The group of four basic radicals, `CO^(2+)` `Nl^(2+)` `Mn^(2+)` and `Żn^(2+)` have group reagent hydrogen sulphide in presence of NH.OH. These radicals are precipitated as sulphides which are insoluble in `NH_(4)OH`. Addition of HS increases the ionization of `NH_4`OH due to reaction of OH and H lon.
`NH_4OHharrNH_(4) + OH`
`H_2S hArr2H+S^(2-)`
Salt (A) of above group cations having black colour treated with above reagent and produced compound (B), (B) on treatment with HCl and KCIO, again converted into (A) with the evolution of HS gas. (A) again treated with KCN gives a buff coloured ppt. (C) which dissolved in excess KCN and produced (D)
QTo avoid the precipitation of hydroxides of Ni*, Co. Mn along with those of the third group cations, the solution should be

Options

A. heated with few drops of conc. `HNO_(3)`
B. heated with excess of `NH_4`CI
C. concentrated to small volume
D. none of the above

Answer

To solve the problem, we need to analyze the behavior of the given cations (Co²⁺, Ni²⁺, Mn²⁺, and Zn²⁺) in the presence of hydrogen sulfide (H₂S) and ammonium hydroxide (NH₄OH). The goal is to avoid the precipitation of hydroxides of Ni²⁺, Co²⁺, and Mn²⁺ along with those of the third group cations.

### Step-by-Step Solution:

1. **Understanding the Group Reagents**:
- The group reagent is hydrogen sulfide (H₂S) in the presence of ammonium hydroxide (NH₄OH).
- H₂S reacts with the metal cations to form insoluble metal sulfides.

2. **Formation of Sulfides**:
- When H₂S is added to the solution containing Co²⁺, Ni²⁺, Mn²⁺, and Zn²⁺, these cations will precipitate as their respective sulfides.
- The sulfides of these metals are generally insoluble in NH₄OH.

3. **Effect of H₂S on NH₄OH**:
- The addition of H₂S increases the ionization of NH₄OH, which can be represented as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]
- The presence of H₂S also shifts the equilibrium, affecting the concentration of hydroxide ions (OH⁻) in the solution.

4. **Black-Colored Salt (A)**:
- The black-colored salt (A) formed from the cations is likely a mixture of their sulfides.
- When treated with HCl and KClO₃, compound (B) is produced, which can be converted back to (A) with the evolution of H₂S gas.

5. **Treatment with KCN**:
- When salt (A) is treated with KCN, a buff-colored precipitate (C) is formed.
- This precipitate can dissolve in excess KCN to produce compound (D).

6. **Avoiding Precipitation of Hydroxides**:
- To avoid the precipitation of hydroxides of Ni²⁺, Co²⁺, and Mn²⁺ along with those of the third group cations, we need to reduce the concentration of OH⁻ ions in the solution.
- This can be achieved by adding excess NH₄Cl to the solution.

7. **Why NH₄Cl?**:
- NH₄Cl dissociates in solution to produce NH₄⁺ and Cl⁻ ions. The NH₄⁺ ions will shift the equilibrium of NH₄OH, reducing the concentration of OH⁻ ions.
- Since the solubility product of the hydroxides of the third group cations is lower than that of the fourth, fifth, and sixth groups, reducing OH⁻ concentration will prevent the precipitation of the hydroxides of Ni²⁺, Co²⁺, and Mn²⁺.

### Final Answer:
To avoid the precipitation of hydroxides of Ni²⁺, Co²⁺, and Mn²⁺ along with those of the third group cations, the solution should be treated with **excess NH₄Cl**.