Transition Elements

Introduction

Across the second and third periods of the Periodic Table, there is a gradation in properties, from the alkali metals to the halogens. The fourth period begins in the same way, with an alkali metal (potassium) and an alkaline earth (calcium). The next ten elements do not show the gradation in properties of previous periods: they are remarkably similar to one another in their properties and are all metals. They are called the first transition series. Periods five and six also contain transition series.

Transition Series	d subshell being filled	Elements
First	3d	Sc- Zn
Second	4d	Y- Cd
Third	5d	La- Hg

The reason for the similarity of the first transition series is that, considering the series from left to right, while each additional electron is entering the 3d shell, the chemistry of the elements continues to be determined largely by the 4s electrons. From one transition element to the next, the nuclear charge increases by 1 unit, and the number of electrons also increases by 1. Since each additional electron enters the 3d shell, 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 series of transition elements. The sizes of the atoms and the magnitudes of the first ionisation energies are therefore very similar and the elements have comparable electropositivities.

The electron configurations of the first transition series are:

Sc	1s²2s²2p⁶3s²3p⁶4s²3d¹	[Ar]4s²3d¹
Ti	1s²2s²2p⁶3s²3p⁶4s²3d²	[Ar]4s²3d²
V	1s²2s²2p⁶3s²3p⁶4s²3d³	[Ar]4s²3d³
Cr	1s²2s²2p⁶3s²3p⁶4s¹3d⁵	[Ar]4s¹3d⁵
Mn	1s²2s²2p⁶3s²3p⁶4s²3d⁵	[Ar]4s²3d⁵
Fe	1s²2s²2p⁶3s²3p⁶4s²3d⁶	[Ar]4s²3d⁶
Co	1s²2s²2p⁶3s²3p⁶4s²3d⁷	[Ar]4s²3d⁷
Ni	1s²2s²2p⁶3s²3p⁶4s²3d⁸	[Ar]4s²3d⁸
Cu	1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰	[Ar]4s¹3d¹⁰
Zn	1s²2s²2p⁶3s²3p⁶4s²3d¹⁰	[Ar]4s²3d¹⁰

You will notice that chromium completes occupation of its d orbitals with unpaired electrons at the expense of its 4s electrons, and copper completes its full d¹⁰ shell at the expense of its 4s electrons. There appears to be a certain measure of stability associated with a full d¹⁰ shell and with a half-filled d⁵ shell.

Definition of a transition element:

Transition elements are often referred to as d-block metals. They are defined as elements which form 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, and copper is only included in copper(II) (3d⁹) compounds. It is convenient to include these metals in a treatment of transition elements, however, on account of the chemical resemblance of their compounds to transition element compounds.

The characteristic properties of transition elements arise from the incomplete d subshell in either the atoms or the ions they form. These characteristics include the ability to:

· form compounds with the transition element in variable oxidation states

· form complex ions

· form coloured ions

· act as catalysts

· exhibit paramagnetism.

Variable Oxidation States

One of the characteristic properties of the transition elements is that they have variable oxidation states. This is because the five inner d orbitals are at a similar energy level to the outer s orbital. In the transition elements, d electrons as well as s electrons are involved in bonding. Ionic bonds form when 4s and then 3d electrons are lost to produce positively charged ions. Covalent bonds are formed as unpaired electrons pair up with those on other atoms.

The table below shows the main oxidation states (bold being the most important) and the electronic structures of the elements titanium to copper.

Element	Ti	V	Cr	Mn	Fe	Co	Ni	Cu
								+1
	+2	+2	+2	+2	+2	+2	+2	+2
Oxidation	+3	+3	+3	+3	+3	+3	+3
states	+4	+4		+4
		+5
			+6	+6	+6
				+7

Complex Ions

Transition metals form compounds containing complex ions, also known as coordination compounds. Such complex ions are formed by the coordination of lone pairs of electrons from a donor (called a ligand) to an atom or cation (called an acceptor) which has empty orbitals to accommodate them. A cation may form a complex with a neutral molecule, e.g.

[Cu(NH₃)₄(H₂O)₂]²⁺

or with an oppositely charged ion, e.g.

[CuCl₄]^2-

An atom may form a complex, e.g., Ni(CO)₄. The charge remaining on the central atom or ion when the ligands are removed together with their lone pairs is the oxidation number of the metal in the complex. The coordination number is the number of atoms forming coordinate bonds with the central atom or ion: 2, 4 and 6 are common. One definition of a complex ion involves these terms: a complex ion is an ion in which the coordination number exceeds the oxidation number of the central atom.

Ligands must possess one or more unshared pairs of electrons. A ligand which can only form one bond to a central atom or ion is called a monodentate (literally, 'one tooth') ligand, e.g., NH₃ , CN^-. A polydentate ('many teeth') ligand can form more than one bond. Examples are the ethanedioate ion,

and ethane-1,2-diamine (known as ‘en’ for short),

in which the lone pairs on the two nitrogen atoms can form coordinate bonds. The two coordinate bonds formed by each ligand are thought to resemble the claws of a crab (Greek: chele), and such compounds are named chelate compounds or chelates. In the polydentate ligand, 1,2-bis[bis(carboxymethyl)amino]ethane

which is called edta (short for its old name, ethylenediaminetetraacetate), there are unshared electron pairs on four oxygen atoms and on the two nitrogen atoms. This ligand forms six coordinate bonds, and its complex ions are very stable.

Nomenclature: how to name transition metal compounds

In the formula of a complex ion, the symbol for the central atom appears first and is followed by the anionic ligands and then by neutral ligands, e.g.

[CoCl₂(NH₃)₄]²⁺

The formula of the complex ion may be enclosed in square brackets.

The name of the complex gives the name and oxidation state of the central metal cation, e.g., cobalt(III), preceded by the name and number of ligands attached to it, e.g.,

[Co(NH₃)₆]³⁺ is the hexaamminecobalt(III) ion

The prefixes di tri tetra penta hexa

are used to show the number of ligands. If several ligands are present, they are listed in alphabetical order, and the prefixes, di, tri, etc., are not allowed to alter this order, e.g.

[CrCl₂(H₂O)₄]⁺ is the tetraaquadichlorochromium(III) ion

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, e.g.

