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)]`, carbonyl (CO) is a neutral ligand. Therefore, the oxidation state of nickel (Ni) can be calculated as follows:

Let the oxidation state of Ni be \( x \).
Since CO is neutral, the overall charge of the complex is 0.

\[ x + 0 = 0 \implies x = 0 \]

**Hint:** Remember that the oxidation state of a neutral ligand is zero.

### Step 2: Write the electron configuration of Ni
Nickel has an atomic number of 28. The electron configuration of Ni in its ground state is:

\[ [Ar] 3d^8 4s^2 \]

**Hint:** Use the periodic table to find the atomic number and corresponding electron configuration.

### Step 3: Analyze the effect of the ligand
Carbonyl (CO) is a strong field ligand. Strong field ligands can cause pairing of electrons in the d-orbitals. In this case, the 4s electrons will be promoted to the 3d orbitals.

Initially, the electron configuration is:

- 3d: ↑ ↑ ↑ ↑ ↑ (5 electrons)
- 4s: ↑ ↑ (2 electrons)

When the strong field ligand CO approaches, one of the 4s electrons is promoted to the 3d orbital, resulting in:

- 3d: ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ (8 electrons)
- 4s: (0 electrons)

**Hint:** Strong field ligands can cause pairing of electrons, so consider how they affect the d-orbitals.

### Step 4: Determine the hybridization
After the promotion of electrons, we have:

- 3d: 8 electrons (fully filled)
- 4s: 0 electrons (empty)

Now, we need to accommodate 4 carbonyl ligands. The hybridization can be determined by the number of orbitals used:

- 1 s orbital (from 4s)
- 3 p orbitals (from 3p)

This gives us a total of 4 hybrid orbitals, which corresponds to \( sp^3 \) hybridization.

**Hint:** Count the number of orbitals involved in bonding to determine the hybridization type.

### Conclusion
The hybridization of Ni in `[Ni(CO)_(4)]` is \( sp^3 \).

**Final Answer:** \( sp^3 \)