Question

The IUPAC name of $K_3\left[Ir{\left(C_2{O}_4\right)}_3\right]$is:

• A. Potassium trioxalatoiridium (III)
• B. Potassium trioxalatoiridate (III)
• C. Potassium tris (oxalato) iridium (III)
• D. Potassium tris (oxalato) iridate (III)

Answer 1

Answer: (B)

Potassium trioxalatoiridate (III)

First we note that the name of the counter ion is potassium, now we write the name of the complex ion.

It contains iridium(III), so it will be called as iridate(III).

Note that we can determine the oxidation number of iridium by charge balance.

Also it contains 3 $[C_2{O}_4{]}^{-2}$ groups so it will be trioxalato.

Combining, we get that the name of the negative ion is trioxalatoiridate(III).

So, the name of the compound becomes Potassium trioxalatoiridate(III).

B is the correct answer.