In the wet tests for identification of various cations by precipitation, which transition element cation doesn’t belong to group IV in qualitative inorganic analysis?
\(Fe^{3+}\)
\(Zn^{2+}\)
\(Co^{2+}\)
\(Ni^{2+}\)
Step 1: Understanding the Grouping of Cations in Qualitative Analysis
In qualitative inorganic analysis, cations are classified into groups based on their precipitation behavior with specific reagents. The classification is as follows:
Group III: Includes \(\text{Fe}^{3+}\), \(\text{Al}^{3+}\), and \(\text{Cr}^{3+}\), which precipitate as hydroxides in the presence of \(\text{NH}_4\text{OH}\) and \(\text{NH}_4\text{Cl}\).
Group IV: Includes \(\text{Zn}^{2+}\), \(\text{Co}^{2+}\), and \(\text{Ni}^{2+}\), which precipitate as sulfides in the presence of \(\text{H}_2\text{S}\) in neutral or slightly acidic medium.
Step 2: Analysis of the Cations
\(\text{Fe}^{3+}\): Belongs to Group III, as it forms \(\text{Fe(OH)}_3\) precipitate with \(\text{NH}_4\text{OH}\).
\(\text{Zn}^{2+}\), \(\text{Co}^{2+}\), \(\text{Ni}^{2+}\): Belong to Group IV, as they precipitate as sulfides (\(\text{ZnS}\), \(\text{CoS}\), \(\text{NiS}\)) with \(\text{H}_2\text{S}\).
Conclusion: \(\text{Fe}^{3+}\) does not belong to Group IV. Therefore, the correct answer is (1) \(\text{Fe}^{3+}\).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)