Question

What is the n-Factor of $KMn{{O}_{4}}$ in neutral medium?
A. $6$
B. $5$
C. $4$
D. $3$

Answer 1

Hint: To answer this question, we should recall the concept of n-factor. Let us see what exactly n factor is. Whenever we talk of redox reaction, n-factor or valency factor is defined as the moles of electrons gained or lost by a mole of species which undergoes oxidation or reduction. So, to solve this question you need to write the reaction of permanganate in a neutral medium, and then just proceed ahead by counting the number of electrons gained or lost.

Complete step-by-step answer:
Having defined the n-factor in the above hint, we can also define n-factor in short as the change in the oxidation number of the species which undergoes either oxidation or reduction in a redox reaction.
Let us first see how exactly potassium permanganate $\left( KMn{{O}_{4}} \right)$ reacts in the neutral medium or weak basic medium.
$Mn{{O}_{4}}^{-}+2{{H}_{2}}O\to Mn{{O}_{2}}+4O{{H}^{-}}$
Thus, we see that purple-colored complex of $Mn{{O}_{4}}^{-}$ is getting reduce to a brown colored complex of $Mn{{O}_{2}}$
For, calculating the n-factor, we will have to see the change in the oxidation number of Mn between $Mn{{O}_{4}}^{-}$ and $Mn{{O}_{2}}$. But for finding out the difference we need to calculate the oxidation state of Mn individually in permanganate ion and manganese dioxide.
Therefore, Oxidation state of Mn in $Mn{{O}_{4}}^{-}$ is calculated as follows
Let the oxidation state of Mn be $x$
$\begin{align}
  & \Rightarrow x+4\left( -2 \right)=-1 \\
 & \Rightarrow x-8=-1 \\
 & \Rightarrow x=-1+8=7 \\
 & \therefore x=+7 \\
\end{align}$
Hence, the oxidation state of Mn in $Mn{{O}_{4}}^{-}$ is +7
Now, let us find the oxidation state on Mn in $Mn{{O}_{2}}$
Let the oxidation state of Mn in this compound be $y$
$\begin{align}
  & \Rightarrow y+2\left( -2 \right)=0 \\
 & \Rightarrow y-4=0 \\
 & \therefore y=+4 \\
\end{align}$
Hence the oxidation state of Mn in $Mn{{O}_{2}}$ is $+4$
Now, the change in the oxidation state is = $x-y=7-4=3$
And we should always remember that change in oxidation state always signifies the number of moles of electrons gained or lost. Since here we see that the oxidation state was getting reduced from $+7$ to $+4$, therefore Mn is getting reduced.
Since, the change in oxidation state is $3$, and we know that Mn is getting reduced, so we can say that Mn in $MnO_{4}^{-}$ gained $3$ electrons to get reduced to $Mn{{O}_{2}}$.
In the definition of n-factor above we learnt that n-factor in simple terms refers to the number of moles of electrons gained or lost.
Here, we saw that $3$ moles of electrons are getting gained by Mn in $MnO_{4}^{-}$ to get reduced to $Mn{{O}_{2}}$.
Therefore, the n-factor of potassium permanganate in neutral medium is $3$.
Hence, the correct option is D.

Note: Sometimes the question asks to find out the equivalent weight of the compound. Then always remember that equivalent weight of compound is given by the formula
$\text{Equivalent weight = }\dfrac{\text{Molecular weight}}{\text{n-factor}}$. So, if we were asked to find the equivalent weight of potassium permanganate in neutral medium, then it would have been equal to $\dfrac{\text{Molecular weight of KMn}{{\text{O}}_{4}}}{3}$. Therefore, the most critical point is the calculation of n-factor and thus its practice is must.