`NiCl_(2) overset(KCN)to "complex"A`
`NiCl_(2) underset("excess") overset(KCl)to "complex"B`
A & B complexes have the co-ordination number 4.
The hybridisation of both complexes are :

Options

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

Answer

To solve the problem regarding the complexes formed from NiCl₂ and KCN, as well as NiCl₂ and KCl, we will follow these steps:

### Step 1: Identify the complexes formed
- **Complex A** is formed when NiCl₂ reacts with KCN. Since KCN is a strong field ligand, it will lead to the formation of a complex with a coordination number of 4.
- **Complex B** is formed when NiCl₂ reacts with KCl. KCl is a weak field ligand, and it will also lead to the formation of a complex with a coordination number of 4.

### Step 2: Determine the oxidation state of Nickel in both complexes
1. **For Complex A (NiCl₂ with KCN)**:
- Nickel (Ni) is in the +2 oxidation state because KCN is neutral and Cl is -1 (2 Cl contribute -2).
- Therefore, the oxidation state of Ni in Complex A is +2.

2. **For Complex B (NiCl₂ with KCl)**:
- Similarly, for NiCl₄²⁻, the oxidation state of Ni is also +2 because Cl is -1 (4 Cl contribute -4).
- Therefore, the oxidation state of Ni in Complex B is +2.

### Step 3: Determine the electronic configuration of Nickel
- The atomic number of Nickel (Ni) is 28, and its ground state electronic configuration is [Ar] 3d⁸ 4s².
- In the +2 oxidation state, Nickel loses 2 electrons from the 4s orbital, resulting in the configuration of 3d⁸.

### Step 4: Determine the hybridization of both complexes
1. **For Complex A**:
- Since KCN is a strong field ligand, it will cause pairing of the 3d electrons.
- The 3d orbitals will be filled as follows: 3d⁶ (paired) and 3d² (unpaired).
- The hybridization involves one d orbital, one s orbital, and two p orbitals, resulting in **DSP² hybridization**.
- The geometry of Complex A is **square planar**.

2. **For Complex B**:
- KCl is a weak field ligand and does not cause pairing of electrons.
- The 3d orbitals remain unpaired: 3d⁸ (unpaired).
- The hybridization involves one s orbital and three p orbitals, resulting in **SP³ hybridization**.
- The geometry of Complex B is **tetrahedral**.

### Final Answer:
- **Complex A**: DSP² hybridization (Square planar)
- **Complex B**: SP³ hybridization (Tetrahedral)