Animated Solution for Chemistry - d and f-Block Elements: $A\stackrel{4{\text{ KOH, O}}_2}{\to }\underset{\text{(Green)}}{2B}+2H_{2}O$ $3B\stackrel{4\text{ HCl}}{\to }\underset{\text{(Purple)}}{2C}+Mn{O}_2+2H_{2}O$ $2C\stackrel{H_{2}O,KI}{\to }2A+2KOH+D$ In the above sequence of reactions, $A$ and $D$, respectively, are

The chemistry of d-block elements is a fascinating journey of vibrant colors and shifting oxidation states. In this problem, we are presented with a sequence of reactions that perfectly encapsulates the industrial preparation and the versatile oxidizing nature of one of the most famous compounds in chemistry: potassium permanganate ($KMn{O}_4$).

Our journey begins with compound A, a dark, unassuming solid that reacts with potassium hydroxide ($KOH$) and oxygen ($O_2$) to yield a striking green compound B. This is the classic first step in the preparation of potassium permanganate.

Compound A is pyrolusite ore, chemically known as manganese dioxide ($Mn{O}_2$). When fused with an alkali like $KOH$ in the presence of an oxidizing agent like atmospheric oxygen, the manganese is oxidized from a $+4$ state to a $+6$ state. The result is potassium manganate ($K_{2}Mn{O}_4$), which is responsible for the characteristic green color.

The reaction is given by:

Next, the green potassium manganate (B) is treated with an acid, such as hydrochloric acid ($HCl$). The manganate ion ($Mn{O}_4^{2-}$) is stable only in highly alkaline solutions. In acidic or even neutral media, it undergoes a fascinating process called disproportionation.

Disproportionation is a specific type of redox reaction where an element in a single oxidation state is simultaneously oxidized and reduced. Here, the manganese in the $+6$ state splits its identity: some of it is oxidized to the $+7$ state, forming the deep purple permanganate ion ($Mn{O}_4^{-}$), while the rest is reduced back to the $+4$ state, precipitating as manganese dioxide ($Mn{O}_2$).

The balanced equation for this transformation is:

Thus, compound C is the iconic purple potassium permanganate ($KMn{O}_4$).

The final step showcases the oxidizing prowess of potassium permanganate. Compound C ($KMn{O}_4$) reacts with potassium iodide ($KI$) in the presence of water ($H_{2}O$). The presence of water indicates a neutral or faintly alkaline medium.

This is where many students fall into a trap! Potassium permanganate is a powerful oxidizing agent, but its behavior changes drastically depending on the pH of the medium. - In a strongly acidic medium, $KMn{O}_4$ oxidizes iodide ($I^{-}$) to iodine gas ($I_2$). - However, in a neutral or faintly alkaline medium, the oxidation is much more profound. The iodide ion is oxidized all the way up to the iodate ion ($I{O}_3^{-}$).

Simultaneously, the permanganate ion is reduced to manganese dioxide ($Mn{O}_2$), which is our original compound A.

The reaction proceeds as follows:

Therefore, the new compound D formed in this reaction is potassium iodate ($KI{O}_3$).

By carefully tracing the oxidation states and remembering the crucial role of the reaction medium, we have successfully decoded the entire sequence. Compound A is $Mn{O}_2$, and compound D is $KI{O}_3$. This elegant cycle not only highlights the synthesis of $KMn{O}_4$ but also its nuanced chemical behavior.