Question

Find the difference in oxidation numbers of sulphur in the $(S{O}_4{)}^{2-}$ ion and $S{O}_3$ compound.

• A. 2
• B. 0.0
• C. -2
• D. 1

Answer 1

The correct option is B 0.0

Let the oxidation number of sulphur be $x$ in $(S{O}_4{)}^{2-}$ and $y$ in $S{O}_3$.

Now the sulphate ion is given to us has a negative charge of -2 on it. This means that the sum of the oxidation numbers of all the elements in this ion is equal to -2 and not zero. The oxidation number of oxygen is -2. So let's calculate the value of $x$.

$x+4\times (-2)=-2$

$x-8=-2$

$x=+6$

Now let's calculate the oxidation number for sulphur in $S{O}_3$.

y + 3 times (-2) = 0

y - 6 = 0

y = +6

So the oxidation number of sulphur in $(S{O}_4{)}^{2-}$ ion is +6.

Hence the difference of oxidation number of sulphur in $(S{O}_4{)}^{2-}$ and $S{O}_3$ is +6 - (+6) = 0

The answer is 0.