Question

The oxidation states of Mn in $$K_{2}Mn{O}_4$$ and $$KMn{O}_4$$ respectively are:
A. +6, +7
B. +6, +6
C. +7, +7
D. +7, +6

Answer 1

Hint: We should know that oxidation state is a number that is assigned to an element in a chemical combination. This number represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element.

Step by step answer:
We should know that the transfer of electrons is described by the oxidation state of the molecule. We use oxidation state to determine the changes in redox reactions and it is numerically similar to valence electrons.
To find the oxidation state of a compound we should know some rules. These rules are as follows:
RULE 1: Any individual atom un-combined with other elements has the oxidation state of 0 (zero). For example: If we take Ag only, its oxidation state will be 0. The oxidation state for O (oxygen) or $O_2$ is 0 as long as it is un-combined with any other element.
RULE 2: We should know that the sum of the oxidation state of all atoms in any given species is equal to the net charge on that species.
a.) In neutral species, the total sum of the oxidation state of all atoms is 0. Example, sum for NaCl is zero because Na=+1 and Cl= -1
b.) In ions, the total sum of the oxidation state is the charge of the ion. Example: $$C{a}^{2+}$$ (Calcium) is = +2. In $$Cr{O}_4^{2-}$$ (Chromate ion), the total sum is -2.
Now, we will find the oxidation state of Mn in $$K_{2}Mn{O}_4$$. So, we should let the oxidation state of Mn be x. So, oxidation state will be:
As we know, the oxidation state of O is -2 and that of H is +1.
$$\begin{array}{rl}& K_{2}Mn{O}_4:2\times (+1)+x+(-2)\times 4=0\\ & :2+x-8=0\\ & :x=+6\\ & KMn{O}_4:1+x+(-2)\times 4=0\\ & :1+x-8=0\\ & :x=+7\end{array}$$
So, from the above calculation we find out the oxidation state of $$K_{2}Mn{O}_4$$ and $$KMn{O}_4$$.
Oxidation state of $$K_{2}Mn{O}_4$$ is +6. And the oxidation state of $$KMn{O}_4$$is +7. So, option C is correct.

Note: We should not confuse formal charge with oxidation states. We should know that oxidation state is commonly used to determine the changes in redox reactions and is numerically similar to valence electrons, but different from formal charge. Formal charge determines the arrangement of atoms and the likelihood of the molecule existing.