To determine which reagent gives a brilliant red precipitate with Nickel ions in a basic medium, let's analyze each option:
When dimethyl glyoxime is added to a solution containing Nickel ions (\(Ni^{2+}\)) in a basic medium (usually achieved by adding ammonia or another base), a red precipitate of nickel(II) dimethylglyoxime complex is formed:
\(Ni^{2+} + 2(C_4H_8N_2O_2) + 2OH^- \rightarrow Ni(C_4H_7N_2O_2)_2 \, (red \, precipitate) + 2H_2O\)
Thus, the correct answer is dimethyl glyoxime because it forms a red precipitate with Nickel ions in a basic medium.
In conclusion, the reagent that gives a brilliant red precipitate with Nickel ions in a basic medium is dimethyl glyoxime.
The reagent that gives a brilliant red precipitate with Nickel ions (\( \text{Ni}^{2+} \)) in a basic medium is dimethyl glyoxime (dmg). When \( \text{Ni}^{2+} \) reacts with dimethyl glyoxime, it forms a complex that produces a bright red precipitate. The reaction is as follows:
\(\text{Ni}^{2+} + \text{dmg} \rightarrow [\text{Ni}(\text{dmg})_2] \, \, (\text{Rosy red/Bright Red precipitate})\)
Thus, the correct answer is dimethyl glyoxime.
The Correct Answer is: dimethyl glyoxime
In the group analysis of cations, Ba$^{2+}$ & Ca$^{2+}$ are precipitated respectively as
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.