
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
545.1k+ views
Hint: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.
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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

