f325 transition elements

F325 Transition Elements

Transition Metals

Electron arrangements of the d-block elements

Recall that the filling order of electron shells and subshells is: 1s 2s 2p 3s 3p 4s 3d 4p 5s etc.

p. 1

Properties deduce the electron configurations of atoms (eg for Fe: 1s22s22p63s23p63d64s2) and ions of the d-block elements of

Period 4 (Sc–Zn), given the atomic number and charge describe the elements Ti–Cu as transition elements, ie d-block elements that have an ion with an incomplete d sub-

shell illustrate:

(i) the existence of more than one oxidation state for each element in its compounds(ii) the formation of coloured ions (no detail of how colour arises is required)(iii) the catalytic behaviour of the elements and/or their compounds

Precipitation reactions describe, including ionic equations, the simple precipitation reactions and the accompanying colour changes of

Cu2+(aq), Co2+

(aq), Fe2+(aq) and Fe3+

(aq) with aqueous sodium hydroxide;

Ligands and complex ions explain the term ligand in terms of coordinate bonding state and use the terms complex ion and coordination number state and give examples of complexes with sixfold coordination with an octahedral shape explain and use the term bidentate ligand (eg NH2CH2CH2NH2, ‘en’) describe the types of stereoisomerism shown by complexes, including those associated with bidentate and

multidentate ligands:(i) cis-trans isomerism, eg Ni(NH3)2Cl2

(ii) optical isomerism, eg [Ni(NH2CH2CH2NH2)3]2+

describe the use of cis-platin as an anticancer drug and its action by binding to DNA in cancer cells, preventing division

Ligands substitution describe the process of ligand substitution and the accompanying colour changes in the formation of:

(i) [Cu(NH3)4(H2O)2]2+ and [CuCl4]2– from [Cu(H2O)6]2+

(ii) [CoCl4]2– from [Co(H2O)6]2+

explain the biochemical importance of iron in haemoglobin, including ligand substitution involving O2 and CO state that the stability constant, Kstab, of a complex ion is the equilibrium constant for the formation of the

complex ion in a solvent from its constituent ions deduce expressions for the stability constant, Kstab, of a ligand substitution,

eg M2+(aq) + 6X–

(aq) MX64–

(aq) Kstab = [MX64–

(aq)]/[M2+(aq)][X–

(aq)]6

relate ligand substitution reactions of complexes to stability constants and understand that a large Kstab results in formation of a stable complex ion

Definition: The elements in between Group 2 and Group 3 are the d-block elements. This is because the highest energy subshell (i.e. the subshell being filled) is a d-subshell in these elements.


Holds: 2 2 6 2 6 2 10 6 2 electrons

The 3d subshell and 4s subshell are very close together in energy. Although the 3d is closer to the nucleus on average, it is slightly higher in energy, so the 4s fills before 3d starts to be filled.

There are some things to consider regarding the filling of a d-subshell: A d-subshell contains five d-orbitals, so can hold up to 10 electrons Electrons go one into each orbital first, before any orbital gets a second electron Having all the d-orbitals half-filled is a stable arrangement. Cr illustrates this because

the atom is more stable with each d-orbital half-filled and a half-filled 4s orbital than with an empty 3d-orbital and a full 4s orbital.

Having all the d-orbitals filled is a stable arrangement. This is illustrated in Cu, where the atom is more stable with each d-orbital filled and a half-filled 4s orbital than with one 3d-orbital only half filled and a full 4s orbital.

Thus we can write the electron arrangements for:Ca (20) 1s2 2s2 2p6 3s2 3p6 4s2

Sc (21) 1s2 2s2 2p6 3s2 3p6 3d1 4s2Ti (22) 1s2 2s2 2p6 3s2 3p6 3d2 4s2V (23) 1s2 2s2 2p6 3s2 3p6 3d3 4s2Cr (24) 1s2 2s2 2p6 3s2 3p6 3d5 4s1Mn (25) 1s2 2s2 2p6 3s2 3p6 3d5 4s2Fe (26) 1s2 2s2 2p6 3s2 3p6 3d6 4s2Co (27) 1s2 2s2 2p6 3s2 3p6 3d7 4s2Ni (28) 1s2 2s2 2p6 3s2 3p6 3d8 4s2Cu (29) 1s2 2s2 2p6 3s2 3p6 3d10 4s1Zn (30) 1s2 2s2 2p6 3s2 3p6 3d10 4s2

Ga (31) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1

Electron arrangements of transition metal ions To do this we start with the electron arrangements of the atoms, then remove electrons corresponding to the charge on the ion.

