Question

Arrange the following in the increasing order of oxidation state of Mn.
(i) $\text{M}{{\text{n}}^{\text{2+}}}$
(ii) $\text{Mn}{{\text{O}}_{\text{2}}}$
(iii) $\text{KMn}{{\text{O}}_{\text{4}}}$
(iv) ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{\text{4}}}$

(A) (i) > (ii) > (iii) > (iv)
(B) (I) < (ii) < (iv) < (iii)
(C) (ii) < (III) < (I) < (iv)
(D) (iii) < (I) < (iv) < (ii)

Answer 1

Hint: In order to answer this question, first of all, we have to find the oxidation number of all the compounds present as options. For every compound present as an option, we need to find the oxidation state of every element. On the basis of the answers, we can rank in order the oxidation state of Mn in the increasing order, as required.

Complete step by step solution:
By oxidation state also known as oxidation number, we mean the number which is assigned to an element in chemical combination which represents the number of electrons lost (or even gained, in case the number is negative), by an atom of that element in the compound.
So, let us first find the oxidation state of Mn in the compound $\text{M}{{\text{n}}^{\text{2+}}}$.
The oxidation state of Mn in $\text{M}{{\text{n}}^{\text{2+}}}$is +2. Which means that Mn loses two electrons to attain the octet configuration of the nearest noble gas.
Now, let us find the oxidation state of Mn in the compound $\text{Mn}{{\text{O}}_{\text{2}}}$. We know that the sum of the oxidation numbers of the elements in a neutral compound is 0. The oxidation state of oxygen in $\text{Mn}{{\text{O}}_{\text{2}}}$. Let us consider the oxidation state of Mn in $\text{Mn}{{\text{O}}_{\text{2}}}$is x. So we can write the equation is:
$\begin{align}
& (x\times 1)+(-2\times 2)=0 \\
& \Rightarrow x=+4 \\
\end{align}
$So, the oxidation number of Mn in $\text{Mn}{{\text{O}}_{\text{2}}}$is +4.
Now, let us find the oxidation state of Mn in the compound $\text{KMn}{{\text{O}}_{\text{4}}}$. We know that the sum of the oxidation numbers of the elements in a neutral compound is 0. The oxidation state of oxygen in $\text{KMn}{{\text{O}}_{\text{4}}}$. The oxidation number of K in $\text{KMn}{{\text{O}}_{\text{4}}}$is +1. Let us consider the oxidation state of Mn in $\text{KMn}{{\text{O}}_{\text{4}}}$is x. So we can write the equation is:
$\begin{align}
& (1\times 1)+(x\times 1)+(-2\times 4)=0 \\
& \Rightarrow 1+x-8=0 \\
& \Rightarrow x=8-1 \\
& \Rightarrow x=+7 \\
\end{align}
$So, the oxidation number of Mn in $\text{KMn}{{\text{O}}_{\text{4}}}$is +7.
Now, let us find the oxidation state of Mn in the compound ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{\text{4}}}$. We know that the sum of the oxidation numbers of the elements in a neutral compound is 0. The oxidation state of oxygen in ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{\text{4}}}$. The oxidation number of K in ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{\text{4}}}$is +1. Let us consider the oxidation state of Mn in ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{\text{4}}}$ is x. So we can write the equation is:
$\begin{align}
& (1\times 2)+(x\times 1)+(-2\times 4)=0 \\
& \Rightarrow 2+x-8=0 \\
& \Rightarrow x=8-2 \\
& \Rightarrow x=+6 \\
\end{align}
$So, the oxidation number of Mn in ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{\text{4}}}$is +6.
So, on the basis of the answers that are given in the above equations, the correct order of the oxidation number of Mn in the increasing order is: (i) < (ii) < (iv) < (iii).

Hence, the correct answer is Option B.

Note: Oxidation state and oxidation number are quantities that commonly equal the same value for atoms in a molecule and are often used interchangeably. Most of the time, it does not matter if the term oxidation state or oxidation number is used. Oxidation state refers to the degree of oxidation of an atom in a molecule.