`ZnCl_(2)` reacts with excess of `NH_(3)` solution to produce

Options

A. A ppt of `Zn(OH)_(2)`
B. A complex ion `Zn(OH)_(4)^(2-)` of tetrahedral geometry
C. A complex ion `Zn(NH_(3))_(4)^(2+)` of tetrahedral geometry
D. A complex ion `Zn(NH_(3))_(4)^(2+)` of square planar geometry

Answer

To solve the question regarding the reaction of ZnCl₂ with excess NH₃ solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this case are zinc chloride (ZnCl₂) and ammonia (NH₃).

2. **Understand the Reaction**: When ZnCl₂ reacts with excess NH₃, it forms a complex. Ammonia acts as a ligand and coordinates with the zinc ion.

3. **Determine the Oxidation State of Zinc**: In ZnCl₂, zinc is in the +2 oxidation state. This can be confirmed as follows:
- Let the oxidation state of zinc be \( x \).
- The total charge of ZnCl₂ is 0: \( x + 2(-1) = 0 \).
- Thus, \( x = +2 \).

4. **Formation of the Complex**: The excess ammonia will coordinate with the Zn²⁺ ion to form a complex ion. The complex formed is [Zn(NH₃)₄]²⁺.

5. **Determine the Geometry of the Complex**:
- Zinc in the +2 oxidation state has the electronic configuration of [Ar] 3d¹⁰ (after losing 2 electrons from the 4s orbital).
- Since it forms a complex with 4 ammonia molecules, we need to determine the hybridization.
- The hybridization for a complex with 4 ligands is typically sp³, which corresponds to a tetrahedral geometry.

6. **Conclusion**: The product of the reaction between ZnCl₂ and excess NH₃ is the complex ion [Zn(NH₃)₄]²⁺, which has a tetrahedral geometry.

### Final Answer:
The correct answer is **a complex ion [Zn(NH₃)₄]²⁺ of tetrahedral geometry**.

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