Question

(i) What is meant by ambidentate ligand? Give examples also. (2)
(ii) Write the IUPAC names of following compounds:
a) [Co(Br)₂(en)₂]⁺
b) K₃[Fe(C₂O₄)₃] (1+1)

Hint:

Ambidentate = one ligand, two possible donor atoms (e.g. NO₂^-, SCN^-). For naming, work out the metal oxidation state from the overall charge, list ligands alphabetically, and add -ate to the metal only when the complex is an anion.

Answer 1

(i) Ambidentate ligand, restated:
Step 1: Recall that a ligand donates a lone pair to the metal. Some anions carry lone pairs on two chemically different atoms. Such a ligand can attach through atom A in one complex and through atom B in another; this switchable behaviour is called ambidentate character (from Latin ambi = both).
Step 2: Quick examples with the linkage isomers they produce: SCN^- (S-bonded thiocyanato vs N-bonded isothiocyanato), NO₂^- (N-bonded nitro vs O-bonded nitrito), and CN^- (C-bonded cyano vs N-bonded isocyano). This is the origin of linkage isomerism.

(ii) Naming by a fixed procedure. Order of writing a coordination name: number-prefix + ligands in alphabetical order + metal (with -ate suffix if the ion is an anion) + oxidation state in Roman numerals; for salts the counter-ion is written first.

a) Deduce the metal charge from the +1 overall charge: neutral en contributes 0, two Br^- contribute -2, so metal \(= +1 - (-2) = +3\). Apply the procedure: dibromido, then bis(ethane-1,2-diamine), then cobalt(III).
Answer: Dibromidobis(ethane-1,2-diamine)cobalt(III) ion.

b) Potassium is the counter-cation, so name it first. The anion has charge -3 (balancing three K⁺); with three oxalate (-2 each = -6), the metal is \(-3 + 6 = +3\). Because the complex is an anion, iron becomes ferrate; three chelating oxalate groups take tris.
Answer: Potassium tris(oxalato)ferrate(III).