[Fe(CN)₆]^4-is the hexacyanoferrate(II) ion

Examples are given in the table below:

Ligand	Type of complex	Example	Name of Complex
Water	Aqua-	[Cr(H₂O)₆]³⁺	Hexaaquachromium(III) ion
Ammonia	Ammine-	[Ag(NH₃)₂]⁺	Diamminesilver(I) ion
Hydroxide ion	Hydroxo-	[Zn(OH)₄]^2-	Tetrahydroxozincate(lI) ion
Chloride ion	Chloro-	[CuCl₄]^2-	Tetrachlorocuprate(lI) ion
Cyanide ion	Cyano-	[Fe(CN)₆]^3-	Hexacyanoferrate(III) ion
Nitrite ion	Nitro-	[CO(NO₂)₆]^3-	Hexanitrocobaltate(III) ion
Carbon monoxide	Carbonyl-	Ni(CO)₄	Tetracarbonylnickel( 0)
Ethane-l,2-diamine	Ethane-I,2-diamine-	[Cr(en)₃]³⁺	Tris(ethane-l,2-diamine)chromium(III) ion
edta	edta-	[Zn(edta)]^2-	edtazincate(lI) ion
Cl^-, NH₃	Mixed	[CoCl₂(NH₃)₄]⁺	Tetraamminedichlorocobalt(III) ion
OH^-, H₂O	Mixed	[Fe(OH)₂(H₂O)₄]⁺	Tetraaquadihydroxoiron(III) ion

The shapes of complex ions

The spatial arrangement of bonds from the ligands to the central atom or ion depends on the identity of the atomic orbitals used by the central atom or ion and on the coordination number of the transition element. The most common shapes are octahedral, tetrahedral, square planar (nickel and platinum) and linear (silver) complex ions.

Octahedral complexes: six-fold coordination

An octahedral complex results when there are six bonds to the central metal ion. This can happen if there are six monodentate ligands, three bidentate ligands, or other combinations that give the six bonds. Three examples of octahedral complexes are shown below.

Tetrahedral complexes: four-fold coordination

A tetrahedral complex has four bonds from the ligands to the central metal ion. The ligands lie at the four corners of a tetrahedron. Three examples are given below.

Square planar complexes: four-fold coordination

Platinum and nickel are unusual in that they form square planar complexes. Two examples are given below. A square planar complex also has four bonds from the ligands to the central metal ion but this time the metal ion and the four ligands all lie in the same plane. A square planar compound, cisplatin, is an anti-cancer drug that has a remarkable success rate in curing testicular cancer. Cisplatin, diamminedichloroplatinum(II), is a complex of platinum(II), Pt(NH₃)₂Cl₂, having chloride ions and ammonia molecules as ligands.

Linear complexes: two-fold coordination

Silver ions are unusual in having a preference for forming linear complexes. In these, the silver ion is bonded to two ligands only and the bond angle is 180°. Examples of these complexes and their uses are:

[Ag(NH₃)₂] ⁺

This is formed when silver chloride or silver oxide dissolves in aqueous ammonia to form the colourless solution known as Tollen's reagent. This reagent is used to distinguish aldehydes from ketones; aldehydes give a silver mirror on the side of the test tube:

[Ag(S₂O₃)₂] ^3-

This ion is formed when silver salts are dissolved in sodium thiosulphate ("hypo") solution. This reaction is important in photographic fixing. The silver bromide which has not been exposed to light is dissolved away from the film leaving the black image of silver as the negative:

[Ag(CN)₂] ^-

This complex ion is formed when silver salts are dissolved in solutions of sodium or potassium cyanides. This solution is used as the electrolyte in silver plating. The deposition potential for silver is greatly raised over that for deposition from aqueous silver solutions so that silver can only be displaced when a sufficiently high potential is applied. Under these circumstances, the silver deposits in a compact and adherent form.

Predicting the shape of transition metal complexes

Most transition elements form both octahedral and tetrahedral complex ions. But when is each formed? The important factors are the size and charge of the ligand. Lots of small, neutral ligands (e.g. NH₃, H₂O) can crowd around a central metal ion, so these ligands tend to form octahedral complexes. On the other hand, fewer large or negatively charged ligands (e.g. OH^-, Cl^-) can fit around the metal ion before repulsions between the electron clouds destabilise the complex. Hence, these ligands tend to favour tetrahedral complexes.

For example, when a pink, aqueous solution of cobalt(II) sulphate, containing [Co(H₂O)₆]²⁺, is treated with an excess of concentrated hydrochloric acid, a deep blue solution is formed, containing [CoCl₄]^2-.

If the blue solution is diluted with water, the pink solution is reformed. Likewise, when concentrated hydrochloric acid is added to an aqueous solution of copper(ll) sulphate which contains the blue complex ion, [Cu(H₂O)₆]²⁺, an olive-green solution containing [CuCl₄]^2- is formed:

If the solution is diluted, then the increased concentration of water forces the equilibrium back to the hexaaqua ion and the solution returns to a blue colour.

Note: if we decrease the size of the ligand to that of F^-, then octahedral complexes become common e.g. AlF₆^3- (compare with AlCl₄^-); if, however, we increase the size of the ligand to that of I^-, we find no octahedral complexes with any of the cations of first three periods.

Isomerism in complex ions

An ion or molecule has isomers if it can exist in two or more different forms with the same formula. Isomers frequently occur in organic compounds, but one of the fascinating things about transition metal chemistry is that many complex ions have isomers. The study of isomerism in complexes was one of the main interests of the German chemist Alfred Werner. We owe much of our understanding of transition metal complexes to him. In his book New Ideas on Inorganic Chemistry published in 1911 he discussed the main types of isomerism. We shall follow some of his examples, although we shall change the names that he used to their modern versions.

Ionisation isomerism

This is the first type of isomerism we shall look at. Werner says:

A peculiar type of isomerism. . . is that in which compounds of the same composition yield different ions in solution.

His example was two compounds with the formula Co(NH₃)₅BrSO₄. Werner noted that in solution:

The one is a reddish-violet colour, and gives in a freshly prepared solution no reaction for bromide ions, but does for the sulphate ion. On the other hand, the solution of the second salt gives the reactions of bromide ion, but not of the sulphate ion.

Well, so much for the facts. How can we explain them? The difference is due to the way the bromide and sulphate are bonded in the compounds. In both of them the five ammonia molecules are ligands. This leaves space for one other ligand. In one of the complexes the bromide is the ligand, in the other it is the sulphate ion. The two possibilities are shown in below. In the first one the bromide is strongly bonded to the cobalt ion. We say that the ammonia molecules and the bromide ion are in the coordination sphere. The sulphate ion is held in the crystal with its two negative charges just balancing the two positive charges on the complex.

When the solid is put in water the sulphate ion floats away, but the bromide is left attached to the cobalt. Therefore, if we test for the presence of sulphate ions we find a positive result, but a test for free bromide ions is negative. With the second variety of the salt, the opposite is true. The sulphate ion is bonded to the cobalt, and free bromide ions can be found in the solution. (The sulphate ion is now in the coordination sphere with the ammonia molecules.)

Hydrate isomerism

This is the second type of isomerism that we need to know about. Werner comments:

The most beautiful example of hydrate isomerism is that furnished by chromium 'chloride. Three different hydrates of the formula CrCI₃ + 6H₂O are known. One is greenish-blue, which dissolves giving a blue-violet solution, and two are green, which dissolve yielding green solutions. ..

