Question

The IUPAC name of $[Ni{(CO)}_4]$ is:
(a)- Tetracarbonyl nickelete(0)
(b)- Tetracarbonyl nickelete(II)
(c)- Tetracarbonyl nickel(0).
(d)- Tetracarbonyl nickel(II)

Answer 1

Hint: $[Ni{(CO)}_4]$ is a coordination compound. In the IUPAC name of coordination compounds, the oxidation state is also mentioned along with the central metal atom in roman form. The oxidation state of nickel in $[Ni{(CO)}_4]$ is zero.

Complete step by step answer:
There are some rules which are followed for the naming of coordination compounds.
Step by step method followed for the naming of the coordination compound is:
The positive ion is named first and then the anion is named. The complex part is written in one word. The numerical prefixes used to indicate the number of ligands is di for 2, tri for 3, tetra for 4, Penta for 5, Hexa for 6, and the name is written without the hyphen.
Ligands are positive, negative, or neutral, the ligands are named according to the alphabetic order.
In the complex, the ligand is named first then the name of the central metal atom is written.
If the complex is a cation, write the ligand name, then the central metal atom with its oxidation state in Romans, and then the name of the anion.
If the complex is an anion, first, write the name of cation and the name of ligand and to the central metal atom add –ate along with its oxidation state.

So, the IUPAC name of $[Ni{(CO)}_4]$ will be,
Here, nickel is the central metal atom and carbonyl is a ligand. Since, it is a neutral complex no need to add –ate to the central metal atom.
In$[Ni{(CO)}_4]$, the oxidation number of nickel is:
The oxidation number of $CO$ is zero because it is a neutral ligand,
$x+4(0)=0$
$x=0$

So, the oxidation state of nickel in $[Ni{(CO)}_4]$is zero.
So, the name will be Tetracarbonyl nickel(0).
So, the correct answer is “Option C”.

Note: In the naming of amines only one ‘m’ is used, but for the ligand ammonia ($N{H}_3$), the name ‘ammine’ is used. We can also use bis and tris at the place if bi and tri, respectively.