The hybridization in `[CO(NH_3)_6]^(3+)` is :

Options

A. `dsp^2`
B. `sp^3d^2`
C. `sp^3`
D. `d^2sp^3`

Explanation

To determine the hybridization of the complex ion \([CO(NH_3)_6]^{3+}\), we can follow these steps:

### Step 1: Determine the Oxidation State of Cobalt
The overall charge of the complex is \(+3\). Ammonia (\(NH_3\)) is a neutral ligand, which means it does not contribute any charge. Therefore, the oxidation state of cobalt in this complex can be calculated as follows:

\[ \text{Oxidation state of Co} + 6 \times 0 = +3 \implies \text{Oxidation state of Co} = +3 \]

### Step 2: Write the Electron Configuration of Cobalt
Cobalt (Co) has an atomic number of 27. The electron configuration of neutral cobalt is:

\[ [Ar] \, 3d^7 \, 4s^2 \]

For cobalt in the \(+3\) oxidation state, we remove three electrons (two from 4s and one from 3d):

\[ \text{Electron configuration of } Co^{3+} = [Ar] \, 3d^6 \]

### Step 3: Analyze the Electron Configuration
In the \(3d^6\) configuration, the electrons are arranged as follows:

- The \(3d\) subshell can hold a maximum of 10 electrons.
- In the presence of a strong field ligand like ammonia, the \(3d\) electrons will pair up.

The arrangement of electrons in the \(3d\) subshell after pairing will be:

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

### Step 4: Determine the Hybridization
Since there are 6 ligands (ammonia) surrounding the cobalt ion, we need to consider the hybridization that accommodates these 6 ligands. The pairing of electrons allows for the involvement of \(d\), \(s\), and \(p\) orbitals in hybridization.

The hybridization can be determined as follows:

- 2 \(s\) orbitals
- 3 \(p\) orbitals
- 2 \(d\) orbitals

This leads to the hybridization type:

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

### Step 5: Conclusion
Thus, the hybridization of the complex \([CO(NH_3)_6]^{3+}\) is \(d^2sp^3\).