Question

What is the oxidation state of copper in $CuS{O}_4$?

Answer 1

Hint: Oxidation state of any element is their ability to gain or lose electrons. The oxidation state of any atom is affected by the bonded atoms in that molecule. Oxidation states are positive and negative depending on their ability to gain or lose electrons and the electronegativities. The sum of all oxidation numbers in a molecule is 0.

Complete answer:
An oxidation state is the ability of atoms to gain or lose electrons to complete their octet. Oxidation state of any atom depends on the nature of bonded atoms in that molecule, whether they are electronegative or electropositive, this gives the atom, the sign of positive and negative, depending on its electronegativity and the ability to gain or lose electrons.
The oxidation state of any atom can be calculated by the valence of that atom. Another method is to calculate by the fact that the sum of all the oxidation numbers in any molecule is 0. Here, $CuS{O}_4$ contains copper as cation and sulphate as anion, we can calculate the individual oxidation state in the sulphate ion and assume it to be added with copper and their sum must be 0. Therefore, we have, -2 oxidation number of oxygen, and +6 as that of sulphur. Now, $(4\times -2)+6+C{u}_{oxidationnumber}=0$ , rearranging this we will get, $C{u}_{oxidationnumber}=+2$.
Hence the oxidation number of copper in $CuS{O}_4$ is +2.

Note:
The name of $CuS{O}_4$ can also be written as copper (II) sulphate, which shows the oxidation number of copper. Also we know that the charge on sulphate ions is -2, through which the oxidation number of copper will be +2. Oxygen has -2 oxidation state due to the absence of 2 electrons to complete its octet, while sulphur possesses a +6 oxidation number due to its less electronegativity.