The complex ion `[Co(NH_(3))_(6)]^(3+)` is formed by `sp^(3) d^(2)` hybridiration. Hence, the ion should posses

Options

A. octahedral geometry
B. tetrahedral geometry
C. square planar geometry
D. tetragonal geometry

Explanation

To solve the question regarding the complex ion \([Co(NH_3)_6]^{3+}\) and its hybridization and geometry, we can follow these steps:

### Step 1: Determine the oxidation state of cobalt in the complex
- The overall charge of the complex ion is \(+3\).
- Amine (\(NH_3\)) is a neutral ligand, contributing \(0\) to the charge.
- Let the oxidation state of cobalt be \(x\).

\[ x + 6(0) = +3 \implies x = +3 \]

### Step 2: Write the electronic configuration of cobalt
- The atomic number of cobalt (Co) is \(27\).
- The ground state electronic configuration of cobalt is:

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

### Step 3: Determine the electronic configuration of \(Co^{3+}\)
- For \(Co^{3+}\), we remove three electrons. The two electrons from the \(4s\) subshell and one from the \(3d\) subshell will be removed:

\[ [Ar] 3d^6 \]

### Step 4: Identify the hybridization
- The complex ion \([Co(NH_3)_6]^{3+}\) has six ligands, which suggests that it will adopt an octahedral geometry.
- In an octahedral geometry, the hybridization involves the \(d\), \(s\), and \(p\) orbitals.
- The hybridization for this complex is \(d^2sp^3\).

### Step 5: Confirm the geometry
- Since the hybridization is \(d^2sp^3\), the geometry of the complex ion \([Co(NH_3)_6]^{3+}\) is octahedral.

### Conclusion
The complex ion \([Co(NH_3)_6]^{3+}\) possesses octahedral geometry.

### Final Answer
- The ion should possess **octahedral geometry**.

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