The difference between the three hydrates is shown by writing their formulae as:

[Cr(H₂O)₆]³⁺(Cl^-)₃ blue-violet in solution

[Cr(H₂O)₅Cl]²⁺(Cl^-)₂ green in solution

[Cr(H₂O)₄Cl₂]⁺Cl^- darker green in solution

You should understand why it is that one mole of each of them give three, two and one moles of chloride ions in solution respectively. When they are crystallised from solution the second takes up one mole and the third takes up two moles of water of crystallisation. This gives them all the overall formula CrCI₃•6H₂O.

Geometrical isomerism

This is the third type of isomerism. Werner points out that a complex ion like [Co(NH₃)₅CI]²⁺ has no isomers. However, two isomers of [Co(NH₃)₄CI₂]⁺ can be identified:

Up to the present, and in spite of much careful work, it has been found impossible to obtain three isomers. These facts are best explained by assuming that the groups are placed around the central atom at the corners of a regular octahedron.

The ion [Co(NH₃)₅CI]²⁺ is an octahedral complex. Because of the symmetry of the octahedron all the diagrams below represent the same complex.

If an ammonia molecule is replaced by a chloride ion we have [Co(NH₃)₄CI₂]⁺. This time there are two possible structures. In one of them the four ammonia molecules and the cobalt ion lie in a plane. The chloride ions lie above and below the plane, in the axial positions. In the second, a plane containing the cobalt ion also contains three ammonia molecules and one of the chloride ions. This leaves the remaining ammonia molecule and chloride ion in the axial positions. The two arrangements are different, as illustrated in below. They are geometrical isomers.

Square planar complexes can also have geometrical isomers. The diagram below shows two isomers that have the formula Ni(NH₃)₂Cl₂. When the two chlorides (or ammonia molecules) are on the same side of the square, they are said to make the cis isomer (left) When they are opposite one another, we have the trans isomer (right).

Optical isomerism

This is the final type of isomerism we shall meet here. An optical isomer rotates the plane of polarised light. If it rotates the plane to the left, it is called an I form (I for laevo, left). If it rotates it to the right, it is called a d form (d for dextro, right). A test to discover if a molecule or ion is optically active is to see if it can be superimposed on its mirror image. Examples that occur in transition metal chemistry are octahedral complexes involving a bidentate ligand such as ethane-1,2-diamine. The three ligand molecules can attach themselves in two ways, shown below, which resemble aeroplane propellers.

Two isomers of [Co(en)₃]³⁺, which are mirror images of each other, are known. They are optical isomers. The top two diagrams show the arrangements of the atoms; the bottom two use curved lines to represent the ligands.

What happens to the d orbitals in a complex ion?

There are two main theories of bonding that are useful for explaining the behaviour of transition metal complexes. The first is crystal field theory, the second is ligand field theory.

Crystal field theory

In crystal field theory we imagine that the ligands act as centres of electric charge. Then we try to work out what effect the field of the ligands has on the electrons in the d orbitals. This was the first theory invented to explain the properties of complexes. Given that it ignores the finer, and even the important, points of bonding, it is remarkably successful. This is what it has to say about he effects on the d orbitals.

[O1]

First, if you bring negative charges near to electrons, the energies of the electrons rise. This has the effect of increasing the energies of all the d orbitals. Now let us look at the effects on the individual orbitals. The diagram below shows the shapes of the d orbitals. Let us make a diagram of an octahedral complex and show the d_z2 and d_x2-y2 orbitals as well. The lobes of the two orbitals point directly at the ligands. Electrons in these orbitals will suffer a significant amount of repulsion. Now look at the next diagram, which shows the arrangements of the d_xy, d_xz and d_yz orbitals.

[O2]

Ligands at the ends of the x, y, and z axes have electron density that will interact strongly with electrons in d_z2 and d_x2-y2 orbitals

These have their lobes pointing between the ligands. As a result, electrons in these orbitals feel less repulsion than those in the other two orbitals. Thus crystal field theory says that the five d orbitals should split into two groups, two of which have a higher energy than the other three.

[O3]

Ligands at the ends of the x, y, and z axes have weaker interactions with d_xy, d_xz and d_yz orbitals than with d_z2 and d_x2-y2 orbitals

We say that the d orbitals are split. Owing to the splitting being caused by the electric field of the ligands, it is called the ligand field splitting (see diagram). The extent of the splitting is given the symbol Δ_o. The two orbitals with the higher energy are called the e_g set; the three with the lower energy are the t_2g set.

[O4]

Ligand field theory

In ligand field theory we take a more realistic view of the bonding in complexes. We use molecular orbital theory to decide how the orbitals on the metal ion and the ligands interact. As you might expect, this is rather complicated. However, the nice thing is that ligand field theory comes up with the same scheme of splitting as crystal field theory, so we do not need to discuss it further. The two theories also agree on the splitting patterns for d orbitals in tetrahedral and square planar complexes which are quite different to the octahedral case, see below.

[O5]

Colour of Transition Metal Complexes

An introduction to colour

The electromagnetic spectrum comprises the total range of physically available frequencies of radiation. High-energy radiations, such as γ-rays and X-rays, are of very short wavelengths whereas low-energy radiations, such as radio waves and electrical waves, are of very long wavelengths. The visible region is situated between the ultra-violet and the infra-red regions and occupies only a very narrow portion of the total spectrum, extending from about 400nm to 750nm in wavelength.

[O6]

Regions of the electromagnetic spectrum

Light has dual characteristics (wave-particle duality). Maxwell first suggested that light was electromagnetic in character, consisting of mutually perpendicular fields whose amplitudes varied in a wave-like manner. The wave front travels with a speed of approximately 3.00 x 10⁸ ms^-1. Light can be characterised either by its wavelength (λ), the distance required for one complete oscillation of the wave, or by its frequency (ν), the number of oscillations occurring in unit time (Hz = s^-1). The speed of the wave front is given by the product of these two quantities:

c = λ ν

Wavelength and frequency are, of course, inversely proportional. The wave theory is most successful in accounting for the propagation of light.

The particle theory of light was put forward in 1905 by Planck and Einstein who regarded light as a stream of discrete particles of energy (photons), travelling at the same speed as that of the wave front in the wave theory. In the particle (quantum) theory, monochromatic radiation is characterised by the energy of each photon. The photon energy (E) is related to the frequency of the radiation by the Planck-Einstein equation:

E = hν

The proportionality constant (h), the Planck constant, has a value of 6.626 x 10^-34 Js. The energy associated with one mole of photons is obtained by multiplying hν by the Avogadro constant (L), which has a value of 6.022 x 10²³ mol^-1. The quantum approach best explains how light interacts with matter.

