The type of hybridization involved in the metal ion of `[Ni(H_2O)_6]^(2+)` complex is

Options

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

Answer

To determine the type of hybridization involved in the metal ion of the complex \([Ni(H_2O)_6]^{2+}\), we will follow these steps:

### Step 1: Identify the oxidation state of Nickel in the complex
The complex \([Ni(H_2O)_6]^{2+}\) has a 2+ charge. Since water (H₂O) is a neutral ligand, the oxidation state of Nickel (Ni) in this complex is +2.

### Step 2: Write the electronic configuration of Nickel
The atomic number of Nickel (Ni) is 28. The electronic configuration of neutral Nickel is:
\[ Ni: [Ar] \, 3d^8 \, 4s^2 \]
When Nickel loses 2 electrons to form \(Ni^{2+}\), the configuration becomes:
\[ Ni^{2+}: [Ar] \, 3d^8 \]

### Step 3: Determine the hybridization
In the case of the complex \([Ni(H_2O)_6]^{2+}\), Nickel is surrounded by six water molecules, which are acting as ligands. This indicates that the geometry of the complex is octahedral.

For an octahedral complex, the hybridization can be determined as follows:
- The \(3d\), \(4s\), and \(4p\) orbitals will mix to form hybrid orbitals.
- In this case, Nickel \(Ni^{2+}\) with the configuration \(3d^8\) will undergo \(sp^3d^2\) hybridization to accommodate six ligands in an octahedral arrangement.

### Step 4: Conclusion
Thus, the type of hybridization involved in the metal ion of the complex \([Ni(H_2O)_6]^{2+}\) is \(sp^3d^2\).

### Final Answer
The type of hybridization involved in the metal ion of \([Ni(H_2O)_6]^{2+}\) complex is \(sp^3d^2\).

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