Question

The n- factor of ${{M}_{0.25}}O$ in the reaction, ${{M}_{0.25}}O\to MO$ is
a. \[1.5\]
b. \[4.5\]
c. \[3\]
d. \[2.5\]

Answer 1

Hint: For bases, the n factor is defined as the number of $O{{H}^{-}}$ (hydroxide) ions replaced by \[1\] mole of base in a given reaction.
\[n\]- Factor is not equal to its acidity.

Complete step by step answer:
Redox reaction can be defined as a chemical reaction in which electron’ are transferred between two reactant participating in it .This transfer of electrons can be determine by the change in oxidation state of reacting specie .In redox reaction one specie undergo oxidation another undergo reduction .the specie is said to be oxidized when electron is removed from it while the specie to which electron is added is said to be reduced .Example of redox reaction is formation of rust , burning fuel etc.
The process of oxidation and reduction occur simultaneously and cannot happen independently of one another similar to acid –base reaction.
\[N\] factor for element is valency of the element and for acid is the number of acidic hydronium ions that the molecule of acid would give dissolve in solvent, n factor for salt is the total charge on cation or anion .
Coming to the solution part: ${{M}_{0.25}}O\to MO$
Oxidation state of \[M\] in ${{M}_{0.25}}O$ can be given as $=+2/0.25$
Oxidation state of \[M\] in $MO$ can be given as \[=+2\].
Then n factor of ${{M}_{0.25}}O=$$\left( +2/0.25 \right)-\left( +2 \right)=6$
Hence the n factor will be \[1.5\] .
Hence, option a is correct.

Note:
Mathematically, In a redox reaction, n factor is defined as the change in their oxidation number or the change in their reduction number on the both right and left side of the chemical reaction. But for a salt, n factor is defined as the number of positive ions in the salt or the number of negative ions of the salt.