Problem 30 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Complete the following equations (no balancing is needed) (i) \(\mathrm{HCO}_{3}^{-}+\mathrm{Al}^{3+} \rightarrow \mathrm{Al}(\mathrm{OH})_{3}+\ldots\) [1981-1 Mark] (ii) \(\mathrm{AlBr}_{3}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{3} \mathrm{PO}_{4}\) \(\rightarrow \mathrm{K}_{3} \mathrm{PO}_{4}+\mathrm{AlPO}_{4}+\mathrm{H}_{2} \mathrm{O}+\ldots+\ldots\)

Step-by-Step Solution

Answer

(i) Add \(\mathrm{CO}_2\); (ii) Add \(\mathrm{K}_2\mathrm{CrO}_4\) and \(\mathrm{Br}_2\).

Step 1: Understanding the Components of Equation (i)

The first part of the equation involves bicarbonate ion \(\mathrm{HCO}_3^-\) and an aluminum ion \(\mathrm{Al}^{3+}\). The products we have so far are aluminum hydroxide \(\mathrm{Al(OH)}_3\), which indicates that hydroxide ions have likely been produced from the bicarbonate during the reaction. To complete the reaction, recognize that \(\mathrm{H_2CO_3}\) will be formed, which can decompose to water \(\mathrm{H_2O}\) and carbon dioxide \(\mathrm{CO}_2\).

Step 2: Completing Equation (i)

Since \(\mathrm{Al(OH)}_3\) forms, carbonate ions must be removed from the bicarbonate. Reactant bicarbonate \(\mathrm{HCO}_{3}^{-}\) transforms to form \(\mathrm{H_2O}\) and \(\mathrm{CO}_2\). Therefore, the complete equation becomes: \[\mathrm{HCO}_{3}^{-}+\mathrm{Al}^{3+}\rightarrow\mathrm{Al(OH)}_{3}+\mathrm{CO}_2\].

Step 3: Understanding the Components of Equation (ii)

In this reaction, aluminum bromide interacts with potassium dichromate \(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7\), and phosphoric acid \(\mathrm{H}_3\mathrm{PO}_4\). The given products \(\mathrm{K}_3\mathrm{PO}_4\), \(\mathrm{AlPO}_4\), and \(\mathrm{H}_2\mathrm{O}\) suggest various ions are precipitating and reducing agents are likely involved. Dichromate thus converts to a chromium-containing product upon reduction.

Step 4: Completing Equation (ii)

The \(\mathrm{K}_3\mathrm{PO}_4\) suggests that potassium chromate \(\mathrm{K}_2\mathrm{CrO}_4\) would be a logical remaining component, resulting from chromium reduction. Thus the completed equation is: \[\mathrm{AlBr}_3+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{3} \mathrm{PO}_{4}\rightarrow \mathrm{K}_3 \mathrm{PO}_4+\mathrm{AlPO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{K}_2 \mathrm{CrO}_4+\mathrm{Br}_2\].

Key Concepts

Bicarbonate Reactions

A bicarbonate reaction involves bicarbonate ions, typically represented as \( \text{HCO}_3^- \). When these ions react with other elements or compounds, they often produce hydroxide ions. In the chemical equation provided, \( \text{HCO}_3^- \) reacts with aluminum ions \( \text{Al}^{3+} \). Here, the bicarbonate ion will dissociate and lead to the formation of aluminum hydroxide \( \text{Al(OH)}_3 \).

• Hydroxide ions from the bicarbonate help in forming aluminum hydroxide.
• This usually results in the liberation of water \( \text{H}_2\text{O} \) and carbon dioxide \( \text{CO}_2 \).

Let's break it into simpler terms. Bicarbonate ion is naturally unstable; it wants to break down. Upon reacting with the positively charged aluminum, it tries to stabilize by forming \( \text{Al(OH)}_3 \). The leftover components, water and carbon dioxide, come from the carbonate group's breakdown process. It's like a rearranging of puzzle pieces to achieve balance. Byproducts like \( \text{CO}_2 \) help indicate the reaction's nature.

Dichromate Reduction

Dichromate reduction involves the conversion of the dichromate ion \( \text{Cr}_2\text{O}_7^{2-} \) into a different chromium species, typically through electron gain. In our case, this concept is vital in reaching the final products of the reaction with aluminum bromide, phosphoric acid, and potassium dichromate.
A key point here is that the dichromate ion acts as an oxidizing agent during the reaction:

• It gains electrons, thus reducing to potassium chromate \( \text{CrO}_4^{2-} \).
• During this conversion, the oxidation state of chromium changes, signaling a reduction process.

In simple words, dichromate is a bit like a collector, gathering extra electrons to change its form. These reactions often produce different chromium-containing species, like in our equation where dichromate's leftover is potassium chromate \( \text{K}_2\text{CrO}_4 \). This conversion marks the critical part of its reduction.

Precipitation Reactions

Precipitation reactions occur when two solutions react to form an insoluble solid, known as a precipitate. In our second equation, several compounds form through such reactions, reflecting how ions reassemble when combined.
When solutions mix, not all ions stay dissolved:

• The insoluble products, like \( \text{AlPO}_4 \) in our example, settle out as a precipitate.
• This settling indicates a chemical change where ions find partners they 'like' more, forming solid compounds instead of staying dissolved.

In plain terms, think of this process as a group dance where everyone is searching for the perfect partner. Once they find it, they leave the dissolving liquid dance floor to sit together. The end results are often pure substances that crystallize out of the solution, significantly changing the chemical landscape of the initial mix.