IntroductionThe first row of transition metals runs from scandium to zinc. All of the elements in this row have outermost electrons in their 3d sub-shells. The 4s sub-shell is at a lower energy than the 3d sub-shells and so the 4s sub shell fills before the 3d sub-shell. A d block element is one which has its highest energy electrons in a d sub-shell. The terms transition metal and d-block element are not the same.

Definition: A transition metal is a d-block element that forms one or more stable ions with incompletely filled d-orbitals.

Student ActivityComplete the electronic arrangements in the following table (in pencil). Remember that Cu and Cr don’t do what you would expect; because half filled and full sub-shells are more stable than partially filled sub-shells. It is worthwhile, in terms of energy, for Cr and Cu to promote electrons from the 4s into the 3d to achieve this extra stability.

Element No of electrons Electronic ConfigurationScandium 21 [Ar] 4s2 3d1

Titanium 22 [Ar] 4s2 3dVanadium 23 [Ar] 4s2 3dChromium* 24 [Ar] 4s 3dManganese 25 [Ar] 4s2 3dIron 26 [Ar] 4s2 3dCobalt 27 [Ar] 4s2 3dNickel 28 [Ar] 4s2 3dCopper* 29 [Ar] 4s 3dZinc 30 [Ar] 4s2 3d10

Chromium has one electron in each orbital of the 4s and 3d sub-shells, giving the configuration [Ar] 4s1 3d5. The 3d is half filled, which is more stable than [Ar] 4s2 3d4. Copper has a full 3d sub-shell giving the configuration [Ar] 4s1 3d10, which is more stable than [Ar] 4s2 3d9.

Formation of ions

When a transition metal forms a positive ion, it loses its valence (outer shell) electrons first. This means that the 4s-electrons are removed before any d-electrons. It’s the other way round from the filling up.

V+ has an electronic configuration of [Ar] 4s1 3d3

Fe3+ has an electronic configuration of [Ar] 4s0 3d5

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Student Activity

1. Complete the table below showing the electronic configurations of the ions listed and the missing ions of elements with given configurations.

Element No of electrons Ion ConfigurationCu 29 Cu2+

Fe 26 Fe2+

Mn 25 Mn4+

Ti 22 Ti3+

Co 27 Co3+

Fe 26 [Ar] 4s0 3d5

Cu 29 [Ar] 4s0 3d10

Ni 28 [Ar] 4s0 3d8

2. What do we call ions/atoms with identical electronic configurations? Give an example

………………………………………………………………………….

………………………………………………………………………….

3. Why is zinc a d block element but not a transition element (Hint: it only forms the Zn2+ ion)?

………………………………………………………………………….

1.Variable oxidation states

Transition metals have more than one stable oxidation state possible and this is seen in their ions. This is because the electrons in the d sub-shells are close together in energy so it is almost as easy to remove several outer electrons as it is to remove just one. There is no easy way to predict the oxidation states which will be displayed by a particular element.

Iron’s common oxidation states are +2 and +3. Copper’s common oxidation states are + 1 and +2.

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Learn how to do!

Iron and Copper are very important

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2. Formation of coloured compounds and ions

Aqueous solutions of the transition metal ions are coloured.. Remember these ions do not occur native but as their hexaaqua metal ions eg [Cu(H2O)6]2+. Examples include:

Fe2+ - green Ni2+ - green Cr6+ - orangeFe3+ - yellow Co2+ - pink Mn2+ - pale pinkCu2+ - blue Cr3+ - green Mn7+ - purple

3. Catalytic behaviour

Transition metals and their ions are very good catalysts, both homogeneous and heterogeneous. One reason for this is that they can have different oxidation states and so can gain or lose electrons easily. Another is that the empty d-orbitals can accept electrons from other molecules/ions at the transition element surface. Intermediate species can be made.

a) Reaction: Haber process Catalyst: Iron (s)

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

b) Reaction : Contact process Catalyst : V2O5(g)

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

Here the V2O5 is believed to react with SO2 to form V2O4 and SO3 and then the V2O4 reacts with the oxygen to regenerate the V2O5.

c) Reaction : Hydrogenation of alkenes (unsaturated oils can be hydrogenated to make margarine) Catalyst : Nickel(s)

Equation:

Student activityWork out the oxidation state of the transition metal species in each of the following compoundsa) KMnO4 ……….. b) Cu2O ………. c) Fe2O3 ……….d) CuSO4 ……….. e) Cr2O7

2- ……….

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Transition metal hydroxides (precipitation reactions)

Transition metal ions react with the hydroxide ion in aqueous solution to give a precipitate. A precipitate is seen as a solid on the inside of the test tube. A precipitate is the insoluble product of a reaction between two aqueous solutions. The colour of the metal hydroxide can be used to identify the metal. Use Experiment 1 : Precipitation Reactions to help complete the following table. You must learn these equations and colours.

