Introduction

A complex consists of a central metal atom or ion surrounded by a set of ligands which have one or more atoms bearing lone-pairs of electrons. These "donor" atom are bound electrostatically and covalently to the metal ion. In non-transition metal complexes such as Na⁺(aq), which can be approximately formulated as [Na(H₂O)₆]⁺, or [Ca(EDTA)]ⁿ⁺, the binding is largely electrostatic, while in transition metal complexes there is significant covalency.

Configuration of Metal Ions

It is most important to be able to correctly arrive at the dⁿ configuration of a particular ion. Remember to place all the remaining valence electrons, after those accounting for the formal charge have been subtracted, in the d-orbitals. This is why those anomalies in configuration for Cr, Cu etc. are of no real importance in describing metals in their chemical compounds. (Unlike the free (gas-phase neutral atoms, the ns subshell always has significantly higher energy than the (n-1)d subshell for the ions.)

Ligand Types (Section 7.2)

Examples of common ligands:

• Monodentate Ligands

  Halide ions, H₂O, OH¯, O²¯ SH¯ S²¯, NH₃, NR₃, NC₅H₅ (pyridene), PR₃, AsR₃, CO
  The following are ambidentate - they can bond via either end, or bridge: CN¯, SCN¯

  Some ambidentate ligands can bind differently in complexes with the same molecular formula. This is called "linkage isomerism". For example, nitrite can bind either through nitrogen or oxygen:

  [Co(NH₃)₅(ONO)]²⁺ (the red "nitrito" complex) and
  [Co(NH₃)₅(NO₂)]²⁺ (the yellow "nitro" complex)

  (Replacement of one NH₃ in [CoNH₃)₆]³⁺ by NO₂¯ in a cold solution yields the kinetically favoured nitrito complex, i.e. in the intimate substitution mechanism, an oxygen binds to the cobalt first. If the solution is then heated, the nitrito complex rearranges to the thermodynamically favoured nitro complex i.e. the cobalt - nitrogen bond is presumably stronger than the cobalt - oxygen bond.)

• Bidentate Ligands

  

• Tridentate Ligands

  

• Other Polydentate Ligands

  The structures below are examples of two important tetradentate and one hexadentate ligand.

  

The Chelate Effect

Polydentate ligands which are flexible enough so that two or more of their "donor" atoms can wrap around an bind to the same metal are called chelating ligands. Their complexes are stabilized by two effects which are basically entropy related:

1. Consider this reaction:

   [Co(NH₃)₆]²⁺   +   3 en       [Co(en)₃]²⁺   +   6NH₃

   Both the reactant and the product complexes contain 6 cobalt-nitogen bonds (the only ones broken in this reaction, so the enthalpy should be very small. On the other hand, there are four molecules to the left and 7 to the right, so there is greater potential for disorder to the right. Therefore the driving force, that is the dominant factor in DG, is the very positive entropy change - look back at the notes on enthalpy/entropy from Chapter 1.

2. If one end of a bidentate ligand becomes detached from the metal ion, there is a strong probability that it will re-attach itself before the other end becomes separated if the chain connecting the two ends is quite short. If the chain is long, the loose end can drift far from the metal ion and the probability of reattachmnet is diminished. The optimum ring size in chelate complexes is found to be four or five members (including the metal ion). This too can be related to related to entropy, but it is in the nature of a kinetic effect since it has to do with order/disorder in the reaction intermediate.

There are a number of possible "defined" geometries for transition metal complexes, together with an infinite range of "in between" cases. Only the most important: octahedral, tetrahedral, and square-planar are covered in this section of the course.

• Coordination Number 6
  • Octahedral

    The principal geometry for six-coordination is octahedral. (Another, which is much much rarer is trigonal-prismatic.) There are a couple of important types of isomerism, and a number of lesser significance. All are relevant only to complexes which are kinetically inert, that is, not subject to ligand exchange processes which will thwart attempts to physically separate the isomers.

    Geometrical Isomerism - cis/trans and fac/mer. These are the named types, but there can be others for more complicated sets of ligands:

    

    Optical isomerism - D/L

    Complexes with two or three chelating ligands can show optical activity. The molecules can exist in two forms which are mirror images of each other. Such isomers, if separated, have the property of rotating the plane of polarization of polarized light.

