Problem 44 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(\mathrm{Ag}^{+}+\mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)\right]^{+} ; k_{1}=6.8 \times 10^{-3}\) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)\right]^{+}+\mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} ; \mathrm{k}_{2}=1.6 \times 10^{-3}\) then the formation constant of \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\)is \([2006-3 \mathrm{M},-1]\) (a) \(6.8 \times 10^{-6}\) (b) \(1.08 \times 10^{-5}\) (c) \(1.08 \times 10^{-6}\) (d) \(6.8 \times 10^{-5}\)

Step-by-Step Solution

Answer

The formation constant is \(1.08 \times 10^{-5}\). The correct option is (b).

Step 1: Understand the Reaction Sequence

We are given two reactions involving silver ions and ammonia:\[ \textrm{Ag}^{+} + \textrm{NH}_3 \rightleftharpoons \left[ \textrm{Ag}(\textrm{NH}_3) \right]^{+} \quad ; \quad k_1 = 6.8 \times 10^{-3} \]\[ \left[ \textrm{Ag}(\textrm{NH}_3) \right]^{+} + \textrm{NH}_3 \rightleftharpoons \left[ \textrm{Ag}(\textrm{NH}_3)_2 \right]^{+} \quad ; \quad k_2 = 1.6 \times 10^{-3} \]The aim is to find the formation constant \(K_f\) for the overall reaction:\[ \textrm{Ag}^{+} + 2\textrm{NH}_3 \rightleftharpoons \left[ \textrm{Ag}(\textrm{NH}_3)_2 \right]^{+} \]

Step 2: Combine the Reaction Steps

To find the overall formation constant \(K_f\), we need to multiply the equilibrium constants of the two steps because the overall reaction is derived from combining these two steps:\[ K_f = k_1 \times k_2 = (6.8 \times 10^{-3}) \times (1.6 \times 10^{-3}) \]

Step 3: Calculate the Formation Constant

Perform the multiplication:\[ K_f = 6.8 \times 1.6 \times 10^{-3} \times 10^{-3} = 10.88 \times 10^{-6} \]Simplify that result to maintain scientific notation:\[ K_f = 1.088 \times 10^{-5} \]

Step 4: Choose the Correct Option

Compare the calculated formation constant with the given options:(a) \(6.8 \times 10^{-6}\)(b) \(1.08 \times 10^{-5}\)(c) \(1.08 \times 10^{-6}\)(d) \(6.8 \times 10^{-5}\)The correct answer is option (b) \(1.08 \times 10^{-5}\) as it is the closest equivalent to our calculated result.

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted by the symbol \(K\), is a fundamental concept in coordination chemistry. It measures the tendency of a chemical reaction to proceed to completion. This constant is calculated based on the concentrations of reactants and products when a reaction reaches equilibrium.
In the case of the reactions discussed here, we have equilibrium constants \(k_1\) and \(k_2\) for two sequential reactions. These values represent how readily each step of complex formation occurs:

• \(k_1 = 6.8 \times 10^{-3}\) for the initial addition of ammonia to the silver ion forming the first silver-ammonia complex.


• \(k_2 = 1.6 \times 10^{-3}\) for the subsequent addition of another ammonia molecule to the complex.

To find the equilibrium constant for the overall reaction forming \([\mathrm{Ag}(\mathrm{NH}_3)_2]^+\), we multiply \(k_1\) and \(k_2\). This reflects that the overall equilibrium constant \(K_f\) can be derived from the product of the individual steps' constants, emphasizing the multiplicative nature of equilibrium conversions in series of reactions.

Complex Formation

Complex formation in coordination chemistry involves the creation of a complex ion or molecule from a central metal atom or ion and ligands. Ligands, which are typically ions or molecules with lone pairs of electrons, donate these electrons to the metal center to form coordinate covalent bonds.
In the reactions involving silver and ammonia:

• \(\mathrm{Ag}^+\) acts as the central metal ion.


• \(\mathrm{NH}_3\) acts as the ligand, binding to the silver ion.

During the process, ammonia molecules coordinate around the silver ion, leading to the formation of complexes such as \([\mathrm{Ag}(\mathrm{NH}_3)]^+\) and \([\mathrm{Ag}(\mathrm{NH}_3)_2]^+\). These complexes represent different stages of ammonia binding to the silver ion, demonstrating how sequential coordination can result in stabilized structures through multiple-step reactions.

Silver-Ammonia Complex

The silver-ammonia complex is a classic example of coordination chemistry in action. It involves the interaction of silver ions with ammonia, which acts as a ligand forming complexes.
The formation of \([\mathrm{Ag}(\mathrm{NH}_3)]^+\) moves further to \([\mathrm{Ag}(\mathrm{NH}_3)_2]^+\) with the additional ammonia molecule. For students learning about these complexes, it is essential to recognize key points:

• Each ammonia molecule donates a pair of electrons to the \(\mathrm{Ag}^+\) central ion.


• This results in formation of a stable coordinate covalent bond.


• The overall stability and formation of these complexes depend on the equilibrium constants of each formation stage.

The silver-ammonia complex series illustrates how transition metals can form different complex ions, which have significant implications in various chemical applications, from catalysis to material science.