Question

The IUPAC name for \[[Ni{{(CO)}_{4}}]\]is:
A. Tetracarbonylnickel (II)
B. Tetracarbonylnickel (0)
C. Tetracarbonylnickel Date (II)
D. Tetracarbonylnickel Date (0)

Answer 1

Hint: In chemistry we have to follow IUPAC to write the names of the compounds.
According to IUPAC for coordination complexes some rules are there to write the names.
We have to split the complex into two parts positive ion and negative ion.
We have to mention the oxidation state of metal in brackets after writing the name of the compound.

Complete step by step answer:
-The given coordination complex is\[[Ni{{(CO)}_{4}}]\].
-First we have to see whether the given coordination complex is going to split into ions or not.
-The given compound \[[Ni{{(CO)}_{4}}]\] split because all the elements are within the coordination complex, then the elements won’t come out from the coordination sphere.
-The metal which is present in the given coordination complex is nickel.
-There is no charge on the given coordination complex then we have to write the metal name is nickel in the IUPAC name.
-Next we have to calculate the oxidation number of the metal in the given coordination complex \[[Ni{{(CO)}_{4}}]\].
\[\begin{align}
& x+4(0)=0 \\
& x=0 \\
\end{align}\]
Here x = oxidation number of the nickel, CO has 0 oxidation number.
-The oxidation number of the metal nickel in the given coordination complex\[[Ni{{(CO)}_{4}}]\] is zero.
-There are four carbonyl ligands (CO) in the given complex. So, we have to write tetra in the IUPAC name.
-Therefore the IUPAC name of the given complex \[[Ni{{(CO)}_{4}}]\]is Tetracarbonylnickel (0).

-So, the correct option is B.


Note: We are not supposed to say four carbonyls, according to IUPAC four means Tetra.
One-mono
Two-bi
Three-tri
Four-tetra
Five-penta
Six-hexa
Seven-septa
Eight-octa