The 4s and 3d orbitals are so close in energy that the filling of the 3d orbitals shields the 4s electrons from the nucleus, and the 3d now becomes lower in energy than the 4s. As a result, when electrons are lost to form an ion, the first electrons to be removed come from 4s, then from 3d.

e.g. iron forms Fe2+ and Fe3+ ionsFe2+ 1s2 2s2 2p6 3s2 3p6 3d6 - 2 x 4s electrons removedFe3+ 1s2 2s2 2p6 3s2 3p6 3d5 - 2 x 4s electrons and 1 x 3d electron removed

e.g. vanadium forms V2+ ions and V3+ ionsV2+ 1s2 2s2 2p6 3s2 3p6 3d3 - 2 x 4s electrons removedV3+ 1s2 2s2 2p6 3s2 3p6 3d2 - 2 x 4s and 1 x 3d electron removed

p. 2

d-block elements in Period 4


BUTZinc only forms 2+ ions

Zn2+ 1s2 2s2 2p6 3s2 3p6 3d10

Scandium only forms 3+ ionsSc3+ 1s2 2s2 2p6 3s2 3p6

This leads us the definition of a transition metal – which is NOT the same as the definition for a a d-block element.

Therefore Zn and Sc are d-block elements but NOT transition elements.

The other d-block elements in Period 4 all form ions with incomplete d-subshells, so ARE transition elements.

The electron arrangements of transition metals, with their incompletely filled d-orbitals, and the ease with which electrons can be removed from, or added to these orbitals, gives them their characteristic properties:

- having coloured compounds, and forming coloured complexes- acting as catalysts- multiple oxidation states

Colour in transition metal compoundsThe aqueous ions of transition metals are coloured.

Cu2+(aq) ions are blue Co2+

(aq) ions are pinkFe2+

(aq) ions are pale green Fe3+(aq) ions are yellow-orange

Cr3+(aq) ions are green

Colour is associated with a metal ion which has incompletely filled d-orbitals, and where surrounding molecules or ions cause these d-orbitals to have slightly different energies, so that wavelengths corresponding to visible light can be absorbed in exciting electrons from one d-orbital to a slightly higher one.

By contrast, compounds of Zn and Sc are white (colourless in solution). This is more evidence that although these are d-block elements, they are not transition metals. Zn2+ ions have a 3d10 electron arrangement, and Sc3+ have no 3d electrons, so neither has incompletely filled d-orbitals.

Yet more evidence that incompletely filled d-orbitals are required for colour is shown when Cu2+ ions ([Ar] 3d9 and blue in colour) are reduced in the reaction with a solution containing iodide ions. The precipitate of copper(I) iodide in which the Cu+ ion is [Ar] 3d10 is white in colour.

p. 3

Definition: A transition element is a d-block element that forms at least one stable ion with a partially filled d sub-shell.


Key ideas in transition metal Chemistry

1) Variable oxidation statesTransition metals exhibit more than one oxidation state in their compounds. For example, both iron(II) and iron(III) compounds are common.

With iron, as with other transition metals in Period 4, this happens because the 3d and 4s energy levels are so close in energy. All transition metals form compounds with ions with +2 oxidation number. In most cases this is due to losing the two electrons from the 4s orbital, however the 3d electrons can also be lost, allowing transition metals to form stable ions with higher oxidation numbers.

Ions of transition metals in Period 4 can also change from one oxidation state to another, by accepting or donating electrons from the 3d subshell. This means that they can act as oxidizing and reducing agents. The change from one oxidation number to another is typically accompanied by a colour change.