[O7]

Electronic excitation (E₁- E₀ = hν)

Because the electrons in atoms and molecules exist in discrete (specific) energy levels, absorption of radiation from the ultra-violet and visible regions of the electromagnetic spectrum leads to the excitation (promotion) of valency electrons in the low-energy ground state to the higher energy excited state. The smallest energy change is that associated with the excitation of an electron from the highest energy level in the ground state to the lowest energy level in the excited state. The same energy difference (Δ) applies in the emission of radiation where an electron in the lowest excited level falls to the highest ground level.

Electronic absorption spectra are usually measured in solution and generally appear as broad bands rather than as sharp peaks. This is because molecular vibrations and rotations very slightly alter the energy of each molecule in a sample.

[O8]

The absorption spectrum of Malachite Green in ethanoic acid

Predicting the colour of substances: the colour circle

All coloured substances absorb radiation selectively from the visible region in a way which is characteristic of the molecular species concerned. It should be realised that absorption in the ultra-violet is characteristic of colourless organic substances. In both cases, electronic transitions are involved. Coloured species usually absorb in the ultraviolet as well as in the visible region.

Wavelength/nm	Colour absorbed	Visible colour
400 - 435	Violet	Yellow-green
435 - 480	Blue	Yellow
480 - 490	Green-blue	Orange
490 - 500	Blue-green	Red
500 - 560	Green	Purple
560 - 580	Yellow-green	Violet
580 - 595	Yellow	Blue
595 - 605	Orange	Green-blue
605 - 750	Red	Blue-green

It is possible to trace a relationship between the colour of light absorbed and the related visible colour when the compound has a narrow single absorption band within the wavelengths listed. When light of a given colour is absorbed, the complementary colour is transmitted or reflected. Thus, for example, a dye or a pigment absorbing light within the yellow region will appear blue to the eye in daylight. The nine regions of the visible spectrum can be depicted in the form of a colour wheel or colour circle. Just as white light can be split into its coloured components by means of a prism, the spectrum colours can be recombined to produce white. White light is also produced by mixing two complementary colours from any pair of opposite sectors in the colour circle. Mixing radiations in this way is called additive mixing. Any colour can be produced by a suitable additive mixture of red, green, and blue light (primary colours).

The colour purple is not duplicated by any single wavelength of light. This so-called non-spectral colour can be produced by mixing red and violet monochromatic radiations.

[O9]

The colour circle of complementary colours

Most coloured substances of interest owe their appearance to a subtractive process. As mentioned above, absorption (removal) of yellow light leads to a blue appearance; the converse is true. The same effect can be achieved by passing white light through appropriate colour filters, as in a colorimeter. The production of green by a subtractive process is of special interest because purple does not exist as monochromatic radiation. However, removal of both red and violet light gives rise to green. The dye Malachite Green is a good example (see the absorption spectrum, above). Thus, absorption of violet (427.5nm) and red (621nm) light results in the appearance of green. Simultaneous absorption in two regions of the visible spectrum also occurs with chlorophyll, the most abundant green pigment, which absorbs at wavelengths close to 430nm and 660nm.

Bright, pure colours are produced by dyes having narrow absorption bands.

Some substances absorb all of the colours of the visible spectrum except for a narrow range of wavelengths which are reflected back to the observer. For example, if copper(II) sulphate is placed in sunlight, i.e. white light, the solution looks a royal blue. If, however, it is placed in a photographic darkroom that has a red safety light, the solution looks black. The reason for this is that copper(II) sulphate reflects blue light very efficiently; but it absorbs light of most other colours, especially light in the red area of the visible spectrum (see absorption spectrum, below). This means that in the dark room, where there is only red light shining on the solution, the photons of red light are absorbed by the copper(II) sulphate. There are no photons of blue light to be reflected by the solution, so no light reaches our eyes and the solution looks black.

[O10]

Substances with broad absorption bands which cover a significant proportion of the visible region tend to give dark or dull hues, such as maroon or brown. Black results from absorption across the whole of the visible spectrum since all the spectral colours are removed.

Why are transition metal compounds coloured?

Transition metal complexes are coloured because visible light has just about the right energy to excite an electron in the lower set of d orbitals into the higher set (a d→ d transition). All substances in which the central metal ion has an incomplete set of d electrons (3d¹ to 3d⁹ for the first transition series) will be coloured, whether in an octahedral or tetrahedral ligand field. For example, the octahedral complex hexaaquanickel(II), [Ni(H₂O)₆]²⁺, is responsible for the green colour of many nickel(II) salts in water (see diagram below). The complex appears green because nickel(II) (3d⁸) absorbs light in the red, orange, yellow and blue regions of the visible spectrum, allowing only the green light to be reflected back to the observer. In this case, the splitting of the d orbitals corresponds to the energy of blue light (400nm), so this wavelength is absorbed when an electron is promoted from the t_2g set of orbitals to the e_g set. The other wavelengths are absorbed by electronic transitions including transfers of electrons between the ligand and the metal ion and by the vibrations of the bonds in the complex.

[O11]

The spectrum (a) of [Ni(H₂O)₆]²⁺ shows a number of strong absorptions. However, the complex absorbs only weakly around 500nm, which is the green and blue-green region of the visible spectrum. The ligand field splitting (b) in [Ni(H₂O)₆]²⁺ matches the frequency of light in the blue region of the spectrum

Obviously, compounds in which the metal has a full set of d electrons (e.g. 3d¹⁰) or, indeed, no d electrons (3d⁰) cannot give rise to d→ d electron transition. However, colour can arise in these species by charge transfer where an electron between the ligand and the metal, resulting in an intense absorption. For example, copper(I) oxide, Cu₂O, in which copper(I) has a full 3d¹⁰ subshell is red due to charge transfer.

The colours of transition metals are directly related to the energies of d→ d transitions, governed by the ligand field strength, Δ. This is affected by:

the number of ligands
the shape of the complex
the oxidation state of the metal
the nature of the ligand

We have already seen how the number of ligands and the shape of the complex affects the colour of copper(II) and cobalt(II) complexes. For example,

The ligand field strength in a tetrahedral field, Δ_tet, is less than that in an octahedral field, Δ_o, so the complex absorbs light of lower energy when an electron undergoes a d→ d transition. The hexaaquacobalt(II) ion absorbs light in the blue-green (490nm) part of the visible spectrum and therefore appears as the complementary colour, pink. The tetrachlorocobaltate(II) ion absorbs light in the lower-energy, yellow region (590nm) of the spectrum and appears as the complementary colour, blue.

