Problem 174 - A Complete Resource Book in Chemistry for JEE Main

Question

A solution containing \(2.675 \mathrm{~g}\) of \(\mathrm{CoCl}_{3} \cdot 6 \mathrm{NH}_{3}\) (molar mass \(=267.5 \mathrm{~g} \mathrm{~mol}^{-1}\) ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of \(\mathrm{AgNO}_{3}\) to give \(4.78 \mathrm{~g}\) of \(\mathrm{AgCl}\) (molar mass \(=143.5 \mathrm{~g} \mathrm{~mol}^{-1}\) ). The formula of the complex is (Atomic mass of \(\mathrm{Ag}=108 \mathrm{u}\) ) [2010] (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{CoCl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}\) (c) \(\left[\mathrm{CoCl}_{3}^{2}\left(\mathrm{NH}_{3}\right)_{3}\right]\) (d) \(\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right] \mathrm{Cl}_{2}\)

Step-by-Step Solution

Answer

The formula of the complex is \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]\mathrm{Cl}_{3}\), corresponding to option (a).

Step 1: Calculate Moles of CoCl3•6NH3

To find the number of moles of \( \mathrm{CoCl}_{3} \cdot 6\mathrm{NH}_{3} \), use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Substitute the given values: \( \text{moles} = \frac{2.675 \text{ g}}{267.5 \text{ g mol}^{-1}} \approx 0.01 \text{ moles} \).

Step 2: Calculate Moles of AgCl Produced

Using the molar mass of \( \mathrm{AgCl} \), calculate the moles produced: \( \text{moles of AgCl} = \frac{4.78 \text{ g}}{143.5 \text{ g mol}^{-1}} \approx 0.0333 \text{ moles} \).

Step 3: Determine Moles of Cl⁻ Ions Released

Since the reaction \( \mathrm{Ag}^+ + \mathrm{Cl}^- \rightarrow \mathrm{AgCl} \) involves a 1:1 molar ratio, the moles of \( \mathrm{Cl}^- \) ions is also approximately \( 0.0333 \text{ moles} \).

Step 4: Find Ratio of CoCl3•6NH3 to Cl⁻ Ions

For every 0.01 moles of the complex, 0.0333 moles of \( \mathrm{Cl}^- \) are released. Calculate the ratio: \( \frac{0.0333}{0.01} = 3.33 \), which indicates 3 moles of \( \mathrm{Cl}^- \) per complex molecule.

Step 5: Identify Correct Complex Formula

Based on the 3 moles of \( \mathrm{Cl}^- \) released, the correct complex produces 3 chloride ions. Thus, the complex formula is \( \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3} \), corresponding to option (a).

Key Concepts

Coordination Complexes

In coordination chemistry, a coordination complex consists of a central metal atom or ion that is bonded to surrounding molecules or ions, known as ligands. These ligands donate electron pairs to form coordinate covalent bonds with the metal center. The structure of these complexes can vary greatly, depending on the nature of the metal and the ligands involved.
For example, in the coordination complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\), cobalt (Co) is the central metal atom. It is bonded to six ammonia (NH₃) ligands. These ligands create a stable coordination compound, in which the central metal ion plays a critical role in determining the structure and properties of the entire complex.
The number of ligands attached to the central atom is referred to as the coordination number, which can significantly affect the geometry of the complex. By understanding these structures, students can better predict compound behavior and reactivity in chemical reactions.

• Coordination complexes are key in catalysis, biological systems, and industrial applications.
• Their stability and reactivity depend on several factors such as ligand type and metal's oxidation state.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and generated in a reaction. This is crucial for experiments and industrial processes to ensure reactions proceed with maximum efficiency and minimal waste.
In the exercise, stoichiometric calculations were used to determine the number of chloride ions released from \(\mathrm{CoCl}_{3} \cdot 6\mathrm{NH}_{3} \), and how they relate to the amount of silver chloride (AgCl) produced. By using molar masses, we were able to convert from mass to moles, a key concept in stoichiometry.
Steps such as these form the backbone of being able to calculate yields, determine limiting reagents, and conducting other critical calculations. Understanding stoichiometry helps in:

• Determining the proportions of reactants and products involved in reactions.
• Calculating theoretical yields and converting between different units of measure.

Practicing stoichiometric calculations can enhance a student's ability to work through complex problems with multiple chemical equations.

Cation Exchange

Cation exchange involves the process by which cations are exchanged between a solid and a liquid phase. It is a fundamental concept in chemistry, playing a vital role in areas such as water purification and soil chemistry.
In our example, the \(\mathrm{CoCl}_{3} \cdot 6\mathrm{NH}_{3}\) solution passed through a cation exchanger. This exchanger helped in facilitating the release of \(\mathrm{Cl}^-\) ions, allowing for the subsequent reaction with silver nitrate (AgNO₃) to form silver chloride (AgCl).
This process can also be key when separating ions from mixtures, as it relies on the preference for certain ions to be retained or exchanged based on size, charge, and other interactions with the exchanging material.

• Cation exchange is used in environmental science to remove excess cations like calcium or magnesium in hard water.
• It helps in soil science to increase soil fertility by exchanging soil nutrients with those in a fertilizer.

Understanding cation exchange mechanisms can provide insights into its practical applications and guide the development of new technologies.