Which of the following correct?

Options

A. `Ti(NO_(3))_(4)` is coloured
B. `[Cr(NH_(3))_(6)]Cl_(3)` is colourless
C. `K_(3)[VF_(6)]` is colourless
D. `[Cu(NC CH_(3))_(4)] [BF_(4)]` is colourless

Explanation

To determine which of the given complexes is colored or colorless, we need to analyze the electronic configurations of the central metal ions in each complex and check for the presence of vacant d orbitals. A compound is colored if it can undergo d-d transitions, which requires vacant d orbitals.

### Step-by-Step Solution:

1. **Analyze the first complex: Ti(NO3)4**
- **Atomic Number of Titanium (Ti)**: 22
- **Electronic Configuration**: [Ar] 3d² 4s²
- **Oxidation State**: +4
- **New Configuration after losing 4 electrons**: [Ar] 3d⁰
- **Conclusion**: No electrons in d orbitals, hence no d-d transition. **Colorless**.

2. **Analyze the second complex: Cr(NH3)6Cl3**
- **Atomic Number of Chromium (Cr)**: 24
- **Electronic Configuration**: [Ar] 3d⁴ 4s¹
- **Oxidation State**: +3
- **New Configuration after losing 3 electrons**: [Ar] 3d³
- **Conclusion**: Electrons are present in d orbitals, and there are vacant d orbitals. **Colored**.

3. **Analyze the third complex: K3VF6**
- **Atomic Number of Vanadium (V)**: 23
- **Electronic Configuration**: [Ar] 3d³ 4s²
- **Oxidation State**: +3
- **New Configuration after losing 3 electrons**: [Ar] 3d²
- **Conclusion**: Electrons are present in d orbitals, and there are vacant d orbitals. **Colored**.

4. **Analyze the fourth complex: Cu(NH3)4BF4**
- **Atomic Number of Copper (Cu)**: 29
- **Electronic Configuration**: [Ar] 3d¹⁰ 4s¹
- **Oxidation State**: +1
- **New Configuration after losing 1 electron**: [Ar] 3d¹⁰
- **Conclusion**: All d orbitals are filled, hence no vacant d orbitals. **Colorless**.

### Final Conclusion:
- **Ti(NO3)4**: Colorless (Incorrectly stated as colored)
- **Cr(NH3)6Cl3**: Colored (Incorrectly stated as colorless)
- **K3VF6**: Colored (Incorrectly stated as colorless)
- **Cu(NH3)4BF4**: Colorless (Correctly stated as colorless)

Thus, the correct answer is **Cu(NH3)4BF4**, which is colorless.