    
    

    Notice that the molecules can be approximated to a propeller shape. If the propeller is "right-handed", that is, it would tend to pull away from you if you rotated it clockwise, then the molecule is the D-isomer. If the propeller would tend to move towards you when rotated clockwise, then it is the L-isomer.

    Complexes with only two chelate rings and two identical monodentate ligands in cis positions could also be isolated as D or L-isomers. The trans isomer has mirror symmetry and is therefore not optically active. Complexes with a single chelate ring and four identical monodentate also have mirror symetry. Make sure you can recognize or draw diagrams which illustrate this!

• Coordination Number 4
  • Tetrahedral

    The tetrahedral geometry is never favoured over the octahedral on the basis of crystal field stabilization energy (CFSE - see below). At best, for d⁰, d⁵ or d¹⁰, the CFSE will be zero for both geometries. The tetrahedral geometry is therefore only found if the ligands are bulky. In addition, tetrahedral complexes are always labile, that is subject to the ligands exchanging, either with the solvent or other dissolved ligands, or exchanging positions. For this reason, optical isomers of tetrahedral transition metal complexes cannot be isolated.

  • Square-planar

    Square-planar complexes exist only when the CFSE (see below) favours this geometry over the alternatives. Most such complexes are formed with d⁸ ions such as Pd²⁺. Pt²⁺, Ni²⁺ (sometimes), Rh⁺, Ir⁺ and Au³⁺. Most square-planar complex ions are inert so cis/trans isomers can be separated:

    

Bonding Theories

The theories included under the general heading Ligand Field Theory directly address the two main properties of transition metal complexes: colour and para/diamagnetism. The theories also give insight into the relative stability of one coordination geometry relative to another, and the properties of inertness and lability which qualify the ease of ligand exchange. This course considers only crystal field theory, where only electrostatic effects are considered. Treatment of the more sophisticated molecular orbital theory of transition metal complexes, ligand field theory, is deferred until Chem 341.

Crystal Field Theory   (Section 7.4 and 7.5)

The d-orbitals are degenerate in the absence of an electrical field (or in a spherically symmetric electric field), that is, for example, in the case of a bare gas phase atom. Their energies are split in ligand fields, i.e. when surrounded by a group of ligands.

• Octahedral

  The orbitals are split into two groups: a set consisting of d_xy, d_xz, and d_yz stabilized by 2/5D_o, known by their symmetry classification as the t_2g set, and a set consisting of the d_x²-y² and d_z², known as the e_g set, destabilized by 3/5D_o where D_o is the gap between the two sets.

  
  For configurations d⁴ to d⁷, there are two ways in which the electrons can be placed into the t_2g and e_g orbitals which depend on the magnitude of D_o. In the weak field case (also called high spin of spin free), the electrons are distributed one at a time, with parallel spins before pairing as if the d-orbitals were still degenerate. In the strong field case (also known as low spin or spin paired), the electrons first fill the t_2g set. For configurations d¹ to d³ and d⁸ to d¹⁰ there is no difference.

  In the diagram above, the arrangements for d⁷ are shown as an example. The configurations would be written t_2g⁵ e_g² for the weak field case and t_2g⁶ e_g¹ for the strong field case.

  A configuration has an associated crystal field stabilization energy (CSFE) calculated by taking a contribution of -2/5D_o for each t_2g electron and +3/5D_o for each e_g electron. Thus for the two possible d7 cases shown, the CFSE is

  for the weak field case:   5(-2/5D_o) + 2(+3/5D_o) = -4/5D_o

  for the strong field case:   6(-2/5D_o) + 1(+3/5D_o) = -9/5D_o

  A particular complex ion will adopt a configuration depending on the balance between to CSFE, which is always negative or zero, and the pairing energy which is always positive.

  As an example consider the complex [Fe(H₂O)₆]²⁺. Iron has a d₆ configuration, the value of D_o is 10,400 cm^-1 and the pairing energy is 17600 cm^-1. (1 kJ mol^-1 = 349.76 cm^-1.) We must compare the total of the CFSE and the pairing energy for the two possible configurations.

