Question

In qualitative analysis, $Cd$ is under.
A) I group.
B) II group.
C) III group.
D) IV group.

Answer 1

Hint: We know that the cations are divided into several groups based on their solubilities of their salts. We all know that the second group cations form the cyano complexes.

Complete step by step answer:
The transition elements form the complex cyanide. The cyanide ion is a monodentate ligand and it donates its electron to the metal atom and results in the formation of a metal-cyanide complex. Cyanide ions have the empty p-orbital, hence it can accept the metal electrons and forms a pi-bond with the metal. This is called back bonding and this stabilizes the metal cyanide complex to a larger extent.
We know that cadmium belongs to group II and the group II metals can be estimated by passing the hydrogen sulfide gas through it. The cadmium can form complexes with the ligands. Cadmium (II) sulfate reacts with the potassium cyanide and forms cyano complexes and the solution is treated with hydrogen sulfide gas and the complex easily dissociates into the cadmium ions and precipitated as $CdS$.
$\left[ {Cd{{\left( {N{H_3}} \right)}_4}} \right]S{O_4} + KCN\xrightarrow{{}}{K_2}\left[ {Cd{{\left( {CN} \right)}_4}} \right]$

Therefore, the option B is correct.

Note:
We must remember that the presence of cadmium can also be confirmed by the reaction of the solution with 4-nitronaphthalene-diazo amino-azobenzene. Take twenty milligram of given salt and dissolved in $100\,cc$ of ethanol and $1cc$ of two normal potassium hydroxide. To the drop of the test solution add a drop of the reagent and then one drop of two normal of potassium hydroxide. A bright pink spot surrounded by a blue circle shows the presence of cadmium ions in the solution.