The IUPAC name of $K_{3} [I r {(C_{2} O_{4})}_{3}]$ is

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potassium trioxalatoiridium (III)

potassium trioxalatoiridate (III)

potassium tris (oxalato) iridate (III)

potassium tris (oxalato) iridium (III)

answer is B.

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Detailed Solution

According to IUPAC nomenclature, while naming the name of the complex-counter ion is written first. In the given complex, potassium is the counter ion. As the oxidation state of Ir in +3, so Iridium is written as iridate(III). There are three oxalate ion (ligands), so trioxalato is written in IUPAC name of the complex.

Finally, name of the complex is $K_{3} [Ir {(C_{2} O_{4})}_{3}]$ becomes Potassium trioxalatoiridate (III).

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