Problem 117 - A Complete Resource Book in Chemistry for JEE Main

Question

A solution when diluted with water and boiled, gives a white precipitate. On addition of excess of \(\mathrm{NH}_{4} \mathrm{Cl} /\) \(\mathrm{NH}_{4} \mathrm{OH}\), the volume of precipitate decreases, leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in \(\mathrm{NH}_{4} \mathrm{OH} / \mathrm{NH}_{4} \mathrm{Cl}\). (a) \(\mathrm{Al}(\mathrm{OH})_{3}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) \(\mathrm{Mg}(\mathrm{OH})_{2}\) (d) \(\mathrm{Zn}(\mathrm{OH})_{2}^{2}\)

Step-by-Step Solution

Answer

The dissolveable precipitate is  Zn(OH)_2.

Step 1: Understanding the Problem

You need to identify the precipitate that dissolves in excess  (NH_4Cl / NH_4OH). There is an initial formation of a white precipitate when the solution is boiled and diluted.

Step 2: Evaluate Precipitate Formation

Precipitates that commonly form white solids include  Al(OH)_3,  Ca(OH)_2,  Mg(OH)_2, and  Zn(OH)_2. They are generally insoluble in water but may react differently with ammonia and ammonium chloride.

Step 3: Consider Reaction with Ammonium Hydroxide and Chloride

When excess  NH_4OH  and  NH_4Cl  are added, we look for the precipitate that dissolves in these reagents. Zn(OH)_2 is known to dissolve in concentrated ammonium hydroxide, forming a complex ion, which aligns with the precipitate dissolving behavior described.

Key Concepts

Inorganic Chemistry

Inorganic chemistry revolves around the study of non-organic compounds, typically those that do not contain carbon-hydrogen (C-H) bonds. An essential part of inorganic chemistry is understanding the behavior and reactions of these compounds.
For example, many inorganic compounds form precipitates, insoluble solids that emerge from a liquid solution. Precipitation reactions are crucial for identifying substances in a solution, as the color, solubility, and reactivity of the precipitate often provide clues about the compound's identity.
In the given exercise, a white precipitate forms when the solution is boiled and diluted with water. This indicates a chemical species that is initially insoluble under these conditions, revealing an important reaction characteristic of inorganic compounds.
Moreover, inorganic chemistry is notable for its exploration of ions and coordination compounds, a key aspect referenced later in amphoteric hydroxides and complex ion formation.

Amphoteric Hydroxides

Amphoteric hydroxides are metal hydroxides that can react with both acids and bases. This dual behavior is due to the compound's ability to either donate or accept a proton (H⁺).
Consider aluminum hydroxide, (\(\text{Al(OH)}_3\), which is amphoteric. It can react with acids to form aluminum salts and with bases to form aluminate ions. This unique property of reacting differently depending on the condition is a defining trait for amphoteric substances.
The exercise involves potential amphoteric hydroxides like (\(\text{Al(OH)}_3\)) and (\(\text{Zn(OH)}_2\)). Aluminum and zinc hydroxides demonstrate this amphoteric nature. (\(\text{Zn(OH)}_2\)) dissolves in both strong acids and concentrated ammonia solutions due to this characteristic, forming soluble complex ions.
Recognizing amphoteric behavior helps us predict how these hydroxides might behave when different reagents are added, like the mix in the solution from the exercise.

Complex Ion Formation

Complex ion formation is a fascinating aspect of inorganic chemistry. It involves creating a charged species consisting of a central metal ion bonded to one or more molecules or ions (ligands).
In the context of our problem, complex ion formation plays a pivotal role. When zinc hydroxide, (\(\text{Zn(OH)}_2\)), dissolves in excess ammonium hydroxide (\(\text{NH}_4\text{OH}\)), a complex ion like (\([\text{Zn(NH}_3)_4]^{2+}\)) can form.
This process involves ammonia acting as a ligand, bonding strongly to the zinc ion and thereby increasing its solubility in the solution. Forming these complex ions is essential for understanding why certain precipitates dissolve in specific reagents.
It's this behavior—the dissolution of (\(\text{Zn(OH)}_2\)) due to complex ion formation—that answers the question from the original exercise, showcasing how knowledge of complex ions elucidates various reactions and transformations in inorganic chemistry.