Question

Find the oxidation number of ${ S }$ in ${ SO }_{ 4 }^{ 2- }$ ion?

Answer 1

Hint: Oxidation number or oxidation state is the total number of electrons that it either gains or loses in order to form a chemical bond with another atom.
In other words, it is the charge acquired by an atom when it accepts or loses electrons.

Complete step-by-step answer:

Let, x be the oxidation number of sulphur.
We know the net charge on the molecule = ${ -2 }$
So, in ${ SO }_{ 4 }^{ 2- }$ oxidation number of sulphur ${ +4\times }$ oxidation number of oxygen = ${ -2 }$
$\Rightarrow { x+4\times (-2)=-2 }$
$\Rightarrow { x-8=-2 }$
$\Rightarrow { x\quad =\quad -2+8 }$
$\Rightarrow { x\quad =\quad +6 }$
Hence, the oxidation number of ${ S }$ in ${ SO }_{ 4 }^{ 2- }$ ion is ${ +6 }$.

Additional Information:
Oxidation number follows following three rules;
The oxidation number of an element in its elemental form is 0. The sum of the oxidation numbers of all the atoms in a species is equal to its total charge.
The oxidation number of an element when it exists as a monatomic ion is the same as its charge. The species which cause oxidation are called Oxidizing agents while the species which cause reduction are called Reducing Agent.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Note: The possibility to make a mistake is that you won't consider the net charge ${ -2 }$. In spite you may take net charge zero.
Oxygen is more electronegative than sulphur. So, the electron polarizes the electron density more towards itself and thus has more negative oxidation state here.