`CuSO_(4).5H_(2)O` is blue in colour while `CuSO_(4)` is colourless. Why ?

Options

A. presence of strong field ligand in `CuSO_(4) * 5H_(2)O`
B. absence of water (ligand) , `d-d` transitions are not possible in `CuSO_(4)`
C. anhydrous `CuSO_(4)` undergoes d-d transitions due to crystal field splitting
D. colour is lost due to loss of unpaired electrons .

Explanation

To understand why `CuSO4.5H2O` (copper(II) sulfate pentahydrate) is blue in color while `CuSO4` (anhydrous copper(II) sulfate) is colorless, we can break down the explanation into several steps:

### Step-by-Step Solution:

1. **Understanding the Compounds**:
- `CuSO4` is an anhydrous form of copper(II) sulfate, which means it does not contain water molecules.
- `CuSO4.5H2O` is the hydrated form, which contains five water molecules associated with each copper sulfate unit.

2. **Role of Water as a Ligand**:
- In `CuSO4.5H2O`, the five water molecules act as ligands. Ligands are ions or molecules that can donate a pair of electrons to the central metal atom (in this case, copper) to form a coordination complex.

3. **Crystal Field Theory**:
- According to crystal field theory, when ligands approach the central metal ion, they cause the d-orbitals of the metal to split into different energy levels. This splitting is essential for the color of the complex.
- In the presence of water ligands, the d-orbitals of copper undergo splitting, allowing for electronic transitions between these split energy levels.

4. **d-d Transitions**:
- The blue color of `CuSO4.5H2O` is due to d-d transitions. When electrons in the lower energy d-orbitals absorb specific wavelengths of light, they can be excited to higher energy d-orbitals. The specific wavelengths absorbed correspond to the blue color observed.

5. **Absence of Ligands in Anhydrous Form**:
- In `CuSO4`, there are no water molecules to act as ligands. Therefore, there is no d-orbital splitting, and consequently, no d-d transitions can occur.
- Without these transitions, `CuSO4` does not absorb visible light and appears colorless.

6. **Conclusion**:
- The presence of water molecules in `CuSO4.5H2O` facilitates d-d transitions due to d-orbital splitting, leading to the blue color, while the absence of water in `CuSO4` results in no color.

### Summary:
- `CuSO4.5H2O` is blue because water acts as a ligand, causing d-orbital splitting and allowing for d-d transitions, while `CuSO4` is colorless due to the absence of ligands and d-d transitions.