Both `[Ni(CO)_4]` and ` [Ni (CN)_4]^(2-) ` are diamagnetic The hybridisations of nickel in these complexes , respectively are :

Options

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

Explanation

To determine the hybridization of nickel in the complexes `[Ni(CO)_4]` and `[Ni(CN)_4]^{2-}`, we can follow these steps:

### Step 1: Identify the oxidation state of nickel in each complex.
- In `[Ni(CO)_4]`, carbon monoxide (CO) is a neutral ligand. Therefore, the oxidation state of nickel (Ni) is 0.
- In `[Ni(CN)_4]^{2-}`, cyanide (CN) is a negatively charged ligand. Since the overall charge of the complex is -2, the oxidation state of nickel must be +2.

**Hint:** Remember that the charge of the complex is equal to the sum of the oxidation states of the metal and the ligands.

### Step 2: Determine the electron configuration of nickel in each oxidation state.
- For Ni in the oxidation state of 0:
- The electron configuration is \( [Ar] 3d^8 4s^2 \).
- For Ni in the oxidation state of +2:
- The electron configuration is \( [Ar] 3d^8 \) (the 4s electrons are removed).

**Hint:** Use the periodic table to find the electron configuration of nickel and adjust it based on the oxidation state.

### Step 3: Analyze the ligands and their field strength.
- CO is a strong field ligand and will cause pairing of electrons in the d-orbitals.
- CN⁻ is also a strong field ligand and will similarly cause pairing of electrons.

**Hint:** Strong field ligands tend to cause pairing of electrons in the lower energy d-orbitals.

### Step 4: Determine the hybridization for `[Ni(CO)_4]`.
- Since Ni is in the 0 oxidation state with the configuration \( 3d^8 4s^2 \), the two 4s electrons will pair with two of the 3d electrons due to the strong field nature of CO.
- After pairing, the configuration becomes \( 3d^{10} \) (with all d-orbitals filled) and the 4s becomes vacant.
- The hybridization will be \( sp^3 \) because there are four equivalent orbitals formed from one s and three p orbitals.

**Hint:** Count the number of ligands and the orbitals involved in bonding to determine the hybridization.

### Step 5: Determine the hybridization for `[Ni(CN)_4]^{2-}`.
- In this case, Ni is in the +2 oxidation state with the configuration \( 3d^8 \).
- The strong field nature of CN⁻ will cause the pairing of electrons in the 3d orbitals, resulting in \( 3d^{8} \) with two vacant orbitals.
- The hybridization here will be \( dsp^2 \) because it involves one d, one s, and two p orbitals.

**Hint:** Again, consider the number of ligands and the orbitals that are hybridized.

### Final Answer:
- The hybridization of nickel in `[Ni(CO)_4]` is **sp³**.
- The hybridization of nickel in `[Ni(CN)_4]^{2-}` is **dsp²**.