The hybridisation of Ni in `[Ni(CO)_(4)]` is

Options

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

Answer

To determine the hybridization of nickel in the complex \([Ni(CO)_{4}]\), we can follow these steps:

### Step 1: Determine the oxidation state of nickel
In the complex \([Ni(CO)_{4}]\), carbon monoxide (CO) is a neutral ligand. Therefore, the oxidation state of nickel (Ni) is 0.

### Step 2: Write the electronic configuration of nickel
The atomic number of nickel is 28. Its electronic configuration is:
\[ \text{Ni: } [Ar] 4s^2 3d^8 \]

### Step 3: Consider the effect of the ligand
CO is a strong field ligand, which means it can cause pairing of electrons in the d-orbitals. In this case, the two electrons from the 4s orbital will pair up with the 3d electrons, resulting in:
\[ 3d^{10} \]

### Step 4: Identify the orbitals involved in hybridization
Since the CO ligand donates electron pairs, it will utilize the empty 4s and 4p orbitals of nickel for hybridization. The hybridization involves:
- 1 4s orbital
- 3 4p orbitals

### Step 5: Determine the hybridization type
The combination of one 4s orbital and three 4p orbitals leads to the formation of four equivalent hybrid orbitals. This type of hybridization is known as \(sp^3\) hybridization.

### Step 6: Confirm the geometry of the complex
The geometry of \([Ni(CO)_{4}]\) is tetrahedral, which is consistent with \(sp^3\) hybridization.

### Conclusion
Thus, the hybridization of nickel in \([Ni(CO)_{4}]\) is \(sp^3\).

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