Question

In the separation of $\text{C}{{\text{u}}^{\text{2+}}}$ and $\text{C}{{\text{d}}^{2+}}$ in the second group of qualitative analysis of cations, tetraamminecopper (II) sulfate and tetraamine cadmium (II) sulfate react with $\text{KCN}$ to form corresponding cyano complexes and their relative stability enables the separation of $\text{C}{{\text{u}}^{\text{2+}}}$ and $\text{C}{{\text{d}}^{2+}}$?
A) ${{\text{K}}_{\text{3}}}\left[ \text{Cu(CN}{{\text{)}}_{\text{4}}} \right]$: less stable and
    ${{\text{K}}_{2}}\left[ \text{Cd(CN}{{\text{)}}_{\text{4}}} \right]$ : more stable

B) ${{\text{K}}_{\text{3}}}\left[ \text{Cu(CN}{{\text{)}}_{\text{4}}} \right]$ : more stable and
     ${{\text{K}}_{2}}\left[ \text{Cd(CN}{{\text{)}}_{\text{4}}} \right]$: Less stable

C) ${{\text{K}}_{2}}\left[ \text{Cu(CN}{{\text{)}}_{\text{4}}} \right]$: less stable and
   ${{\text{K}}_{2}}\left[ \text{Cd(CN}{{\text{)}}_{\text{4}}} \right]$: More stable

D) ${{\text{K}}_{2}}\left[ \text{Cu(CN}{{\text{)}}_{\text{4}}} \right]$ : more stable and
    ${{\text{K}}_{2}}\left[ \text{Cd(CN}{{\text{)}}_{\text{4}}} \right]$ : Less stable

Answer 1

Hint: The second group cations, copper $\text{C}{{\text{u}}^{\text{2+}}}$ and cadmium $\text{C}{{\text{d}}^{2+}}$ forms the cyano complexes: ${{\text{K}}_{\text{3}}}\left[ \text{Cu(CN}{{\text{)}}_{\text{4}}} \right]$ and ${{\text{K}}_{2}}\left[ \text{Cd(CN}{{\text{)}}_{\text{4}}} \right]$. The second ionization of copper in the cyano complex is $+1$ and for cadmium it is $+2$. Due to low ionization, the copper forms very stable complexes that do not dissociate easily in solution.

Complete step by step answer:
The transition metals form the complex cyanides. The cyanide $\text{C}{{\text{N}}^{-}}$ is a monodentate ligand. It donates its electron to the metal atom and results in the formation of metal –cyanide complex. $\text{C}{{\text{N}}^{-}}$ 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. This stabilizes the metal cyanide complex to a larger extent.

We know that the copper $\text{C}{{\text{u}}^{\text{2+}}}$ and $\text{C}{{\text{d}}^{2+}}$ are the group II metal ions. The group II metal ions can be estimated by passing the hydrogen sulphide gas through it. Here we are provided with the solution which contains the $\text{C}{{\text{u}}^{\text{2+}}}$ and $\text{C}{{\text{d}}^{2+}}$ ions in it. Therefore, we use the concept of masking.
Masking is defined as the process in which the substance, without its physical separation, is transformed into a form that does not interfere in the particular reaction.

Here, the copper and cadmium can form complexes with the ligands. The tetra amine copper (II) sulfate and tetraamine cadmium (II) sulfate reacts with the potassium cyanide $\text{KCN}$ .The $\text{KCN}$ is the masking agent. The copper and cadmium form the cyano complexes. The complex is as follows:
$\left[ \text{Cu(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}} \right]\text{S}{{\text{O}}_{\text{4}}}\text{ + KCN }\to \text{ }{{\text{K}}_{\text{3}}}\left[ \text{Cu(CN}{{\text{)}}_{\text{4}}} \right]$

Similarly, the cadmium forms the cyano complex as follows:
$\left[ \text{Cd(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}} \right]\text{S}{{\text{O}}_{\text{4}}}\text{ + KCN }\to \text{ }{{\text{K}}_{2}}\left[ \text{Cd(CN}{{\text{)}}_{\text{4}}} \right]$
The copper forms a very stable complex with cyanide ions. The secondary ionization power of copper ${{\text{K}}_{\text{3}}}\left[ \text{Cu(CN}{{\text{)}}_{\text{4}}} \right]$ is $+1$ . On the other hand, the secondary ionization power of cadmium ${{\text{K}}_{2}}\left[ \text{Cd(CN}{{\text{)}}_{\text{4}}} \right]$ is $+2$ . The second ionization of potassium cadmi cynanide is higher than that of the potassium cupro cyanide $\text{C}{{\text{d}}^{2+}}\rangle \text{ C}{{\text{u}}^{\text{1+}}}$.
Therefore, the copper forms a stable complex with the cyanide than the cadmium.

When $\text{KCN}$ is introduced in the solution, the cyano complexes are formed. When the solution is treated with the ${{\text{H}}_{\text{2}}}\text{S}$ gas, the unstable ${{\text{K}}_{2}}\left[ \text{Cd(CN}{{\text{)}}_{\text{4}}} \right]$ easily dissociates into the $\text{C}{{\text{d}}^{2+}}$ ions and precipitated as the $\text{CdS}$ but ${{\text{K}}_{\text{3}}}\left[ \text{Cu(CN}{{\text{)}}_{\text{4}}} \right]$ is very stable complex therefore very few $\text{C}{{\text{u}}^{\text{2+}}}$ ions are present in the solution, therefore, it does not get precipitated as $\text{CuS}$ .
$\text{C}{{\text{d}}^{\text{2+}}}\text{+}{{\text{H}}_{\text{2}}}\text{S}\rightleftharpoons \text{CdS (}\downarrow \text{) + 2}{{\text{H}}^{\text{+}}}$
Therefore, here ${{\text{K}}_{\text{3}}}\left[ \text{Cu(CN}{{\text{)}}_{\text{4}}} \right]$ is a more stable complex and ${{\text{K}}_{2}}\left[ \text{Cd(CN}{{\text{)}}_{\text{4}}} \right]$ is less stable complex.
So, the correct answer is “Option B”.

Note: Masking is a technique to reduce or prevent the interference of ion in the estimation. Here, the potassium cyanide acts as the masking agent. It forms a stable complex with the copper and prevents it from undergoing further reaction with sulphide. This eases the determination of cadmium.