Question

Find the oxidation number of elements in each case: $Mn$ in ${{K}_{2}}Mn{{O}_{4}}$ , ${{K}_{2}}Mn{{O}_{3}}$ , $M{{n}_{3}}{{O}_{4}}$ , $MnS{{O}_{4}}$ , ${{K}_{3}}Mn{{F}_{6}}$ .
A. $6,4,\dfrac{8}{3},2,3$
B. $10,5,\dfrac{8}{3},3,3$
C. $6,5,8,2,3$
D. None of the above

Answer 1

Oxidation is defined as a term which is used for the addition of oxygen to an element or any compound. In the atmosphere, there is a presence of dioxygen, that results in a combination of elements with it. And this is the principle reason why they are present in the form of their oxide on the earth. In these chemical reactions, transfer of electrons takes place.

Complete step by step answer:
Oxidation number is also referred to as an oxidation state that helps in describing the degree of oxidation of an atom in a compound. Oxidation state can be positive, negative or zero. It is considered as a hypothetical charge that an atom must have, if all the bonds formed are ionic. When an oxidation state increases in a chemical reaction, it is known as oxidation, whereas if an oxidation decreases in a chemical reaction, it is known as reduction. In these chemical reactions, transfer of electrons takes place. The oxidation state is zero for pure elements.
Let us suppose the oxidation number of $Mn$ be $x$ for all the cases.
$Mn$ in ${{K}_{2}}Mn{{O}_{4}}$
Oxidation number of $K=+1$
Oxidation number of $O=-2$
Therefore, $2(K)+1(Mn)+4(O)=0$
On substituting the values, we get,
$2+x-8=0$
On further solving, we get,
$\Rightarrow x-6=0$
$\Rightarrow x=+6$
$Mn$ in ${{K}_{2}}Mn{{O}_{3}}$
Oxidation number of $K=+1$
Oxidation number of $O=-2$
Therefore, $2(K)+1(Mn)+3(O)=0$
On substituting the values, we get,
$2+x-6=0$
On further solving, we get,
$\Rightarrow x=+4$
$Mn$ in $M{{n}_{3}}{{O}_{4}}$
Oxidation number of $O=-2$
$3(Mn)+4(O)=0$
On substituting the values, we get,
$3x-4\times 2=0$
On further solving, we get,
$\Rightarrow x=+\dfrac{8}{3}$
$Mn$ in $MnS{{O}_{4}}$
Oxidation number of $S{{O}_{4}}=-2$
$1(Mn)+1(S{{O}_{4}})=0$
On substituting the values, we get,
$x-2=0$
On further solving, we get,
$\Rightarrow x=+2$
$Mn$ in ${{K}_{3}}Mn{{F}_{6}}$
Oxidation number of $K=+1$
Oxidation number of $F=-1$
$3(K)+1(Mn)+6(F)=0$
On substituting the values, we get,
$3\times 1+1\times x+6\times (-1)=0$
On further solving, we get,
$\Rightarrow 3+x-6=0$
$\Rightarrow x=+3$
Oxidation number of element in each case will be $6,4,\dfrac{8}{3},2,3$

So, the correct answer is Option A.

Note: The oxidation state of an atom does not represent a real charge on the atom.
Monatomic ion contains an oxidation number equal to the charge of the ion. For example, $H$ having oxidation number equal to $+1$
Oxidation number of a polyatomic ion can be calculated as a sum of all the oxidation numbers of atoms equal to the net charge of the polyatomic ion.