Question

Assertion: Oxidation number of \[Mn\] in \[MnS{{O}_{4}}\] and \[Mn{{C}_{2}}{{O}_{4}}\] is same.
Reason: Oxidation number of \[Mn\] is different in $M{{n}_{3}}{{O}_{4}}$ and $MnO_{4}^{-2}$
A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
C) Assertion is correct but Reason is incorrect.
D) Assertion is incorrect but Reason is correct.

Answer 1

Hint: Oxidation number is the charge that is acquired by gain or loss of electrons, the oxidation numbers are integers. It can be positive, negative or zero. To solve this question, we must know the oxidation number of oxygen and ${{C}_{2}}{{O}_{4}}$.

Complete step-by-step answer:
A charge is formed on an atom when ionic bonds are formed with other atoms. This charge is known as oxidation number. If an atom shows high electronegativity, then it will show a negative oxidation state. As we know that atoms have multiple valence electrons, and it forms multiple bonds with all the atoms.
If we want to calculate the oxidation number of an atom, we must follow these steps:
First, we will add up the constant oxidation state of an atom.
Then, we will equate the total oxidation state of a molecule to the total charge of the molecule.
Now, we will calculate the oxidation number of \[Mn\] in \[MnS{{O}_{4}}\]
 \[MnS{{O}_{4}}\]
Let us take oxidation number of \[Mn\] equal to $x$
The oxidation number of sulphate ion $(S{{O}_{4}})$ is $-2$
Now, we will substitute the value and calculate the oxidation number of Manganese
 $x+(-2)=0$
 $\Rightarrow x=2$
Now, we will calculate the oxidation of \[Mn\] in \[Mn{{C}_{2}}{{O}_{4}}\]
 \[Mn{{C}_{2}}{{O}_{4}}\]
Let us take oxidation number of \[Mn\] equal to $x$
The oxidation number of oxalate ion $({{C}_{2}}{{O}_{4}})$ is $-2$
Now, we will substitute the value and calculate the oxidation number of Manganese
 $x+(-2)=0$
 $\Rightarrow x=2$
Therefore, this proves that the oxidation number of \[Mn\] in \[MnS{{O}_{4}}\] and \[Mn{{C}_{2}}{{O}_{4}}\] is same.
In both these compounds, the manganese is present as \[M{{n}^{+2}}\] ion.
Now, we will calculate the oxidation number of \[Mn\] in $M{{n}_{3}}{{O}_{4}}$
 $M{{n}_{3}}{{O}_{4}}$
Let us take oxidation number of \[Mn\] equal to $x$
The oxidation number of oxygen is $-2$
Now, we will substitute the value and calculate the oxidation number of Manganese
 $3x+4(-2)=0$
 $\Rightarrow 3x-8=0$
On further solving, we get
 $x=\dfrac{8}{3}$
The oxidation number of \[Mn\] in $M{{n}_{3}}{{O}_{4}}$ is $+\dfrac{8}{3}$
Now, we will calculate the oxidation number of \[Mn\] in $MnO_{4}^{-2}$
 $MnO_{4}^{-2}$
Let us take oxidation number of \[Mn\] equal to $x$
The oxidation number of oxygen is $-2$
Now, we will substitute the value and calculate the oxidation number of Manganese
 $x+4(-2)=-2$
 $\Rightarrow x-8=-2$
On further solving, we get
 $x=6$
The oxidation number of \[Mn\] in $MnO_{4}^{-2}$ is $+6$
Therefore, the reason is incorrect as the oxidation number of \[Mn\] in \[MnS{{O}_{4}}\] and \[Mn{{C}_{2}}{{O}_{4}}\] is $+2$ and the oxidation number in $M{{n}_{3}}{{O}_{4}}$ and $MnO_{4}^{-2}$ is $+\dfrac{8}{3}$ and $+6$ respectively.
So, we can say that both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.

Therefore, the correct answer is option (B).

Note:
i) In this question, we have calculated the oxidation number of manganese from different manganese complexes.
ii) The oxidation number of alkali metal is always $+1$
iii) The oxidation number of alkaline earth metals is always $+2$
iv) The oxidation number of alkali metal is always $-2$