Hybridization shape and magnetic moment of `K_(3)[Co(CO)_(6)]` is

Options

A. `d^2sp^3`. Octahedral , 4.9 BM
B. `sp^3d^2`, octahedral, 4.9 BM
C. `dsp^2`, square planer, 4.9 BM
D. `sp^3`, tetrahedral , 4.9 BM

Answer

To determine the hybridization, shape, and magnetic moment of \( K_3[Co(CO)_6] \), we can follow these steps:

### Step 1: Determine the Oxidation State of Cobalt
In the complex \( K_3[Co(CO)_6] \), potassium (K) has a +1 oxidation state. Since there are three potassium ions, their total contribution to the charge is +3. The carbonyl (CO) ligands are neutral. Therefore, the oxidation state of cobalt can be calculated as follows:

\[ \text{Oxidation state of Co} + 3 \times (+1) = 0 \]
\[ \text{Oxidation state of Co} = -3 \]

### Step 2: Determine the Electronic Configuration of Cobalt
Cobalt has an atomic number of 27. The ground state electronic configuration of cobalt is:

\[ \text{Co: } [Ar] 4s^2 3d^7 \]

When cobalt is in the +3 oxidation state, it loses two electrons from the 4s subshell and one from the 3d subshell:

\[ \text{Co}^{3+}: [Ar] 3d^6 \]

### Step 3: Identify the Hybridization
Since carbonyl (CO) is a weak field ligand, it does not cause pairing of electrons in the 3d orbitals. Thus, the 3d electrons remain unpaired. The arrangement of the 3d electrons will be:

\[ 3d: \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \]

For hybridization, cobalt will use one s orbital and three p orbitals along with two d orbitals, leading to:

\[ \text{Hybridization: } sp^3d^2 \]

### Step 4: Determine the Geometry
The hybridization \( sp^3d^2 \) corresponds to an octahedral geometry. Thus, the shape of the complex is octahedral.

### Step 5: Calculate the Magnetic Moment
To find the magnetic moment, we need to determine the number of unpaired electrons. In \( Co^{3+} \) with a \( 3d^6 \) configuration and weak field ligands, there are 4 unpaired electrons.

The formula for magnetic moment (\( \mu \)) is given by:

\[ \mu = \sqrt{n(n + 2)} \]

Where \( n \) is the number of unpaired electrons. Substituting \( n = 4 \):

\[ \mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.9 \, \text{Bohr magnetons} \]

### Summary
- **Hybridization**: \( sp^3d^2 \)
- **Shape**: Octahedral
- **Magnetic Moment**: Approximately \( 4.9 \, \text{Bohr magnetons} \)