Identify the colourless complexes

Options

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

Answer

To identify the colorless complexes among the given options, we need to analyze each complex based on the oxidation state of the central metal ion and its electronic configuration. The presence or absence of unpaired electrons in the d-orbitals will determine if a complex is colorless or colored. Colorless complexes typically have either a fully filled d-orbital configuration or no d-electrons at all, which prevents d-d transitions that are responsible for color.

### Step-by-Step Solution:

1. **Identify the first complex: Ti(NO3)4**
- **Oxidation State Calculation**: The oxidation state of titanium (Ti) can be calculated as follows:
\[ X + 4(-1) = 0 \implies X = +4 \]
- **Electronic Configuration**: The atomic number of titanium is 22. Thus, its electronic configuration is:
\[ \text{Ti: } [\text{Ar}] 4s^2 3d^2 \]
For Ti in +4 oxidation state, it loses 4 electrons:
\[ \text{Ti}^{4+}: [\text{Ar}] 3d^0 \]
- **Conclusion**: Since there are no d-electrons, d-d transitions are not possible. Therefore, Ti(NO3)4 is colorless.

2. **Identify the second complex: Cu(CH3)4PF6**
- **Oxidation State Calculation**: The oxidation state of copper (Cu) is calculated as follows:
\[ X + 4(0) + (-1) = +1 \implies X = +1 \]
- **Electronic Configuration**: The atomic number of copper is 29. Thus, its electronic configuration is:
\[ \text{Cu: } [\text{Ar}] 4s^1 3d^{10} \]
For Cu in +1 oxidation state, it loses 1 electron:
\[ \text{Cu}^{+1}: [\text{Ar}] 3d^{10} \]
- **Conclusion**: Since the d-orbitals are fully filled (10 electrons), d-d transitions are not possible. Therefore, Cu(CH3)4PF6 is colorless.

3. **Identify the third complex: Cr(NH3)6Cl3**
- **Oxidation State Calculation**: The oxidation state of chromium (Cr) is:
\[ X + 6(0) + 3(-1) = +3 \implies X = +3 \]
- **Electronic Configuration**: The atomic number of chromium is 24. Its electronic configuration is:
\[ \text{Cr: } [\text{Ar}] 4s^2 3d^4 \]
For Cr in +3 oxidation state, it loses 3 electrons:
\[ \text{Cr}^{3+}: [\text{Ar}] 3d^3 \]
- **Conclusion**: Since there are unpaired electrons in the d-orbitals (3 electrons), d-d transitions are possible. Therefore, Cr(NH3)6Cl3 is colored.

4. **Identify the fourth complex: K3VF6**
- **Oxidation State Calculation**: The oxidation state of vanadium (V) is:
\[ X + 3(-1) = +3 \implies X = +3 \]
- **Electronic Configuration**: The atomic number of vanadium is 23. Its electronic configuration is:
\[ \text{V: } [\text{Ar}] 4s^2 3d^3 \]
For V in +3 oxidation state, it loses 3 electrons:
\[ \text{V}^{3+}: [\text{Ar}] 3d^2 \]
- **Conclusion**: Since there are unpaired electrons in the d-orbitals (2 electrons), d-d transitions are possible. Therefore, K3VF6 is colored.

### Final Conclusion:
The colorless complexes among the given options are:
- **Ti(NO3)4**
- **Cu(CH3)4PF6**