The oxidation states of iron atoms in compounds A , B and C , respectively, are x ,   y and z . Then…

Oxidation state of $\mathrm{Fe}$ in ${\mathrm{Na}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_5\left(\mathrm{NOS}\right)\right]$:

Charge of $\mathrm{CN}$ =$-1$

Charge of $\mathrm{NOS}=-1$

Suppose oxidation state of $\mathrm{Fe} = \mathrm{x}$

$$\begin{gathered} 4+\mathrm{x}-5-1=0 \\ \Rightarrow \mathrm{x}=+2 \end{gathered}$$

Oxidation state of $\mathrm{Fe} \mathrm{in} {\mathrm{Na}}_4\left[{\mathrm{FeO}}_4\right]$:

$$\begin{gathered} 4+\mathrm{x}-8=0 \\ \Rightarrow \mathrm{x}=+4 \end{gathered}$$

Oxidation state of $\mathrm{Fe} \mathrm{in} \left[{\mathrm{Fe}}_2{\left(\mathrm{CO}\right)}_9\right]$ $=0$ because carbonyl is neutral ligand.

The oxidation states of iron in these compounds will be

The sum of oxidation states will be $= 6$.