Question

In the complex $F{e_4}{[Fe{(CN)_6}]_3}$ is:
A.Both Fe atoms are in the same oxidation state.
B.Both Fe atoms are in different oxidation states.
C.The coordination number of ions is 4.
D.The complex is a high spin complex.

Answer 1

Ferric ferrocyanide is also known as Berlin blue or Prussian blue or ferric hexacyanoferrate (II). It has two iron atoms having oxidation states +3 and +2. The coordination number of the central iron atom is the number of cyanide ions surrounding it. Cyanide is a strong feel ligand and thus the complex will be low spin.

Complete step by step answer:
In this complex, we can see both iron atoms have different oxidation states. The first iron needs three groups of ferrocyanide to make the compound. The ferrocyanide group has -4 charge from 6 cyanide ions which have –1 charge and an atom of iron having +2 charge. Mathematically, we can combine these atoms and their oxidation states which gives us this neutral complex.
Let us see the method through which we can synthesise this complex.
$4F{e^{3 + }}(aq) + 3{[Fe{(CN)_6}]^{4 - }}(aq) \to F{e_4}{[Fe{(CN)_6}]_3}(s)$
The innermost iron sphere ${[Fe{(CN)_6}]^{4 - }}$ has an oxidation state +2. This is because the overall charge on the ion is -4 and each cyanide ion is contributing -1 charge. Thus, $x + 6\left( { - 1} \right) = - 4$ , so $\;x = + 2$ .
Similarly, we can find the oxidation state of outer iron atoms. Four atoms of outer Fe and three of inner ${[Fe{(CN)_6}]^{4 - }}$ complex ions need to be balanced. 3 ${[Fe{(CN)_6}]^{4 - }}$ will have total charge of $3 \times ( - 4) = - 12$ , so outer four iron atoms must have +12 charge in total to balance the ions on complex. Therefore, each outer iron atom will have +3 charge on it and total it will be $4 \times ( + 3) = + 12$ .
Both Fe atoms are in different oxidation states i.e. +2 and +3. Therefore, option (A) is incorrect and (B) is correct. Let us look at the third option. Coordination number represents the number of ligands surrounding the central metal atom. Here, the central metal atom is Fe(II) and the ligands attached to it are cyanide ions. We can see from the complex formula that six cyanide ions surround the central Fe atom to form an octahedral geometry. Thus, its coordination number will be six and not four. This marks option (C) as incorrect.
Cyanide is a strong field ligand as per spectrochemical series. It will have only a few unpaired electrons and thus the complex will be low spin and not high spin. This marks option (D) as incorrect as well.

Hence, the correct option is (B).

Note:
The main difference between high spin and low spin complexes is that high spin complexes have unpaired electrons but low spin complexes tend to contain paired electrons. Ligands like CO, $C{N^ - }$ and ammonia form low spin complexes whereas $N{O^{2 - }}$ forms high spin complexes. If the field is strong, it will have only few unpaired electrons and thus low spin. If it is a weak field, it will have more unpaired electrons and thus high spin.