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Hint: We must understand that the oxidation state of $[Ni{(CO)_4}]$ is 0 in 1st case. And the oxidation state of \[{K_2}[Hg{I_4}]\]is 2 in case 2.
Complete step by step solution:
First, let’s go through the steps which are used for naming the complex compounds.
When naming, Ligand is to be named first.
While writing the names of ligand it should be in the following order: neutral, negative, positive. If multiple ligands have the same charge then they should be named in alphabetical order.
If a ligand is having multiple occurrences (monodentate) in that case we use the prefixes like – di, tri, tetra, penta, hexa respectively. For polydentate ligands we use –bis-, tris-, tetrakis-, etc.
Anions ends with –ido. Like anion ending with e becomes ato example, sulphate becomes sulphato and e becomes ‘ido’, example, cyanide becomes cyanide.
Neutral ligands are given their usual names.
After that write the name of the central atom/ion. If the complex is an anion then it will end with –ate and its Latin name is to be used except in case of mercury.
If it is required to specify the oxidation state(in case of atoms with multiple oxidation states) us Roman numerals.
End with “cation” or “anion” as separate (if applicable).
Now start with naming the complex compound
$[Ni{(CO)_4}]$
In this complex compound the oxidation state of both carbonyl and nickel is 0 and 4 carbonyl groups are being attached with central atom nickel so, the nomenclature will be tetracarbonyl nickel (0)
\[{K_2}[Hg{I_4}]\]
In this the complex is \[{\left[ {Hg{I_4}} \right]^{2 - }}\] and since it is having a charge which means the oxidation state is 2. Also, 4 iodine are connected with central atom Hg so, the nomenclature will be potassium tetraiodomercurate (II).
Note: While answering such questions the student should follow the steps for IUPAC nomenclature for complex compounds. IUPAC full form is International Union of Pure and Applied Chemistry; it is a world authority for giving chemistry based terminologies, units, nomenclatures, etc.
Complete step by step solution:
First, let’s go through the steps which are used for naming the complex compounds.
When naming, Ligand is to be named first.
While writing the names of ligand it should be in the following order: neutral, negative, positive. If multiple ligands have the same charge then they should be named in alphabetical order.
If a ligand is having multiple occurrences (monodentate) in that case we use the prefixes like – di, tri, tetra, penta, hexa respectively. For polydentate ligands we use –bis-, tris-, tetrakis-, etc.
Anions ends with –ido. Like anion ending with e becomes ato example, sulphate becomes sulphato and e becomes ‘ido’, example, cyanide becomes cyanide.
Neutral ligands are given their usual names.
After that write the name of the central atom/ion. If the complex is an anion then it will end with –ate and its Latin name is to be used except in case of mercury.
If it is required to specify the oxidation state(in case of atoms with multiple oxidation states) us Roman numerals.
End with “cation” or “anion” as separate (if applicable).
Now start with naming the complex compound
$[Ni{(CO)_4}]$
In this complex compound the oxidation state of both carbonyl and nickel is 0 and 4 carbonyl groups are being attached with central atom nickel so, the nomenclature will be tetracarbonyl nickel (0)
\[{K_2}[Hg{I_4}]\]
In this the complex is \[{\left[ {Hg{I_4}} \right]^{2 - }}\] and since it is having a charge which means the oxidation state is 2. Also, 4 iodine are connected with central atom Hg so, the nomenclature will be potassium tetraiodomercurate (II).
Note: While answering such questions the student should follow the steps for IUPAC nomenclature for complex compounds. IUPAC full form is International Union of Pure and Applied Chemistry; it is a world authority for giving chemistry based terminologies, units, nomenclatures, etc.
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