  CFSE (high spin) = 4 x -2/5 x 10400 + 2 x 3/5 x 10400 = -4160cm^-1 (-11.89 kJ mol^-1)
  Pairing energy (1 pair) = 1 x 17600 = 17600 cm^-1 (50.32 kJ mol^-1
  Total = +13440 cm^-1 (38.43 kJ mol^-1)

  CFSE (low spin) = 6 x -2/5 x 10400 = -24960 cm^-1 (-71.36 kJ mol^-1)
  Pairing energy (3 pairs) = 3 x 17600 = 52800 (151.0 kJ mol^-1)
  Total = +27840 cm^-1 (79.60 kJ mole^-1

  The high spin configuration is about 41 kJ mol^-1 more stable than the low spin configuration, so it is preferred. This is in accordance with the experimental observation of the paramagnetism of [Fe(H²O)₆]²⁺. (Note that it looks as if both configurations are unstable, but remember that what happens to the d-electron energies is only part of the picture. The ligands are held to the central ion (in crystal field theory) by electrostatic attractions which provide a substantial exothermic contribution to the total heat of formation.

• Tetrahedral

  The orbital splitting is the inverse of octahedral. The lower group of two orbitals, known as the e set, are stabilized by 3/5D_T and the upper group of three, known as the t₂ set, are destabilized by 2/5D_T where D_T would equal 4/9D_o if the metal ion and ligands were the same.

  Note that there are no known complexes where D_T is great enough to cause spin pairing. Tetrahedral 4-coordination is never favoured over octahedral 6-coordination by the CFSE, and because there are only four metal - ligand bonds vs six for octahedral, it is also disfavoured on electrostatic (or covalent) grounds. Tetrahedral is found where there is zero CFSE difference i.e. d⁰, d⁵ (high spin O_h), and d¹⁰, and where bulky ligands preclude 6-coordination.

  
  

• Square-planar

  The orbitals are split into four sets. From the lowest in energy: d_xz and d_yz (known as e₁), d_z² (a₁), d_xy (b₂) and d_x²-y² (b₁). The gap between the b₁ and b₂ sets is the only gap of importance and would be equivalent to D_o if the metal ion and ligands were the same.

  This geometry is favoured if the value of D_o is sufficently high and if the configuration is d⁸. This includes some Ni²⁺ complexes and many 4-coordinate complexes of Pd²⁺, Pt²⁺, Rh⁺, Ir⁺, and Au³⁺

  
  

The actual value of D depends on both the metal ion and the nature of the ligands:

• The splitting increases with the metal ion oxidation state. For example, it roughtly doubles going from II to III.

• The splitting increases by 30 - 50% per period down a group.

• Tetrahedral splitting would be 4/9 of the octahedral value if the ligands and metal ion were the same. Of course, generally only the octahedral or tetrahedral complex can actually be prepared for a particular combination, so a direct comparison is rarely feasible.

• It is possible to arrange representative ligands in an order of increasing field strength called the spectrochemical series:

  I¯ < Br¯ < -SCN¯ < Cl¯ < F¯ < OH¯ < C₂O₄²¯ < H₂O < -NCS¯
  < py < NH₃ < en < bipy < o-phen < NO₂¯ < CN¯ < CO

  In comparing goups of similar ligands, for example the halogens, it is possible to rationalize the order. (The negative charge on F¯ is more localized that that on the larger Cl¯ and so on, so it has a larger effect on the metal ion.) In general, ligands e.g. CN¯ or CO which are capable of p-bonding to the metal appear high in the series. Molecular orbital theory (called ligand field theory for transition metal complexes) is required to explain this in more detail, but this is beyond the scope of this course.

Ligand f - Factor Br¯ 0.72 -SCN¯ 0.73 Cl¯ 0.78 N3¯ 0.83 F¯ 0.90 C2O42¯ 0.99 H2O 1.00 -NCS¯ 1.02 NH2CH2CO2¯ 1.18 py 1.23 NH3 1.25 en 1.28 bipy 1.33 CN¯ 1.70

Metal Ion g - Factor Mn(II) 8.0 Ni(II) 8.7 Co(II) 9.0 V(II) 12.0 Fe(III) 14.0 Cr(III) 17.4 Co(III) 18.2 Ru(II) 20.0 Mn(IV) 23.0 Mo(III) 24.6 Rh(III) 27.0 Tc(IV) 30.0 Ir(III) 32 Pt(IV) 36

Molecular Orbital Theory   (Section 7.6)

The variety of molecular orbital theory applied to transition metal complexes is called ligand field bond theory. It is not covered in Chem 241. (It wll be covered in Chem 341.)

