Amongst `TiF_6^(2-), CoF_6^(3-), Cu_2Cl_2" and " NiCl_4^(2-)`, which are the colourless species ? (atomic number of Ti = 22 , Co = 27, Cu = 29, Ni = 28)

Options

A. 1. `CoF_6^(3-)and NiCl_4^(2-)`
B. 2. `TiF_6^(2-)and Cu_2Cl_2`
C. 3. `Cu_2Cl_2 and NiCl_4^(2-)`
D. 4. `TiF_6^(2-)and CoF_6^(2-)`

Explanation

To determine which of the given species are colorless, we need to analyze the electronic configurations of the metal ions in each complex and see if they have any d-electrons. Color arises in transition metal complexes due to the presence of unpaired d-electrons. If there are no d-electrons, the species will be colorless.

### Step-by-Step Solution:

1. **Identify the complexes and their metal ions**:
- \( \text{TiF}_6^{2-} \)
- \( \text{CoF}_6^{3-} \)
- \( \text{Cu}_2\text{Cl}_2 \)
- \( \text{NiCl}_4^{2-} \)

2. **Determine the oxidation states of the metal ions**:
- For \( \text{TiF}_6^{2-} \): Titanium (Ti) is in the +4 oxidation state.
- For \( \text{CoF}_6^{3-} \): Cobalt (Co) is in the +3 oxidation state.
- For \( \text{Cu}_2\text{Cl}_2 \): Copper (Cu) is in the +1 oxidation state (each Cu is +1).
- For \( \text{NiCl}_4^{2-} \): Nickel (Ni) is in the +2 oxidation state.

3. **Write the electronic configurations of the metal ions**:
- **Titanium (Ti)**: Atomic number = 22
- Electronic configuration: \( [Ar] 4s^2 3d^2 \)
- In \( \text{TiF}_6^{2-} \), Ti loses 4 electrons (2 from 4s and 2 from 3d), resulting in \( 3d^0 \) (no d-electrons).
- **Cobalt (Co)**: Atomic number = 27
- Electronic configuration: \( [Ar] 4s^2 3d^7 \)
- In \( \text{CoF}_6^{3-} \), Co loses 3 electrons (2 from 4s and 1 from 3d), resulting in \( 3d^6 \) (6 d-electrons).
- **Copper (Cu)**: Atomic number = 29
- Electronic configuration: \( [Ar] 4s^2 3d^{10} \)
- In \( \text{Cu}_2\text{Cl}_2 \), each Cu loses 1 electron (from 4s), resulting in \( 3d^{10} \) (10 d-electrons).
- **Nickel (Ni)**: Atomic number = 28
- Electronic configuration: \( [Ar] 4s^2 3d^8 \)
- In \( \text{NiCl}_4^{2-} \), Ni loses 2 electrons (from 4s), resulting in \( 3d^8 \) (8 d-electrons).

4. **Determine the color of each species**:
- \( \text{TiF}_6^{2-} \): Colorless (no d-electrons).
- \( \text{CoF}_6^{3-} \): Colored (6 d-electrons).
- \( \text{Cu}_2\text{Cl}_2 \): Colorless (10 d-electrons, but fully paired).
- \( \text{NiCl}_4^{2-} \): Colored (8 d-electrons).

5. **Conclusion**:
The colorless species among the given options are \( \text{TiF}_6^{2-} \) and \( \text{Cu}_2\text{Cl}_2 \).

### Final Answer:
- The colorless species are: **\( \text{TiF}_6^{2-} \) and \( \text{Cu}_2\text{Cl}_2 \)**.