Question

How can you find the oxidation state of polyatomic ions?

Answer 1

Oxidation state is also called an oxidation number which can be positive or negative. It is used by an atom to display its degree of reduction or oxidation. If the oxidation number is positive, this means that there are more protons in the atom than electrons thus it has an ability to gain electrons.

Complete answer:
It is helpful to know the oxidation numbers of atoms in each compound during the reaction to understand the changes that follow in a chemical reaction. It also plays an important role in the systematic nomenclature of chemical compounds. By definition, the oxidation number of an atom is the charge that atom would have if the compound was composed of ions.
When an atom is in a neutral state, it will have its oxidation state as zero. This is found when the atoms contain only one element. Example- aluminium, gold copper metals, ${O_2}$, etc
When it comes to simple ions, the oxidation number is equal to the charge on the ion. As we know, $N{a^ + }$ ion, it will have oxidation number $+ 1$. And the oxidation number of $C{l^ - }$ ion will be $- 1$
When in compounds of nonmetal such as $HCl,N{H_3},{H_2}O$ then the oxidation number of hydrogen will be $+ 1$.
When it comes to polyatomic ions, the sum of oxidation numbers is equal to the charge on an ion. This means that when it comes to $S{O_4}^{2 - }$ ion then the oxidation number of sulfur atom must be $+ 6$, for examples, because the sum of oxidation numbers of atoms in this ion must be equal $- 2$
$S{O_4}^{2 - }:\left( { + 6} \right){\text{ }} + {\text{ }}4\left( { - 2} \right) = {\text{ }} - 2$

Note: Elements which have positive oxidation numbers are present on the bottom left corner of periodic table which is more likely to that of those towards the upper right corner of the table. Sulfur has a positive oxidation number in $S{O_2}$ as it is below oxygen in the periodic table. The oxidation number of oxygen in most compounds is $- 2$.