Give reason for the statement `[Ni(CN)_(4)]^(2-)` is diamagnetic while `[NiCl_(4)]^(2-)` is paramagnetic in nature .

Options

A. In `[NiCl_(4)]^(2-)` no unpaired electrons are present while in `[Ni(CN)_(4)]^(2-)` two unpaired electrons are present
B. In `[Ni(CN)_(4)]^(2-)` , no unpaired electrons are present while in `[NiCl_(4)]^(2-)` two unpaired electrons are present .
C. `[NiCl_(4)]^(2-)` shows `dsp^(2)` hybridisation hence it is paramagnetic .
D. `[Ni(CN)_(4)]^(2-)` shows `sp^(3)` hybridisation hence it is diamagnetic .

Explanation

To explain why \([Ni(CN)_{4}]^{2-}\) is diamagnetic while \([NiCl_{4}]^{2-}\) is paramagnetic, we will analyze the electronic configurations and the effects of the ligands involved.

### Step 1: Determine the oxidation state of Nickel in both complexes
- In both \([Ni(CN)_{4}]^{2-}\) and \([NiCl_{4}]^{2-}\), Nickel is in the +2 oxidation state.

### Step 2: Write the electronic configuration of Nickel
- The electronic configuration of Nickel (Ni) in its ground state is \([Ar] 3d^8 4s^2\).
- When Nickel is in the +2 oxidation state, it loses two electrons, which are removed from the 4s orbital first. Therefore, the electronic configuration becomes:
\[ Ni^{2+}: [Ar] 3d^8 \]

### Step 3: Analyze the ligand field strength
- **Cyanide (CN\(^-\))** is a strong field ligand, while **Chloride (Cl\(^-\))** is a weak field ligand.
- Strong field ligands like CN\(^-\) cause greater splitting of the d-orbitals and promote electron pairing.
- Weak field ligands like Cl\(^-\) do not cause significant splitting, resulting in less pairing of electrons.

### Step 4: Determine the arrangement of electrons in \([Ni(CN)_{4}]^{2-}\)
- In the presence of the strong field ligand CN\(^-\), the 3d electrons will pair up. The arrangement will be:
- Pairing occurs in the 3d orbitals:
\[ 3d: \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \]
- All 8 electrons are paired, resulting in **no unpaired electrons**.

### Step 5: Determine the arrangement of electrons in \([NiCl_{4}]^{2-}\)
- In the presence of the weak field ligand Cl\(^-\), the 3d electrons will not pair up:
- The arrangement will be:
\[ 3d: \uparrow \uparrow \uparrow \uparrow \uparrow \downarrow \downarrow \]
- This results in **two unpaired electrons**.

### Conclusion
- Since \([Ni(CN)_{4}]^{2-}\) has no unpaired electrons, it is **diamagnetic**.
- Since \([NiCl_{4}]^{2-}\) has two unpaired electrons, it is **paramagnetic**.

### Summary
- \([Ni(CN)_{4}]^{2-}\) is diamagnetic due to the absence of unpaired electrons (strong field ligand causing pairing).
- \([NiCl_{4}]^{2-}\) is paramagnetic due to the presence of two unpaired electrons (weak field ligand allowing unpaired electrons).