Problem 80 - A Complete Resource Book in Chemistry for JEE Main

Question

The product formed on oxidation of \(\mathrm{I}^{-}\)with \(\mathrm{MnO}_{4}^{-}\)in alkaline medium is (a) \(\mathrm{IO}_{4}^{-}\) (b) \(\mathrm{I}_{2}\) (c) \(\mathrm{IO}^{-}\) (d) \(\mathrm{IO}_{3}^{-}\)

Step-by-Step Solution

Answer

The product is \(\mathrm{IO}_{4}^{-}\). (Option a)

Step 1: Understand the Reaction

In an alkaline medium, the oxidation reaction typically proceeds with the permanganate ion \(\mathrm{MnO}_{4}^{-}\) acting as the oxidizing agent. Here, \(\mathrm{I}^{-}\) is oxidized by \(\mathrm{MnO}_{4}^{-}\) to form a new iodine species.

Step 2: Identify Possible Products

Oxidation of \(\mathrm{I}^{-}\) in an alkaline medium can lead to the formation of different iodine oxyanions. Given the options, we need to predict which iodine oxoanion forms based on the alkaline condition.

Step 3: Consider Permanganate's Role

In a strongly alkaline medium, permanganate ion is reduced to \(\mathrm{MnO}_{2}\). Correspondingly, \(\mathrm{I}^{-}\) is oxidized to an oxyanion of iodine. Typically, in alkaline solutions, the higher oxidation state product, \(\mathrm{IO}_{4}^{-}\), is favored.

Step 4: Check Oxidation State Changes

\(\mathrm{I}^{-}\) initially has an oxidation state of -1. Upon oxidation, it reaches +7 in \(\mathrm{IO}_{4}^{-}\), which is the highest possible oxidation state for iodine. This supports the formation of \(\mathrm{IO}_{4}^{-}\) in the reaction.

Step 5: Verify with the Process

Taken all observations, in an alkaline medium, \(\mathrm{MnO}_{4}^{-}\) oxidizes \(\mathrm{I}^{-}\) to \(\mathrm{IO}_{4}^{-}\):\[ \mathrm{MnO}_4^- + \mathrm{I}^- \rightarrow \mathrm{IO}_4^- + \mathrm{MnO}_2 \]

Key Concepts

Oxidation Reactions

Oxidation reactions are a type of chemical reaction where a substance loses electrons. This results in an increase in the oxidation state of the substance. In our example, iodine ions (\(\mathrm{I}^-\)) are oxidized. Oxidation indicates that these ions lose electrons and transform into a different form with a higher oxidation state.

In an oxidation reaction, there is usually an oxidizing agent. This is the substance that facilitates the electron loss. In our case, the oxidizing agent is the permanganate ion (\(\mathrm{MnO}_4^-\)). During the reaction, \(\mathrm{MnO}_4^-\) gains electrons and is itself reduced to \(\mathrm{MnO}_2\). The key points in oxidation reactions include:

• Electron transfer from one substance to another.
• Substance losing electrons is oxidized.
• Substance gaining electrons is reduced.
• Changes in oxidation states are used to identify what has been oxidized or reduced.

Alkaline Medium Chemistry

In chemistry, the environment, or "medium" in which a reaction takes place, can significantly affect the reaction's course. An alkaline medium means the reaction happens in a basic environment, often containing bases like \(\mathrm{OH}^-\) ions. Such environments can alter both the nature and speed of chemical reactions.

In an alkaline medium, certain ions or molecules are made more reactive. This is due to the presence of high concentrations of \(\mathrm{OH}^-~ \)ions, which can affect the available species in the reaction. The change in reactive conditions can determine which product forms. In our reaction, the presence of \(\mathrm{OH}^-~ \) ions helps drive the oxidation of \(\mathrm{I}^-\) to produce \(\mathrm{IO}_4^-\), promoting a higher oxidation state. Some key points about alkaline medium effects in chemistry include:

• Favors the formation of higher oxidation states of some elements.
• Influences solubility of compounds.
• Alters the balance between reactants and products.
• Can stabilize certain reaction intermediates.

Iodine Species

Iodine can exhibit multiple oxidation states, leading to a variety of iodine species. Each species corresponds to a specific oxidation state of iodine. When iodine exists in a reduced form like \(\mathrm{I}^-\), it can be oxidized to form various oxyanions, depending on the reaction conditions.

In an alkaline medium, the final product of the oxidation of \(\mathrm{I}^-\) is \(\mathrm{IO}_4^-\). \(\mathrm{IO}_4^-\) represents iodine in its highest oxidation state of +7. Unlike \(\mathrm{I}_2\), which is neutral, or \(\mathrm{IO}^-\) and \(\mathrm{IO}_3^-\) which involve lower oxidation states, \(\mathrm{IO}_4^-\) is a stable species in highly alkaline environments. The transformations of iodine are influenced by:

• The oxidizing agents, such as \(\mathrm{MnO}_4^-\)
• The medium of the reaction (acidic, neutral, or alkaline).
• The concentration and presence of certain ions, like \(\mathrm{OH}^-\).

Understanding these species and their behavior helps in predicting the products of iodine-related reactions.