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Which of the following reactions are kin...

Which of the following reactions are kinetically favourable ?
I. `[Cu(H_(2)O)_(4)]^(2+)+4NH_(3)rarr[Cu(NH_(3))_(4)]^(2+)+4H_(2)O`
II. `[Cu(H_(2)O)_(4)]^(2+)+4Cl^(-)rarr[CuCl_(4)]^(2-)+4H_(2)O`
III. `[Co(H_(2)O)_(6)]^(3+)+6Cl^(-)rarr[CoCl_(6)]^(3-)+6H_(2)O`

A

I and II

B

II and III

C

I and III

D

I, II and III

Text Solution

AI Generated Solution

The correct Answer is:
To determine which of the given reactions are kinetically favorable, we need to analyze the stability of the complexes involved and the nature of the ligands. ### Step 1: Analyze Reaction I **Reaction I:** \[ [Cu(H_2O)_4]^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} + 4H_2O \] - In this reaction, copper is in the +2 oxidation state in both the reactant and product. - Ammonia (NH₃) is a stronger field ligand compared to water (H₂O), which means it can stabilize the copper complex more effectively. - The transition from water to ammonia ligands is favorable because it leads to a more stable complex. Thus, this reaction is kinetically favorable. ### Step 2: Analyze Reaction II **Reaction II:** \[ [Cu(H_2O)_4]^{2+} + 4Cl^{-} \rightarrow [CuCl_4]^{2-} + 4H_2O \] - Here, copper is again in the +2 oxidation state. - Chloride (Cl⁻) is a weaker field ligand compared to water and ammonia. The formation of the tetrahedral complex \([CuCl_4]^{2-}\) is less stable than the octahedral complex \([Cu(H_2O)_4]^{2+}\). - Therefore, this reaction is not kinetically favorable because the product is less stable than the reactant. ### Step 3: Analyze Reaction III **Reaction III:** \[ [Co(H_2O)_6]^{3+} + 6Cl^{-} \rightarrow [CoCl_6]^{3-} + 6H_2O \] - In this case, cobalt is in the +3 oxidation state in both the reactant and product. - The complex \([Co(H_2O)_6]^{3+}\) is relatively stable, and the formation of \([CoCl_6]^{3-}\) involves a change in coordination number and geometry. - However, cobalt in the +3 state is more stable with strong field ligands, and chloride is a weaker field ligand. Thus, this reaction is not kinetically favorable because the reactant is more stable than the product. ### Conclusion Based on the analysis: - **Kinetically favorable reactions:** I - **Not kinetically favorable reactions:** II and III ### Summary of Results - **Reaction I:** Kinetically favorable - **Reaction II:** Not favorable - **Reaction III:** Not favorable

To determine which of the given reactions are kinetically favorable, we need to analyze the stability of the complexes involved and the nature of the ligands. ### Step 1: Analyze Reaction I **Reaction I:** \[ [Cu(H_2O)_4]^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} + 4H_2O \] - In this reaction, copper is in the +2 oxidation state in both the reactant and product. - Ammonia (NH₃) is a stronger field ligand compared to water (H₂O), which means it can stabilize the copper complex more effectively. ...
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