Question

Match the reactions given in column I with average oxidation numbers given in column II.

Answer 1

The oxidation number of Fe has been calculated below:
${Fe}_4^{3+}{\left[{Fe}^{2+}{\left(CN\right)}_6\right]}_3$
Average oxidation number of $Fe$ is $\frac{3\times 4+2\times 3}{7}=\frac{18}{7}$
${Fe}_2^{2+}\left[{Fe}^{2+}{\left(CN\right)}_6\right]$
Average oxidation number of $Fe$ is $\frac{2\times 2+2}{3}=\frac{6}{3}=2$
${Fe}_3^{2+}{\left[{Fe}^{3+}{\left(CN\right)}_6\right]}_2$
Average oxidation number of $Fe$ is $\frac{2\times 3+3\times 2}{5}=\frac{12}{5}$
${Fe}^{3+}\left[{Fe}^{3+}{\left(CN\right)}_6\right]$
Average oxidation number of $Fe$ is $\frac{3+3}{2}=\frac{6}{2}=3$
${Na}_2^{2+}{\left[{Fe}^{2+}{\left({CN}^{5-}\right)}_{5}N{O}^{1+}\right]}^{2-}$
Oxidation state of $Fe$ is $+2$.
${\left[{Fe}^{1+}\left({NO}^{1+}\right){\left(\stackrel{0}{H_{2}O}\right)}_5\right]}^{2+}S{O}_4^{2-}$
Oxidation state of $Fe$ is $+1$.
${Fe}_3{O}_4$:
$3x-8=0$
$\Rightarrow x=\frac{8}{3}$
Therefore, oxidation state of $Fe$ is $\frac{8}{3}$.