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# n-factor of ${{\text{P}}_4}{{\text{O}}_6}$ in the change, ${{\text{P}}_4}{{\text{O}}_6} \to {{\text{H}}_3}{\text{P}}{{\text{O}}_4}$

Last updated date: 20th Jun 2024
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Hint:n-factor is the change in oxidation number per molecule or ion. It depends on the type of reaction. Reactants react or products are formed in their equal number of equivalents.

Complete step by step answer:n-factor can be calculated for three different types of reactions:
Acid-base neutralization
Redox reactions
Precipitation and double decomposition reactions.
There is only a chance of redox reaction to occur. Oxidation state has also changed. So here only one atom changes its oxidation number and transforms to three different compounds with different oxidation numbers. There is a chance of either oxidation or reduction.
There are three cases in which different formulae are used to calculate the n-factor. They are:
One atom changes its oxidation number and converts into one compound only.
One atom changes its oxidation number and converts into two compounds of the same oxidation state.
One atom changes its oxidation number and converts into two or more compounds of different oxidation states.
Here balancing the chemical reaction is necessary. Therefore the complete balanced chemical equation is given below:
$\mathop {6{{\text{P}}_4}{{\text{O}}_6}}\limits^{ + 3} + 24{{\text{H}}_2}{\text{O}} \to 8\mathop {\text{P}}\limits^0 + 15\mathop {{{\text{H}}_3}{\text{P}}{{\text{O}}_4}}\limits^{ + 5} + \mathop {{\text{P}}{{\text{H}}_3}}\limits^{ - 3}$
$\mathop {\text{P}}\limits^{ + 3}$ has changed to $\mathop {\text{P}}\limits^0 ,\mathop {\text{P}}\limits^{ + 5} ,\mathop {\text{P}}\limits^{ - 3}$
The reactant side has $24{\text{P}}$ atoms and the product side has $8 + 15 + 1 = 24{\text{P}}$ atoms. Thus the reactant side and product side have equal numbers of phosphorus atoms.
n-factor is the change in oxidation per molecule. We should calculate the effect per molecule.
Therefore n-factor
$= \dfrac{{8 \times \left| {3 - 0} \right| + 15 \times \left| {3 - 5} \right| + 1 \times \left| {3 - \left( { - 3} \right)} \right|}}{6} = \dfrac{{8 \times 3 + 15 \times 2 + 1 \times 6}}{6} = \dfrac{{24 + 30 + 6}}{6} = \dfrac{{60}}{6} = 10$
i.e., the number of atoms in each phosphorus atom in the product side is multiplied with the change in the oxidation state and then it is divided by the number of phosphorus atoms in the reactant side.

Note: In acids, n-factor is termed for the number of protons displaced by $1{\text{mole}}$ of acid. Since ${{\text{P}}_4}{{\text{O}}_6}$ does not have any proton in it, this does not belong to acid-base neutralization. Double decomposition also does not occur here.