The hybridization of `[CoF_(6)]^(3-)` and `[Co(C_(2)O_(4))}^(3-)` are :

Options

A. both `sp^(3)d^(2)`
B. both `d^(2)sp^(3)`
C. `sp^(3)d^(2)` and `d^(2)sp^(3)` respectively
D. `d^(2)sp^(3)` and `sp^(3)d^(2)` respectively

Answer

To determine the hybridization of the complexes \([CoF_6]^{3-}\) and \([Co(C_2O_4)]^{3-}\), we will follow these steps:

### Step 1: Determine the oxidation state of cobalt in both complexes.

1. **For \([CoF_6]^{3-}\)**:
- Let the oxidation state of cobalt be \(x\).
- The oxidation state of fluorine (F) is \(-1\).
- The overall charge of the complex is \(-3\).
- Therefore, the equation is:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]

2. **For \([Co(C_2O_4)]^{3-}\)**:
- The oxalate ion \((C_2O_4)^{2-}\) has an overall charge of \(-2\).
- Let the oxidation state of cobalt again be \(x\).
- The overall charge of the complex is \(-3\).
- Therefore, the equation is:
\[ x + (-2) = -3 \implies x - 2 = -3 \implies x = +3 \]

### Step 2: Identify the ligand types and their coordination numbers.

- **For \([CoF_6]^{3-}\)**:
- Fluoride (F) is a weak field ligand and is monodentate.
- The coordination number is \(6\) (since there are 6 fluoride ligands).

- **For \([Co(C_2O_4)]^{3-}\)**:
- The oxalate ion \((C_2O_4)^{2-}\) is a bidentate ligand.
- The coordination number is also \(6\) (since it can bind through two sites).

### Step 3: Determine the hybridization based on the ligands and oxidation states.

1. **For \([CoF_6]^{3-}\)**:
- Cobalt in the +3 oxidation state has the electronic configuration of \([Ar] 3d^7\).
- In the presence of weak field ligands (like F), the electrons will fill the higher energy orbitals before pairing occurs.
- The hybridization will be \(sp^3d^2\) because the 4s orbital is empty and the 4p orbitals are not involved in bonding. The 3d orbitals will be used for hybridization.
- Therefore, the hybridization is \(sp^3d^2\).

2. **For \([Co(C_2O_4)]^{3-}\)**:
- The oxalate ion is a strong field ligand, which causes pairing of electrons.
- The 3d orbitals will fill first, leading to pairing before filling the higher energy orbitals.
- The hybridization will be \(d^2sp^3\) because the inner d orbitals are used for hybridization.
- Therefore, the hybridization is \(d^2sp^3\).

### Final Answer:
- The hybridization of \([CoF_6]^{3-}\) is \(sp^3d^2\).
- The hybridization of \([Co(C_2O_4)]^{3-}\) is \(d^2sp^3\).