Problem 126 - A Complete Resource Book in Chemistry for JEE Main

Question

Mixture of \((\mathrm{X})=0.02\) moles of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) and \(0.02\) mole of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}\) was prepared in 2 litre of solution 1 litre of mixture \((\mathrm{X})+\mathrm{excess} \mathrm{AgNO}_{3} \longrightarrow(\mathrm{Y})\) 1 litre of mixture \((\mathrm{X})+\) excess \(\mathrm{BaCl}_{2} \longrightarrow(\mathrm{Z})\) (a) \(0.01,0.01\) (b) \(0.02,0.01\) (c) \(0.01,0.02\) (d) \(0.02,0.02\)

Step-by-Step Solution

Answer

The correct answer is (a) 0.01, 0.01.

Step 1: Determine Ion Release from Mixture

The compound \( [\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4]\mathrm{Br} \) releases one \( \mathrm{Br}^{-} \) ion, and the compound \( [\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}]\mathrm{SO}_4 \) releases one \( \mathrm{SO}_4^{2-} \) ion into the solution. Since both compounds are mixed in equal amounts, i.e., both have 0.02 moles in 2 liters of solution, they will contribute equally to ion concentrations in the solutions.

Step 2: Calculate Ions in Reaction with AgNO3

When 1 liter of the mixture reacts with excess \( \mathrm{AgNO}_3 \), \( \mathrm{Br}^{-} \) ions will form a precipitate of \( \mathrm{AgBr} \). The amount from \( [\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4]\mathrm{Br} \) is 0.01 moles because it's half the total volume. Thus, \( 0.01 \) moles of \( \mathrm{Br}^{-} \) will react with \( \mathrm{Ag}^{+} \) ions to form \( 0.01 \) moles of \( \mathrm{AgBr} \).

Step 3: Calculate Ions in Reaction with BaCl2

When 1 liter of the mixture reacts with excess \( \mathrm{BaCl}_2 \), \( \mathrm{SO}_4^{2-} \) ions will precipitate as \( \mathrm{BaSO}_4 \). From the compound \( [\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}]\mathrm{SO}_4 \), 0.01 moles of \( \mathrm{SO}_4^{2-} \) are contributed, similar to the \( \mathrm{Br}^{-} \) ions. Thus, \( 0.01 \) moles of \( \mathrm{SO}_4^{2-} \) will react with \( \mathrm{Ba}^{2+} \) ions to form \( 0.01 \) moles of \( \mathrm{BaSO}_4 \).

Step 4: Confirm Options and Select the Correct One

From the calculations, it's clear that \( 0.01 \) moles of \( \mathrm{AgBr} \) and \( 0.01 \) moles of \( \mathrm{BaSO}_4 \) are formed in their respective reactions. Therefore, the correct answer is \((a)\) \(0.01,0.01\).

Key Concepts

Complex Compounds

Complex compounds are fascinating structures in chemistry formed between metal ions and other molecules. These molecules, or ligands, have lone pairs of electrons they can share with the metal ion to form coordinate covalent bonds. For example, in this case, we have the complex ions \([\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4]\) and \([\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}]\).

• The cobalt ion is the central metal ion.
• Ammonia (\(\mathrm{NH}_3\)) acts as the ligand providing the electron pairs.
• Sulfate (\(\mathrm{SO}_4^{2-}\)) and bromide (\(\mathrm{Br}^{-}\)) ions act as counterions, balancing out the charge of the whole complex.

Understanding these complexes involves recognizing their structure and the role of each component: the metal, ligands, and counterions. Depending on the ligands and the metal, these complexes can vary greatly in properties and uses. Complex compounds are essential in fields like catalysis, biochemistry, and material science.

Stoichiometry

Stoichiometry is the mathematical backbone of chemistry that helps in calculating the amounts of reactants and products in a chemical reaction. When we have a balanced reaction, stoichiometry enables us to determine the quantities needed for a reaction to occur fully without excess of any component.
In the given exercise, stoichiometry allows us to calculate how much of each product forms when the mixture interacts with different reactants like \(\mathrm{AgNO}_3\) and \(\mathrm{BaCl}_2\). By knowing:

• Both compounds contribute equally (each has \(0.02\) moles in a \(2\) liter solution).
• Each reaction uses only half of the mixture (\(1\) liter of solution).
• The exact stoichiometric amount necessary to react fully with \(\mathrm{Br}^{-}\) or \(\mathrm{SO}_4^{2-}\) from the mixture.

Stoichiometry ensures that we know precisely \(0.01\) moles of \(\mathrm{AgBr}\) and \(\mathrm{BaSO}_4\) are produced, answering the exercise accurately.

Precipitation Reactions

Precipitation reactions happen when two solutions are mixed, and an insoluble solid, called a precipitate, forms. In this exercise, when \(\mathrm{AgNO}_3\) is added to the solution containing \(\mathrm{Br}^{-}\), \(\mathrm{AgBr}\), a yellowish solid, precipitates. Similarly, adding \(\mathrm{BaCl}_2\) leads to the formation of \(\mathrm{BaSO}_4\), a white solid. Precipitation plays a crucial role in:

• Analytical chemistry for detecting ions in a solution.
• Water treatment processes.
• Forming new materials in lab settings.

By understanding how specific ions combine to form solids, chemists gain tools to control reactions and separate compounds from mixtures efficiently. The exercise illustrates how specific halide ions interact under precipitation conditions, highlighting their role in making certain substances visible and measurable.

Ion Exchange Reactions

Ion exchange reactions involve the exchange of ions between two compounds, resulting in the formation of new compounds. These are significant in various chemical processes including water purification and in the functioning of many biological systems.
In the given problem, the reactions with \(\mathrm{AgNO}_3\) and \(\mathrm{BaCl}_2\) are examples of ion exchange:

• \(\mathrm{AgNO}_3\) exchanges its \(\mathrm{NO}_3^{-}\) ion for \(\mathrm{Br}^{-}\), resulting in \(\mathrm{AgBr}\) formation.
• \(\mathrm{BaCl}_2\) exchanges \(\mathrm{Cl}^{-}\) for \(\mathrm{SO}_4^{2-}\), forming \(\mathrm{BaSO}_4\).

These exchanges are guided largely by the tendency to form a more stable or less soluble product, driving the reaction forward to the completion. Through ion exchange reactions, we can remove unwanted ions from a system and selectively recover or produce desired compounds. These reactions are fundamental in developing chemical processes and treatments for practical applications.