`underset(("Coloured"))(P("soln.")) overset(Air) to underset(("Coloured"))(Q("soln.")) overset(KOH)to underset((ppt.)" (Insoluble in both excess NaOH and excess "NH_(3)" solution)")(Rdarr)`
then P contains:

Options

A. `Cu^(2+)(aq.)`
B. `Fe^(2+)(aq.)`
C. `Cr^(2+)(aq.)`
D. `Ni^(2+)(aq.)`

Answer

To solve the question, we need to analyze the information given about the colored solution P, its reaction with air to form Q, and its reaction with KOH to form a precipitate R that is insoluble in both excess NaOH and excess NH3.

### Step-by-Step Solution:

1. **Identify the Nature of P**:
- We know that P is a colored solution. Common colored solutions in chemistry include those containing transition metals, such as Fe²⁺ (iron), Cu²⁺ (copper), and Cr³⁺ (chromium).

2. **Reaction with Air**:
- The question states that P reacts with air to form Q. This suggests an oxidation reaction. For transition metals, oxidation typically involves an increase in oxidation state.
- For example, Fe²⁺ can be oxidized to Fe³⁺ in the presence of air.

3. **Formation of Q**:
- If we assume P is Fe²⁺, then upon oxidation in the presence of air, it would convert to Fe³⁺. Therefore, Q would be a solution of Fe³⁺, which is also colored.

4. **Reaction with KOH**:
- The next part of the question states that when P reacts with KOH, it forms a precipitate R that is insoluble in both excess NaOH and excess NH3.
- When Fe²⁺ reacts with KOH, it forms a precipitate of iron(II) hydroxide, Fe(OH)₂, which is a green precipitate. However, Fe³⁺ reacts with KOH to form iron(III) hydroxide, Fe(OH)₃, which is a brown precipitate.

5. **Insolubility in NaOH and NH3**:
- The precipitate formed from Fe³⁺ (Fe(OH)₃) is indeed insoluble in excess NaOH and NH3, confirming that R is likely Fe(OH)₃.

6. **Conclusion**:
- Given that P is Fe²⁺, it reacts with air to form Q (Fe³⁺) and with KOH to form R (Fe(OH)₃), which is insoluble in both excess NaOH and NH3.

Thus, the colored solution P contains **Fe²⁺**.