MnO2(s) 2H2O2(aq) → O2(g) + 2H2O(l) Under normal conditions, this reaction is very slow. It can be speeded up by using more concentrated hydrogen peroxide or by heating it. A much easier way of speeding it up is to add manganese(IV) oxide. MnO2 is a fine, black powder. Even it is cold and dilute, the hydrogen peroxide starts to decompose rapidly in the presence of MnO2(s). The catalytic process occurs at the interface between the solid phase of MnO2 and the aqueous phase of H2O2. Once the reactants H2O2 molecules are adsorbed on the active surface of MnO2, they can be decomposed via a new pathway that proceeds at a much higher rate. Active catalyst surfaces are regenerated once the product molecules of O2 and H2O diffuse away. The energy profile for the decomposition of hydrogen peroxide in the presence of MnO2 and in the absence of a heterogeneous catalyst is shown as follows : Potential energy

Reaction Co-ordinate

Unit 10 - 20 （3）Homogeneous catalysis In homogeneous catalysis, the transition metals and their compounds can exhibit variable oxidation states, thus allowing electron transfer between reactants and products by means of the catalyst changing between two oxidation states. Use of Fe3+ or Fe2+ in the reaction between iodide and peroxodisulphate(VI) ions The oxidation of I-

(aq) by peroxodisulphate(VI) ions, S2O82-

(aq) , is catalysed homogeneously by small concentrations of aqueous Fe(III), Fe(II) or Cu(II) ions. 2I-

(aq) + S2O82-

(aq) → I2(aq) + 2SO42-

(aq) The rate of this reaction can be measured conveniently by adding a little starch solution and a fixed amount of sodium thiosulphate. As soon as sufficient iodine has been formed to react with the thiosulphate present, a blue colour is formed with the starch, and the time for the appearance of the colour is inversely proportional to the average reaction rate. Alternatively the rate of the reaction can be followed by a colorimeter, as the colour intensity of I2(aq) will increase when iodine is being formed during the reaction. In the catalytic process, the Fe3+

(aq) ion takes part in the reaction by oxidizing the I-(aq) ions to I2(aq),

itself being reduced to Fe2+(aq) ion :

The Fe2+(aq) ion is subsequently oxidized by the S2O8

2-(aq) ion, so that the original Fe3+

(aq) ion is regenerated. The overall process is the sum of these two stages : 2I-

(aq) + S2O82-

(aq) → I2(aq) + 2SO42-

(aq) The energy profile for the reaction in the presence of iron(III) ion and in the absence of a homogeneous catalyst is shown as follows : Potential energy

Reaction Co-ordinate Fe2+

(aq) ions can also catalyse the same reaction by another mechanism, the Fe2+(aq) ions being

regenerated unchanged at the end of the reaction :

Fe2+(aq) + S2O8

2-(aq) → Fe3+

(aq) + 2SO42-

(aq) + Fe3+

(aq) + 2I-(aq) → Fe2+

(aq) + I2(aq) 2I-

(aq) + S2O82-

(aq) → I2(aq) + 2SO42-

(aq)

Unit 10 - 21

Section 10.2 D Coloured ions （1）Simple theory for the appearance of colour

In general, a substance appears coloured because it absorbed some of the light which falls on it. The light which is then reflected or transmitted to the observer’s eyes is not a complete spectrum of the wavelengths which make up white light, and appears to have a colour complementary to that of the absorbed light. For example, copper(II) sulphate solution appears blue because it absorbs red light, as shown in figure :

white light white minus red appears blue

To account for the particular colour of a transition metal complex, Ligand Field Theory is

proposed. Under the influence of a ligand, the 3d orbitals of the central metal ion split into two groups with slightly different energy levels, and the energy difference that it causes depends on the nature of the ligand and the shape of the complex. The promotion of an electron from the lower to the higher of these d orbitals just happens to require energy within the range of visible light. For instance, the energy difference between the two groups of d orbitals for the [Cu(H2O)]2+ ion is of just the right magnitude for promotion of an electron by the absorption of red light. That is why the ion appears to be blue.

Energy ___ ___ two groups of split 3d orbitals ___ ___ ___ In [Cu(NH3)4]2+ , the energy between two groups of d orbitals is larger, the absorbed light is orange,

and the ion appears deep blue in colour. I.R. Red Orange Yellow Green Indigo Blue Violet U.V. increasing frequency

Why do you think that the following compounds are colourless - ScCl3, ZnSO4, CuCl ?

