Question

What is the IUPAC name of $Ca[PtC{l_4}]$ .

Answer 1

Hint: Coordination compounds consist of an array of ligands that are bounded to a central metal via coordinate covalent bonds. $Ca[PtC{l_4}]$ is a coordination compound, where central metal platinum is bound to chloride ligands.

Complete answer:
In order to name coordination complexes, there are some points that we need to consider;
In coordination compounds, the name of cation is always written before anion.
If the complex is anionic, the name of the metal ends with the suffix $'ate'$
Ligands (Lewis bases) are named first in alphabetical order.
Metal is named after ligands.
The oxidation state of metal is written in Roman numerals.
In the coordination complex $Ca[PtC{l_4}]$ , Calcium is a cation with $ + 2$ charge whereas the complex ion is the anion with $ - 2$ charge.
We need to find the oxidation state of Platinum in order to name the coordination compound.
${[PtC{l_4}]^{ - 2}}$ = $x + 4( - 1) = - 2$ where, chloride ligand bears $ - 1$ charge
   $x = 2$ ( The oxidation state of Platinum is $ + 2$ )
The IUPAC name of the coordination compound will be;
$Ca[PtC{l_4}]$ = Calcium tetrachloroplatinate(II)
Here;
Calcium is cation due to which its name is written first.
In the complex ion, the name of ligand ( $C{l^ - }$) is written first. Prefix tetra is added to denote four chloride ligands.
Since the complex is anionic, the name of the metal ended with $'ate'$ along with the oxidation state.

Note:
The IUPAC name was found to be Calcium tetrachloroplatinate (II). It should be noted that while naming the complex, there should be no space in between. For example- There is no space between tetrachloroplatinate(II).