According to valence bond theory the geometry of `[Ni(CO)_(4)]` is________ .

Answer

To determine the geometry of the coordination compound \([Ni(CO)_{4}]\) according to valence bond theory (VBT), we can follow these steps:

### Step 1: Determine the Oxidation State of Nickel
- Let the oxidation state of nickel be \(X\).
- Carbonyl (CO) is a neutral ligand, which means it has a charge of 0.
- The overall charge of the coordination complex is 0.

Thus, we have:
\[ X + 0 \cdot 4 = 0 \implies X = 0 \]
So, the oxidation state of nickel in \([Ni(CO)_{4}]\) is 0.

**Hint:** Remember that the charge of neutral ligands like CO is zero when calculating oxidation states.

### Step 2: Write the Electron Configuration of Nickel
- Nickel has an atomic number of 28.
- The electron configuration of nickel in its ground state is:
\[ [Ar] 4s^2 3d^8 \]

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

### Step 3: Draw the Orbital Diagram
- The orbital diagram for nickel will show:
- 3d orbitals: 8 electrons
- 4s orbital: 2 electrons
- 4p orbital: empty

**Hint:** Visualizing the electron distribution in orbitals can help in understanding hybridization.

### Step 4: Consider the Ligand Field Strength
- Carbon monoxide (CO) is a strong field ligand, which means it will cause pairing of electrons in the d-orbitals.
- The pairing will occur as follows:
- The two electrons in the 4s orbital will be promoted to the 3d orbital, leading to a configuration of \(3d^{10} 4s^0\).

**Hint:** Strong field ligands can cause electron pairing, which is crucial for determining hybridization.

### Step 5: Determine the Hybridization
- After the pairing of electrons, we have:
- 3d: 10 electrons (fully filled)
- 4s: 0 electrons (empty)
- 4p: empty
- Since we need four orbitals to accommodate the four CO ligands, we use the \(sp^3\) hybridization.

**Hint:** The number of bonds formed by the central atom can indicate the type of hybridization.

### Step 6: Predict the Geometry
- The \(sp^3\) hybridization corresponds to a tetrahedral geometry.
- Therefore, the geometry of \([Ni(CO)_{4}]\) is tetrahedral.

**Hint:** Remember that the hybridization type directly correlates with the molecular geometry.

### Final Answer
The geometry of \([Ni(CO)_{4}]\) is **tetrahedral**.