Example of variable oxidation states:A solution containing vanadate(V) ions is yellow, but when a reducing agent such as zinc is added, the oxidation state can be reduced from +5 in steps down to +2 with a corresponding change in colour at each stage:

VO2+

(aq) Oxidation state of V: +5 Colour: yellowVO2+

(aq) Oxidation state of V: +4 Colour: blue* an intermediate green solution is usually seen before the blue, with a mixture of VO2

+ and VO2+ ions

V3+(aq) Oxidation state of V: +3 Colour: green

V2+(aq) Oxidation state of V: +2 Colour: lilac

In each stage, the vanadium has been progressively reduced. At the same time, the zinc has been oxidized, Zn(s) Zn2+

(aq) so this is a series of redox reactions.

2) Redox reactionsTransition metal ions are easily oxidized or reduced (lose or gain electrons) with an accompanying change in oxidation state. As a result they can also be used to oxidize or reduce other species, acting as reducing agents, or oxidizing agents.

Examples:1) Purple manganate(VII) ions in acidic solution are decolourised by iron(II) ions. The pale green solution turns pale yellow.

Oxidation state changes:MnO4

-(aq) becomes Mn2+

(aq)

+7 +2 reduction

Fe2+(aq) becomes Fe3+

(aq)

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+2 +3 oxidation

2) When blue copper(II) ions (e.g. in copper sulphate solution) react with colourless iodide ions, an off-white precipitate of copper(I)iodide is formed, and a brown solution of iodine.

2Cu2+(aq) + 4I-

(aq) 2CuI(s) + I2(aq)

Oxidation states: +2 -1 +1 -1 0

reduction oxidation

3) Transition metals as CatalystsHeterogeneous transition metal catalysts tend to be solid, with the reactants being in the liquid or gas phase. The solid transition metal catalyst adsorbs reactants onto its surface by using the partly-filled d-orbitals to form weak chemical bonds. This has the effect of weakening the bonds within the reactants and holding them close together on the catalyst surface and in the correct orientation to react. The product molecules can then be desorbed and leave the catalyst surface.

A copper strip (or coin) heated and suspended over propanone continues to glow brightly. The copper is acting as a heterogeneous catalyst for the reaction between propanone vapour and the oxygen in the air – an exothermic reaction. This is a laboratory demonstration of the reaction which takes place in a motor vehicle exhaust, where unburned hydrocarbon fuel molecules are oxidized to carbon dioxide and water on a transition metal catalyst surface.

Industrial examples of heterogeneous transition metal catalysts:Iron in the Haber ProcessNickel in the hydrogenation of ethene/alkenes

Homogeneous transition metal catalysts tend to be transition metal ions in solution, with the reactants also in solution. They act as a temporary store for electrons. In a first step, the substance being oxidized passes electrons to the transition metal ion which takes on a lower oxidation number. In the second step the transition metal ion donates electrons to the substance being reduced, and in the process is restored to its original oxidation number. Because of the ease with which transition metals can accept and donate electrons from the d-subshell, this may be a lower energy route than the direct transfer of electrons between the oxidized and reduced substance.

The reaction between zinc and sulphuric acid is speeded up by the addition of a little copper sulphate solution.

Cu2+

Zn(s) + 2H+(aq) Zn2+

(aq) + H2(g)

Here the Cu2+ ions are acting as a heterogeneous catalyst. In the first step, zinc is oxidized by the Cu2+ ions, losing two electrons to become Zn2+ ions. The Cu2+ ions gain two electrons and reduced to copper atoms. The catalyst is regenerated a second step where two H+ ions each

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Definition: A precipitate is formed when a solution containing a positively-charged ion is mixed with a solution containing a negatively-charged ion which combine to create a compound which is insoluble in the solvent used (usually water).


gain an electron from the copper atoms, and are reduced to form hydrogen gas. The copper atoms are oxidized back to Cu2+ ions.

4) Precipitates Transition metal hydroxides are insoluble, and coloured. These precipitates are commonly used to help identify the transition metal ion. The colour of the ion in solution can be rather pale, but the colour of the precipitate tends to be more distinctive.

