Question

What is the oxidation number of $\mathrm{Fe}$ in ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$?

Answer 1

Oxidation number: Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to a more electronegative element.

Calculating the oxidation number of Iron(Fe) in ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$:

The oxidation state of Potassium(K) = +1

The charge on Cyanide ion(CN^-) = -1

Consider the oxidation state of Iron(Fe) = x

Now,

$$\begin{gathered} \left(+1\times 4\right)+x+\left(-1\times 6\right)=0 \\ 4+x-6=0 \\ x=+2 \end{gathered}$$

Hence, the oxidation number of Iron in ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ is $+2$.