Question

What is the IUPAC name of $Xe[Pt{F_6}]$
A) Hexafluoridoplatinate $(VI)$ xenon
B) Xenon Hexafluoroplatinate $(V)$
C) Xenon hexafluoridoplatinate $(VI)$
D) Xenonium hexafluoridoplatinum $(V)$

Answer 1

The given compound is a coordination complex compound in which Platinum is the metal, Fluorine is the ligand and Xenon is the cation. Calculate the oxidation state of the metal and use IUPAC rules of coordination compounds to name the given molecule.

Complete solution:
We can use the IUPAC nomenclature rules for coordination compounds to name the given compound.
In the given compound the ligand Fluorine is a monodentate.
1. According to the IUPAC rules, multiple occurring monodentate ligands should receive a prefix according to the number of occurrences such as di, tri, tetra etc.
Here, we have six Fluorines in the compound so we name the ligand as ‘hexafluoro’.
2. The ligand should be named first and if the complex is an anion, then the central atom’s name would end with ‘ate’.
Therefore, the name becomes ‘hexafluoroplatinate’. We still have Xenon in our complex that hasn’t been named.
3. The cation should be named first followed by the anion. Since Xenon is the cation here and the rest of the complex is the anion, we write Xenon first while naming the compound.
Therefore the name becomes ‘Xenon hexafluoroplatinate’.
Now, the oxidation state of Platinum is Five and this should be written in Roman numbers and in brackets.
Therefore, the IUPAC name of $Xe[Pt{F_6}]$ is Xenon hexafluoroplatinate $(V)$ i.e. option B.

Note: In order to name a coordination compound, one should be able to identify the metal, ligand and the cation/anion. One should also be able to calculate the oxidation state of the metal taking the oxidation state of cation/anion into consideration.