Adding hydroxide ions to a solution containing transition metal ions will form a precipitate of the transition metal hydroxide. Suitable sources of hydroxide ions include:

sodium or potassium hydroxide solution (NaOH(aq) or KOH(aq))ammonia solution - NH3(g) + H2O(l) ⇌ NH4

+(aq) + OH-

(aq)

We need to know four examples.1) Cu2+

(aq) + 2 OH-(aq) Cu(OH)2(s)

Observation: The blue colour of the copper ions in solution fades as a precipitate of pale blue copper(II) hydroxide is formed.

2) Co2+(aq) + 2OH-

(aq) Co(OH)2(s)

Observation: The pink colour of the cobalt ions in solution fades as a precipitate of dark blue-green cobalt(II) hydroxide is formed

3) Fe2+(aq) + 2OH-

(aq) Fe(OH)2(s)

Observation: The pale green aqueous Fe2+ ions react with OH- ions to produce a dark green precipitate of iron(II) hydroxide

4) Fe3+(aq) + 3OH-

(aq) Fe(OH)3(s)

Observation: The aqueous Fe3+ ions are yellow if the solution is dilute, and can appear orange in more concentrated solutions. They react with OH- ions to create a rust-coloured precipitate of iron(III) hydroxide.

It can be difficult to tell Fe2+(aq) ions (pale green) from Fe3+

(aq) ions (yellow) in very dilute solutions. The formation of the very different coloured iron hydroxide precipitates is a good test to show which iron ion was present.

p. 6


Transition metal complexes

A transition metal complex is formed when a transition metal ion is bonded to one or more ligands by dative covalent bonds. If the resulting complex has an overall charge, it is referred to as a complex ion.

Because transition metal ions are small they have a strong electric field around them which attracts electron-rich species. Species which have a lone pair available to donate can form dative covalent bonds with the transition metal ion. We call such species LIGANDS.

Examples of ligands:H2O: :NH3 :Cl- :CN- :SCN- :OH-

water ammonia chloride cyanide thiocyanate hydroxide

The electron pairs bonding the ligands to the central metal ion repel one another, and so are arranged around the transition metal ion in geometric arrangements where they get as far apart from each other as possible (VSEPR Theory). The number of dative covalent bonds arranged around a transition metal is called the co-ordination number.

Complex ions with a co-ordination number of 6 tend to be octahedral Complex ions with a co-ordination number of 4 are tetrahedral or occasionally square

planar

Drawing complexes, and writing formulaeComplex ions can be drawn in 3D using shaded and wedge bonds to represent bonds into and out of the plane of the paper respectively. The bonds between ligands and the transition metal ion are dative covalent bonds, so we should use an arrow to indicate the direction in which the electron pair has been donated (towards the transition metal ion), although this is sometimes omitted in drawings of complexes to help clarity.

The complex is drawn in square brackets, and the overall charge written at the top right.

Examples:1) [Cu(H2O)6]2+ octahedral

pale blue

Name: hexa aqua copper II ion

p. 7

Definition: A ligand is a molecule or ion which can donate a lone pair to form a dative covalent bond to a transition metal ion.


2) [CuCl4]2- tetrahedralyellow

Name: tetra chloro cuprate II ion

3) [Fe(H2O)5SCN]2+ octahedralblood red

Name: penta aqua thiocyanato iron III ion

It is the size of the ligands which has the most significant effect on the geometry of the complex ions – chloride ions are much larger than water molecules, so only four chloride ions will pack around the small transition metal ion, where six water molecules will fit.

Aqueous solutions of all the transition metal ions (with no other ligands present) have a co-ordination number of 6 and are octahedral.

Crystallising complexesComplex ions are not found on their own – there is always an oppositely-charged ion to balance the charge on the complex ion, although we ignore it much of the time. We may focus on the [Cu(H2O)6]2+ ion when we dissolve copper(II) sulphate in water, forgetting the SO4

2- ion is also present in equal concentrations. Similarly when we form [CuCl4]2- by the addition of hydrochloric acid to copper sulphate solution, we have two H+ ions present for every [CuCl4]2- ion.