An increase in oxidation state of the central metal tends to increase the ligand field strength, Δ_o. For example, the blue hexaaquachromium(II) is readily oxidised to the ruby-red hexaaquachromium(III) in air:

the hexaaquachromium(II) ion absorbs light in the yellow part of the visible spectrum (590nm) and so it appears as the complementary colour, blue. However, the hexaaquachromium(III) ion absorbs blue-green light in a higher energy part of the spectrum (490nm) so it appears as the complementary colour, red

In general, the ligand field strength, Δ_o, for d→ d transitions in octahedral complexes increases in the order:

Mn²⁺ < Ni²⁺ < Co²⁺ < Fe²⁺ < V²⁺ < Fe³⁺ < Cr³⁺ < V³⁺ Co³⁺

As might be expected, the higher the charge on the metal ion, the stronger the interactions with ligands in the crystal field. This is the spectrochemical series for metal ions.

Ligand substitution can have a very dramatic effect on the colour of a compound. Treating the ruby-red hexaaquachromium(III) ion with concentrated ammonia solutions leads to the formation of the yellow hexaamminechromium(III) ion:

The hexaaquachromium(III) ion absorbs light in the blue-green region (490nm) -because of a d→ d transition and so it appears red. The hexaamminechromium(III) ion has a stronger ligand field, so it absorbs higher energy blue light (460nm) and appears as the complementary colour, yellow.

A spectrochemical series for ligands, in order of increasing ligand field strength is:

I^- < Br^- < Cl^- < S^2- < F^- < OH^- < CH₃COO^- < C₂O₄^{2- ≈}O^2- < H₂O < NH₃ < en < CO

Why are compounds of transition elements often paramagnetic?

Most people know that iron is magnetic. In fact, iron shows a very special type of magnetism called ’ferromagnetism’. The two metals following it in the first transition series, cobalt and nickel, also show this type of magnetism. Ferromagnetism is a variety of magnetism known as paramagnetism. A paramagnetic substance put close to a magnetic field will be attracted into the field. Many transition metal compounds are paramagnetic.

It is the magnetic field associated with the spin of an electron that can give rise to paramagnetism. However, only compounds that have unpaired electrons are paramagnetic. When two electrons occupy the same orbital, they have their spins paired, and we can think of the magnetic fields cancelling out. To sum up: transition metals are paramagnetic when they have one or more unpaired electrons.

If we measure the amount of their paramagnetism we can work out how many unpaired electrons are present. This can be done using a Gouy balance (see diagram). An empty glass tube is suspended between the jaws of an electromagnet. Then the tube is filled with crystals of the chemical being investigated. The magnet is switched on and, if the substance is paramagnetic, the tube is pulled down into the field. Masses are placed, on the balance pan until the balance is zeroed again. The mass used is a measure of the amount of paramagnetism, and therefore the number of unpaired electrons. If we are to understand how these unpaired electrons come about, we need to look carefully at what happens to the d orbitals when a complex ion is made.

[O12]

The principle of the Gouy balance: the sample is balanced with the electromagnet off, and again with it on. The difference between the two weighings is a measure of the number of unpaired electrons in the sample.

The two complexes of iron(II), hexaaquaferrate(II), [Fe(H₂O)₆]²⁺ and hexacyanoferrate(II), [Fe(CN)₆]^4-, are both octahedral. The first has paramagnetism corresponding to four unpaired electrons. The second is not paramagnetic; it has no unpaired electrons. Now, the electron structure of iron is [Ar]3d⁶4s². When it loses two electrons, it is left with the argon core and six 3d electrons. We have to arrange these six electrons on the energy level diagram for an octahedral complex. We know that electrons will go into separate orbitals whenever possible so that the repulsion among them is minimised. If we follow this pattern we shall have one electron in each of the five orbitals, and one left over. This will go into the orbital of lowest energy. Making sure that we stick to the Pauli principle, the two electrons in the. same orbital will have their spins paired. This is the arrangement shown below. You can see that there are four unpaired electrons, just as found from experiment.

[O13]

Cyanide ions cause a much larger splitting than do water molecules. One result is that cyanide complexes are often ‘low spin’.

In the case of [Fe(CN)₆]^4- there are no unpaired electrons. The only way of achieving this, and keeping the energy of the electrons to a minimum, is to put them as three pairs in the t_2g set. This is the arrangement shown below. The reason why they take up this arrangement is that cyanide ions have a larger effect on the splitting of the d orbitals than do water molecules. The ligand field splitting is larger in [Fe(CN)₆]^4-than [Fe(H₂O)₆]²⁺. When the gap Δ_o becomes large, it takes more energy to put an electron into one of the e_g set than it does to pair electrons in the t_2g orbitals. For fairly obvious reasons, [Fe(H₂O)₆]²⁺ is called a high spin complex, and [Fe(CN)₆]^4- a low spin complex.

Finding the formula of a complex by colorimetry

The colour of a transition metal ion can change as the ligand is changed, as we have seen above. We can use this change in colour to find the ratio of ligand to metal in a complex, using a colorimeter. In a colorimeter a narrow beam of light passes through the solution under test towards a sensitive photocell (see diagram). The wavelength of this light can be selected by using an appropriate filter. The current generated in the photocell is proportional to the amount of light transmitted by the solution, which depends on the colour of the solution. The colorimeter is usually calibrated to show the fraction of light absorbed by the solution - the most intensely coloured solution absorbs the most light, and the faintest coloured solution absorbs the least light.

[O14]

Let's use an example to see how this method works. A solution of Fe³⁺ ions, such as aqueous FeCl₃, is yellow. The complex ion is actually [Fe(H₂O)₆]³⁺. When aqueous sodium thiocyanate, NaSCN, is added the complex turns blood-red as the thiocyanate ligands (SCN^-) displace the water ligands. How do we work out the formula of the Fe³⁺/ SCN^- complex? Ten tubes are prepared which contain different amounts of Fe³⁺ and SCN^-. The tube with the most intense colour, as read by the colorimeter, contains the maximum amount of ligand molecules to metal ions. The table contains details of one method you can follow. The principles of the method are always the same, although the volumes and concentrations of the metal ion and ligand solutions can change. What is extremely important is to use the same volume of solution in each tube.

Tube	1	2	3	4	5	6	7	8	9	10
vol. in cm³ of 0.1 moldm^-3 Fe³⁺(aq)	10	10	10	10	10	10	10	10	10	10
vol. in cm³ of 0.1 moldm^-3 SCN^-(aq)	2	4	6	8	10	12	14	16	18	20
vol. in cm³ of H₂O added to make final volume 40cm³	28	26	24	22	20	18	16	14	12	10

The Fe³⁺ and SCN^- solutions are measured as accurately as possible into the tubes. The final solution from each tube in turn is then placed in a test-tube or cuvette and placed in the colorimeter. The complex is red so it absorbs in the blue region of the spectrum, so a blue filter is used, showing that the complementary colour to the complex is used in the filter. The absorbance reading is taken. The results are shown below.

