Question

If the equivalent weight of S in $S{{O}_{2}}$ is 8, the equivalent weight of S in $S{{O}_{3}}$ is:
(A)- $\dfrac{8\times 2}{3}$
(B)- $\dfrac{8\times 3}{2}$
(C)- $8\times 2\times 3$
(D)- $\dfrac{2\times 3}{8}$

Answer 1

Equivalent weight of an element depends on the valency or oxidation state of the element in a compound. It is given as
$\text{Equivalent weight = }\dfrac{\text{Atomic weight}}{\text{valency}}\text{=}\dfrac{\text{Atomic weight}}{\text{oxidation state}}$

Complete step by step answer:
Let us first find the oxidation state of S in $S{{O}_{2}}$. Let the oxidation number of S be ‘$x$’. Oxidation state of O is -2. $S{{O}_{2}}$ is a neutral compound, so the total charge on it is taken as zero.
$\begin{aligned} & x+2(-2)=0 \\\ & x-4=0 \\\ & x=4 \\\ \end{aligned}$

The oxidation state of S in $S{{O}_{2}}$ is 4.
Given an equivalent weight of S in $S{{O}_{2}}$ is 8. Therefore, we have equivalent weight of S
$\begin{aligned} & \text{Atomic weight}=\text{Equivalent weight (in S}{{\text{O}}_{2}}\text{) }\times \text{oxidation state} \\\ & \text{Atomic weight}=\text{8 }\times 4=32 \\\ \end{aligned}$

Thus, 32 g of S is combined with $2\times 16$ = 32 g of O in $S{{O}_{2}}$.
Let the oxidation number of S in $S{{O}_{3}}$ be ‘$x$’. Taking the oxidation state of O as -2, we get
$\begin{aligned} & x+3(-2)=0 \\\ & x-6=0 \\\ & x=6 \\\ \end{aligned}$

Thus, the oxidation state of S in $S{{O}_{3}}$ is +6.
If 32 g of S is combined with $3\times 16$= 48g of O in $S{{O}_{3}}$. Therefore, we can write
$\text{Atomic weight}=\text{Equivalent weight (in S}{{\text{O}}_{3}}\text{)}\times 6$

Atomic weight of remains the same in both $S{{O}_{2}}$ and $S{{O}_{3}}$, so we can equate the two equations as
$\text{Equivalent weight (in S}{{\text{O}}_{3}}\text{)}\times 6=\text{Equivalent weight (in S}{{\text{O}}_{2}}\text{) }\times \text{4 }$

Given an equivalent weight of S in $S{{O}_{2}}$is 8. Thus,
$\text{Equivalent weight (in S}{{\text{O}}_{3}}\text{)}\times 6=8\text{ }\times \text{4 }$

Therefore, the equivalent weight of S in $S{{O}_{3}}$ is
$\text{ Equivalent weight (in S}{{\text{O}}_{3}}\text{) =}\dfrac{\text{8 }\times 4}{6}\text{ =}\dfrac{\text{8 }\times 2}{3}$
So, the correct answer is “Option A”.

Note: Do not get confused between the options. We can simply calculate the equivalent weight of S in $S{{O}_{3}}$ by dividing the atomic mass of S by the oxidation state of S in $S{{O}_{3}}$, i.e. $\dfrac{32}{6}=\dfrac{16}{3}$.
Now, $\dfrac{16}{3}=\dfrac{8\times 2}{3}$ which is the correct option.