Question

What is the IUPAC name of ${\mathrm{Fe}}_4{\left[(\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}_3$?

Answer 1

Step 1: Rules for naming coordination compounds

• Cation is named first
• Ligands are named in alphabetical order followed by the central metal atom.
• Names on anionic ligands will end in -o and for neutral or cationic ligands the name remains the same except ${\mathrm{H}}_2\mathrm{O}$(aqua), ${\mathrm{NH}}_3$(ammine), $\mathrm{CO}$ (carbonyl), and $\mathrm{NO}$ (nitrosyl).
• Prefixes like mono, di, tri, etc. are used to indicate the number of each ligand. But if the names of ligands itself contain a numerical prefix then the used terms are bis, tris, tetrakis, etc. and those ligands will be enclosed in paranthesis.
• The oxidation state of metal in the coordination entity is represented using the Roman number in parenthesis after the metal name.
• If the central metal atom is involved in a cation then the name is written as it is. But if it is in an anion then the name of the central metal atom end with the suffix -ate. For some metals, Latin names are used in anions (e.g. ferrate is used for $\mathrm{Fe}$).
• There will be a space between anionic and cationic part naming. No other extra spaces are given.

Step 2: Naming of the given compound ${\mathbf{Fe}}_{\mathbf{4}}{\left[(\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}_{\mathbf{3}}$

• Following the rules, here cation is ${\mathrm{Fe}}^{3+}$. So the name starts with iron$\left(\mathrm{III}\right)$. Note that iron will be in $+3$ oxidation state in the cation because there is total $4$ irons in cation and $3$ ions in the anion. So in order to balance the neutrality the iron in cation has to be charged $+3$.

$$\begin{gathered} chargeoncation=(+3\times 4) \\ =+12 \\ chargeonanion=(-4\times 3) \\ =-12 \\ Chargeonnetcompound=+12-12 \\ =0 \end{gathered}$$

• Here there are six cyanide ligands present. Since cyanide ligand is an anionic ligand we have to name it as cyanido. Thus next comes hexacyanido.
• Now we have to name central metal atom. Here $\mathrm{Fe}$is in anionic part so we have to name it ferrate.
• The oxidation number of $\mathrm{Fe}$: Charge on anionic part is $-4$. Let,s assume the charge of $\mathrm{Fe}$ is $x$. Thus,

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

• So the Roman number after metal will be $\left(\mathrm{II}\right)$.

Therefore the name of given compound ${\mathrm{Fe}}_4{\left[(\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}_3$ will be iron$\mathbf{\left(}\mathbf{III}\mathbf{\right)}$ hexacyanidoferrate$\mathbf{\left(}\mathbf{II}\mathbf{\right)}$.