`[Co(H_(2)O)_(6)]^(3+)` complex is

Options

A. a.High spin complex
B. b.Having `d^(2)sp^(3)`-hybridization
C. c.Low spin complex
D. d.Having octahedral structure .

Answer

To analyze the complex `[Co(H₂O)₆]³⁺`, we will follow these steps:

### Step 1: Determine the oxidation state of cobalt.
The complex is `[Co(H₂O)₆]³⁺`. Since water (H₂O) is a neutral ligand, it does not contribute to the oxidation state. Therefore, we can set up the equation:

\[ \text{Oxidation state of Co} + 0 = +3 \]

This implies that the oxidation state of cobalt (Co) is +3.

### Step 2: Write the electronic configuration of cobalt in the +3 oxidation state.
The atomic number of cobalt (Co) is 27. The ground state electronic configuration of cobalt is:

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

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

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

### Step 3: Analyze the ligand and its field strength.
The ligand in this complex is H₂O, which is generally considered a weak field ligand. However, in the case of cobalt in the +3 oxidation state, H₂O behaves as a strong field ligand. Strong field ligands tend to cause pairing of electrons in the lower energy d orbitals.

### Step 4: Fill the d orbitals.
For the 3d⁶ configuration, we fill the d orbitals considering that H₂O is a strong field ligand, leading to pairing of electrons:

1. The first two electrons will occupy the first d orbital (pairing).
2. The next two electrons will occupy the second d orbital (pairing).
3. The last two electrons will occupy the third d orbital (pairing).

The resulting configuration will be:

\[ \text{3d: } \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \]

This indicates that all six electrons are paired, confirming that it is a low spin complex.

### Step 5: Determine the hybridization.
For a coordination number of 6, the hybridization involves two d orbitals, one s orbital, and three p orbitals:

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

### Step 6: Identify the geometry of the complex.
With a coordination number of 6 and d²sp³ hybridization, the geometry of the complex is octahedral.

### Conclusion:
The complex `[Co(H₂O)₆]³⁺` is a low spin complex with d²sp³ hybridization and has an octahedral structure.

### Summary of Findings:
- Oxidation state of Co: +3
- Electronic configuration: 3d⁶
- Type of complex: Low spin
- Hybridization: d²sp³
- Geometry: Octahedral