During the race of complex compounds, finish line is fixed by achieving $\mathrm{EAN}$value equal to the atomic number of corresponding inert gas. Then which of the following complex is present at the finish line?

$K_4{\left[Fe\right(\left(N{)}_6\right]}^{-}$ will be present at the finish line because its $\mathrm{EAN}$value is 36.
If n, Z, and x are the numbers of ligands, then the metal ion's atomic number and oxidation state are as follows:

$\mathrm{EAN}=\mathrm{z}-\mathrm{x}+2\mathrm{nL}$ $[\mathrm{L}=1$ for mono di tetra ligand $\mathrm{L}=2$for bi di tetra ligand] 

For,  $K_4{\left[Fe\right(\left(N{)}_6\mathit{\right]}}^{\mathit{-}}$

$\mathrm{x}=+2\mathrm{z}=26, \mathrm{N}=6, \mathrm{L}=1$

$\mathrm{EAN} =26-2+(2\times 1\times 6)$

$\mathrm{EAN}=24+12=36$