For `[FeF_(6)]^(3-) and [CoF_(6)]^(3-)`, the correct statement is

Options

A. both are coloured
B. both are colourless
C. `[FeF_(6)]^(3-)` is coloured and `[CoF_(6)]^(3-)` is colourless
D. `[FeF_(6)]6(3-)` is colourless and `[CoF_(6)]^(3-)` is coloured

Answer

To determine the correct statement regarding the complexes \([FeF_6]^{3-}\) and \([CoF_6]^{3-}\), we will analyze the oxidation states, electronic configurations, and magnetic properties of both complexes.

### Step 1: Determine the oxidation state of Iron in \([FeF_6]^{3-}\)

1. Let the oxidation state of Iron (Fe) be \(x\).
2. The fluorine (F) has an oxidation state of \(-1\).
3. The equation for the oxidation state can be set up as:
\[ x + 6(-1) = -3 \]
\[ x - 6 = -3 \]
\[ x = +3 \]

### Step 2: Find the electronic configuration of \([FeF_6]^{3-}\)

1. The atomic number of Iron (Fe) is 26, so its ground state electronic configuration is:
\[ [Ar] 3d^6 4s^2 \]
2. In the +3 oxidation state, Iron loses 2 electrons from the 4s orbital and 1 electron from the 3d orbital:
\[ [Ar] 3d^5 4s^0 \]

### Step 3: Analyze the d-orbital filling for \([FeF_6]^{3-}\)

1. The d-orbital configuration for \([FeF_6]^{3-}\) is \(3d^5\).
2. Since F is a weak field ligand, the electrons will fill the d-orbitals singly before pairing. The configuration will be:
\[ \uparrow \uparrow \uparrow \uparrow \uparrow \]
This means there are 5 unpaired electrons.

### Step 4: Determine the magnetic property of \([FeF_6]^{3-}\)

1. The number of unpaired electrons is 5, which means \([FeF_6]^{3-}\) is paramagnetic.
2. However, since all the d-orbitals are filled in a way that does not allow for d-d transitions (due to the same energy level), it is colorless.

### Step 5: Determine the oxidation state of Cobalt in \([CoF_6]^{3-}\)

1. Let the oxidation state of Cobalt (Co) be \(y\).
2. Setting up the equation:
\[ y + 6(-1) = -3 \]
\[ y - 6 = -3 \]
\[ y = +3 \]

### Step 6: Find the electronic configuration of \([CoF_6]^{3-}\)

1. The atomic number of Cobalt (Co) is 27, so its ground state electronic configuration is:
\[ [Ar] 3d^7 4s^2 \]
2. In the +3 oxidation state, Cobalt loses 2 electrons from the 4s orbital and 1 electron from the 3d orbital:
\[ [Ar] 3d^6 4s^0 \]

### Step 7: Analyze the d-orbital filling for \([CoF_6]^{3-}\)

1. The d-orbital configuration for \([CoF_6]^{3-}\) is \(3d^6\).
2. Again, since F is a weak field ligand, the configuration will be:
\[ \uparrow \uparrow \uparrow \uparrow \uparrow \downarrow \]
This means there are 4 unpaired electrons.

### Step 8: Determine the magnetic property of \([CoF_6]^{3-}\)

1. The number of unpaired electrons is 4, which means \([CoF_6]^{3-}\) is also paramagnetic.
2. Additionally, since there are unpaired electrons, d-d transitions can occur, making the complex colored.

### Conclusion

Based on the analysis:
- \([FeF_6]^{3-}\) is colorless and paramagnetic.
- \([CoF_6]^{3-}\) is colored and paramagnetic.

Thus, the correct statement is:
**\([FeF_6]^{3-}\) is colorless and \([CoF_6]^{3-}\) is colored.**

### Final Answer
**Option D: \([FeF_6]^{3-}\) is colorless and \([CoF_6]^{3-}\) is colored.**