Magnetic Properties   (Section 7.4)

• Molecules or ions containing no unpaired electrons are diamagnetic. A sample of such a compound is very very weakly repelled out of a magnetic field.

• Compounds containing unpaired electrons will be attracted into a magnetic field very weakly if the electrons are acting independently. Such compounds are paramagnetic.

• A substance, usually a metal e.g. iron or nickel with unpaired electrons acting in a concerted way will be strongly attracted into a magnetic field. Such materials are ferromagnetic.

• (Some substances have unpaired electrons (that is one electron per orbital) which are nevertheless organized with their spins oriented in opposite directions in approximately equal numbers. They are called anti-ferromagnetic.)

A paramagnetic substance is characterised experimentally by its (molar) magnetic susceptibility, c_m. This is measured by suspending a sample of the compound under a sensitive balance between the poles of a powerful electro-magnet, but above the regions where the field is strongest. The weight of the sample is measured with the field off, and then with the field turned, on which pulls the paramagnetic sample downwards. From the weight increase c_m can be determined (usually by comparison with a known standard).

The magnetic moment of the substance is given by the Curie Law:

m   =   2.54(c_m-corr.T)^½   (in units of Bohr magnetons - don't ask!)

m   =   2(S(S+1))^½

Even a rather approximate measurement of the magnetic moment of a complex can allow the assignment of geometry and, in the case of an octahedral species, a discrimination between the high or low spin configuration.

Examples

Consider the complexes [Ni(CN)₄]²¯ and [NiCl₄]²¯. The former is observed to be diamagnetic, and the latter is paramagnetic. What are their structures?

1. Work out the oxidation state and d-configuration of the nickel in both complexes. You have to know that chloride and cynanide are anions (both -1) and that neutral nickel has 10 valence electrons. This leads to the conclusion that both complexes contain Ni²⁺, a d⁸ ion.

2. Choose the possible geometries and draw the configurations. Both compounds contain four ligands so the only likely geometries are square-planar or tetrahedral. Remember, square-planar is normally spin paired and tetrahedral is always spin free. The diagrams below show the configurations:
   
3. Clearly, [Ni(CN)₄]²¯ fits the square-planar geometry, and [NiCl₄]²¯ fits the tetrahedral geometry. This makes sense because it requires a strong field ligand such as CN¯ to produce the spin paired configuration which stabilizes the square-planar geometry for a lighter transition metal like nickel (relative to the spin free tetrahedral equivalent that is).

The complex [Fe(CN)₆]⁴¯ is diamagnetic while [Fe(H₂O)₆]²⁺ is paramagnetic. Why?

1. Work out the oxidation state and the d-electron configuration of the iron in both complexes. You have to know that cyanide has a charge of -1 and water is neutral and that neutral iron has 8 valence electrons. This leads to the conclusion that both complexes contain Fe²⁺, a d⁶ ion.

2. There are six ligands in both complexes, so they are virtually certainly both octahedral. Therefore the only possible difference must be weak field vs strong field. The two configurations are illustrated below.
   
3. The strong field configuration has no unpaired electrons, so it goes with the cyanide complex. The weak field configuration has four unpaired electrons.

Electronic Absorption Spectroscopy   Chapter 13

This is quite a complex topic which is not covered in any real detail in Chem 241. The text is rather too detailed to be very useful. The d-d transitions in complex ions correspond to absorptions which are often, though not always, the cause for their colour. The position of absorption peaks in the spectra allow the direct measurement of D. This is particularly straightforward for ions with a d¹ or d¹⁰ configurations.

Some complexes, usually very intensely coloured, owe their colour to charge-transfer transitions which involve the excitation of an electron from a molecular orbital largely centered on the metal to one largely centered on the ligands or vice versa. Such transitions often result in a big dipole change for the molecule which is a factor which is associated with a highly probable transition and hence an intense colour.

Selection Rules for d-d Transitions and Colour Intensity (Section 13.5)

• The Laporte Rule. In a molecule or ion possessing a centre of symmetry, transitions are not allowed between orbitals of the same type, for example d to d. The geometries affected by this rule include octahedral and square-planar depending on the ligands and isomers involved. The rule never applies to tetrahedral complexes.

  In cases where the rule applies, the colours of the complexes are usually relatively pale. The reason transitions are observed at all is because the symmetry centre is transiently destroyed by vibrations of the molecules or ions.

