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) IO (d) \(\mathrm{IO}_{3}^{-}\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Reaction

We are dealing with the oxidation of iodide ions (\(\mathrm{I}^{-}\)) using permanganate ions (\(\mathrm{MnO}_{4}^{-}\)) in an alkaline medium. The permanganate ion is a strong oxidizing agent.

Step 2: Identify the Oxidation States Involved

Iodide (\(\mathrm{I}^{-}\)) has an oxidation state of -1. In the oxidation reaction occurring in an alkaline medium with \(\mathrm{MnO}_{4}^{-}\), it will be oxidized to a higher oxidation state.

Step 3: Determine Possible Oxidation Products

In an alkaline medium, permanganate ions can oxidize iodide ions to \(\mathrm{IO}_{4}^{-}\) (where iodine is in the +7 oxidation state). Other potential products (\(\mathrm{I}_2\), \(\mathrm{IO}\), \(\mathrm{IO}_{3}^{-}\)) do not reach the highest oxidation state possible for iodine.

Step 4: Select the Correct Answer

Given the conditions (oxidation by \(\mathrm{MnO}_{4}^{-}\) in an alkaline medium), the product will be \(\mathrm{IO}_{4}^{-}\). This is consistent with the behavior of permanganate as a strong oxidizer converting iodide to its highest oxidation state.

Key Concepts

oxidation of iodide

The oxidation of iodide involves the transformation of \(\mathrm{I}^{-}\), or iodide ions, into a compound where iodine has a higher oxidation state. Oxidation is essentially the process where a molecule, atom, or ion loses electrons. In the context of iodide ions, each \(\mathrm{I}^{-}\) ion can undergo this process under certain conditions which result in an increase in the oxidation state of iodine. To put it simply:

• Iodide has an initial oxidation state of -1 because it carries an extra electron.
• Through oxidation, it can lose this electron and be converted to various iodine-containing species, where iodine displays higher oxidation states.
• This change depends on the reaction conditions and the strength of the oxidizing agent used.

It means that not only do the conditions have to be right, but there also must be a sufficiently strong agent present to promote such a transfer of electrons.

oxidizing agents

An oxidizing agent is a substance that can bring about oxidation by accepting electrons. These agents are critical players in redox reactions like the oxidation of iodide ions. In these reactions, oxidizing agents themselves are reduced since they gain electrons. Let's break it down:

• The oxidizing agent accepts electrons from another substance, thereby reducing its own oxidation state.
• Common oxidizing agents include substances like oxygen, ozone, and halogens, along with ions such as \(\mathrm{MnO}_{4}^{-}\) and \(\mathrm{CrO}_4^{2-}\).

When iodide ions are oxidized by a strong oxidizing agent like the permanganate ion, they serve as an electron source that gets accepted by the agent. The choice and strength of the oxidizing agent greatly impact the end product of a redox reaction. Essentially, it dictates how far the oxidation process will go, as seen with iodide ions being oxidized to \(\mathrm{IO}_{4}^{-}\) in alkaline mediums.

permanganate ion oxidation

Permanganate ions, \(\mathrm{MnO}_{4}^{-}\), are well-known for their potent oxidizing power, particularly in alkaline conditions. Here’s why they are so effective:

• Permanganate ions can accept electrons easily and undergo reduction. Specifically, the manganese atom in \(\mathrm{MnO}_{4}^{-}\) can be reduced from the +7 oxidation state to lower states, primarily \(\mathrm{MnO}_{2}\) in alkaline media.
• In oxidative actions like the one involving iodide, permanganate ions facilitate significant electron transfer, leading to a substantial change in the chemical state of the substance being oxidized.

In the specific context of oxidizing iodide, permanganate ions oxidize iodine from the -1 state to the +7 state, forming \(\mathrm{IO}_{4}^{-}\). These ions are vital because their high oxidation state allows them to 'pull' electrons from various substances, inducing dramatic chemical changes.

iodine oxidation states

Iodine is unique as it can exhibit multiple oxidation states, ranging from -1 to +7. Understanding these states is key to comprehending iodine’s chemical behavior during oxidation reactions. Let’s take a look:

• -1: Iodine starts as \(\mathrm{I}^{-}\) when it gains an electron, suited for reactions in reducing environments.
• 0: When iodine forms molecules like \(\mathrm{I}_2\), each iodine is neutral.
• +1 to +7: High oxidation states occur when iodine is part of compounds such as \(\mathrm{IO}\), \(\mathrm{IO}_3^{-}\), and \(\mathrm{IO}_4^{-}\). Here, iodine shares or loses electrons progressively, reaching the highest state in \(\mathrm{IO}_4^{-}\).

The flexibility in the oxidation states allows iodine to be part of various chemical reactions, influencing its reactivity. When oxidized by strong agents like permanganate ions, iodine moves towards its higher oxidation states, showing its versatile chemistry influenced strongly by the medium and oxidizing conditions.