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(C) The $E.A.N.$ (Effective Atomic Number) is calculated as: $E.A.N. = Z - O.S. + 2 	imes C.N.$,where $Z$ is the atomic number,$O.S.$ is the oxidatio

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$K_4[Fe(CN)_6]$

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(C) The $E.A.N.$ (Effective Atomic Number) is calculated as: $E.A.N. = Z - O.S. + 2 	imes C.N.$,where $Z$ is the atomic number,$O.S.$ is the oxidation state,and $C.N.$ is the coordination number.For $K_4[Fe(CN)_6]$:$Z$ for $Fe = 26$.$O.S.$ of $Fe = +2$.$C.N.$ of $Fe = 6$.$E.A.N. = 26 - 2 + 2(6) = 24 + 12 = 36$.Since $36$ is the atomic number of the inert gas Krypton $(Kr)$,this complex satisfies the condition.

During the race of complex compounds,the finish line is fixed by achieving an $E.A.N.$ value equal to the atomic number of the corresponding inert gas. Which of the following complexes is present at the finish line?

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