Question

If the molecular weight of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$and ${{I}_{2}}$are ${{M}_{1}}$and ${{M}_{2}}$respectively then what will be the equivalent weight of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$and ${{I}_{2}}$in the following reaction?
$\text{2}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}^{\text{2}-}\text{ + }{{\text{I}}_{2}}\text{ }\xrightarrow{{}}\text{ }{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}^{\text{2}-}\text{ + 2}{{\text{I}}^{-}}$
A) ${{M}_{1}}$, ${{M}_{2}}$
B) ${{M}_{1}}$, $\dfrac{{{M}_{2}}}{2}$
C) ${2{M}_{1}}$, ${{M}_{2}}$
D) ${{M}_{1}}$, ${2{M}_{2}}$

Answer 1

Hint: To answer this question, we should be having an idea about how to find the oxidation number of the required element in the compound it is present. In this case, it is S. So, at first we should be beginning to answer this question by finding the oxidation number of S. Once we know the oxidation number, we can move on to find the equivalent weight of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$and ${{I}_{2}}$as required.

Complete step by step answer:
From the given problem, we need to find the oxidation state of each compound which can be seen in the following equation. In the given equation, the oxidation states of the compounds are given below the respective compounds.
 $\begin{align}
  & \text{2}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}^{\text{2}-}\text{ + }{{\text{I}}_{2}}\text{ }\xrightarrow{{}}\text{ }{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}^{\text{2}-}\text{ + 2}{{\text{I}}^{-}} \\
 & +2\text{ }-2\text{ 0 +2}\text{.5 }-2\text{ }-1 \\
\end{align}$
Oxidation number of S in $\text{2}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}^{\text{2}-}$ is +2 as can be obtained from below.
$\begin{align}
  & 2\text{x }-\text{ }6\text{ = }-2 \\
 & \therefore \text{ x = +2} \\
\end{align}$
Oxidation number of S in ${{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}^{\text{2}-}$ is +2.5 as can be obtained from below.
$\begin{align}
  & \text{4x }-\text{ 12 = }-2 \\
 & \therefore \text{ x = +2}\text{.5} \\
\end{align}$

We know that, the change in Oxidation Number of S per mole can be written as
$0.5\text{ }\times \text{ 2 = 1}$
Similarly, change in oxidation change of Iodine (I) per mole is
$1\text{ }\times \text{ 2 = 2}$
Therefore, equivalent mass of the compound $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ is
$\dfrac{{{\text{M}}_{\text{1}}}}{1}\text{ = }{{\text{M}}_{\text{1}}}$
And equivalent mass of ${{\text{I}}_{\text{2}}}$is $\dfrac{{{\text{M}}_{2}}}{2}$.
Therefore, the correct answer is Option B.

Note: As we have been mentioning the concept of equivalent weight in the answer, we should be defining it as well. Equivalent weight is defined as the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance. Equivalent weight has the dimensions and units of mass, unlike atomic weight which is dimensionless.