Question

Find the n-factor in the following chemical changes:
\[KMn{O_4}\xrightarrow{{{H^ + }}}M{n^{2 + }}\]
A. \[3\]
B. \[2\]
C. \[4\]
D. \[5\]

Answer 1

Hint:The n-factor is related to the change in number of number of electrons while converting reactant to product. It is also related to the oxidation state of the central metal ion.

Complete answer:
The given reaction is an example of redox reaction. The change of oxidation state of the central metal ion i.e. manganese is taking place during the reaction.
Manganese is an element in the periodic table with atomic number \[25\] . Its electronic configuration is \[\left[ {Ar} \right]3{d^5}4{s^2}\] . Thus manganese can attain a highest oxidation state of \[ + 7\] . The potassium permanganate \[KMn{O_4}\] is a neutral compound having no net charge.
In \[KMn{O_4}\], the oxidation state of \[Mn\] = valency of \[K\] + oxidation state of \[Mn\] + \[4{\text{ }} \times \] valency of \[O\] = \[0\]
$ = + 1 + x + 4 \times ( - 2) = 0$ or $x = + 7$.
In the presence of acid, \[M{n^{7 + }}\] is converted to \[M{n^{2 + }}\] and thus there is a gain of five electrons which is the n-factor.
In an acidic medium, the ionic reaction is shown as
\[Mn{O_4}^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O\]
As electrons are gained by the \[M{n^{7 + }}\] ion so it makes it a strong oxidizing agent and it gets reduced to \[M{n^{2 + }}\]. Hence the n factor is \[5\] for the given transformation, i.e. option D is the correct answer.​

Note: n-factor is also applicable in case of acids and bases. For acids n factor is equal to the number of \[{H^ + }\] ions substituted by one mole of acid during a reaction. Hence the n-factor for acid is not equal to its basicity. For bases n-factor is equal to the number of \[O{H^ - }\] ions substituted by one mole of base during a reaction. Hence n-factor is not equal to its acidity.