Question

Which compound is zero valent metal complex?
(A) $[Cu{(N{H_3})_4}]S{O_4}$
(B) $[Pt{(N{H_3})_2}C{l_2}]$
(C) $[Ni{(CO)_4}]$
(D) ${K_3}[Fe{(CN)_6}]$

Answer 1

Hint:The given compound is a coordination compound. Coordination compounds are the one which contain one or more metal atoms which are bonded to one or more ligands. We know that all the given compounds do not contain any charge so the overall charge of this compound will be zero.

Complete step-by-step answer:According to Werner coordination theory, two types of valency are present which are known as the primary valency and the secondary valency. The primary valency is the number of negative ions which are equal to the charge on the metal ion. The primary valency is basically the oxidation number of the metal ion. On the other hand, the secondary valency is the number of ions or molecules(ligands) that are coordinated to the metal ion. This is also known as the coordination number of the metal ion.
Here we need to find the oxidation number of the central atom of each of the following given compounds to find the zero valent compound.
$[Cu{(N{H_3})_4}]S{O_4}$
We know that $N{H_3}$ is a neutral ligand so its oxidation state will be $0$ and the oxidation state of $S{O_4}$ is $ - 2$
The oxidation state of $Cu$will be:
Let the oxidation state of $Cu$ be $x$
So,
$
x + 0(4) + ( - 2) = 0 \\
\Rightarrow x + 0 - 2 = 0 \\
\Rightarrow x - 2 = 0 \\
\therefore x = + 2 \\
$
The oxidation of $Cu$ is $ + 2$
$[Pt{(N{H_3})_2}C{l_2}]$
As we already know that $N{H_3}$ is a neutral ligand so its oxidation state will be $0$ and $Cl$ is a negatively charge ligand so it will have an oxidation state of $ - 1$
Hence the oxidation state of platinum will be:
let the oxidation state of $Pt$ be $x$
So,
$
x + 0(2) + ( - 1 \times 2) = 0 \\
\Rightarrow x + 0 - 2 = 0 \\
\Rightarrow x - 2 = 0 \\
\therefore x = + 2 \\
$
Hence the oxidation state of $Pt$ is $ + 2$
$[Ni{(CO)_4}]$
We know that $CO$ is a neutral ligand, so the charge on it will be $0$
Let the oxidation state of $Ni$ be $x$
So,
$
x + 0 \times (4) = 0 \\
\Rightarrow x + 0 = 0 \\
\therefore x = 0 \\
$
Hence the oxidation number of $Ni$ in $[Ni{(CO)_4}]$ is $0$
${K_3}[Fe{(CN)_6}]$
We know that the oxidation number potassium is $ + 1$ and $CN$ has an oxidation state of $ - 1$
Let the oxidation state of $Fe$ be $x$
So,
$
+ 1(3) + x + ( - 1 \times 6) = 0 \\
\Rightarrow + 3 + x - 6 = 0 \\
\Rightarrow x - 6 + 3 = 0 \\
\Rightarrow x - 3 = 0 \\
\therefore x = + 3 \\
$
Hence the oxidation state of $Fe$ is $ + 3$
There only one of the following compounds has an oxidation state of zero which is $[Ni{(CO)_4}]$.

Therefore, the correct answer is Option (C).

Note:Generally, coordination compounds are neutral compounds. By looking at the number of ligands attached to the central metal atom we can easily determine the coordination number of the central metal. So, the coordination numbers are $4,4,4\& 6$ in $[Cu{(N{H_3})_4}]S{O_4}$, $[Pt{(N{H_3})_2}C{l_2}]$, $[Ni{(CO)_4}]$ and ${K_3}[Fe{(CN)_6}]$ respectively.