Question

The IUPAC name of $$K_4[Fe{(CN)}_6]$$ is:
a.) Potassium ferricyanide
b.) Potassium ferrocyanide
c.) Potassium hexacyanoferrate (II)
d.) Potassium hexacyanoferrate (III)

Answer 1

Hint: In chemistry we have to follow IUPAC to write the names of the compounds.
According to IUPAC for coordination complexes some rules are there to write the names.
We have to split the complex into two parts positive ion and negative ion.
We have to mention the oxidation state of metal in brackets after writing the name of the compound.


Complete step by step solution:
The given coordination complex is $$K_4[Fe{(CN)}_6]$$.
Split the molecule into two parts based on the ions generated after adding the complex to water.
The given complex is splits in to two ions in water $$4K^{+}$$and $${[Fe{(CN)}_6]}^{4-}$$.
The metal which is present in the given complex is iron or ferrous. The metal is present in anion part, so we should write as ferrate.
We have to calculate the oxidation state of the metal in $$K_4[Fe{(CN)}_6]$$.
$$\begin{array}{rl}& x+6(-1)=-4\\ & x=6-4\\ & x=2\end{array}$$, here x = oxidation state of the iron, cyanide has -1 charge.
The oxidation state of iron or ferrous in the given complex is +2.
Cyanide (CN) is also present in the negative part of the ligand so, it is called cyano, the number of cyanide ligands are six (Hexa).
Potassium is present as cation (positive charge).
Therefore, the name of the given complex ($$K_4[Fe{(CN)}_6]$$) is Potassium hexacyanoferrate (II)

So, the correct option is C.


Note: We are not supposed to say six cyano, according to IUPAC six means hexa.
One-mono
Two-bi
Three-tri
Four-tetra
Five-penta
Six-hexa
Seven-septa
Eight-octa.