Question

Which of the following can act as oxidising as well as reducing agent?
A. ${H_2}{O_2}$
B. $S{O_3}$
C. ${H_2}S{O_4}$
D. $HN{O_3}$

Answer 1

To solve this question, we should have knowledge about the stable and possible oxidation states of an element. We will find the oxidation states of each of the constituents in each compound. Only those compounds in which the constituents or one of the constituents is present in their intermediate oxidation state can oxidise as well as reduce itself and act as both oxidising as well as reducing agent.

Complete step by step answer:
In option A (${H_2}{O_2}$), we know that the stable oxidation state of hydrogen with electronegative elements is +1 and with electropositive elements like most of the metals is -1. Oxygen is mostly found at its stable -2 oxidation state and sometimes in -1 oxidation state. In this compound, if oxygen is present at -2 oxidation state, it would imply that hydrogen is in +2 oxidation state which is not possible. Thus, we can conclude that oxygen is present in its less stable but intermediate -1 oxidation state and hydrogen is in +1 oxidation state.
In option B ($S{O_3}$), both the elements are from group 16, but oxygen being more electronegative than sulphur will be present at -2 oxidation state. This implies that sulphur will be present in +6 oxidation state which is its highest oxidation state.
In option C (${H_2}S{O_4}$), oxygen being the most electronegative element will be present in -2 oxidation state and hydrogen being the most electropositive element will be in +1 oxidation state. This implies that sulphur will be present in +6 oxidation state which is its highest oxidation state.
In option D ($HN{O_3}$), oxygen being electronegative will be in -2 oxidation state while hydrogen being electropositive will be in +1 oxidation state. This implies that nitrogen will be at +5 oxidation state, highest for a group 15 element.
$\therefore$ It can be concluded that among the four options, only Option A (${H_2}{O_2}$) has one of its constituent, i.e. oxygen present at intermediate oxidation state and can be converted to ${O^{2 - }}$ on reduction and ${O_2}$ on oxidation.

So, the correct answer is Option A .

Note:
Though oxygen is mostly present as -2 oxidation state, in case of peroxides, it is present in -1 oxidation state. The ability to act as both oxidising and reducing agents are very common among d and f-block elements due to their variable oxidation state and presence of d and f-orbitals. Most elements like vanadium, chromium, manganese, iron, cobalt, lanthanides etc. can all form compounds in their intermediate oxidation states and can act as oxidising as well as reducing agents.