The complex ion `[Ni(CN)_4]^(2-)` is :

Options

A. Square planar and diamagnetic
B. Tetrahedral and paramagnetic
C. Square planar and paramagnetic
D. Tetrahedral and diamagnetic

Explanation

To determine the characteristics of the complex ion \([Ni(CN)_4]^{2-}\), we will follow these steps:

### Step 1: Determine the Oxidation State of Nickel
1. Let the oxidation state of nickel be \(x\).
2. The cyanide ion \((CN^-)\) has a charge of \(-1\) and there are four cyanide ligands, contributing a total charge of \(-4\).
3. The overall charge of the complex ion is \(-2\).
4. Set up the equation:
\[ x + 4(-1) = -2 \]
5. Simplifying gives:
\[ x - 4 = -2 \implies x = +2 \]

### Step 2: Write the Electronic Configuration of Nickel
1. The atomic number of nickel (Ni) is 28.
2. The ground state electronic configuration of nickel is:
\[ [Ar] 4s^2 3d^8 \]

### Step 3: Determine the Configuration of the Nickel Ion
1. For \(Ni^{2+}\), we remove two electrons from the outermost shell:
\[ Ni^{2+} : [Ar] 3d^8 \]

### Step 4: Consider the Ligand Field
1. The cyanide ion \((CN^-)\) is a strong field ligand, which means it will cause pairing of electrons in the \(3d\) orbitals.
2. The \(3d^8\) configuration will pair up the electrons:
- Before pairing: \(3d: \uparrow \uparrow \uparrow \uparrow \uparrow\)
- After pairing: \(3d: \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow\)

### Step 5: Identify the Hybridization and Geometry
1. With four ligands, the hybridization can be determined:
- The \(3d\), \(4s\), and \(4p\) orbitals are involved, leading to \(dsp^2\) hybridization.
2. The geometry of the complex ion with \(dsp^2\) hybridization is square planar.

### Step 6: Determine the Magnetic Properties
1. Since all the electrons in the \(3d\) orbitals are paired, the complex is **diamagnetic**.

### Conclusion
The complex ion \([Ni(CN)_4]^{2-}\) is:
- **Square planar** in geometry
- **Diamagnetic** in nature