Problem 60 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following on reaction with \(\mathrm{H}_{2} \mathrm{~S}\) does not produce metallic sulphide? (a) \(\mathrm{CdCl}_{2}\) (b) \(\mathrm{ZnCl}_{2}\) (c) \(\mathrm{CoCl}_{2}^{2}\) (d) \(\mathrm{CuCl}_{2}^{2}\)

Step-by-Step Solution

Answer

\(\mathrm{CoCl}_{2}\) with \(\mathrm{H}_{2} \mathrm{~S}\) can fail to produce a detectable metallic sulfide in dilute solutions.

Step 1: Identify the compounds and their common reactions

We need to consider how hydrogen sulfide (\(\mathrm{H}_{2} \mathrm{~S}\)) reacts with different metal chlorides to form metal sulfides. The reaction is typically of the form \( \text{Metal chloride} + \mathrm{H}_{2} \mathrm{~S} \rightarrow \text{Metal sulfide} + \text{hydrochloric acid} \).

Step 2: Consider solubility rules for sulfides

Not all metal chlorides will react with \(\mathrm{H}_{2} \mathrm{~S}\) to form sulfides. Generally, transition metal sulfides are insoluble while certain groups, like alkaline earth metals, have soluble sulfides. Cadmium (\(\mathrm{Cd}\)), Zinc (\(\mathrm{Zn}\)), Cobalt (\(\mathrm{Co}\)), and Copper (\(\mathrm{Cu}\)) typically form insoluble sulfides.

Step 3: Analyze each given option

- \(\mathrm{CdCl}_{2}\) and \(\mathrm{ZnCl}_{2}\): Both will form insoluble sulfides when reacted with \(\mathrm{H}_{2} \mathrm{~S}\), i.e. \(\mathrm{CdS}\) and \(\mathrm{ZnS}\) respectively. - \(\mathrm{CoCl}_{2}\) and \(\mathrm{CuCl}_{2}\): These too typically form \(\mathrm{CoS}\) and \(\mathrm{CuS}\); however, \(\mathrm{CoS}\) is slightly more soluble under certain conditions compared to \(\mathrm{CuS}\).

Step 4: Determine the exception based on experimental conditions

While all options potentially lead to sulfide formation, \(\mathrm{CoCl}_{2}\)'s sulfide (\(\mathrm{CoS}\)) can be more soluble or unstable depending on excess \(\mathrm{H}_{2} \mathrm{~S}\) presence and aqueous conditions, and less precipitate may form compared to other chlorides, particularly in very diluted solutions.

Key Concepts

Reaction with Hydrogen Sulfide

When metal chlorides encounter hydrogen sulfide (\( \mathrm{H}_{2} \mathrm{S} \)), an interesting chemical process takes place. Generally, these reactions result in the transformation of metal chlorides into metal sulfides alongside the production of hydrochloric acid. The generic reaction follows the form:

• \( \text{Metal chloride} + \mathrm{H}_{2} \mathrm{S} \rightarrow \text{Metal sulfide} + \text{hydrochloric acid} \)

This seemingly simple exchange involves the sulfide ion (\( \mathrm{S}^{2-} \)) replacing the chloride ion (\( \mathrm{Cl}^- \)) in the metal chloride structure. This reaction serves to illustrate the fascinating nature of chemistry, where the exchange of ions alters the compound's properties entirely, resulting in substances with markedly different features.

Solubility Rules for Sulfides

Understanding the solubility of sulfides is foundational in predicting reactions involving metal sulfides. Most transition metal sulfides are characteristically insoluble in water. This means that when such sulfides are formed in aqueous solutions, they tend to precipitate out as solids. Here are some critical points to remember regarding sulfide solubility:

• Most transition metal sulfides, including cadmium (\( \mathrm{CdS} \)) and zinc (\( \mathrm{ZnS} \)), are quite insoluble in water.
• Sulfides of alkaline earth metals, like calcium, are an exception as they usually remain soluble.
• Solubility can sometimes vary under specific conditions, such as temperature or concentration, which may affect the precipitate's formation and stability.

These rules help anticipate the outcome of reactions involving sulfides, particularly when hydrogen sulfide is present, facilitating predictions about whether a precipitate will form.

Transition Metal Chlorides and Sulfides

Transition metals like cadmium (\( \mathrm{Cd} \)), zinc (\( \mathrm{Zn} \)), cobalt (\( \mathrm{Co} \)), and copper (\( \mathrm{Cu} \)) often form complex chlorides that interact distinctively with \( \mathrm{H}_{2} \mathrm{S} \). This interaction often produces sulfides, which appear as precipitates. However, the extent of solubility varies:

• Cadmium chloride (\( \mathrm{CdCl}_{2} \)) will react with \( \mathrm{H}_{2} \mathrm{S} \) to render \( \mathrm{CdS} \), an insoluble sulfide.
• Zinc chloride (\( \mathrm{ZnCl}_{2} \)) similarly forms \( \mathrm{ZnS} \), which is also insoluble.
• Copper chloride (\( \mathrm{CuCl}_{2} \)) turns into \( \mathrm{CuS} \), generally forming a stable precipitate.
• Cobalt chloride (\( \mathrm{CoCl}_{2} \)), however, forms \( \mathrm{CoS} \) which under certain conditions, such as the presence of excess \( \mathrm{H}_{2} \mathrm{S} \) or very dilute solutions, may not form a solid precipitate as readily compared to the others.

This variability underscores the nuanced nature of chemical reactions, where even subtle changes in conditions can lead to different outcomes in the resulting compounds.