Which of the following complexes may be coloured due to d-d transition ?

Options

A. `Ni(CO)_(4)`
B. `K_(3)[Fe(CN)_(6)]`
C. `Cu(H_(2)O)_(4)^(+2)`
D. `K_(3)[Cu(CN)_(4)]`

Answer

To determine which of the given complexes may be colored due to d-d transitions, we need to analyze the oxidation states of the metal ions in each complex and their electronic configurations. Let's go through each complex step by step.

### Step 1: Analyze Ni(CO)₄
- **Complex:** Ni(CO)₄
- **Oxidation State Calculation:** Let the oxidation state of Ni be \( x \). Since CO is a neutral ligand, the overall charge is 0.
\[ x + 0 = 0 \implies x = 0 \]
- **Electronic Configuration of Ni:** The atomic number of Ni is 28, so its electronic configuration is [Ar] 3d⁸ 4s².
- **d-D Transition:** In Ni(CO)₄, Ni is in the zero oxidation state, meaning it has a full 3d orbital (3d⁸). Since there are no unpaired electrons, it does not undergo d-d transitions and is colorless.

### Step 2: Analyze K₃[Fe(CN)₆]
- **Complex:** K₃[Fe(CN)₆]
- **Oxidation State Calculation:** Let the oxidation state of Fe be \( x \). The charge of CN is -1, and there are 6 CN ligands.
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- **Electronic Configuration of Fe³⁺:** The atomic number of Fe is 26, so its configuration is [Ar] 3d⁶ 4s². For Fe³⁺, it becomes [Ar] 3d⁵.
- **d-D Transition:** The 3d⁵ configuration has unpaired electrons, allowing for d-d transitions. Therefore, K₃[Fe(CN)₆] is colored.

### Step 3: Analyze Cu(H₂O)₄²⁺
- **Complex:** Cu(H₂O)₄²⁺
- **Oxidation State Calculation:** Let the oxidation state of Cu be \( x \). H₂O is neutral, and the overall charge is +2.
\[ x + 0 = +2 \implies x = +2 \]
- **Electronic Configuration of Cu²⁺:** The atomic number of Cu is 29, so its configuration is [Ar] 3d¹⁰ 4s¹. For Cu²⁺, it becomes [Ar] 3d⁹.
- **d-D Transition:** The 3d⁹ configuration has unpaired electrons, allowing for d-d transitions. Therefore, Cu(H₂O)₄²⁺ is colored.

### Step 4: Analyze K₃[Cu(CN)₄]
- **Complex:** K₃[Cu(CN)₄]
- **Oxidation State Calculation:** Let the oxidation state of Cu be \( x \). The charge of CN is -1, and there are 4 CN ligands.
\[ x + 4(-1) = -3 \implies x - 4 = -3 \implies x = +1 \]
- **Electronic Configuration of Cu⁺:** The atomic number of Cu is 29, so its configuration is [Ar] 3d¹⁰ 4s¹. For Cu⁺, it becomes [Ar] 3d¹⁰.
- **d-D Transition:** The 3d¹⁰ configuration is fully filled, meaning there are no unpaired electrons. Therefore, K₃[Cu(CN)₄] does not undergo d-d transitions and is colorless.

### Conclusion:
The complexes that may be colored due to d-d transitions are:
- K₃[Fe(CN)₆] (Option B)
- Cu(H₂O)₄²⁺ (Option C)

### Final Answer:
The complexes that may be colored due to d-d transitions are K₃[Fe(CN)₆] and Cu(H₂O)₄²⁺.

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