“n-factor” of $M{n_2}{O_7}$ in the change, $2M{n_2}{O_7} \to 4Mn{O_2} + 3{O_2}$, is:

Hint: In the given reaction, manganese (VII) oxide is decomposed to give manganese oxide and oxygen. In this reaction reduction is taking place as electrons are accepted by the reactant to form the product and change in oxidation state is seen.

Complete step by step answer:
The redox reaction is defined as the reaction where oxidation and reduction reaction takes place simultaneously where transfer of electrons takes place.
When the electrons are released during the reaction to form the product, then it is oxidation. When the electrons are accepted during the reaction to form the product, then it is reduction.
The given reaction is shown below.
$2M{n_2}{O_7} \to 4Mn{O_2} + 3{O_2}$
In this reaction, two manganese (VII) oxide decompose to give four moles of manganese oxide and three mole of oxygen.
In this reaction decomposition takes place where one reactant compound broke to two or more products.
In this reaction the oxidation state of manganese changes from +7 in manganese (VII) oxide to +4 in manganese oxide. In this reaction reduction is taking place.
The n-factor is defined as the conversion factor by which molecular weight is divided to calculate the equivalent weight.
The change in oxidation state is 7 – 4 =3.
In $M{n_2}{O_7}$, two manganese atoms are present so the n-factor is calculated as shown below.
$\Rightarrow n - factor = 2 \times 3$
$\Rightarrow n - factor = 6$
Therefore, the n-factor of $M{n_2}{O_7}$ is 6.

Note:
During oxidation the oxidation state increases from the reactant to product. During reduction the oxidation state decreases from the reactant to product. In the given reaction oxidation state is decreased as reduction is taking place.