This is of no consequence when we are considering solutions – the "other" ions are spectators, but when we evaporate the water and form crystals, these other ions become part of the giant ionic lattice along with the complex ions.

e.g. K3Fe(CN)6 contains the complex ion [Fe(CN)6]3- and three K+ ions to balance the charges. Name: potassium hexacyanoferrate(III)

Typical ions we may find in this role include Na+, K+ or NH4+.These other ions are NOT ligands

and are NOT bonded to the central metal ion.

p. 8


Ligand Substitution

When a new ligand is added to a solution containing a complex, an equilibrium can be set up.

The new ligand can replace the original ligand to form a new complex. This is known as ligand substitution. The position of equilibrium will tend to favour the ligand whch forms a stronger dative covalent bond with the transition metal ion, resulting in a more stable complex.

The position of the equilibrium will be able to be shifted by- changing the concentration of either of the ligands in the solution

(more of the complex which "uses up" that ligand will be formed")- or changing the temperature

(position of equilibrium will shift to oppose the temperature change)

Examples:Addition of concentrated ammonia solution to Cu2+

(aq) results initially in the formation of a copper hydroxide precipitate because ammonia solution contains OH- ions. This redissolves in excess ammonia to produce a solution containing a dark blue complex ion.

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

blue dark blue

octahedral octahedral

If concentrated hydrochloric acid is added to a solution containing Cu2+

(aq), the solution turns green and progressively more yellow as chloride ligands replace the water molecules. The new complex is yellow, but the equilibrium solution containing both complexes appears green.

[Cu(H2O)6]2+ + 4Cl- ⇌ [CuCl4]2- + 6H2O blue yellowoctahedral tetrahedral

A similar set of reactions takes place with aqueous cobalt ions. If concentrated hydrochloric acid is added to a solution containing pale pink Co2+

(aq) ions, the water ligands are replaced by chloride ions, forming a blue complex ion. The solution initially turns violet as both complex ions are present, becoming blue as the chloride complex predominates.

[Co(H2O)6]2+ + 4Cl- ⇌ [CoCl4]2- + 6H2O ΔH +ve pale pink blueoctahedral tetrahedral

It is possible to shift the position of equilibrium in these reactions, causing the colour of the complex to change as one or other ion predominates, according to Le Chatelier's principle. For

p. 9


example if a violet solution containing similar concentrations of [Co(H2O)6]2+ and [CoCl4]2- ions is made:

Increasing the concentration of water shifts the equilibrium to the left, reducing the concentration of water as it forms the pink complex

Increasing the concentration of chloride ions shifts the equilibrium to the right

Increasing the temperature shifts the position of equilibrium in the endothermic direction, and more of the blue complex is formed, whereas decreasing the temperature causes the pink complex to predominate.

Multidentate ligandsAll the ligands we have seen so far have formed a single dative covalent bond to the transition metal ion. We refer to these as monodentate ligands.

Larger molecules may have more than one site which is able to donate an electron pair, and so may be able to form more than one dative covalent bond to the transition metal ion.

Example: salicylate (2-hydroxybenzoate) ions

Salicylic acid (a benzene ring with an adjacent –OH and –COOH group) does not act as a ligand, but the salicylate ion does.

The O of the –OH group and the –O- of the –COO- group each have a lone pair which can form a bond to the transition metal ion, so two dative covalent bonds are formed. This as a bidentate ligand.

When this is added to a solution containing yellow aqueous iron(III) ions, the water ligands are replaced with salicylate ligands to form a deep purple complex of iron(III) salicylate. (Note that this complex has no charge – it is not a complex ion, just a complex).

[Fe(H2O)6]3+(aq) + 3HOC6H4COO-

(aq) [Fe(HOC6H4COO)3](s) + 6H2O(l)

yellow purple

Example: EDTA (ethylenediaminetetraacetic acid)

This ligand exists in complexes as the EDTA4- ion. It has six lone pairs which can form dative covalent bonds, and is hence a hexadentate ligand.

EDTA is used to bind metal ions, removing them from solution, and is referred to as a chelating agent. Uses include binding calcium and magnesium ions to reduce water hardness, or being added to blood to treat patients suffering from lead or mercury poisoning.

p. 10


Example: ethane-1,2-diamine NH2CH2CH2NH2

This has two N atoms each with a lone pair which can form a bond to the transition metal ion. We would refer to this as a bidentate ligand.

