The hybrdisation of the complex `[NiCl_4]^(-2)?` is

Options

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

Explanation

To determine the hybridization of the complex \([NiCl_4]^{2-}\), we can follow these steps:

### Step 1: Determine the oxidation state of Nickel (Ni)
In the complex \([NiCl_4]^{2-}\), we need to find the oxidation state of Nickel.

- Chlorine (Cl) has an oxidation state of -1.
- Since there are 4 chlorine atoms, their total contribution to the oxidation state is \(4 \times (-1) = -4\).
- The overall charge of the complex is -2.

Using the formula for oxidation states:
\[ \text{Oxidation state of Ni} + \text{Total oxidation state of Cl} = \text{Charge of the complex} \]
Let the oxidation state of Ni be \(x\):
\[ x + (-4) = -2 \]
Solving for \(x\):
\[ x = -2 + 4 = +2 \]
Thus, the oxidation state of Nickel in \([NiCl_4]^{2-}\) is +2.

### Step 2: Determine the electronic configuration of Ni
Nickel (Ni) has an atomic number of 28. The electronic configuration of neutral Nickel is:
\[ [Ar] 3d^8 4s^2 \]
When Nickel is in the +2 oxidation state, it loses 2 electrons, which are removed from the 4s orbital first:
\[ \text{Configuration of } Ni^{2+} = [Ar] 3d^8 \]

### Step 3: Analyze the ligand and its effect
Chlorine is a weak field ligand, which means it does not cause pairing of electrons in the d-orbitals. Therefore, the 3d orbitals will remain unpaired.

### Step 4: Determine the hybridization
Since Nickel in the +2 state has the configuration of \(3d^8\) and does not have any paired electrons, we need to find the number of orbitals required for bonding with the 4 chlorine atoms.

- To accommodate 4 ligands, Nickel will require 4 hybrid orbitals.
- The hybridization will involve the 4s and 3p orbitals because there are no paired d electrons available.

The hybridization can be determined as follows:
- 1 orbital from 4s and 3 orbitals from 3p will hybridize to form 4 equivalent sp³ hybrid orbitals.

### Conclusion
The hybridization of the complex \([NiCl_4]^{2-}\) is **sp³**.