`[NiCl_(4)]^(2-)` has geometry

Options

A. square planar
B. tetrahedral
C. square bipyramidal
D. trigonal bipyramidal

Answer

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

### Step 1: Identify the oxidation state of Nickel (Ni)
- Nickel (Ni) typically has an oxidation state of +2 in its complexes. In \([NiCl_4]^{2-}\), the overall charge is -2, and since each chloride ion (Cl) has a charge of -1, we can set up the equation:
\[ x + 4(-1) = -2 \]
where \(x\) is the oxidation state of Ni. Solving for \(x\):
\[ x - 4 = -2 \implies x = +2 \]
- Therefore, the oxidation state of Nickel in \([NiCl_4]^{2-}\) is +2.

### Step 2: Determine the electronic configuration of Ni
- The atomic number of Nickel (Ni) is 28. The electronic configuration of Ni is:
\[ [Ar] 3d^8 4s^2 \]
- In the +2 oxidation state, Nickel loses 2 electrons, typically from the 4s orbital first. Thus, the configuration becomes:
\[ [Ar] 3d^8 \]

### Step 3: Identify the nature of the ligands
- Chloride ions (Cl\(^-\)) are considered weak field ligands. This means they do not cause significant splitting of the d-orbitals in transition metal complexes.

### Step 4: Determine the hybridization
- Since we have a total of 4 chloride ligands coordinated to the Nickel ion, we can use the VSEPR theory to predict the hybridization. For 4 ligands, the hybridization is:
\[ sp^3 \]
- This is because the coordination number is 4.

### Step 5: Determine the geometry
- The geometry associated with \(sp^3\) hybridization is tetrahedral. Therefore, the geometry of the complex \([NiCl_4]^{2-}\) is tetrahedral.

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

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