[O15]

As you can see, the absorbance readings taken on the colorimeter go up as the red colour gets more intense, and finally the readings level out. The point where the horizontal line is reached shows the number of moles of Fe³⁺ and SCN^- which form the complex. In this case it is at tube 5. In this experiment, this solution contains 10 cm³ of 0.1 moldm^-3 Fe³⁺ and 10 cm³ of 0.1 moldm^-3 SCN^-. From these values you can work out the amount in moles of each species: 0.001 mol Fe³⁺ and 0.001 mol SCN^-. So the ratio is 1 Fe³⁺ : 1 SCN^-and the formula of the complex is therefore [Fe(H₂O)₅(SCN)]²⁺.

Transition elements in paints

Transition metal complexes are often used in paints to make many coloured pigments. In fact, for brilliance and clarity of colour, transition metal complexes with organic ligands cannot be bettered. Examples of such pigments include Monastral Blue, in which copper(II) ions (d⁹) are complexed by phthalocyanine ligands. If hydrogen atoms in phthalocyanine are replaced by bromine or chlorine atoms, the colour of the pigment can be changed to other blue shades or various shades of green.

Titanium(IV) oxide, TiO₂, includes a titanium(IV) ion with the electronic configuration [Ar]3d⁰4s⁰. Obviously, no d electrons are present and so d→ d transitions are impossible. Titanium(IV) oxide therefore reflects all incident light and is white. It is used as the pigment for most white paints.

How do the transition elements act as catalysts?

Transition elements can act as catalysts for two reasons. First, because they have different oxidation states, they can take part in electron transfer reactions. One example that demonstrates this is the effect of adding iron(II) or iron(III) ions to a mixture of peroxodisulphate(VI) ions, S₂O₈^2-, and iodide ions, I^-. Normally the reaction:

S₂O₈^2-(aq) + 2I^-(aq) → 2SO₄^2-(aq) + I₂(s) (A)

takes place at a convenient rate, which we can follow by one of the usual methods. If a little iron(III) sulphate solution is added, the rate of production of iodine increases markedly. This is because the Fe³⁺ ions react with the iodide ions:

2Fe³⁺(aq) + 2I^-(aq) → 2Fe²⁺(aq) + I₂(s) (B)

If this were all that happened, we would not be dealing with true catalysis: the Fe³⁺ ions appear to be used up in the reaction. However, the Fe²⁺ ions are oxidised back to Fe³⁺ by S₂O₈^2- ions:

2Fe²⁺(aq) + S₂O₈^2-(aq) → 2Fe³⁺(aq) + 2SO₄^2-(aq) (C)

If you combine equations (B) and (C) you will obtain the original equation, (A). The Fe³⁺ ions have taken part in the reaction, but they are regenerated.

[O16]

The key point is that by giving a different path by which electron transfer can take place, the ions have provided a route with lower activation energy. Reaction (A) takes place between ions of with the same charge which repel each other. Reactions (B) and (C) are between ions of opposite charge which attract each other and the reactions are therefore faster.

Transition metals and their ions can also act as catalysts by providing sites at which reactions can take place. They can bond to a wide range of ions and molecules, e.g. those with lone pairs or π electrons. Also they show the ability to make different numbers of bonds, often four or six, but sometimes two, three and five as well. They can show catalytic behaviour when dissolved in solution or as solids, i.e. as homogeneous or heterogeneous catalysts.

An important example of a catalytic reaction in solution is the Ziegler-Natta polymerisation of alkenes (see below). A typical Ziegler-Natta catalyst is a mixture of an organic aluminium compound, e.g. Al(C₂H₅)₃, and titanium(III) chloride. One of the ethyl groups from the aluminium compound bonds to the titanium ion; but being a transition metal ion, it has further sites available for bonding. The π cloud of an alkene can fill one of these sites, with the result that the alkene and ethyl group are held very close to one another. For reasons that we do not have to consider, the alkene inserts itself between the ethyl group and the titanium ion. Now there is room for a further alkene to bond to the titanium ion, which in turn swaps its position. In this way a hydrocarbon chain that originally had only two carbon atoms in it now has six of them joined together. Provided the supply of alkene molecules is kept up, the chain will grow to great length. In other words, a polymer is made. The discovery of this type of reaction by Ziegler and Natta has revolutionised the manufacture of polymers, partly because the method can give polymers of consistent quality.

[O17]

These diagrams show the type of change that takes place with Ziegler-Natta catalysts. The first step is an alkene bonding to a titanium complex through its π cloud of electrons. Then there is a rearrangement of the atoms, which leaves a complex, A, that can repeat the process. The result is growth of the polymer chain.

The chemistry of selected transition elements

Vanadium

The chemistry of vanadium provides us with one of the most striking illustrations of variable oxidation states. The highest oxidation states occur when the element combines with the highly electronegative element oxygen. Vanadium shows its oxidation number +5 in the vanadate(V) ion. When zinc reduces ammonium vanadate(V) in concentrated hydrochloric acid, the solution changes from yellow to blue-green, and eventually violet as the oxidation number changes from +5 through to +2:

VO₂²⁺(aq) → VO²⁺(aq) → V³⁺(aq) → V²⁺(aq)

V(V) V(IV) V(III) V(II)

yellow blue blue-green violet

A stereotypical transition element, vanadium also displays catalytic properties. It is used to catalyse the oxidation of sulphur(IV) oxide in the Contact Process.

Vanadium and the manufacture of sulphuric acid (the Contact Process)

Sulphuric acid is manufactured by oxidising sulphur dioxide to sulphur trioxide. This intermediate product is then converted into sulphuric acid.

(i) Sulphur dioxide is obtained from one of the following sources:

(a) burning sulphur in air:

S(l) + O₂(g) SO₂(g) DH^ø= -298kJmol^-1

(b) burning H₂S from crude oil in air.

(ii) Sulphur dioxide combines with oxygen in an exothermic reaction (the Contact Process):

According to Le Chatelier’s Principle, the exothermic reaction will be favoured by low temperatures and high pressures. At pressures above 2atm, however, the sulphur dioxide liquefies; at low temperatures, the rate of attainment of equilibrium is slow because the forward and reverse reactions have a large activation energy. The use of a catalyst (V₂O₅) allows compromise conditions (450°C and 2atm) to be used whilst obtaining an economically viable yield.

(iii) Once sulphur trioxide, SO₃, has been prepared, it is reacted with water produce sulphuric acid:

SO₃(g) + H₂O(l) H₂SO₄(l)

In fact, this reaction is not used much in industry in Britain because it would produce a dangerous mist of sulphuric acid. Instead, the reaction is controlled by absorbing the sulphur trioxide in sulphuric acid to form oleum, H₂S₂O₇:

SO₃(g) + H₂SO₄(l) H₂S₂O₇(l)

The oleum is then converted into more sulphuric acid by adding water:

H₂S₂O₇(l) + H₂O(l) 2H₂SO₄(aq)

The contact process is used to make about 156 million tonnes each year of sulphuric acid (2.3 million tonnes in the UK).

