In sodium nitroprusside the oxidation number, coordination number and $\mathrm{EAN}$ of iron are respectively.

$$\begin{gathered} [{\mathrm{Na}}_2 [\mathrm{Fe}{\left(\mathrm{CN}\right)}_5\mathrm{NO}] \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{NO} = +1 \\ \mathrm{Oxidation} \mathrm{state} \mathrm{of} \mathrm{Fe} = +2 \\ \mathrm{Coordination} \mathrm{number} = 6 \\ \mathrm{Effective} \mathrm{atomic} \mathrm{number} = \mathrm{Atomic} \mathrm{number} – \mathrm{Oxidation} \mathrm{state} + 2 \times \mathrm{Coordination} \mathrm{number} \\ = 26 – 2 + 2 \times 6 = 36] \end{gathered}$$