This ligand is often shortened to 'en' in formulae.

For example, nickel forms an octahedral complex with co-ordination number 6 when it bonds with three en ligands:

[Ni(en)3]2+ or [ Ni(NH2CH2CH2NH2)3]2+

Example: ethanedioate ions C2O42-

The ethanedioate ion is a bidentate ligand. Each O- can form a dative covalent bond to the transition metal ion.

Cr3+ forms an octahedral complex with two ethanedioate ions and two water molecules.

[Cr(C2O4)2(H2O)2]-

p. 11

Pt

O

OO

H

C OO

C

C

O

O

O

H

C OO O

C

O C

O

OPt

O

H

C OO

C

C

O

O O

O

O

H

C O OC

C

O

O O

O


We might be given the structure of a complex and asked to work out and draw the ligands present, indicating how they act as ligands. To do this we need to break the dative covalent bonds to the transition metal ion and restore the lone pair to the ligand atoms which formed the bond:

e.g. [Pt(C2O4)( HOC6H4OCOO)]-

1) Take the complex ion apart:

2) Look for atoms whose valency is not satisfied with bonds – one too few bonds indicates a negative ion.

3) Add in the lone pairs which formed the dative covalent bonds to the metal ions

Biological role of ligand substitutionHaemaglobin in blood is responsible for the transport of oxygen to cells for respiration, and the transport away of carbon dioxide.

Haemaglobin is a complex protein, containing four non-protein components called haem groups, which have an Fe2+ ion at their centre, and four dative covalent bonds between the iron and four N-atoms in the haem structure, and a further dative covalent bond to the protein globin.

p. 12

-


Oxygen can bind to the haem group as a ligand, giving a co-ordination number of 6, in order to be transported. The colour of the complex when oxygen is bound as a ligand is the rich red of oxygenated blood.

Carbon monoxide can also bind to the iron at the same binding site as the oxygen would, the stability constant for this complex is greater (see later) so the bond between the iron and the CO is stronger and much less likely to break. CO will therefore bind preferentially if both carbon monoxide and oxygen are present in

the lungs; the binding is irreversible, so that haemaglobin becomes useless.

Since tobacco smoking produces carbon monoxide, and this is one of the reasons why smokers become short of breath.

Isomerism in complex ionsStereoisomerism arises when the ligands can be arranged in different spatial arrangements. In complexes, as with organic substances, we will see both cis-trans isomerism and optical isomerism.

Cis – Trans isomerismFor example when a complex contains four of one ligand and two of another, we can have cis and trans isomers.

For example, cobalt forms an octahedral complex of formula [Co(NH3)4Cl2]+. The cobalt is in oxidation state +3 here. It was found that a green salt and a purple salt both with the same formula could be crystallized. This was clear evidence for two different isomers:

cis- isomer trans- isomerpurple green

Cl at 90° to one another Cl at 180° to one another(adjacent = cis) ("opposite" / "across" = trans)

Cis-trans isomerism is also possible in some complexes with co-ordination number 4 if they have a square-planar configuration.

p. 13


Nickel forms a square planar complex [NiCl2(NH3)2]. This has no overall charge, as the nickel is in the +2 oxidation state.

cis- isomer trans- isomer

A similar complex of platinum [PtCl2(NH3)2] is one of the most effective drugs in cancer treatment. It is the cis-form of the complex which is biologically active. The drug is a liquid, usually administered intravenously as a drip, and goes under the name of cis-platin.

The importance of the exact shape and structure of the molecule is emphasized by the fact that the trans-form is ineffective.

It is thought that cis-platin may work by binding to the DNA of fast-growing cancer cells, changing the DNA structure and preventing them from reproducing. A new generation of cancer-treatment drugs with lower doses and fewer side effects than cis-platin have been developed, e.g. carboplatin.

Optical isomerismOptical isomerism in transition metal chemistry arises with octahedral complexes containing multidentate ligands, where these can be arranged around the central transition metal atom as non-superimposable mirror images.

