Question

Calculate effective atomic number of Fe^2⊕ in \[\ce{[Fe(CN)6]^{4\ominus}}\]

[Given: (Z = 26)]

Answer 1

For \[\ce{[Fe(CN)6]^{4\ominus}}\]:

Let the oxidation state of Fe = x

Each CN^– ligand has a charge = –1

There are 6 CN^– ligands → total charge = –6

Overall charge on complex = –4

x − 6 = −4 

x = +2

So, Fe is in the +2 oxidation state.

EAN = Z − oxidation state + electrons donated by ligands

Atomic number (Z) of Fe = 26

Oxidation state = +2

Each CN^– donates 2 electrons.

6 CN^– ligands → 6 × 2 = 12 electrons

EAN = 26 − 2 + 12

= 36