The coordination number of `Ni^(2+)` is 4
`NiCI_(2) + KCN` (excess)rarr`A` (cyano complex)
The hybridisation of `A` and `B` are .

Options

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

Explanation

To solve the question regarding the hybridization of the complexes formed by `Ni^(2+)`, we will follow these steps:

### Step 1: Identify the Complexes
We start with the reaction of `NiCl2` with excess `KCN`. This will lead to the formation of a cyano complex, which we will denote as complex A. The formula for this complex is `Ni(CN)4^(2-)`.

Next, we consider the reaction of `NiCl2` with excess `KCl`, which will lead to the formation of a chloro complex, denoted as complex B. The formula for this complex is `NiCl4^(2-)`.

### Step 2: Determine the Coordination Number
The coordination number of `Ni^(2+)` in both complexes is 4. This is because there are four ligands (either CN or Cl) surrounding the nickel ion.

### Step 3: Determine the Oxidation State of Nickel
In both complexes, the oxidation state of nickel is +2. This can be determined from the overall charge of the complexes and the charges of the ligands.

### Step 4: Electronic Configuration of Nickel
The electronic configuration of `Ni` is `[Ar] 3d^8 4s^2`. When `Ni` is in the +2 oxidation state, it loses two electrons, resulting in the configuration `3d^8`.

### Step 5: Hybridization of Complex A (Ni(CN)4^(2-))
- The ligands in complex A are CN^-, which are strong field ligands. This means they will cause pairing of the d-electrons.
- In the presence of strong field ligands, the `3d` electrons will pair up, and the hybridization will involve the `3d`, `4s`, and `4p` orbitals.
- Since the coordination number is 4, the hybridization for complex A is `DSP2`.

### Step 6: Hybridization of Complex B (NiCl4^(2-))
- The ligands in complex B are Cl^-, which are weak field ligands. They do not cause pairing of the d-electrons.
- The `3d` electrons remain unpaired, and the hybridization will involve the `4s` and `4p` orbitals.
- Since the coordination number is also 4, the hybridization for complex B is `SP3`.

### Final Answer
- The hybridization of complex A (Ni(CN)4^(2-)) is `DSP2`.
- The hybridization of complex B (NiCl4^(2-)) is `SP3`.