The oxidation number of ‘Cu’ in ${\mathrm{CuSO}}_4 \mathrm{is}$

In ${\mathrm{CuSO}}_4$ the oxidation number of oxygen “O” is −2 and sulphur “S” is +6 So, the oxidation number of ‘Cu’ can be calculated as below.

${\mathrm{CuSO}}_4\Rightarrow 1\times \left(\mathrm{x}\right)+1\times (+6)+4\times (-2)=0$

x=8-6

x=+2

Therefore, the oxidation number of ‘Cu’ in ${\mathrm{CuSO}}_4 \mathrm{is} +2$