Question

The oxidation state of sulphur in Caro’s and Marshall’s acid are:
\[
  A.\;\; + 6,{\text{ }} + 6 \\
  B.\;\; + 4,{\text{ }} + 6 \\
  C.\;\; + 6,{\text{ }} - 6 \\
  D.\;\; + 6,{\text{ }} + 4 \\
 \]

Answer 1

Hint: We have to draw the structure of both Caro’s and Marshall’s acid and then only we can check the oxidation state in which sulphur is connected. We always remember that when both sulphur and oxygen are together sulphur always takes a positive oxidation state. For example, in the case, of \[S{O_2}\], the oxidation state of sulphur is +4.

Complete step by step solution:
Case 1: We know the chemical formula for Caro’s acid is \[{H_2}{O_5}S\] which also known as Peroxymonosulfuric acid and we can draw the structure as

Now we are going to calculate the oxidation number of Caro’s acid by oxidation calculation.
We can write the equation as
$(2 \times H) + (5 \times O) + (1 \times S) = 0$
We know that the oxidation number of hydrogen and oxygen is +1 and -2 respectively.
If we consider the oxidation number of sulphur as X. therefore, we can write
$(2 \times H) + (5 \times O) + (1 \times X) = 0$
Where, X is the oxidation number of sulphur atoms.
Now, we substitute the oxidation number of hydrogen and sulphur in this equation.
Therefore,
 $0 = (2 \times ( + 1)) + (5 \times ( - 2)) + (1 \times X)$
$0 = 2 + ( - 10) + X$
$0 = - 8 + X$
$X = - 8$
But the S cannot exceed the oxidation number is +6. Since, S has only 6 electrons in the valence shell. This exceptional value of oxidation state for sulphur in H2SO5 is due to the peroxy linkage shown by two atoms.
Therefore,
$0 = (2 \times ( + 1)) + (1 \times X) + (3 \times ( - 2)) + (2 \times ( - 1)[{\text{for peroxy linkage which has - 1 oxidation value }}])$$0 = 2 + X + ( - 6) + ( - 2)$
$0 = 2 + X + ( - 8)$
$X = + 6$
Therefore, the oxidation of sulphur in \[{H_2}{O_5}S\] is +6.
Case 2: Also, the chemical formula for Marshall’s acid is \[{H_2}{S_2}{O_8}\] which can be drawn as

Calculate the oxidation of a number of Marshall’s acids by oxidation calculation.
So, here also we can see that both the sulphur atoms are connected with oxygens.
And also both these acids have peroxy links. Therefore,
$0 = (2 \times H) + (2 \times S) + (6 \times O) + (2 \times O{\text{ }}[{\text{for peroxy linkage which has - 1 oxidation value }}])$
We know that the oxidation number of hydrogen and oxygen is +1 and -2 respectively.
$0 = (2 \times ( + 1)) + (2 \times X) + (6 \times ( - 2)) + (2 \times ( - 1))$
$0 = (2) + (2X) + ( - 12) + ( - 2)$
$0 = (2X) + ( - 12)$
$0 = (2X) - 12$
$ + 12 = (2X)$
\[ + \dfrac{{12}}{2} = X\]
$X = + 6$
Therefore, the oxidation number of sulphur in Marshall’s acid \[{H_2}{S_2}{O_8}\]is +6

We can conclude that the answer will be A. +6, +6

Additional Information:
We must understand that the sulphur has many oxidation states which are -2, −1, 0, +1, +2, +3, +4, +5, and +6. Out of these, the most common one is -2.
When we compare to oxygen, sulphur can also accept electrons which leads to its positive oxidation states.

Note: We must know that the peroxy linkage is \[O - O\] linkage. Therefore, both the oxygen has -1 oxidation state. Oxygen also attains other oxidative states such as +2 and +1, but the most common is -2. Example of +2 is \[O{F_2}\] since fluorine is more electronegative the oxygen has to attain +2 oxidative state and example of +1 are peroxides (\[{H_2}{O_2}\]).