Question

What is the oxidation state of Copper in$\;{\left[ {Cu{{\left( {CN} \right)}_4}} \right]^{3 - }}$

Answer 1

Hint: Since the charge on cyanide ion as it acts as a ligand is -1 as there are four cyanide ligands so the total charge makes it to – 4 and the charge on the complete complex is -3. By taking the oxidation state of copper in ligand as ‘x’, we can find out the oxidation state of Copper in $\;{\left[ {Cu{{\left( {CN} \right)}_4}} \right]^{3 - }}$

Complete answer:
Oxidation number, also called oxidation state, the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom. Each atom that participates in an oxidation-reduction reaction is assigned an oxidation number that reflects its ability to acquire, donate, or share electrons.
The charge on cyanide ion as it acts as ligand is -1 as there are four cyanide ligands so the total charge is
$- 1 \times 4 = - 4$
Let the oxidation state of copper in ligand is ‘x’
The charge on the complete complex is -3.
So, the sum of total anionic and cationic charge should be equal to -3.
So, we have
$ \Rightarrow x + \left( { - 4} \right) = - 3$
$ \Rightarrow x = - 3 + 4$
$ \Rightarrow x = + 1$
Hence, the oxidation state of copper in $\;{\left[ {Cu{{\left( {CN} \right)}_4}} \right]^{3 - }}$ is $+ 1$.

Note: Oxidation state shows the total number of electrons which have been removed from an element i.e. a positive oxidation state or added to an element i.e. a negative oxidation state. To get to its present state. Oxidation involves obviously an increase in the oxidation state and similarly reduction involves a decrease in the oxidation state. Recognising the simple pattern is the single most important thing about the concept of oxidation states. If we people know how the oxidation state of an element changes during a reaction, we can obviously firstly tell whether it is being oxidised or reduced without having to work in terms of electrons half equations and electron transfers.