Vanadium is able to catalyse the reaction because it has variable oxidation states. Firstly, vanadium(V) oxide reacts with sulphur dioxide to form sulphur trioxide and sulphur trioxide:

V₂O₅ + SO₂ → V₂O₄ + SO₃

The vanadium(V) oxide is then regenerated (a requirement for catalysis) by reaction with oxygen:

2V₂O₄ + O₂ → 2V₂O₅

These two steps each have a much smaller activation energy than the direct, uncatalysed oxidation of sulphur dioxide (see diagram). The reaction is therefore much faster.

[O18]

The course of reaction involving V₂O₅ as a catalyst

Chromium

Element 24, chromium, is hard but brittle, and has an attractive silvery appearance which has led to it coating the bumpers and other bits of bicycles and cars.

Chromium forms compounds in the +2, +3, +4, +5 and +6 oxidation states, out of which +3 and +6 are the two most often seen in chromium compounds.

An oxyanion is a negatively charged ion containing oxygen. In the +6 state chromium forms two different oxyanions. These are the orange dichromate(VI) ion, Cr₂O₇^2- and the yellow chromate(VI) ion, CrO₄^2-. They are commonly used as potassium salts, potassium dichromate(VI), K₂Cr₂O₇, and potassium chromate(VI), K₂CrO₄. In solution each of these oxyanions can be converted into the other by altering the pH, since the following equilibrium exists:

Cr₂O₇^2- + H₂O 2CrO₄^2- + 2H⁺

Cr₂O₇^2- is formed in acidic solution, while CrO₄^2- the ion formed in alkaline solution (le Chatelier).

Note: The interconversion of Cr₂O₇^2- and CrO₄^2- can be shown as an equilibrium in three different ways:

Cr₂O₇^2- + H₂O 2CrO₄^2- + 2H⁺

Cr₂O₇^2- + OH^- 2CrO₄^2- + H⁺

Cr₂O₇^2- + 2OH^- 2 CrO₄^2- + H₂O

All three versions are valid and predict the same outcome: CrO₄^2- is formed in alkaline solution, Cr₂O₇^2- in acidic solution.

Potassium dichromate(VI) in acid solution is a strong oxidising agent, being itself reduced to Cr³⁺ ions which are very stable:

Cr₂O₇^2- + 14H⁺ 2Cr³⁺ + 7H₂O E^ø = +1.33 V

In aqueous solution Cr³⁺ forms the violet [Cr(H₂O)₆]³⁺ complex ion. However, when potassium dichromate(VI) in acid solution is used as an oxidising agent, the colour change observed is from orange to green. Particularly when hot, [Cr(H₂O)₆]³⁺ undergoes ligand substitution reactions forming green species. This is the reason why most solutions and solids containing Cr³⁺ are green. For example, in hydrochloric acid the dark-green complex ion [Cr(H₂O)₄Cl₂]⁺ is formed.

Uses of chromium

In 1913, Harry Brearley, a Sheffield chemist, was trying to find new steel alloys for making gun barrels. One particular alloy, containing about 13% chromium, proved useless for this purpose, so he threw his test piece outside on to his ever-growing scrap heap. Some months later he noticed that while all the other bits of scrap had rusted, this piece hadn't.

After further investigation and scientific fiddling around, Brearley settled on an alloy consisting of 70% iron, 20% chromium and 10% nickel as the ideal 'stainless steel'. The chromium and nickel in the alloy quickly form a layer composed of their oxides on the surface of the steel when it first comes in contact with the air. Unlike iron oxide, which is flaky and porous, this oxide layer is tenacious and impervious to air and water - it protects the steel underneath.

If smaller amounts of chromium are added to steel, the resulting alloy is very hard. Steel hardened in this way is used for making many tools.

Cobalt

Element 27, cobalt, is a hard, unreactive, white metal with a slightly blue appearance. It forms a wide range of compounds in the +2 and +3 oxidation states, of which the +2 state is usually the most stable. Co²⁺ forms the stable [Co(H₂O)₆]²⁺ ion, which is pink - as are most octahedral Co²⁺ complexes. Tetrahedral complexes of Co²⁺, such as [CoCl₄]^2-, are mostly blue.

Cobalt chloride paper is used as a test for water. When dry, the paper is blue which is the colour of anhydrous cobalt chloride, CoCl₂. Following the addition of water, the [Co(H₂O)₆]²⁺ ion is formed and the paper turns pink.

Co³⁺ forms the aqueous complex [Co(H₂O)₆]³⁺ which is blue. However, it is so easily reduced to [Co(H₂O)₆]²⁺ that there is no simple aqueous chemistry involving the [Co(H₂O)₆]³⁺ ion. There are, in fact, very few simple compounds containing Co³⁺, and those that do exist, such as CoF₃, instantly react with water to give a Co²⁺ compound. The water is oxidised in the reaction, producing oxygen gas. In complexes with ligands other than water, Co³⁺ is stable, however, and this leads to the 'wide range' of compounds already referred to, exhibiting an equally wide range of colours.

[Co(H₂O)₆]²⁺ is a good example of a complex ion that changes colour due to ligand substitution. The endothermic change

[Co(H₂O)₆]²⁺ + 4Cl^- [CoCl₄]^2- + 6H₂O

takes place in CoCl₂(aq) if either a high concentration of CI^- or a high temperature is applied.

An example of the enhanced stability of Co³⁺ with different ligands is the stability of [Co(NH₃)₆]³⁺. Compare the E^ø values:

[Co(H₂O)₆]³⁺ + e^- [Co(H₂O)₆]²⁺ ; E^ø = +1.81 V

[Co(NH₃)₆]³⁺ + e^- [Co(NH₃)₆]²⁺ ; E^ø = +0.11 V

(The magnitude of the standard electrode potential, E^ø, is an indication of the reactivity. The larger the magnitude, the more reactive.)

A wide variety of reducing agents can reduce [Co(H₂O)₆]³⁺ to [Co(H₂O)₆]²⁺, including water, iodide ions and silver metal. The reduction of [Co(NH₃)₆]³⁺ to [Co(NH₃)₆]²⁺ requires much stronger reducing agents, however, like iron metal or V²⁺ ions. The enhanced stability of [Co(NH₃)₆]³⁺ is due to ammonia being a stronger ligand than water i.e. ammonia forms stronger dative bonds.

Copper

Element 29, copper, is a reddish-gold colour. It is soft and ductile, and its ability to conduct heat and electricity is second only to silver. It was first smelted from its ores over 5500 years ago. Early metallurgists in Cyprus produced very pure copper, so the metal was named after the island.

Copper forms a wide range of stable compounds in the +2 oxidation state but, unlike most other transition metals, copper also forms compounds in the +1 oxidation state. While compounds and complexes containing Cu²⁺ are virtually all green or blue, compounds containing Cu⁺ are white solids.

