Question

The oxidation number of Mn is $+ 8$ in the compound
(A) ${K_2}Mn{O_4}$
(B) $Mn{O_4}$
(C) $KMn{O_4}$
(D) $M{n_2}{O_7}$

Answer 1

Oxidation number or oxidation state is the number of electrons gained or lost to form a chemical bond with another atom. It can be positive, negative or zero. The oxidation number of metal is found if the oxidation state of all other entities of a compound are known and the overall charge is also known.

Complete step-by-step solution: We know that for a stable and uncharged compound the total of all oxidation numbers is zero. Let the oxidation number of manganese(Mn) in the given compound be x.
Then we have the compound ${K_2}Mn{O_4}$ . We know that oxidation of alkali metals like sodium, potassium is always one.
The oxidation state of oxygen is $- 2$
Then oxidation number of manganese in ${K_2}Mn{O_4}$ is calculated in the following way:
Let the oxidation state of manganese(Mn) in ${K_2}Mn{O_4}$ be x
We know that the sum of the oxidation numbers of the constituent elements of a neutral compound is zero. Then
$1 \times 2 + x + 2 \times ( - 4) = 0$
$\Rightarrow 2 + x - 8 = 0$
$\therefore x = + 6$
Hence the oxidation state of Mn in ${K_2}Mn{O_4}$ is $+ 6$
Now we have the compound $Mn{O_4}$. Then,
$x + ( - 2) \times (4) = 0$
$\Rightarrow x - 8 = 0$
$\therefore x = 8$
Hence the oxidation state of Mn in $Mn{O_4}$ is $+ 8$
Now we have the compound potassium permanganate. The value of x will be as follows:
$1 + x + ( - 2) \times 4 = 0$
$\Rightarrow 1 + x - 8 = 0$
$\therefore x = + 7$
Hence the oxidation state of manganese in $KMn{O_4}$ is $+ 7$.
Now we have the compound $M{n_2}{O_7}$ . The value of x is as follows:
$2 \times x - 2 \times 7 = 0$
$\Rightarrow 2x - 14 = 0$
$\therefore x = + 7$
Hence the oxidation state of manganese in $M{n_2}{O_7}$ is $+ 7$.

Therefore the correct answer is option (B).

Note: Although oxidation state of manganese in manganese tetroxide is $+ 8$ , this compound does not exist. This is because manganese has only seven valence electrons.
Also, manganese acts as a self -indicator in titration reactions.