Problem 3 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Complex A has a composition of \(\mathrm{H}_{12} \mathrm{O}_{6} \mathrm{Cl}_{3} \mathrm{Cr}\). If the complex on treatment with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) loses \(13.5 \%\) of its original mass, the correct molecular formula of \(\mathrm{A}\) is : [Given: atomic mass of \(\mathrm{Cr}=52 \mathrm{amu}\) and \(\mathrm{Cl}=35 \mathrm{amu}]\) (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}\) (c) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) (d) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The correct formula is \(\left[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_4\mathrm{Cl}_2\right]\mathrm{Cl} \cdot 2 \mathrm{H}_2\mathrm{O}\). (Option d)

Step 1: Determine the Molecular Mass of Complex A

Start by calculating the molecular mass of complex \( A \). The given composition is \( \mathrm{H}_{12} \mathrm{O}_{6} \mathrm{Cl}_{3} \mathrm{Cr} \). Calculate each component's contribution: \(\mathrm{H}_{12} = 12 \times 1 = 12\, \text{amu}\), \(\mathrm{O}_{6} = 6 \times 16 = 96\, \text{amu}\), \(\mathrm{Cl}_{3} = 3 \times 35 = 105\, \text{amu}\), and \(\mathrm{Cr} = 52\, \text{amu}\). Add these values together: \(12 + 96 + 105 + 52 = 265\, \text{amu}\).

Step 2: Calculate the Mass Loss from Treatment

The complex loses 13.5% of its original mass when treated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Calculate this mass loss: \( 0.135 \times 265 = 35.775\, \text{amu} \) is lost. This loss corresponds to \(\mathrm{H}_{2} \mathrm{O}\) molecules being removed since \(\mathrm{H}_{2} \mathrm{O} = 18\, \text{amu} \) each.

Step 3: Evaluate the Options Based on Mass Loss

Identify which structure's removal of \(\mathrm{H}_{2} \mathrm{O}\) will result in the calculated mass loss of approximately \(35.775\, \text{amu}\). Begin evaluating options:- Option (a): \([\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_3\) - If \(6\) water molecules are lost completely, it would be \(6 \times 18 = 108\, \text{amu}\).- Option (b): \([\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_3\mathrm{Cl}_3] \cdot 3 \mathrm{H}_2\mathrm{O}\) - Loss of 3 water molecules is \(3 \times 18 = 54\, \text{amu}\).- Option (c): \([\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_5\mathrm{Cl}] \mathrm{Cl}_2 \cdot \mathrm{H}_2\mathrm{O}\) - Loss of 1 external \(\mathrm{H}_2\mathrm{O}\) is \(18\, \text{amu}\).- Option (d): \([\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_4\mathrm{Cl}_2]\mathrm{Cl} \cdot 2 \mathrm{H}_2\mathrm{O}\) - Loss of 2 external \(\mathrm{H}_2\mathrm{O}\)'s is \(2 \times 18 = 36\, \text{amu}\).

Step 4: Determine the Correct Molecular Formula

The calculated mass loss of \(35.775\, \text{amu}\) best matches the option losing 2 water molecules (36 amu). Thus, the correct molecular formula of complex \(A\) is \( \left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} \cdot 2 \mathrm{H}_{2} \mathrm{O} \), which corresponds to option (d).

Key Concepts

Molecular Mass Calculation

Understanding how to calculate molecular mass is crucial when dealing with coordination compounds. This process starts by identifying each element within a complex and its corresponding atomic mass. In the case of complex compound \( \mathrm{H}_{12} \mathrm{O}_{6} \mathrm{Cl}_{3} \mathrm{Cr} \), each element contributes to the compound's total mass.

To find the molecular mass, multiply the atomic mass of each element by the number of atoms of that element in the formula. Specifically:

• The twelve hydrogen atoms each have an atomic mass of 1 amu, contributing a total of \(12 \times 1 = 12\, \text{amu}\).
• Six oxygen atoms with atomic mass 16 amu add \(6 \times 16 = 96\, \text{amu}\).
• Three chlorine atoms, each 35 amu, contribute \(3 \times 35 = 105\, \text{amu}\).
• One chromium atom at 52 amu contributes \(52\, \text{amu}\).

Adding these together gives a total molecular mass of \(12 + 96 + 105 + 52 = 265\, \text{amu}\). This sum represents the collective mass of atoms in the given coordination compound. Learning this cornerstone technique aids in analyzing more complex coordination structures and reactions.

Water of Crystallization

Water of crystallization refers to the specific amount of water molecules that are integral to the crystallization of a compound. This water is bound within the structure, impacting both the physical and chemical properties of a compound. When dealing with coordination compounds, identifying the correct number of water molecules, or the 'waters of crystallization,' is essential.

In the provided example, the complex originally includes water in its formula, explained by its total composition before any treatment. Through absorption or release of water, these compounds can change masses significantly, which is pivotal in problems like the one analyzed here. When the compound is treated with concentrated \( \mathrm{H}_2\mathrm{SO}_4 \), it loses its water of crystallization. This mass loss quantified at about 13.5% corresponds to specific water molecules being removed from the system.

Determining the exact number of water molecules typically involves comparing calculated mass losses with the known mass of water. Each mole of water is 18 amu, so removal affects the compound’s total mass notably. By analyzing changes in mass, you can confirm the precise structure and formula of coordination compounds, acknowledging the water of crystallization as a fundamental concept.

Complex Formation

Complex formation is at the heart of coordination chemistry. It entails the combination of a central metal atom with various ligands, like water or chloride ions, to form a complex. This can impact the stability, composition, and reactivity of the compound.

In coordination compounds like the one given, chromium serves as the central metal. It's surrounded by different ligands that donate lone pairs of electrons, forming coordinate bonds. Understanding how these complexes form helps explain why particular configurations stabilize.

The complex formation also dictates how substances interact or dissociate under certain treatments. For instance, when treated with specific reagents, ligands in the coordination sphere or water molecules outside the primary sphere may be removed, changing the mass and possibly the coordination structure.

This section of the compound denotes these changes, such as in option (d), where water loss from the exterior results in confirming the structure [\(\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_4\mathrm{Cl}_2\)]\( \mathrm{Cl} \cdot 2\mathrm{H}_2\mathrm{O} \). Mastery in evaluating complex formation helps derive accurate formulations of coordination compounds, critical for further chemical synthesis or applications.