It is non experimental fact that `Cs_(2)[CuCl_(4)]` is orange coloured but `(NH_(4))_(2)[CuCl_(4)]` is yellow. It is further known that total paramagnetic moment of a unpaired electron is due to spin as well as due to nature of orbital , 'd' orbital contributing more than 's' or 'p'. Thus the total paramgentic moment of orange compound is found to be more than that of yellow compound. Then which of the following is correct?

Options

A. Anion of orange compound is tetrahedral and that of yellow is square planar
B. Anion of orange compound is square planar and that of yellow is tetrahedral
C. Both the anions are tetrahedral
D. Both the anions are square planar.

Explanation

To solve the problem, we need to analyze the two coordination compounds given: \( Cs_2[CuCl_4] \) (orange) and \( (NH_4)_2[CuCl_4] \) (yellow). We will look into their electronic configurations, oxidation states, geometries, and paramagnetic properties.

### Step-by-Step Solution:

1. **Identify the oxidation state of Copper in both compounds**:
- In both \( Cs_2[CuCl_4] \) and \( (NH_4)_2[CuCl_4] \), copper is in the +2 oxidation state.

2. **Determine the electronic configuration of Copper**:
- The electronic configuration of Copper (Cu) in its elemental state is \( [Ar] 3d^{10} 4s^1 \).
- In the +2 oxidation state, the configuration becomes \( [Ar] 3d^9 \) (since two electrons are removed, typically from the 4s orbital first).

3. **Analyze the geometry of \( Cs_2[CuCl_4] \)**:
- \( Cs_2[CuCl_4] \) is orange in color and exhibits tetrahedral geometry.
- In tetrahedral complexes, the hybridization is \( sp^3 \).
- The unpaired electron in \( 3d^9 \) configuration remains in the \( d \) orbital, contributing to its paramagnetism.

4. **Analyze the geometry of \( (NH_4)_2[CuCl_4] \)**:
- \( (NH_4)_2[CuCl_4] \) is yellow in color and exhibits square planar geometry.
- In square planar complexes, the hybridization is \( dsp^2 \).
- Here, the unpaired electron from the \( 3d^9 \) configuration is promoted to a higher energy level (either \( s \) or \( p \)), leading to the formation of a vacant \( d \) orbital.

5. **Compare the paramagnetic moments**:
- The orange compound \( Cs_2[CuCl_4] \) has a higher total paramagnetic moment due to the presence of an unpaired electron in the \( d \) orbital.
- The yellow compound \( (NH_4)_2[CuCl_4] \) has a lower total paramagnetic moment because the unpaired electron is in a higher energy orbital, leading to less contribution from the \( d \) orbital.

6. **Conclusion**:
- The orange compound \( Cs_2[CuCl_4] \) is tetrahedral, while the yellow compound \( (NH_4)_2[CuCl_4] \) is square planar. Therefore, the correct statement is that the orange compound has tetrahedral geometry and the yellow compound has square planar geometry.

### Final Answer:
The correct option is that the orange compound is tetrahedral and the yellow compound is square planar.

---