Problem 56 - A Complete Resource Book in Chemistry for JEE Main

Question

Mixture \(\mathrm{X}=0.02 \mathrm{~mol}\) of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) and \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Br}\right] \mathrm{SO}_{4}\) 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 \(\mathrm{BaCl}_{2} \longrightarrow Z\) Number of moles of \(\mathrm{Y}\) and \(\mathrm{Z}\) are (a) \(0.02,0.01\) (b) \(0.01,0.01\) (c) \(0.01,0.02\) (d) \(0.02,0.02\)

Step-by-Step Solution

Answer

(b) 0.01, 0.01

Step 1: Define the Mixture Components

The mixture consists of two different complexes: \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) and \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Br}\right] \mathrm{SO}_{4}\). We need to evaluate how these will react with \(\mathrm{AgNO}_{3}\) and \(\mathrm{BaCl}_{2}\).

Step 2: Reaction with AgNO₃

When the mixture \(\mathrm{X}\) is reacted with excess \(\mathrm{AgNO}_3\), the Br⁻ ions in \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) will precipitate as \(\mathrm{AgBr}\). Since this mixture is prepared in 2 litres and divided into 1 litre for reaction, initially it had 0.02 moles of the complex, so in 1 litre, it has 0.01 moles of \(\mathrm{Br}^-\). Therefore, \(\mathrm{Y}\) will form 0.01 moles.

Step 3: Reaction with BaCl₂

When \(\mathrm{X}\) is reacted with excess \(\mathrm{BaCl}_2\), the \(\mathrm{SO}_{4}^{2-}\) ions in both complexes will precipitate as \(\mathrm{BaSO}_4\). Since both complexes contribute \(\mathrm{SO}_4^{2-}\), and the combined initial moles are 0.02, in 1 litre it corresponds to 0.01 moles. Thus, \(\mathrm{Z}\) will also form 0.01 moles.

Key Concepts

Complex Formation

Complex formation in coordination chemistry involves the combination of a metal ion with ligands to form a complex ion. These ligands are molecules or ions that donate a pair of electrons to the metal. In our example, the mixture consists of two cobalt complexes:

• The first is \[\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\]where cobalt is coordinated by five ammonia \((\mathrm{NH}_3)\) ligands and one sulfate ligature.
• The second complex is \[\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Br}\right] \mathrm{SO}_{4}\]where cobalt is surrounded by three ammonia ligands and one bromide ion.

Each complex ion acts as a single unit. The charged parts of these complexes determine their behavior in reactions and the kinds of ions they release in solutions. Understanding the composition and structure of these complexes helps predict how they will interact with other chemicals.

Precipitation Reactions

Precipitation reactions involve the formation of an insoluble solid, called a precipitate, when two aqueous solutions are mixed. In the given problem, precipitation reactions occur when the complexes react with \(\text{AgNO}_3\) and \(\text{BaCl}_2\).

• With \(\text{AgNO}_3\), bromide ions from\[ \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br} \] are converted into \(\text{AgBr}\), an insoluble compound.
• With \(\text{BaCl}_2\), sulfate ions from both complexes form \(\text{BaSO}_4\), another insoluble compound.

These reactions illustrate the different ways ions can leave a solution. Precipitation tells us not only about the ions present but also about their quantities when reacted with specific reagents. Observing these results helps in calculating moles and understanding the overall reaction behavior.

Mole Calculations

Mole calculations are essential for quantifying reactants and products in chemical reactions. Given our reaction setup, calculating the number of moles involved helps predict the amount of precipitate formed.When the reaction was divided into each 1 liter:

• For \(\text{AgNO}_3\), we focus on the bromide ions. As initial moles for the complex containing \(\text{Br}^-\) ions were 0.02 in 2 liters, this is halved in 1 liter resulting in 0.01 moles of \(\text{AgBr}\).
• For \(\text{BaCl}_2\), we account for the sulfate ions present in both complexes. Overall, 0.01 moles of \(\text{BaSO}_4\) were formed due to the similar halving process.

Grasping these calculations is crucial for accurately predicting the reaction outputs, and it lays the foundation for more complex stoichiometry in further chemistry studies.