A complex in which the oxidation number of the metal is zero
A.${K_4}[Fe{(CN)_6}]$
B.${K_3}[Fe{(CN)_6}]$
C.$[Ni{(CO)_4}]$
D.$[Pl{(N{H_3})_4}]C{l_2}$

Hint: A species is said to be diamagnetic when all the electrons present in are paired, and it is not attracted to a magnetic field. If a particle has unpaired electrons then they are paramagnetic.

Complete step by step answer:
To calculate oxidation state of a metal in the complex, net charge of the complex will be equal to the total charge on the complex.
Let $x$ be the oxidation state of the metal.
${K_4}[Fe{(CN)_6}]$
The net charge on the complex is $ - 4$ as we have an anionic complex.
$\therefore \quad - 4 = x + 6 \times - 1 \Rightarrow x = 2$
${K_3}[Fe{(CN)_6}]$
The net charge on the complex is $ - 3$as we have an anionic complex.
$\therefore \quad - 3 = x + 6 \times - 1 \Rightarrow x = 3$
$[Ni{(CO)_4}]$
The net charge on the complex is $0$as we have a neutral complex.
$\therefore 0 = x + 0 \Rightarrow x = 0$
$[Pl{(N{H_3})_4}]C{l_2}$
The net charge on the complex is $ + 2$ as we have a cationic complex.
$\therefore + 2 = x + 0 \times 4 \Rightarrow x = 2$
Hence, The only Complex which has metal oxidation state as zero is $[Ni{(CO)_4}]$.

Additional Information: Metals in zero oxidation state undergo bond formation by the exchange of Lone pair on ligands. Hence, Ligands with Lone pairs such as ${H_2}O,N{H_3},CO$ can form complexes with metals in zero oxidation state. Any ligand which is forming a bond because of it has gained an extra electron such as $C{l^ - },C{N^ - }$ cannot form complexes with metal in the zero oxidation state. Even the counter ion for such metals needs to be neutral and cannot be a charged species such as potassium.

Note:
Draw the structure of the ligand to determine whether they are neutral or charged species. This will also help in determining the number of donor sites in one molecule, which will affect the oxidation number.