Aqueous solutions containing Cu²⁺ ions are stable; however, aqueous solutions containing Cu⁺ ions are not. They disproportionate to give Cu atoms and Cu²⁺ ions:

2Cu⁺ Cu + Cu²⁺

The disproportionation of Cu⁺ ions only takes place in aqueous solution. Dry copper(I) compounds and those that are insoluble in water are therefore stable.

All the copper(I) halides are extremely insoluble in water. If a solution containing Cu²⁺ ions is mixed with a solution containing an excess of iodide ions, the iodide ions will reduce the Cu²⁺ ions to Cu⁺ ions. The iodide ions are oxidised to iodine.

Cu²⁺ + I^- Cu⁺ + ½ I₂

The Cu⁺ ions then combine with more iodide ions to form a precipitate of copper(I) iodide:

Cu⁺ + I^- CuI

The overall equation is therefore:

Cu²⁺(aq) + 2I^- CuI(s) + ½I₂(aq)

Although the CuI precipitate that forms is actually white, it always appears a brown colour due to the iodine that forms at the same time. The Cu⁺ ions in CuI are stable and do not disproportionate because they are not in aqueous solution.

Copper alloys

Copper is a constituent of many useful alloys.

Copper and zinc make brass. Its resistance to corrosion, pleasant ringing tones and attractive gold-coloured appearance all explain why brass is used to make musical instruments. Brass is harder than copper. Both metals are used in central heating systems. Being more ductile, copper is drawn out to make the pipes: Being harder, brass is more suitable for the joints, which are made by casting or machining.

Copper and tin make bronze. Bronze is harder than copper and is used to make statues and medals.

Bronze (in this case 97% copper, 2.5% zinc and 0.5% tin) is used for 1p and 2p coins. However, since September 1992, the price of copper has made it too expensive to make 1p and 2p coins out of bronze, so steel coated in a thin layer of copper has been used instead - except for some of the 2p coins issued in 1998. You can distinguish the copper-plated steel coins from the older ones with a magnet.

Cupro-nickel is one of the alloys used for UK coinage; it is soft enough for the coins to be made by stamping but hard enough to resist wear. The more 'silvery' looking the coin, the higher the percentage of nickel in the alloy. For example, 5p, l0p and 50p coins are 75% copper and 25% nickel, while 20p coins are 84% copper and 16% nickel.

Analysing the amount of copper in an alloy

If a piece of brass is dissolved in nitric acid, the solution obtained is a mixture of copper(II) nitrate and zinc nitrate. Analysing this mixture can tell us how much copper it contains. If the mass of the original piece of brass is also known, then the composition of the original alloy can be calculated.

The analysis is done by adding an excess of KI(aq) to the mixture containing Cu²⁺. As has already been seen, this causes the precipitation of all the copper as copper(I) iodide:

2Cu²⁺ + 4I^- 2CuI + I₂

The amount of iodine produced can then be found by titrating the resulting mixture with a sodium thiosulphate (Na₂S₂O₃) solution of known concentration. The thiosulphate ions reduce the iodine back to iodide ions:

2S₂O₃^2- + I₂ S₄O₆^2- + 2I^-

As the end-point of this titration is neared, the brown colour due to the iodine becomes faint. Adding a little starch solution produces a dark blue colour, which in turn disappears when enough sodium thiosulphate solution has been added to react with all the iodine. The disappearance of a dark blue colour is much easier to judge accurately than the disappearance of a pale brown colour.

The two equations above show us that two Cu²⁺ ions produce one iodine molecule, which then reacts in the titration with two thiosulphate ions. The number of moles of thiosulphate added in the titration is therefore the same as the number of moles of copper originally present. From this the original mass of copper in the alloy can be calculated. This measurement is known as 'estimating the composition of the alloy'. The word 'estimating' in this context does not suggest inaccuracy!

Precipitating transition metal hydroxides

When sodium hydroxide is added to a solution of a transition metal, a precipitate of the metal hydroxide always appears.

Let us consider why this happens. When a metal aqua ion is placed in water, an equilibrium is established:

[M(H₂O)₆]ⁿ⁺ [M(H₂O)₅(OH)]^(n-1)+ + H⁺

This reaction is called a ‘hydrolysis’ because a water molecule has undergone a reaction in which it has split into OH^- and H⁺.

When alkali is added to a solution of a transition metal, they attack the strongest acid present, i.e. H⁺:

OH^- + H⁺ H₂O

This upsets the hydrolysis equilibrium and when the equilibrium has been completely pushed over to the right hand side, a new equilibrium is established:

[M(H₂O)₅(OH)]^(n-1)+ [M(H₂O)₄(OH)₂]^(n-2)+ + H⁺

For example, when sodium hydroxide is added to copper(II) sulphate solution, a gelatinous (jelly-like) blue precipitate appears. The hydroxide ions firstly upset the hydrolysis reaction:

[Cu(H₂O)₆]²⁺ [Cu(H₂O)₅(OH)]⁺ + H⁺

When this equilibrium is pushed fully to the right hand side by addition of excess hydroxide ions, the second equilibrium is established:

[Cu(H₂O)₅(OH)]⁺ Cu(H₂O)₄(OH)₂ + H⁺

The product, copper(II) hydroxide, Cu(H₂O)₄(OH)₂, is uncharged. In this case, there is insufficient solvation energy to overcome the lattice enthalpy, so a solid (i.e. precipitate) is formed. Usually, copper(II) hydroxide is written in its simpler form, Cu(OH)₂. However, it should be remembered that Cu(H₂O)₄(OH)₂ and Cu(OH)₂ represent the same thing. The co-ordinated water molecules are present in both.

A similar sequence occurs whatever the metal ion so that we can summarise the reactions of sodium hydroxide with metal ions:

Cu²⁺(aq) + 2OH^-(aq) Cu(OH)₂(s) pale (Cambridge!) blue gelatinous precipitate

Fe²⁺(aq) + 2OH^-(aq) Fe(OH)₂(s) green gelatinous precipitate

Fe³⁺(aq) + 3OH^-(aq) Fe(OH)₃(s) brown gelatinous precipitate

Mn²⁺(aq) + 2OH^-(aq) Mn(OH)₂(s) cream gelatinous precipitate

Cr³⁺(aq) + 3OH^-(aq) Cr(OH)₃(s) grey-green gelatinous precipitate

Co²⁺(aq) + 2OH^-(aq) Co(OH)₂(s) blue-green gelatinous precipitate

Note: (i) Mⁿ⁺(aq) is a shorthand for [M(H₂O)₆]ⁿ⁺

(ii) Fe(OH)₃ is unstable and rapidly dehydrates to form rust (Fe₂O₃).

(iii) These precipitates will form with any source of hydroxide ion. For example, addition of dilute aqueous ammonia to copper(II) sulphate solution will give a gelatinous blue solution.