Question: Select the correct IUPAC name for \[[{\text{Pt(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}{\te...

Question

Select the correct IUPAC name for $[{\text{Pt(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}{\text{][PtC}}{{\text{l}}_4}]$ .
(A) Tetraammineplatinum(II) tetrachloroplatinate(II)
(B) Tetraammineplatinum(II) tetrachloroplatinate(II)
(C) Tetraammineplatinum(II) tetrachloride platinum(II)
(D) All are correct

Answer 1

Solution

By knowing the cationic and anionic entities, and simple IUPAC nomenclature rules, we can easily narrow down to our correct option.

Complete step by step solution:
In a coordination compound, cation is always written first and then followed by anion. The usual oxidation numbers of platinum are $+ 2$ and $+ 4$ .
Oxidation number of ammonia molecules is $= 0$ .
Oxidation number of chlorine is $= - 1$ .
If the oxidation number of platinum is $+ 4$ then the entity ${\text{[PtC}}{{\text{l}}_4}]$ will become neutral, but we know that it is an anion. Therefore, the oxidation number of platinum must be $+ 2$ .
Thus, charge on cation, $[{\text{Pt(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}{\text{]}}$ is $+ 2$ and charge on anion, ${\text{[PtC}}{{\text{l}}_4}]$ is $- 2$ .
The given coordination compound can be written as ${[\mathop {{\text{Pt}}}\limits^{{\text{II}}} {{\text{(N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{4}}}{\text{]}}^{2 + }}{{\text{[}}\mathop {{\text{Pt}}}\limits^{{\text{II}}} {\text{C}}{{\text{l}}_4}]^{2 - }}$ .
According to IUPAC rules, the name of anion entity to end with the suffix –ate. Also, note the following nomenclature rules:
For four atoms/molecules- tetra
For positive/ negative charge- roman numeral in parentheses as: (II)
For ammonia- amine
Name of anionic ligand, e.g., $- {\text{Cl}}$ ends in –o, as chlorido.
Therefore, the correct IUPAC name of the given coordination compound is:
Tetraammineplatinum(II) tetrachloroplatinate(II)
Hence, the correct option is A.

Note: Any molecule is a neutral ligand. The total oxidation number of any molecule must be equal to zero as in this case for ammonia.
To form a stable coordination compound, the complex entities $[{\text{Pt(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}{\text{]}}$ and ${\text{[PtC}}{{\text{l}}_4}]$ cannot be neutral. One should be cation and the other should be anion.