JEE MAIN - Chemistry (2024 - 27th January Evening Shift - No. 13)
Explanation
The ability of a species to function as an oxidizing agent depends on its ability to gain electrons (be reduced). For a species to be an oxidizing agent, it must contain an element that has a positive oxidation state and is capable of being reduced (gain electrons), thereby oxidizing another species.
Let's look at each of the options :
Option A: $$\mathrm{SO}_4^{2-}$$ (sulfate ion) has sulfur in a +6 oxidation state. Sulfate is the fully oxidized form of sulfur in aqueous solution, and while it can undergo certain reductions under specific conditions (to form, for example, $\mathrm{SO}_3^{2-}$ or sulfite), it is generally stable and not commonly considered a strong oxidizing agent.
Option B: $$\mathrm{MnO}_4^{-}$$ (permanganate ion) has manganese in a +7 oxidation state. This ion is a very strong oxidizing agent and is well known for its ability to oxidize a wide range of organic and inorganic substances, with manganese being reduced to a lower oxidation state (often to +2 as $\mathrm{Mn}^{2+}$ in acidic solution).
Option C: $$\mathrm{N}^{3-}$$ is the nitride ion, with nitrogen in a -3 oxidation state. Since nitrogen is in its lowest oxidation state, it cannot be further reduced (gain more electrons), and hence cannot act as an oxidizing agent. In fact, it would more likely act as a reducing agent because it has room to be oxidized (lose electrons).
Option D: $$\mathrm{BrO}_3^{-}$$ (bromate ion) has bromine in a +5 oxidation state. This ion can act as an oxidizing agent because the bromine can be reduced to a lower oxidation state (often to -1 as $\mathrm{Br}^{-}$).
Therefore, the species that cannot function as an oxidizing agent due to its inability to undergo reduction is Option C, the $$\mathrm{N}^{3-}$$ ion.
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