Nickel (Z=28) combines with a uninegative monodenatate ligands to form a diamagnetic complex `[NiL_(4)]^(2-)`. The hybridisation involved and the number of unpaired electrons present in the complex are respectively:

Options

A. `dsp^(2)` , zero
B. `sp^(3)`, zero
C. `dsp^(2),` .one
D. `sp^(3)`, two

Answer

To solve the problem, we need to determine the hybridization and the number of unpaired electrons in the complex \([NiL_4]^{2-}\) where nickel (Ni) has an atomic number of 28 and combines with uninegative monodentate ligands.

### Step 1: Determine the oxidation state of Nickel
- Nickel (Ni) has a neutral charge, and when it forms a complex with four uninegative ligands (L), the total charge contributed by the ligands is \(-4\) (since each ligand has a charge of \(-1\)).
- The overall charge of the complex is \(-2\).
- Let the oxidation state of Ni be \(x\). The equation can be set up as:
\[ x + 4(-1) = -2 \]
Simplifying this gives:
\[ x - 4 = -2 \implies x = +2 \]

### Step 2: Determine the electronic configuration of \(Ni^{2+}\)
- The ground state electronic configuration of neutral nickel (Ni) is \( [Ar] 3d^8 4s^2 \).
- Upon losing two electrons to form \(Ni^{2+}\), the configuration becomes:
\[ [Ar] 3d^8 \]
(The \(4s\) electrons are removed first.)

### Step 3: Analyze the nature of the ligands
- The ligands are uninegative and monodentate. Since the complex is stated to be diamagnetic, the ligands must be strong field ligands, which means they will cause pairing of electrons in the \(d\) orbitals.

### Step 4: Fill the \(d\) orbitals
- The \(3d\) subshell can hold a maximum of 10 electrons. In \(Ni^{2+}\), we have 8 electrons in the \(3d\) orbitals.
- The filling of the \(3d\) orbitals will look like this:
- \(3d\) orbitals: ↑↓ ↑↓ ↑↓ ↑ ↑
- Since the complex is diamagnetic, all electrons are paired.

### Step 5: Determine the hybridization
- The complex \([NiL_4]^{2-}\) involves 4 ligands. The hybridization can be determined based on the number of orbitals involved:
- 1 \(d\) orbital, 1 \(s\) orbital, and 2 \(p\) orbitals are used to accommodate the 4 ligands.
- Therefore, the hybridization is \(dsp^2\).

### Step 6: Count the number of unpaired electrons
- Since all the \(d\) electrons are paired, the number of unpaired electrons is \(0\).

### Conclusion
- The hybridization involved in the complex \([NiL_4]^{2-}\) is \(dsp^2\) and the number of unpaired electrons present is \(0\).

### Final Answer
The hybridization involved and the number of unpaired electrons present in the complex are respectively: **\(dsp^2\) and \(0\)**.

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