Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Mixture \(X=0.02 \mathrm{~mol}\) of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) and \(0.02 \mathrm{~mol} \mathrm{o}\) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}\) was prepared in 2 litre of solution. [2003S] 1 litre of mixture \(X+\) excess \(\mathrm{AgNO}_{3} \rightarrow Y\) 1 litre of mixture \(X+\) excess \(\mathrm{BaCl}_{2} \rightarrow Z\) No. of moles of \(Y\) and \(Z\) are (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

(a) 0.01, 0.01

Step 1: Analyze the Compounds

Mixture \( X \) consists of two coordination compounds: \([\text{Co}(\text{NH}_3)_5 \text{SO}_4] \text{Br}\) and \([\text{Co}(\text{NH}_3)_5 \text{Br}] \text{SO}_4\). Each compound contains different ions that can precipitate with \(\text{AgNO}_3\) and \(\text{BaCl}_2\):* \([\text{Co}(\text{NH}_3)_5 \text{SO}_4] \text{Br}\) releases \(\text{Br}^-\)* \([\text{Co}(\text{NH}_3)_5 \text{Br}] \text{SO}_4\) releases \(\text{SO}_4^{2-}\)

Step 2: Determine Precipitate with AgNO3

\(\text{AgNO}_3\) reacts with \(\text{Br}^-\) ions to form \(\text{AgBr}\) precipitate.In 1 L of the mixture:- From \([\text{Co}(\text{NH}_3)_5 \text{SO}_4] \text{Br}\), we have \(0.01\) mol \(\text{Br}^-\).Therefore, the moles of \(Y = \text{AgBr}\) are \(0.01\) mol.

Step 3: Determine Precipitate with BaCl2

\(\text{BaCl}_2\) reacts with \(\text{SO}_4^{2-}\) ions to form \(\text{BaSO}_4\) precipitate.In 1 L of the mixture:- From \([\text{Co}(\text{NH}_3)_5 \text{Br}] \text{SO}_4\), we have \(0.01\) mol \(\text{SO}_4^{2-}\).Thus, the moles of \(Z = \text{BaSO}_4\) are \(0.01\) mol.

Step 4: Conclusion

From Steps 2 and 3:- The number of moles of \(Y\) is \(0.01\)- The number of moles of \(Z\) is \(0.01\)Option (a) which is \(0.01, 0.01\) is the correct answer.

Key Concepts

Coordination Compounds

Coordination compounds are fascinating chemical entities formed from a central metal atom surrounded by non-metal ions or molecules, known as ligands. In the given exercise, we have two coordination compounds:

• \([\text{Co}(\text{NH}_3)_5 \text{SO}_4] \text{Br}\)
• \([\text{Co}(\text{NH}_3)_5 \text{Br}] \text{SO}_4\)

The central cobalt metal ion (\(\text{Co}\)) can coordinate with surrounding ligands like \(\text{NH}_3\), \(\text{SO}_4\), and \(\text{Br}^-\). The coordination sphere includes the metal and the surrounding ligands, while ions outside this sphere act as counter ions. This coordination leads to fascinating reactivity and properties, which are utilized in identifying the resulting precipitates in reactions.

Precipitation Reactions

Precipitation reactions occur when two solutions are mixed, and an insoluble solid, known as a precipitate, forms. These reactions are crucial in analytical chemistry for identifying ions in solution.
In our scenario, two different reactions are used:

• When \(\text{AgNO}_3\) combines with the \(\text{Br}^-\) ions from the first compound, it forms \(\text{AgBr}\), a solid precipitate.
• Similarly, \(\text{BaCl}_2\) interacts with \(\text{SO}_4^{2-}\) ions from the second compound to create the \(\text{BaSO}_4\) precipitate.

These reactions are predictably based on the specific insolubility of the resulting compounds. Understanding which ions will precipitate with which reagent is essential for correctly determining the amounts of product formed.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It revolves around the conservation of mass and energy in reactions. In this exercise, stoichiometry is applied to ensure that the number of moles of reactants corresponds accurately to the precipitates formed.
From 1 litre of the given mixtures:

• The \([\text{Co}(\text{NH}_3)_5 \text{SO}_4] \text{Br}\) releases \(0.01\) mol of \(\text{Br}^-\)
• The \([\text{Co}(\text{NH}_3)_5 \text{Br}] \text{SO}_4\) releases the same amount of \(\text{SO}_4^{2-}\)

This calculated conversion demonstrates the effectiveness of stoichiometry in predicting the amounts of substances in products, reaffirming the solution's accuracy in deducing the moles of \(Y\) and \(Z\).

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction, causing no overall change in the concentrations of reactants and products over time. Within the scope of the given exercise, it informs us about the conditions under which precipitation reactions may be reversed or shifted.
Considering an equilibrium scenario:

• If enough \(\text{Ag}^+\) or \(\text{Ba}^{2+}\) was not added, the reaction may not go to completion, preventing all \(\text{Br}^-\) or \(\text{SO}_4^{2-}\) from precipitating.
• Excess presence of these ions ensures full reaction, thus equilibrium is driven toward complete precipitation without reversal.

In this context, an understanding of equilibrium helps ensure that the theoretical predictions align with practical laboratory outcomes.