The complex `[Ni(CN)_(4)]^(2-)` is diamagnetic and the complex `[NiCl_(4)]^(2-)` is paramagnetic . What can you conclude about their molecular geometries ?

Options

A. Both complexes have square planar geometries
B. Both complexes have tetrahedral geometries
C. `[NiCl_(4)]^(2-)` has a square planar geometry while `[Ni(CN)_(4)]^(2-)` has a tetrahedral geometry .
D. `[NiCl_(4)]^(2-)` has a tetrahedral geometry while `[Ni(CN)_(4)]^(2-)` has a square planar geometry .

Answer

To determine the molecular geometries of the complexes \([Ni(CN)_{4}]^{2-}\) and \([NiCl_{4}]^{2-}\), we will analyze their oxidation states, electronic configurations, hybridizations, and the presence of unpaired electrons.

### Step 1: Determine the oxidation state of Nickel in both complexes

For \([Ni(CN)_{4}]^{2-}\):
- Let the oxidation state of Ni be \(x\).
- The charge of cyanide \((CN^{-})\) is \(-1\), and there are 4 cyanide ligands.
- The overall charge of the complex is \(-2\).

Thus, we can set up the equation:
\[ x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2 \]

For \([NiCl_{4}]^{2-}\):
- Let the oxidation state of Ni be \(x\).
- The charge of chloride \((Cl^{-})\) is \(-1\), and there are 4 chloride ligands.
- The overall charge of the complex is \(-2\).

Setting up the equation:
\[ x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2 \]

### Step 2: Determine the electronic configuration of Ni²⁺

The atomic number of Nickel (Ni) is 28, so its electronic configuration is:
\[ [Ar] 3d^{8} 4s^{2} \]
For Ni²⁺, we remove 2 electrons from the 4s orbital:
\[ Ni^{2+} = [Ar] 3d^{8} 4s^{0} \]

### Step 3: Analyze the complexes for unpaired electrons

**For \([Ni(CN)_{4}]^{2-}\)**:
- CN⁻ is a strong field ligand, which causes pairing of electrons.
- The 3d orbitals will have all electrons paired:
\[ 3d: \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \]
- Since there are no unpaired electrons, this complex is **diamagnetic**.

**For \([NiCl_{4}]^{2-}\)**:
- Cl⁻ is a weak field ligand, which does not cause pairing of electrons.
- The 3d orbitals will have some unpaired electrons:
\[ 3d: \uparrow \uparrow \uparrow \uparrow \uparrow \]
- Since there are unpaired electrons, this complex is **paramagnetic**.

### Step 4: Determine the hybridization and geometry

**For \([Ni(CN)_{4}]^{2-}\)**:
- The hybridization is \(dsp^{2}\) (due to pairing of electrons).
- With a coordination number of 4 and \(dsp^{2}\) hybridization, the geometry is **square planar**.

**For \([NiCl_{4}]^{2-}\)**:
- The hybridization is \(sp^{3}\) (due to the presence of unpaired electrons).
- With a coordination number of 4 and \(sp^{3}\) hybridization, the geometry is **tetrahedral**.

### Conclusion
- \([Ni(CN)_{4}]^{2-}\) has a **square planar** geometry and is **diamagnetic**.
- \([NiCl_{4}]^{2-}\) has a **tetrahedral** geometry and is **paramagnetic**.

### Final Answer
The complex \([Ni(CN)_{4}]^{2-}\) is square planar, while \([NiCl_{4}]^{2-}\) is tetrahedral.

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