Question

What is the oxidation number of K in ${K_2}C{r_2}{O_2}$?
(A) $1$
(B) $2$
(C) $3$
(D) $4$

Answer 1

$K$ is a highly electropositive element, it is an alkali metal and commonly has $+ 1$ oxidation state in all ionic salts. It shows $+ 1$ oxidation state due to its tendency to lose one electron to attain noble gas configuration.

Complete step by step answer:
Oxidation number also known as the oxidation state is defined as the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.
The most common oxidation state of ${\text{potassium}}$ is $+ 1$ such as in $KCl$ or $KMn{O_4}$.
So, in ${K_2}C{r_2}{O_2}$ the oxidation number of K will also be $+ 1$.

Some facts regarding oxidation number are:
(1) The Oxidation number of all ${\text{alkali metals}}$ ions is always $+ 1$.
(2) The Oxidation number for all ${\text{alkaline earth metals}}$ ions is always $+ 2$.
(3) Oxidation number of all ${\text{boron}}$ family metal ions is $+ 3$ always.
(4) Oxidation number of hydrogen in proton is $+ 1$ & in hydride it is $- 1$.
(5) Oxidation number of oxygen in oxide ion $\left( {{O^{2 - }}} \right)$ is $- 2$ & in peroxide is $- 1$.
The Oxidation number of K in ${K_2}C{r_2}{O_2}$ is $+ 1$.

So, Option $A$ is correct.

Note: As you know, oxidation number in general describes the transfer of electrons. But one thing should be noted that it is different from formal charge which actually defines the arrangement of atoms.