Reaction Colour of Precipitate

Co2+(aq) + 2OH-(aq) → Co(OH)2(s)

pale blue precipitate

……….(aq) + ………(aq) → Fe(OH)2(s)

Fe3+(aq) + 3OH-(aq) → ……………foxy red precipitate

Complex ions: structure and bonding

Transition metals form complex ions. A Complex consists of a central metal ion/metal surrounded by ligands which are bonded co-ordinately. The co-ordination number is the number of co-ordinate bonds in the complex (not the number of ligands).

A ligand is a molecule or an ion, which donates a lone pair of electrons to the transition metal ion by co-ordinate bonding

Example : [Cu(H2O)6]2+

Copper is the central metal ion

Water is the ligand

Common Ligands include: H2O, Cl-, NH3 and SCN- , note that some are neutral ligands and others are charged, all have a lone pair of electrons. Complex ions can be drawn in full as shown below when their shape is also displayed.

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Square brackets go round the complex ion and the total charge on the complex sits outside the square brackets

Student Activity 1State (i) the formula and (ii) the charge of the following complex ions:

(a) one titanium (III) ion and six water molecules;

……………………………………………

(b) one copper (II) ion and four chloride ions;

……………………………………………

(c) one cobalt (II) ion and four chloride ions;

……………………………………………

(d) one copper(II) ion, four ammonia molecules and two water molecules;

……………………………………………

(e) one iron(III) ion, one thiocyanate ion and five water molecules.

……………………………………………

Shapes of complex ions

There are three shapes that are required at this level.

1. Octahedral 2. Tetrahedral[CuCl4]

2-

Bond Angles : 90O and 180O Bond Angle : 109.5O

Co-ordination Number = 6 Co-ordination Number = 4

Bond angles around the oxygen of the water ligand(H-O-H)= ………………

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3. Square Planar - cis platin (anti-cancer drug) [PtCl2(NH3)2]

Bond Angle(Cl-Pt-N): 90 O; Co-ordination Number = 4

This is a very effective chemotherapy drug which works by binding to the DNA in cancer cells and causing the shape of the DNA to change such that it can no longer replicate, so eventually these cells die. However, there are unpleasant side effects for the patient. You can assume that any 4 co-ordinate complexes are tetrahedral unless they are cis-platin or [Ni(CN)4]2-, [Ni(NH3)2Cl2] or [AuCl4]- which are square planar.

Student Activity 2

(i) Using 3 dimensional diagrams, draw each of the shapes for the complexes listed below and predict the bond angles.

(ii) For each complex, give the oxidation number of the transition metal and the coordination number.

a) [Co(NH3)5Cl]2+ b) [Cr(NH3)6]3+

c) [Fe(SCN)(H2O)5]2+ d) [CoCl4]2-

Bidentate or polydentate ligands

Some ligands can donate two (or more electron pairs) and are called bidentate (or poly/multidentate). Three bidentate ligands will make a six coordinate complex. The atoms/species which donate the electrons in a ligand are usually: N, O, O-,P, Br-. A neutral oxygen of a carboxylic acid or an alcohol (bonded to an H) can act as a ligand, but a carbonyl oxygen will not act as a ligand.

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Student activity 31. State whether each ligand below is monodentate, bidentate or polydentate.

(a) Glycine (aminoethanoic acid) (e) Br-

(b) (f) 1,2-diaminoethane (en)

(c) (g)

(d) 1,1-cyclobutane dicarboxylate (h)

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Student Activity 4Write the formulae of the complexes formed between:

(a) nickel (II) ions and three 1,2-diaminoethane molecules;

………………………………………………(b) copper (II) ions and three ethanedioate ions;

………………………………………………(c) cobalt (III) ions, two chloride ions and two 1,2-diaminoethane molecules.

……………………………………………...Stereoisomerism

Many complexes can exist as stereoisomers; they have the same formulae but different spatial arrangements of their atoms. There are two types – cis/trans isomers and optical isomers.

Cis/Trans isomers

These only happen in square planar or octahedral complexes.For example, in [Pt(NH3)Cl2] which is square planar, there are two possible arrangements of the two different ligands:

or

cis trans Only the cis isomer has anti-cancer activity. Note that in the cis the bond angles Cl-Pt-Cl are 90 degrees whereas in the trans it is 180 degrees.

Student Activity 5 In pencil draw the cis and trans isomers of

a) diamminedichloronickel (II);

b) tetraamminedichlorocobalt (III).

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The latter is particularly striking because the cis crystals are violet whereas the trans crystals are green.

Optical isomerism Octahedral complexes which have at least two bidentate ligands also show optical isomerism. As a result of their 3D shape, there are two possible mirror image isomers, which are non-super imposable and will rotate the plane of polarised light in opposite directions. The central ion is not described as chiral; this term is reserved for organic molecules only. On the next page is a diagram showing the two isomers from a complex with three ethanediamine (en) ligands around an M+ metal ion

Student Activity 6In pencil, draw stereoisomers of the following:

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

(ii) [CoCl2(en)2] (there are three structures; draw the two optical isomers first, with a mirror plane between them).

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Ligand Substitution Reactions

Electron pair donors in one complex can be replaced by other ligands to form more stable complexes. This is known as ligand substitution. Often the reactions are reversible.

