Question

In the given reaction of formation of prussian blue, the value of $m$ + $n$ is :
$K^{+}$ + ${Fe}^{m+}$ + $\left[{{Fe}^{n+}{\left(CN\right)}_6}^{4-}\right]\to \text{prussian blue (colloidal solution)}$

Answer 1

$K^{+}+F{e}^{3+}+\left[F{e}^{2+}\right(CN{)}_6{]}^{4-}\to Prussianblue$

Therefore, Value of (m+n) is 5