Question

In which of the following reactions of $ {H_2}S{O_4} $ , sulphur retains its oxidation state?
(A) Reaction of dil. $ {H_2}S{O_4} $ with sulphur
(B) Reaction of conc. $ {H_2}S{O_4} $ with sulphur
(C) Reaction of conc. $ {H_2}S{O_4} $ with glucose
(D) Reaction of dil. $ {H_2}S{O_4} $ with zinc

Answer 1

Hint :The cumulative number of electrons that an atom absorbs or loses in order to form a chemical bond with another atom is referred to as the oxidation number, also known as the oxidation state.
The word "oxidation" was coined to describe the reaction of a product with oxygen. Far further, it was discovered that the compound loses electrons when oxidised, and the meaning was expanded to include all reactions in which the loss of electrons take place, regardless of whether oxygen is present.

Complete Step By Step Answer:
Sulphuric acid which has a molecular formula $ {H_2}S{O_4} $ , is a mineral acid which is made up of three elements: sulphur, oxygen, and hydrogen. It is commonly used in the production of chemicals, such as hydrochloric acid, pigments, dyes, nitric acid, sulphate salts and drugs. The oxidation of sulphur in $ {H_2}S{O_4} $ is $ + 6 $ .
The above question has multiple correct answers:
As dilute sulphuric acid interacts with Sulphur, the following happens:
 $ 2{H_2}S{O_4} + S \to 3S{O_2} + 2{H_2}O $
The oxidation state of sulphur changes from $ + 6 $ to $ + 4 $ in $ {H_2}S{O_4} $ . So, the oxidation state differs from sulphur , and it is a reduction which takes place. Similarly, when sulphur reacts with concentrated $ {H_2}S{O_4} $ , the oxidation state of sulphur in $ {H_2}S{O_4} $ differs.
As Conc. $ {H_2}S{O_4} $ interacts with glucose, the following reaction takes place:
 $ 6{H_2}S{O_4} + {C_6}{H_{12}}{O_6} \to 6C + 6{H_2}S{O_4} + {H_2}O $
In this case, $ {H_2}S{O_4} $ participates indirectly and is obtained as a combination of water and acid. In this combination, the oxidation state remains the same.
As dil. $ {H_2}S{O_4} $ interacts with Zinc, the following reaction takes place:
 $ Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2} $
Since, the oxidation state of sulphur in $ ZnS{O_4} $ is $ + 6 $ . The oxidation state of sulphur remains the same.
Thus, the correct options are options $ C $ and $ D $ .

Note :
The loss of electrons or an increase in the oxidation state of an atom, an ion, or any atoms in a molecule is referred to as oxidation. Reduction is the process by which an atom, an ion, or some atoms in a molecule gain electrons or decrease in their oxidation state.