Question

Stronger reducing agent
(1) $S{O_3}$
(2) $S{O_2}$
(3) $Se{O_3}$
(4) $Te{O_2}$

Answer 1

All the given oxides in options are formed by the elements of group 16. Reducing agent is the one which reduces the other elements and itself undergoes oxidation. Thus, a stronger reducing agent will be the one who will readily lose its electrons to attain higher oxidation state. You must also know that stability of higher oxidation state i.e. 6 decreases down the group 16 and stability of +4 oxidation state increases.

Complete step by step answer:
Oxygen (O), sulphur (S), selenium (Se), tellurium (Te) and polonium (Po) constitute group 16 of the periodic table. All these elements form oxides of the type $E{O_2}$ and $E{O_3}$ where E= S, Se, Te or Po.
On moving down the group 16, stability of higher oxidation state i.e. 6 decreases down the group and stability of lower oxidation state i.e., +4 increases down the group. This is attributed to inert pair effects down the group. Now, we know that reducing agent is the one which reduces the other elements and itself undergoes oxidation. Thus, a reducing agent loses electrons and consequently, attains a higher oxidation state. Let us discuss each given oxide one by one:

In $S{O_3}$, sulphur is in +6 oxidation state, which is its highest oxidation state after losing its outer 6 electrons. It will not further lose its inner electrons. Thus, it will not behave as a reducing agent. Similarly, in $Se{O_3}$, Se is in the highest +6 oxidation state and thus, will not be a reducing agent.

In $S{O_2}$ , sulphur is in +4 oxidation state and is the lighter element of group 16. But we know that the higher oxidation state (+6) is stable is the lighter elements of group 16. Thus, sulphur will try to attain +6 oxidation state which is more stable than +4. Hence, $S{O_2}$ will behave as a strong reducing agent after losing its 2 electrons. Now in $Te{O_2}$, tellurium (Te) is in +4 oxidation state and is the heavier element of the group. Down the group, stability of +4 oxidation state decreases due to inner pair effect. Thus, Te in $Te{O_2}$ will not attain higher oxidation state and hence, will not be a reducing agent.
So, the correct answer is “Option 2”.

Note: Heavier elements of group 16 are more stable in lower oxidation states that is, +4 (two less than +6 oxidation states in lighter elements) due to inert pair effect. The reluctance or inertness of two electrons of the outermost s-orbital towards the bond formation is called inner pair effect.