`Mg_(3)N_(2)(s)+H_(2)O overset(R.T.)to P darr+Q uarr`
Excess 'Q' gas does not form coloured complex with:

Options

A. `Ni^(2+)(aq.)`
B. `Zn^(2+)(aq.)`
C. `Cr^(3+)(aq.)`
D. `Cu^(2+)(aq.)`

Explanation

To solve the problem, we need to analyze the reaction of magnesium nitride (Mg₃N₂) with water and identify the products formed. We will then determine which ion does not form a colored complex with ammonia (Q gas).

### Step-by-Step Solution:

1. **Identify the Reaction**:
- The reaction involves magnesium nitride (Mg₃N₂) reacting with water (H₂O).
- The balanced equation for the reaction is:
\[ Mg_3N_2 (s) + 6H_2O (l) \rightarrow 3Mg(OH)_2 (s) + 2NH_3 (g) \]
- Here, magnesium hydroxide (Mg(OH)₂) is a white precipitate (P), and ammonia (NH₃) is a colorless gas (Q).

2. **Identify the Products**:
- From the reaction, we have:
- P = Magnesium hydroxide (Mg(OH)₂) - a white precipitate.
- Q = Ammonia (NH₃) - a colorless gas.

3. **Determine the Complex Formation**:
- We need to check which metal ions do not form colored complexes with ammonia (NH₃).
- Common metal ions and their behavior with ammonia:
- Nickel (Ni²⁺): Forms a blue complex (hexamine nickel(II)).
- Zinc (Zn²⁺): Forms a colorless complex (tetraamine zinc(II)).
- Chromium (Cr³⁺): Forms a yellow complex (hexaamine chromium(III)).
- Copper (Cu²⁺): Forms a deep blue complex (tetraamine copper(II)).

4. **Identify the Ion that Does Not Form a Colored Complex**:
- Among the ions mentioned:
- Nickel, chromium, and copper form colored complexes with ammonia.
- Zinc does not form a colored complex; it forms a colorless complex with ammonia.

5. **Conclusion**:
- The excess Q gas (ammonia) does not form a colored complex with zinc ion (Zn²⁺).
- Therefore, the correct answer is that ammonia does not form a colored complex with **Zinc ion (Zn²⁺)**.

### Final Answer:
Excess 'Q' gas (ammonia) does not form a colored complex with **Zinc ion (Zn²⁺)**.