Scandium(III) ions, Sc3+, have no electrons in the 3d orbitals. Therefore, electronic transitions

involving such electrons are not possible. Zinc ions, Zn2+, and copper(I) ions, Cu+, have completely filled 3d subshell (3d10). As the

electronic transitions responsible for the colour always involve partly filled orbitals, such electronic transitions cannot occur in Zn2+ and Cu+ ions.

Unit 10 - 22 （2）Hydrated ions of Fe(II) and Fe(III)

The iron(II) ion is hydrated in aqueous solution. It exists as the pale green hexaaquairon(II) ion, [Fe(H2O)6]2+.

Upon addition of alkali, hydrated iron(II) ions give a dirty green precipitate of hydrated iron(II)

hydroxide : [Fe(H2O)6]2+

(aq) + 2OH-(aq) [Fe(OH)2(H2O)4](s) + 2H2O(l)

green dirty green precipitate The hexaaquairon(II) ion is stable in acidic solution. In neutral or alkaline solution, it is unstable

and is oxidized by oxygen to form the yellow hexaaquairon(III) ion, [Fe(H2O)6]3+(aq) .

Upon addition of alkali, hydrated iron(III) ions give a reddish brown precipitate of hydrated iron(III)

hydroxide : [Fe(H2O)6]3+

(aq) + 3OH-(aq) [Fe(OH)3(H2O)3](s) + 3H2O(l)

yellow reddish brown precipitate Hydrated iron(III) ions form a blood-red complex with thiocyanate ions, SCN-

(aq) : [Fe(H2O)6]3+

(aq) + SCN-(aq)

yellow The blood-red colour only fades when excess NaF(aq) solution is added : [FeSCN(H2O)5]2+

(aq) + 6F-(aq)

blood-red

（3）Hydrated ions of Co(II)

Cobalt(II) ions form complex ions with the ligands H2O, Cl-, NH3 and CN-. If cobalt(II) compound is dissolved in aqueous solution, pink colour hexaaquacobalt(II) ions [Co(H2O)6]2+

(aq) are formed. Upon addition of excess Cl- ions (from conc. HCl(aq) ) , the colour of solution turns blue, because the Cl- ion displaces the water ligand to form the complex ion [CoCl4]2-

(aq) : [Co(H2O)6]2+

(aq) + 4Cl-(aq)

pink octahedral This reaction is reversed by adding water.

Unit 10 - 23 （4）Hydrated ions of Cu(II)

Hydrated copper(II) ions are pale blue in aqueous solution : [Cu(H2O)6]2+

(aq) pale blue If concentrated hydrochloric acid is added to aqueous solution of Cu(II), water molecules in

[Cu(H2O)6]2+(aq) are displaced by Cl- ions, with the formation of a yellow solution containing [CuCl4]2- :

conc. HCl [Cu(H2O)6]2+

(aq) + 4Cl-(aq)

pale blue water On dilution the above equilibrium shifts to the left, the solution turns green and finally blue due to

the formation of hydrated Cu(II) ions, [Cu(H2O)6]2+(aq). The green colour results from the presence of

both the yellow [CuCl4]2-(aq) ions and the blue [Cu(H2O)6]2+

(aq) ions in solution. When ammonia solution or NaOH(aq) is added to a solution containing hydrated Cu(II) ions, a blue

precipitate of hydrated copper(II) hydroxide is produced : [Cu(H2O)6]2+

(aq) + 2OH-(aq) [Cu(OH)2(H2O)4](s) + 2H2O(l)

pale blue blue ppt. Copper(II) hydroxide dissolves in excess ammonia, forming the deep blue tetraamminecopper(II)

ion : [Cu(OH)2(H2O)4](s) + 4NH3(aq) [Cu(NH3)4]2+

(aq) + 2 OH-(aq) + 4 H2O(l)

blue ppt. deep - blue

Copper(II) hydroxide does not dissolve in excess NaOH(aq).

The following tests can be used to distinguish a hydrated iron(II) ion from a hydrated iron(III) ion. Write the observations for the tests.

Test Fe2+

(aq) Fe3+(aq)

Add NaOH(aq) Add potassium thiocyanate

The following tests can be used to distinguish a hydrated Zn(II) ion from a hydrated Cu(II) ion. Write the observations for the tests.

Test Zn2+

(aq) Cu2+(aq)

Add NaOH(aq) Add excess NaOH(aq) Add NH3(aq) Add excess NH3(aq)