Question

The ionizable valency of $Ni$ in $Ni{\left( {CO} \right)_4}$ is __________.
A.$2$
B.$4$
C.$0$
D.$1$

Answer 1

In the given question firstly we have to define the chemical formula and the structure of $Ni{\left( {CO} \right)_4}$ . Now when we get the structure of the complex we will get that it has $4{\text{ }}CO$ components bonded to a central nickel atom and that it is having no charge over the central atom. Now this thus gives us the ionizable valency of a complete zero.

Complete answer:
-Here according to the question statement we have to find the ionizable valency for the case of the molecule of the compound $Ni{\left( {CO} \right)_4}$ .
-This $Ni{\left( {CO} \right)_4}$ molecule is molecule with $4{\text{ }}CO$ molecules bonded to the central nickel atom in such a way that it forms $4$ single bonds. The entire structure is a balanced tetrahedral which gives it a stability and thus no polarity towards any direction.
-Here we should be aware of the fact that the given phenomenon of Ionizable valency is the one valency which is in such a way linked or associated with the net charge, and this net charge should be on any complex component or molecule.
-Now as for the case of $Ni{\left( {CO} \right)_4}$ we can clearly see that in this case it is no charge or zero charge, therefore it should have no ionizable valency.
Hence, the correct option is (C).

Note:
The given compound is named as Nickel carbonyl whereas the given IUPAC name for the compound is tetracarbonylnickel. It is an organonickel compound. This colorless liquid is the principal carbonyl of nickel. It is a widely obtained and bonding complex.