Question

Find out the oxidation state of sodium in $\text{N}{{\text{a}}_{2}}{{\text{O}}_{2}}$.

Answer 1

Oxidation state of an element is defined as the charge which occurs on the after the gain or loss of the electron. According to the rules of the oxidation state, elements which are found free such as H, He, ${{\text{O}}_{2}}$ have the oxidation state of zero.

Complete step by step answer:
-To calculate the oxidation state of the sodium we will let it as 'x'.
-We know that the oxidation state of the oxygen is -2 because according to the electronic configuration of the element i.e. $\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{4}}$.
-So, we can see that the oxygen needs to gain only two electrons to complete its octet. So, the oxidation state of the oxygen will be -2.
-But in case of the peroxide oxygen, the bond is formed between two oxygen atoms the oxidation state of oxygen is -1.
-Whereas in case of superoxide, the oxidation state of oxygen is -1/2.

-Now, we will calculate the oxidation of sodium in the given molecule i.e.
$\begin{aligned} & \text{N}{{\text{a}}_{2}}{{\text{O}}_{2}} \\\ & \text{2x + (2 }\times \text{ -1) = 0} \\\ \end{aligned}$
$\begin{aligned} & 2\text{x = 2} \\\ & \text{x = +1} \\\ \end{aligned}$
-So, the oxidation state of sodium is +1 which tells us that sodium can easily lose one electron to form a bond and complete its octet.
-Therefore, the oxidation state of sodium is +1.

Note: One should remember that the oxidation of a monatomic ion is equal to the charge present on it. The electronic configuration of the sodium is $\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{1}}$ because of its tendency to lose a electron it forms cation and it is considered as a metal. The given is widely used as a bleaching agent, laboratory chemical and as disinfectant.