The geometry, hybridization and magnetic moment of `[Ni(CO)_(4)]` are …. .. And … respectively.

Options

A. tetrahedral, `sp^(3)`, zero
B. aquare planar `sp^(3),sqrt(3)`
C. tetrahedral `sp^(3),2sqrt(2)`
D. square planar `dsp^(2)`. Zero

Explanation

To solve the question regarding the geometry, hybridization, and magnetic moment of `[Ni(CO)_(4)]`, we can follow these steps:

### Step 1: Determine the oxidation state of Nickel in `[Ni(CO)_(4)]`
- The overall charge of the complex is zero, and since CO is a neutral ligand, the oxidation state of Ni is also 0.

**Hint:** Remember that the charge of neutral ligands like CO is zero.

### Step 2: Write the electronic configuration of Nickel
- Nickel has an atomic number of 28, so its electronic configuration is:
\[ \text{Ni: } [Ar] 4s^2 3d^8 \]

**Hint:** Use the atomic number to find the electronic configuration, and remember that the 4s orbital is filled before the 3d orbital.

### Step 3: Analyze the effect of the ligand on the electronic configuration
- CO is a strong field ligand, which means it can cause pairing of electrons. In the presence of CO, the 4s electrons can be excited to the 3d orbital.
- This results in the 4s orbital being empty and the 3d orbitals being fully filled:
\[ 3d: 10 \text{ electrons (fully filled)} \]

**Hint:** Strong field ligands like CO can cause electron pairing, leading to a change in the electron configuration.

### Step 4: Determine the number of unpaired electrons
- Since all the electrons in the 3d orbitals are paired, there are no unpaired electrons.

**Hint:** Count the number of unpaired electrons to determine the magnetic properties.

### Step 5: Calculate the magnetic moment
- The magnetic moment (μ) can be calculated using the formula:
\[ μ = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons. Since \(n = 0\):
\[ μ = \sqrt{0(0 + 2)} = 0 \]

**Hint:** The magnetic moment is directly related to the number of unpaired electrons.

### Step 6: Determine the hybridization
- The hybridization of `[Ni(CO)_(4)]` can be determined by the number of sigma bonds formed. Since there are 4 CO ligands, the hybridization is:
\[ \text{Hybridization: } sp^3 \]

**Hint:** Count the number of ligands to determine the hybridization.

### Step 7: Determine the geometry
- The geometry corresponding to sp³ hybridization is tetrahedral.

**Hint:** Remember the common geometries associated with different hybridizations.

### Final Answer
- The geometry of `[Ni(CO)_(4)]` is tetrahedral, the hybridization is sp³, and the magnetic moment is 0.

**Summary:**
- Geometry: Tetrahedral
- Hybridization: sp³
- Magnetic Moment: 0