Question

Calculate the oxidation number of $Fe$ in ${{K}_{3}}[Fe{{(CN)}_{6}}]$.

Answer 1

Coordinate bonds are the type of covalent bonds in which pairs of electrons are given by one atom and shared by both atoms. Unidentate ligands are those ligands which contain one donor site. Oxidation number describes the degree of oxidation of metal ions in a complex .

Complete answer:
The given complex is ${{K}_{3}}[Fe{{(CN)}_{6}}]$ where cationic species is potassium $(K)$and anionic species is coordinate species. Potassium $(K)$is alkali metal and thus has oxidation state $+1$. Therefore, anionic species has charge $-3$. Cyanide ion $(C{{N}^{-}})$is an anionic and unidentate ligand which has charge $-1$. Therefore,
$\text{(charge on Fe)}+6\times \text{(charge on }C{{N}^{-}})=-3$
$\text{(charge on Fe)}+6\times (-1)=-3$
Therefore, charge on $Fe$ is given as
$\text{Charge on Fe}=+3$
$\Rightarrow \text{Oxidation number of Fe}=+3$
Hence, the oxidation number of $Fe$ in ${{K}_{3}}[Fe{{(CN)}_{6}}]$is $+3$.

Additional information: There are two types of valencies present in coordinate complexes which are primary valency and secondary valency. Primary valency is ionisable whereas secondary valency is non-ionisable. The species present in square brackets represents the non-ionisable species. On the above basis, the number of ionisable species from ${{K}_{3}}[Fe{{(CN)}_{6}}]$ is $4$. In contrast to oxidation number, coordination number of central metal ions describes the number of donor ligands attached to it. In the given example, $Fe$ is attached to six cyanide ions $(C{{N}^{-}})$. Therefore, the coordination number of $Fe$ in ${{K}_{3}}[Fe{{(CN)}_{6}}]$ is $+3$. Instead of unidentate ligands, bidentate ligands are those which contain two donor sites for the formation of coordinate bonds. Ethylene 1,2-diamine is an example of bidentate ligand.

Note:
It is important to note that the oxidation number of $Fe$ in ${{K}_{3}}[Fe{{(CN)}_{6}}]$ is $+3$. Cyanide ion $(C{{N}^{-}})$ is an anionic ligand. Potassium $(K)$ is alkali metal and thus has oxidation state $+1$.