An ion `M^(2+)` forms the complexes. `[M(H_(2)O)_(6)]^(2+)" "[M(en)_(3)]^(2+)" "[MBr_(6)]^(4-)`
Match the complex with the appropriate colour respectively :

Options

A. Green, blue and red
B. Blue, red and green
C. Green, red and blue
D. Red, blue and green

Explanation

To solve the problem of matching the complexes with their respective colors, we will follow these steps:

### Step 1: Identify the complexes and their ligands
The complexes given are:
1. \([M(H_2O)_6]^{2+}\)
2. \([M(en)_3]^{2+}\)
3. \([MBr_6]^{4-}\)

The ligands in these complexes are:
- \(H_2O\) (water) - a weak field ligand
- \(en\) (ethylenediamine) - a strong field ligand
- \(Br^-\) (bromide) - a weak field ligand

### Step 2: Determine the field strength of the ligands
- Weak field ligands (like \(H_2O\) and \(Br^-\)) typically produce larger splitting of the d-orbitals, resulting in higher energy transitions and longer wavelengths of light absorbed.
- Strong field ligands (like \(en\)) produce smaller splitting of the d-orbitals, resulting in lower energy transitions and shorter wavelengths of light absorbed.

### Step 3: Analyze the color emitted by each complex
- For \([M(H_2O)_6]^{2+}\): As a weak field ligand, it absorbs light in the green region of the spectrum, thus it emits red light.
- For \([M(en)_3]^{2+}\): As a strong field ligand, it absorbs light in the red region, thus it emits green light.
- For \([MBr_6]^{4-}\): As a weak field ligand, it absorbs light in the orange region, thus it emits blue light.

### Step 4: Match the complexes with their respective colors
Based on the analysis:
1. \([M(H_2O)_6]^{2+}\) emits **red** light.
2. \([M(en)_3]^{2+}\) emits **green** light.
3. \([MBr_6]^{4-}\) emits **blue** light.

### Final Matching
- \([M(H_2O)_6]^{2+}\) → Red
- \([M(en)_3]^{2+}\) → Green
- \([MBr_6]^{4-}\) → Blue

### Summary of the solution:
- \([M(H_2O)_6]^{2+}\) → Red
- \([M(en)_3]^{2+}\) → Green
- \([MBr_6]^{4-}\) → Blue