`[NiCl_(4)]^(2-)` is paramagnetic and therefore its geometry is :

Options

A. pyramidal
B. bi-pyramidal
C. tetrahedral
D. square planar

Answer

To determine the geometry of the complex ion \([NiCl_4]^{2-}\) and confirm its paramagnetism, we can follow these steps:

### Step 1: Determine the oxidation state of nickel (Ni) in the complex.
- The formula for the complex is \([NiCl_4]^{2-}\).
- Let the oxidation state of Ni be \(x\).
- Each Cl has an oxidation state of \(-1\), and there are 4 Cl atoms.
- The equation for the oxidation state is:
\[ x + 4(-1) = -2 \]
\[ x - 4 = -2 \]
\[ x = +2 \]

### Step 2: Write the electronic configuration of Ni in the +2 oxidation state.
- The atomic number of Ni is 28, and its ground state electronic configuration is:
\[ [Ar] 3d^8 4s^2 \]
- In the +2 oxidation state, Ni loses 2 electrons from the 4s orbital:
\[ Ni^{2+}: [Ar] 3d^8 \]

### Step 3: Determine the nature of the ligand and its effect on electron pairing.
- Chlorine (Cl) is a weak field ligand, which means it does not cause significant pairing of electrons in the d-orbitals.
- Therefore, in the presence of weak field ligands, the electrons in the d-orbitals remain unpaired.

### Step 4: Count the number of unpaired electrons.
- The \(3d\) subshell can accommodate a maximum of 10 electrons. In \(Ni^{2+}\) (which has \(3d^8\)):
- The configuration can be represented as:
\[ \uparrow\downarrow \quad \uparrow \quad \uparrow \quad \uparrow \quad \uparrow \]
- There are 2 unpaired electrons in the \(3d\) orbitals.

### Step 5: Determine the hybridization and geometry of the complex.
- Since there are 4 chloride ligands surrounding the nickel ion, we can determine the hybridization:
- The hybridization for 4 ligands is \(sp^3\).
- The geometry corresponding to \(sp^3\) hybridization is tetrahedral.

### Conclusion:
- Therefore, the geometry of the complex \([NiCl_4]^{2-}\) is tetrahedral, and it is paramagnetic due to the presence of unpaired electrons.

### Final Answer:
The geometry of \([NiCl_4]^{2-}\) is tetrahedral.

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