When two compounds `Acl_(3) and DCl_(3)` of two elements A and D are mixed together a compound `ADCl_(6)` is formed. Structural analysis showed that `AdCl_(6)` is an ionic compound. Given that `DCl_(3)` is trigonal planar and `Acl_(3)` is trigonal pyramidal, predict the shape of the anion `DCl_(4)^(-)`.

Options

A. See - saw
B. Perfect tetrahedral
C. Square planar
D. None of these

Answer

To solve the problem, we need to analyze the information given about the compounds ACl₃ and DCl₃, and then predict the shape of the anion DCl₄⁻.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Structures**:
- ACl₃ is described as trigonal pyramidal.
- DCl₃ is described as trigonal planar.

2. **Determine the Hybridization of A and D**:
- Since ACl₃ is trigonal pyramidal, the central atom A has a hybridization of sp³, which implies it has one lone pair and three bond pairs.
- DCl₃ being trigonal planar indicates that the central atom D has a hybridization of sp², which means it has no lone pairs and three bond pairs.

3. **Understand the Formation of ADCl₆**:
- When ACl₃ and DCl₃ are mixed, they form the compound ADCl₆. This compound is ionic in nature.
- The likely ionic forms are DCl₂⁺ (cation) and ACl₄⁻ (anion) or DCl₄⁻ (anion) and ACl₂⁺ (cation).

4. **Focus on the Anion DCl₄⁻**:
- We need to predict the shape of the anion DCl₄⁻. Since D is from the boron family (group 13), it has three valence electrons. When it forms DCl₄⁻, it gains an additional electron, resulting in four bond pairs and no lone pairs.

5. **Determine the Shape of DCl₄⁻**:
- With four bond pairs and no lone pairs, the molecular geometry of DCl₄⁻ will be tetrahedral. This is due to the arrangement of the four bonding pairs around the central atom D, which minimizes the repulsion between them.

6. **Conclusion**:
- The shape of the anion DCl₄⁻ is tetrahedral.

### Final Answer:
The shape of the anion DCl₄⁻ is tetrahedral.