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

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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.