Question

The equivalent weight of $MnS{{O}_{4}}$is half its molecular weight when it is converted to:
(A)$M{{n}_{2}}{{O}_{3}}$
(B)$Mn{{O}_{2}}$
(C)$MnO_{4}^{-}$
(D)$MnO_{4}^{2-}$

Answer 1

To solve this question first we should know how we can calculate the equivalent weight relating the terms molecular weight and the n- factor.
The equation with which we can calculate the equivalent weight is as follows:
$\text{Equivalent}\,\text{weight}\,\text{=}\,\dfrac{\text{Molecular}\,\text{weight}}{\text{ valency factor}}$

Complete step-by-step answer: In the question, it is given that, from the given compounds as options which will have its molecular weight half the time with the equivalent weight of$MnS{{O}_{4}}$.
From the lower classes, we are very familiar with the terms like molecular weight and equivalent weight. Now let's briefly discuss these terms for better understanding and to correlate the two terms to find the answer to the above questions.
Equivalent weight: it is also called the combining quantity. The equivalent weight is the quantity of the substance that reacts or combines with the arbitrarily fixed value of the other substance.
Molecular weight: It is the total sum of masses of the atoms present in the molecule.
Now we have to know how we can find the equivalent mass if the molecular mass is given.
When the molecular mass of a compound is divided with the n-factor or the valency factor of the compound then it will yield the equivalent weight of the compound.
$\text{Equivalent}\,\text{weight}\,\text{=}\,\dfrac{\text{Molecular}\,\text{weight}}{\text{ valency factor}}$
So here when $MnS{{O}_{4}}$is converted to a compound, then the compound obtained will have molecular weight half the weight of the equivalent weight of$MnS{{O}_{4}}$.
Therefore, $\text{Equivalent}\,\text{weight}\,\text{=}\,\dfrac{\text{Molecular}\,\text{weight}}{\text{2}}$
The n-factor involved is 2 i.e. the change in oxidation number will be 2 or we can say that the number of electrons involved in the reaction will be 2.
In$MnS{{O}_{4}}$, Mn has an oxidation number +2. So in the converted compound, the oxidation state of Mn should be +2.
Now let's check the oxidation state of Mn in the given options.
In$M{{n}_{2}}{{O}_{3}}$, the oxidation state of Mn is +3.
In$Mn{{O}_{2}}$, the oxidation number of Mn is +4.
In$MnO_{4}^{-}$, the oxidation number of Mn is +7.
In$MnO_{4}^{2-}$, the oxidation number of Mn is +6.

So the correct answer for the above question is option (B).

Note: The valency factor or n-factor is the change in the oxidation state or oxidation number per molecule in a chemical reaction.
We can find the oxidation state of the atom of consideration by assigning it with any variable and calculate the valence number of other atoms and do the appropriate mathematical calculation.
For example, let's calculate the oxidation state of Mn in$M{{n}_{2}}{{O}_{3}}$
Assign a value x for Mn and the valence electrons of O is -2.
So, $\overset{x}{\mathop{M{{n}_{2}}}}\,\overset{-2}{\mathop{{{O}_{3}}}}\,$
$2x+3\left( -2 \right)=0$
$2x=6$
And $x=3$