Exercise practice: Coordination Compounds

4.1 How the exercise set is arranged

The chapter has 31 exercises, numbered 5.1 to 5.31. They fall into five groups. Werner's theory and terminology cover 5.1 to 5.3. Oxidation states, formulas and names cover 5.4 to 5.7 and 5.23, 5.24. Isomerism covers 5.8 to 5.12. Bonding, crystal field theory, colour and magnetism cover 5.13 to 5.22, 5.25 and 5.29 to 5.31. Stability, chelates, applications and counting ions cover 5.14, 5.26 to 5.28. The worked items below are the patterns that repeat.

4.2 Oxidation states, formulas and names

Exercise 5.5 asks for the oxidation numbers of the metals. Take the complex charge, subtract the ligand charges and solve for the metal. For [Co(H2O)(CN)(en)2]2+ the ligands contribute zero from water, minus one from cyanide and zero from two neutral en, so cobalt is +3. Exercise 5.6 reverses the process: tetrahydroxidozincate(II) is [Zn(OH)4]2- and hexaamminecobalt(III) sulphate is [Co(NH3)6]2(SO4)3. Exercise 5.7 asks for systematic names, so list ligands alphabetically, give the metal its Roman numeral and finish with the counter ion.

Exercise 5.24 combines naming with the electronic configuration and coordination number. For each complex, find the oxidation state, count the d electrons, decide the hybridisation from the geometry and then write the name. The answer is complete only when all four pieces are present, because the question asks for the electronic configuration and coordination number as well as the name.

4.3 Isomerism

Exercise 5.8 lists the types: ionisation, hydration, linkage, coordination, geometrical and optical. Exercise 5.10 asks for the optical isomers of [Cr(C2O4)3]3-, which exists as two non-superimposable mirror images because the three bidentate oxalate ligands form a propeller around the metal. Exercise 5.11 asks for all isomers of [CoCl2(en)2]+, which has cis and trans geometrical forms, and the cis form has two optical isomers. Exercise 5.12 on [Pt(NH3)(Br)(Cl)(py)] gives three geometrical isomers and shows that square planar complexes can be optically active only when the arrangement has no mirror plane.

4.4 Bonding and crystal field theory

Exercise 5.15 asks for the bonding in [Fe(CN)6]3- and related complexes on the basis of valence bond theory. Fe3+ is d5, cyanide is a strong field ligand, so the electrons pair in t2g and one inner d orbital is free for d2sp3 hybridisation; the complex is octahedral and has one unpaired electron. Exercise 5.18 asks how Δo decides the configuration: if Δo exceeds the pairing energy the low spin arrangement is preferred, otherwise the high spin arrangement follows Hund's rule. Exercise 5.20 and 5.21 ask why the same metal gives different colours with different ligands; the answer is the change in Δo and the resulting change in the wavelength absorbed.

4.5 Stability and counting ions

Exercise 5.26 asks for the chelate effect with an example: [Ni(en)3]2+ is more stable than [Ni(NH3)6]2+ because the bidentate ligands form rings and the entropy change favours the chelated product. Exercise 5.28 counts the ions from Co(NH3)6Cl2, which gives three ions because two chloride ions are outside the coordination sphere. Exercise 5.30 asks for the most stable complex and is answered by comparing formation constants and chelation.

4.6 Two worked numericals

Magnetic moment of [Fe(CN)6]3-. Iron in the complex is +3, so the ion is d5. Cyanide is strong field, so the configuration is t2g5 with one unpaired electron. The moment is the square root of 1 times 3, which is 1.73 BM.

CFSE of [CoF6]3-. Cobalt is +3, so the ion is d6. Fluoride is a weak field ligand, so the configuration is t2g4 eg2, which is high spin. The CFSE is minus 0.4Δo times four plus 0.6Δo times two, which is minus 1.6Δo plus 1.2Δo, equal to minus 0.4Δo. A low spin d6 complex such as [Co(CN)6]3- has all six electrons in t2g and a CFSE of minus 2.4Δo, which is much more stabilised.

4.7 Common mistakes in the numericals

Using the formula for the octahedral splitting in a tetrahedral complex without the factor four ninths is the most frequent slip. A second slip is to count the electrons in the wrong shell: the crystal field splits only the d orbitals of the metal, so the d count must come from the oxidation state of the metal in that complex. A third is to state a magnetic moment without the unit BM, or to confuse the number of d electrons with the number of unpaired electrons.

A final check on any coordination answer: the oxidation state, the d count, the coordination number and the geometry should all be consistent with each other. If the geometry needs six hybrid orbitals but the answer states four, the hybridisation line is wrong even if the name is right.

When the answer asks for the number of ions produced, count only the species outside the coordination sphere. A complex such as [Co(NH3)6]Cl2 gives three ions, not two, because the complex cation and both chlorides are separate particles in solution.