  As examples, consider [Cu(H₂O)₆]²⁺ which is a rather pale blue colour vs [Cu(NH₃)₄]²⁺ which is an intense dark blue.

• Spin Allowed - Spin Forbidden Transitions which would require an electron to change its spin are strongly forbidden. Consider the case of the high spin d⁵ complex [Mn(H₂O)₆]²⁺. This complex is sometimes very pale pink due to the presence of very small amounts of impurities.

Recommended Questions Shriver and Atkins - Chapter 7:

	Exercises
7.1, 7.2	You should be able to do these.
7.3	Only the common geometry, octahedral, was covered.
7.4	You should be able to draw them, but do not worry about the naming.
7.5	You should be able to do this.
7.6	Don't bother.
7.7	You should be able to draw structures from their names...
7.8	...but don't worry about naming them yourself. Can you draw them, though?
7.9	You should be able to do this.
7.10 - 7.14	These are all good questions.
7.15	Not difficult, if you think about it.
7.16 - 7.19	Not covered.
7.20	You should be able to do this.
7.21	Not covered.
7.22	Covered, but in the following section of the notes.
7.23	You should be able to do this.
7.24 - 7.29	Not covered
	Problems
7.1 - 7.10	These questions are beyond the scope of Chem 241.
7.11	You could take a shot at this one. NO2- is ambidentate.
7.12	You should be able to do this question.

Recommended Questions from Shriver and Atkins - Chapter 13:

	Exercises
13.1 - 13.9	Beyond th scope of Chem 241.
13.10, 13.11	You could give a brief answers to these questions.
13.12 - 13.14	Beyond the scope of Chem 241.
13.15	You could give a brief answer to this question.
13.16	Beyond the scope of Chem 241.
13.17	You could give a brief answer to this question.
13.18 - 13.20	Beyond the scope of Chem 241.
	Problems
All	Beyond the scope of Chem 241.

Recommended Questions from Cotton, Wilkinson and Gaus - Chapter 6:

	"Study Questions" (Chapter 6)
	"A. Review"
1	Just do 4 and 6.
2	You should know these.
3	Not covered.
4	You should know this, particularly how to distinguish between these terms. (Not all polydentate ligands can chelate; they are useful for bridging.)
5	Know them.
6	Do this - be able to clearly illustrate the three-dimensionality of the species.
7	You should know what coordination isomers are. The others are less important. Remember to give real examples for this type of question.
8 - 10	Not covered.
11	Know this.
12	You might know this - it was mentionned, and is in these notes.
13	These terms were mentionned - check your notes.
14 - 22	Not covered.
	"B. Additional Questions"
1	Not covered.
2	Not explicitly covered, but you should be able to attempt this problem.
3	Not explicitly covered, but you might be able to attempt this problem. Look up trigonal prismatic coordination which was not covered.
4	Not covered in this part of the course, but review electron deficient compounds in Chapter 3.
5	You should be able to do this.
6	Not covered.
7	Not explicitly covered but you might be able to tackle this problem.
8 - 13	Not covered.
14, 15	You should be able to do these questions.
16 - 18	Not covered.
19	You should be able to do this.
20 - 37	Not covered.

Recommended Questions from Cotton, Wilkinson and Gaus - Chapter 23:

	"Study Questions"
	"A. Review"
1	You should know the answer to the first part and be able to figure out the second from your familiarity of the periodic table.
2	You should be able to answer this.
3, 4	You should be able to answer these.
5	Not covered, but if you look up the answer it might help you remember the splitting for the square-planar geometry.
6	Although ligand field theory (M.O.) was not covered, you should know what it is as opposed to crystal field theory.
7	You must be able to answer this.
8	You might be able to tackle this though no examples were worked in lectures.
9	You should be able to answer this.
10	Not covered.
11	You should be able to answer this.
12	A bit of a trick question, but you should be able to get the answer.
13	You should be able to answer this.
14, 15	Not covered.
16	You should be able to do this (LFSE º CFSE)
	"B. Additional Questions"
1 - 3	Beyond the scope of Chem 241.
4, 5	You should be able to answer these
6	A bit of a trick question - the answer is simpler than you might think!
7, 8	You should be able to do these with some "educated guesses".
9 - 12	Not covered.