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
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Hint: 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.
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.
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