For optical isomers we need:- a complex with three bidentate ligands

e.g. [Ni(en)3]2+

p. 14


- OR a complex with two bidentate and two monodentate ligands* NOTE this will also have cis- and trans- isomers. The optical isomers are of the cis- form (the trans form is symmetrical)

e.g. [Co(en)2(H2O)2]2+- (two optical isomers of the cis-isomer)

The trans- isomer, which is not optically active

- OR a complex with one hexadentate ligande.g. [Cu(EDTA)]2- … we're not even going to attempt to draw this ourselves!

p. 15


Stability ConstantsWe can express the position of an equilibrium in terms of an equilibrium constant, Kc and the concentrations of the species present at equilibrium. We know that the more stable the complex, the more the position of equilibrium shifts in that direction, so we can use the equilibrium constant to compare the stability of complexes. To give a meaningful comparison, we measure the stability of a complex relative to the complex which has just water as a ligand.

We define the stability constant of a complex, Kstab, for the equilibrium:

Example:[Co(H2O)6]2+ + 4Cl- ⇌ [CoCl4]2- + 6H2O ΔH +ve pale pink blue

Note also that water is left out. This is common practice because all the species are dissolved in water, so it is in great excess and its concentration is therefore virtually constant.

Kstab = [[CoCl 4] 2- ] . [[Co(H2O)6]2+] [Cl-]4

Note the potentially confusing use of two sets of square brackets; one for the complex ion and one to mean 'concentration in mol dm-3'. The more important use is that denoting concentration, and you may sometimes see the square brackets denoting complex ions omitted:

Kstab = [CoCl 42- ] .

[Co(H2O)62+] [Cl-]4

A large value of stability constant denotes a stable complex – i.e. that the position of equilibrium lies to the right, and that the complex ion is easily formed.

Note that the when the values of Kstab are larger than 1, which they almost always are, this indicates that the aqua complex is less stable than the complex containing ligands other than water. This is typically the case.

You may well be asked to compare the stability of two complexes, based on Kstab values, or to work out Kstab values given appropriate data.

A very stable complex is very unlikely to undergo further ligand substitution reactions.

p. 16

Definition: The stability constant, Kstab,, of a complex ion is the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions


Practice:1a) Write the expression for the stability constant for the reaction below, and give the units for Kstab:

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

1b) At 25°C the equilibrium mixture contained 2.0 x 10-9 mol dm-3 of [Cu(H2O)6]2+ and 0.10 mol dm-3 of NH3. Given that the value for Kstab at 25°C is 1.2 x 1013 mol-4 dm12, calculate the concentration of the [Cu(NH3)4(H2O)2]2+ complex in the solution at equilibrium:

1c) When chloride ions substitute for water as a ligand, the following equilibrium is set up:[Cu(H2O)6]2+ + 4Cl- ⇌ [CuCl4]2- + 6H2O

At 25°C the equilibrium mixture contained the following concentrations:[Cu(H2O)6]2+ 1.17 x 10-5 mol dm-3

Cl- 0.800 mol dm-3

[CuCl4]2- 2.00 mol dm-3

Calculate the value of Kstab and give the units.

1d) Comment on the relative stabilities of [CuCl4]2- and [Cu(NH3)4(H2O)2]2+ compared to [Cu(H2O)6]2+, and to each other.

p. 17


Answers to Practice Questions:

1a) Kstab = [Cu(NH3)4(H2O)22+ ] Units mol-4 dm12

[Cu(H2O)62+] [NH3]4

1b) [Cu(NH3)4(H2O)2]2+ = Kstab x [Cu2+] x [NH3]4

= 1.2x1013 x 2.0x10-9 x (0.1)4

= 2.4 mol dm-3

1c) Kstab = 4.17 x 105 mol-4 dm12

1d) Both [CuCl4]2- and [Cu(NH3)4(H2O)2]2+ have Kstab values which are much greater than 1, so the equilibrium lies far to the right and both complexes are much more stable than [Cu(H2O)6]2+

The much larger value of Kstab for the formation of [Cu(NH3)4(H2O)2]2+ indicates that this is a much more stable complex than [CuCl4]2-.

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f325 transition elements

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