Problem 71 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one is the most likely structure of \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is \(1 / 3\) of total chlorine of the compound is precipitated by adding \(\mathrm{AgNO}_{3}\) to its aqueous solution? (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \cdot\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\) (b) \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) (c) \(\left[\mathrm{CrCl}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) (d) \(\left[\mathrm{CrCl}_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right] . \mathrm{Cl} .{ }_{2} \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The correct structure is option (d): \([\text{CrCl}_2(\text{H}_2\text{O})_4] \cdot \text{Cl} \cdot 2\text{H}_2\text{O}\).

Step 1: Understanding the Problem

We need to determine which structure of \(\text{CrCl}_3 \cdot 6 \text{H}_2\text{O}\) corresponds to \(\frac{1}{3}\) of its total chlorine being precipitated by adding \(\text{AgNO}_3\) to the solution.

Step 2: Define Precipitation Concept

Silver nitrate (\(\text{AgNO}_3\)) precipitates free chloride ions as \(\text{AgCl}\). Thus, the precipitated \(\text{Cl}^-\) ions must be those that are outside the coordination sphere.

Step 3: Analyze Each Option

Examine each given structure to identify how many chlorine atoms are free and can be precipitated:- **Option (a)**: \([\text{Cr}(\text{H}_2\text{O})_3\text{Cl}_3] \cdot (\text{H}_2\text{O})_3\) - All \(\text{Cl}\) atoms are inside the coordination sphere.- **Option (b)**: \(\text{CrCl}_3 \cdot 6 \text{H}_2\text{O}\) - This representation is ambiguous without a clear structure indicated.- **Option (c)**: \([\text{CrCl}(\text{H}_2\text{O})_5] \text{Cl}_2 \cdot \text{H}_2\text{O}\) - There is 1 \(\text{Cl}\) in the coordination sphere and 2 outside.- **Option (d)**: \([\text{CrCl}_2(\text{H}_2\text{O})_4] \cdot \text{Cl} \cdot 2\text{H}_2\text{O}\) - There are 2 chlorines in the coordination sphere and 1 outside.

Step 4: Calculate Fraction of Precipitated Chlorine

For precipitation of \(\frac{1}{3}\) of chlorine to occur, 1 of every 3 chlorine atoms must be free. From analysis:- **Option (c)**: 2 out of 3 total \(\text{Cl}\) atoms are free, precipitating 2/3.- **Option (d)**: 1 out of 3 total \(\text{Cl}\) atoms are free, precipitating 1/3.Therefore, **Option (d)** satisfies the requirement.

Key Concepts

Crystal Field Theory

The concept of crystal field theory is fundamental in understanding the electronic structure of coordination compounds, particularly those containing transition metals. This theory helps to explain the splitting of degenerate orbitals (orbitals with the same energy) that occur when ligands, like water and chlorine ions, approach the central metal ion, in this case, chromium (Cr).

When ligands coordinate with a transition metal, their electric fields repel the metal's electrons, leading to an energy gap between the resulting split orbitals. This process is known as 'crystal field splitting'. In an octahedral complex, which is common for many transition metals, the \(d\) orbitals split into two sets:

• The higher energy \(e_g\) orbitals (consisting of \(d_{x^2-y^2}\) and \(d_{z^2}\))
• The lower energy \(t_{2g}\) orbitals (consisting of \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\))

This splitting impacts the electronic transitions and, therefore, the color of the compounds, as well as their magnetic properties.

In \([\mathrm{CrCl}_2(\mathrm{H}_2\mathrm{O})_4] \cdot \mathrm{Cl} \cdot 2\mathrm{H}_2\mathrm{O}\), the presence of water and chlorine as ligands will each have a unique effect on the crystal field splitting. Water, being a weaker field ligand compared to chloride, will be less effective in causing greater splitting, contributing to the overall stability and reactivity of the complex.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two soluble salts in aqueous solution react to form one or more insoluble products. These insoluble products are called precipitates. An essential application of precipitation reactions is in detecting and analyzing ion presence in solutions.

For the compound \(\mathrm{CrCl}_3 \cdot 6 \mathrm{H}_2\mathrm{O}\), understanding precipitation reactions plays a crucial role when determining which chlorine atoms can be precipitated using \(\mathrm{AgNO}_3\). Silver nitrate reacts specifically with free \(\mathrm{Cl}^-\) ions to form insoluble silver chloride (AgCl), which appears as a white precipitate.

In the exercise, one of the structures discussed contains one free chloride ion, allowing exactly one-third of the total chlorine to precipitate when treated with \(\mathrm{AgNO}_3\). This selectively precipitated \(\mathrm{Cl}^-\) confirms the solution's ionic content and helps infer the compound's composition. It highlights the transient nature of ionic species within a coordination complex, particularly those not tightly bound within the coordination sphere.

Transition Metal Chemistry

Transition metal chemistry focuses on the unique properties and reactions of metals found in the d-block of the periodic table, specifically those with partially filled d orbitals. Chromium is a classic example, displaying multiple oxidation states and forming distinct compounds with various ligands. These characteristics are largely due to their d orbital electrons.

Chromium, a critical transition metal, has the ability to form coordination complexes like \(\mathrm{CrCl}_3 \cdot 6 \mathrm{H}_2\mathrm{O}\). Due to its variable oxidation states, chromium can adapt to form multiple complexes, characterized by different numbers and types of ligand attachments.

• Oxidation States: Chromium commonly exhibits oxidation states of +2, +3, and +6. In coordination compounds, it often appears in the +3 state, balancing the charge among the metal center and the ligands.
• Ligand Coordination: Chromium can coordinate with varying numbers and types of ligands, leading to diverse chemical properties and structural arrangements.

The example compound highlights these unique features. It shows how chromium, in its +3 oxidation state, can coordinate with both water and chloride ligands. This capability affects the overall geometry, solubility, and reactivity of the compound, providing a basis for understanding more complex reactions and applications in transition metal chemistry.