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)
QWhich of the following pairs of ions cannot be separated by `H_(2)S`in dilute hydrochloric acid?

Options

A. `Bi^(3+) Sn^(4+)`
B. `Al^(3+) Hg^(2+)`
C. `Zn^(2+). Ca^(2+)`
D. `Ni^(2+)Cu^(2+)`

Answer

To solve the question regarding which pairs of ions cannot be separated by `H₂S` in dilute hydrochloric acid, we will follow a systematic approach based on the information provided in the question and the video transcript.

### Step-by-Step Solution:

1. **Understanding the Group of Radicals**:
The question mentions four basic radicals: `Co^(2+)`, `Ni^(2+)`, `Mn^(2+)`, and `Zn^(2+)`. These radicals can form precipitates when treated with hydrogen sulfide (`H₂S`) in the presence of ammonium hydroxide (`NH₄OH`).

**Hint**: Identify the radicals and their behavior with `H₂S`.

2. **Role of `H₂S` in Acidic Medium**:
When `H₂S` is added to a solution containing cations in an acidic medium (like dilute HCl), it can lead to the formation of insoluble sulfides. The presence of HCl increases the concentration of `H+` ions, which can affect the dissociation of `H₂S`.

**Hint**: Consider how the acidic environment influences the solubility of sulfides.

3. **Common Ion Effect**:
The video transcript mentions the common ion effect, which states that the addition of a common ion (in this case, `H+` from both `H₂S` and `HCl`) decreases the solubility of a salt. This means that if two cations can form precipitates in the presence of `H₂S`, they may not be separated effectively due to this effect.

**Hint**: Recall the common ion effect and its implications on solubility.

4. **Identifying Cations that Form Precipitates**:
The cations that form precipitates with `H₂S` in acidic medium include bismuth (`Bi^(3+)`) and tin (`Sn^(4+)`). Both of these cations form insoluble sulfides (bismuth sulfide is black, and tin sulfide is yellow).

**Hint**: Focus on the specific cations mentioned in the context of `H₂S` and their precipitate colors.

5. **Conclusion**:
Since both bismuth and tin form precipitates in the presence of `H₂S` in dilute hydrochloric acid, they cannot be separated from each other. Thus, the answer to the question is that the pair of ions that cannot be separated by `H₂S` in dilute hydrochloric acid is:

**Answer**: Option A: `Bi^(3+)` and `Sn^(4+)`.

### Summary:
The ions `Bi^(3+)` and `Sn^(4+)` cannot be separated by `H₂S` in dilute hydrochloric acid due to the formation of precipitates and the common ion effect, which affects their solubility in the acidic medium.