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

Explanation

To determine which of the species \( TiF_6^{2-}, CoF_6^{3-}, Cu_2Cl_2, \) and \( NiCl_4^{2-} \) are colorless, we need to analyze their electronic configurations and the presence of unpaired electrons. Colorless species are typically diamagnetic, meaning they have no unpaired electrons.

### Step-by-Step Solution:

1. **Identify the oxidation states and electronic configurations:**
- **For \( TiF_6^{2-} \):**
- The oxidation state of Titanium (Ti) can be calculated as follows:
\[ x + 6(-1) = -2 \implies x - 6 = -2 \implies x = +4 \]
- The ground state electronic configuration of Titanium (atomic number 22) is \( [Ar] 3d^2 4s^2 \).
- In the +4 oxidation state, Titanium loses 4 electrons:
\[ 3d^0 4s^0 \implies \text{All electrons are paired.} \]
- Therefore, \( TiF_6^{2-} \) is colorless.

2. **Analyze \( CoF_6^{3-} \):**
- The oxidation state of Cobalt (Co) is calculated as:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- The ground state configuration of Cobalt (atomic number 27) is \( [Ar] 3d^7 4s^2 \).
- In the +3 oxidation state, Cobalt loses 3 electrons:
\[ 3d^6 4s^0 \]
- Fluoride is a weak field ligand, so there will be no pairing of electrons in the \( 3d \) subshell:
- The configuration will be \( 3d^6 \) with 4 unpaired electrons.
- Therefore, \( CoF_6^{3-} \) is paramagnetic and not colorless.

3. **Evaluate \( Cu_2Cl_2 \):**
- The oxidation state of Copper (Cu) is:
\[ 2x + 2(-1) = 0 \implies 2x - 2 = 0 \implies x = +1 \]
- The ground state configuration of Copper (atomic number 29) is \( [Ar] 3d^{10} 4s^1 \).
- In the +1 oxidation state, Copper loses 1 electron:
\[ 3d^{10} 4s^0 \implies \text{All electrons are paired.} \]
- Therefore, \( Cu_2Cl_2 \) is colorless.

4. **Check \( NiCl_4^{2-} \):**
- The oxidation state of Nickel (Ni) is:
\[ x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2 \]
- The ground state configuration of Nickel (atomic number 28) is \( [Ar] 3d^8 4s^2 \).
- In the +2 oxidation state, Nickel loses 2 electrons:
\[ 3d^8 4s^0 \]
- Chlorine is also a weak field ligand, so there will be no pairing of electrons:
- The configuration will be \( 3d^8 \) with 2 unpaired electrons.
- Therefore, \( NiCl_4^{2-} \) is paramagnetic and not colorless.

### Conclusion:
The colorless species among the given options are:
- \( TiF_6^{2-} \) (colorless)
- \( Cu_2Cl_2 \) (colorless)

### Final Answer:
- The colorless species are \( TiF_6^{2-} \) and \( Cu_2Cl_2 \).