(a) Copper (II) ions and excess ammonia

Observations : Dropwise addition: a pale blue precipitate is observed which dissolves in excess ammonia to give a royal blue solution.There are two reactions. The first is where ammonia acts a base.The second is where the ammonia takes part in ligand substitution. Write an equation to show how ammonia behaves as a base with water. …………………………………………………………………………..

1. Cu2+(aq) + 2OH-

(aq)→ Cu(OH)2(s)

pale blue ppt

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

(aq) + 4H2O(l)

Royal/deep blue solution

(b) Copper(II) ions and concentrated HCl (aq)

Observations: On addition of the HCl(aq) the blue solution turns yellow. H+ is spectator ion so does not appear in the equation.

[Cu(H2O)6]2+ (aq) + 4Cl-(aq) [CuCl4 ]2- (aq) + 6H2O(l)

Octahedral Tetrahedral

The above reaction is reversible, it is an example of an equilibrium. How could you change the yellow solution to blue?

(c) Iron (III) and thiocyanate ions (SCN-)- watch your teacher demo this!Observations: On addition of the SCN- ions to the solution of Iron (III) the yellow solution turns blood red.

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

+ SCN- (aq) → [Fe(H2O)5(SCN)]2+

(aq) + H2O(l)

Yellow Solution Blood Red Solution

(Note: this is not on the specification but is a valid example of ligand substitution)

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(d) Cobalt (II) ions and concentrated HCl (aq)

When excess chloride ions are added to the hexaaquacobalt (II) ion, there is a colour change and coordination number change as a substitution reaction occurs.

Write an equation for the reaction here and explain how you would reverse the colour change using Le Chatelier’s principle.

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

(e) Demonstrations: Your teacher will demonstrate the following reactions. For each one you should note down the colour change and write the equation. You do not need to learn these colour changes but could be asked to write an equation. These colour changes are valid to use in essay questions.

i) Copper (II) ions and 1,2 – diaminoethane

…………………………………………………………………………………………………………

…………………………………………………………………………………………………………

ii) Copper (II) ions and Sodium ethanedioate

…………………………………………………………………………………………………………

…………………………………………………………………………………………………………

iii) Copper (II) ions and EDTA

…………………………………………………………………………………………………………

…………………………………………………………………………………………………………

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Complete the following table for the reactions of copper (II) ions:

Reagent Formula of ligand

Formula of complex

Colour of complex

Shape of complex

Co-ordination number

Water H2O Octahedral

Conc HCl [CuCl4 ]2- 4

Ammonia NH3 6

1,2 -diaminoethane

en [Cu(en)3]2+

Sodium ethanedioate

[Cu(C2O4)3]4- 6

EDTA EDTA4- [CuEDTA]2- Octahedral

Haemoglobin

Ligand substitution is very important in the transport of oxygen by haemoglobin(Hb). The active bit is the Fe (II) ion which forms four co-ordinate bonds to the four nitrogen atoms in the haem ligand and to one nitrogen atom in the globin. There is also a weak bond to a water molecule which can be reversibly substituted by the oxygen.

Hb(aq) + O2 (g) HbO2 (aq)

This allows haemoglobin to give up oxygen where it is needed in the body. Sadly, CO can also act as a ligand with the iron (II) ions and can not only bind more strongly but also irreversibly with the globin which prevents the oxygen from binding. This can result in death – hence the importance of CO detectors!

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Stability constants We can write an expression to show how far the equilibrium lies to the left or right of a reversible reaction. This quantifies equilibria, this is whole topic in itself (more in Sep/Oct)

[Cu(H2O)6]2+ (aq) + 4Cl-(aq) [CuCl4 ]2-(aq) + 6H2O(l)

If we divide concentrations products/reactant in an equation we will get a measure of how far to the right the equilibrium lies. We omit the water as its value is huge in comparison to the others and it does not change much during the establishment of equilibrium.

Kstab = [[CuCl4 ]2- ] _________________________ You must use two sets of square brackets [[Cu(H2O)6] 2+] x [Cl- ] 4 for a complex

The value is called the stability constant.You should be able to deduce expressions for Kstab for any ligand substitution reaction.

Student Activity 7Write expressions for Kstab for

(i) Hb(aq) + O2 (g) HbO2 (aq)

(ii) [Fe(H2O)6]3+(aq) +SCN-(aq) [Fe(H2O)5(SCN)]2+ +H2O

For a complex ion, Kstab is the equilibrium constant for the formation of the complex ion from its constituent ions in a solvent (usually water). A large value means that the equilibrium lies to the right hand side and that the new complex is very stable. In order to avoid awkward powers of 10 we often take logarithms.

Student Activity 8Compare the stabilities of these three complex ions from the data below and put them in decreasing order of stability (1 is the most stable 3 is the least).

Ligand Complex ion Kstab Log Kstab

Cl- [CuCl4 ]2- 4.0 x 105 5.60

edta [Cu(edta)]2- 6.3 x 1018 18.8

NH3 [Cu(NH3)4(H2O)2]2+ 1.3 x 1013 13.1

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June 2011: Ques 6b

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