n-factor of $KMn{{O}_{4}}$in neutral medium is:
A. 6
B. 5
C. 4
D. 3

VerifiedVerified
139.8k+ views
Hint: n- factor is also known as valence factor and for neutral medium it can be defined as the change in oxidation state of an atom in a compound per molecule. $KMn{{O}_{4}}$is neither an acid nor a base, it is actually neutral. It is a purplish black solid which when dissolved in water gives intensely pink or purple solutions.
Step by step solution:
- $KMn{{O}_{4}}$is a salt because it is produced from KOH which is a strong base and $MnO_{4}^{-}$ion are produced from $HMn{{O}_{4}}$, which is an acid.
- The reaction in neutral medium is:
\[MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O\]
- We can calculate the oxidation state of $MnO_{4}^{-}$. Here, the charge present on oxygen is -2 and the overall charge is -1.So,
\[\begin{align}
  & x+\left( 4\times -2 \right)=-1 \\
 & x+\left( -8 \right)=-1 \\
 & x-8=-1 \\
 & x=+7 \\
\end{align}\]
-Now we will calculate the oxidation state of $Mn{{O}_{2}}$. Here, the charge present on oxygen is -2 and the overall charge is 0. So,
\[\begin{align}
  & x+\left( 2\times -2 \right)=0 \\
 & x+\left( -4 \right)=0 \\
 & x=+4 \\
\end{align}\]

- We can see from the reaction that, In neutral medium, $MnO_{4}^{-}$ is reduced to $Mn{{O}_{2}}$.Here the oxidation state of Mn changes from+7 to +4. For which 3 electrons are taken, hence we can say that the n factor here is 3.
- Therefore we can conclude that the option(d)that is 3 is correct.
Additional information:
 - We can also find the n factor for acid, here the n factor in case of acid can be defined as the number of ions replaced by one mole of acid in a reaction.
- In case of base, n factor can be defined as the number of ions replaced by one mole of base in a reaction.
Note:
n-factor value can be further used to determine the equivalent weight of a compound. We can see that n-factor methods can be used to solve complicated reactions including redox reactions.