Find the oxidation state, type of hybridization and geometry of the following complexes.
`[Ni(CN)_4]^(2-)`

Explanation

To solve the problem, we will find the oxidation state, type of hybridization, and geometry of the complex \([Ni(CN)_4]^{2-}\).

### 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 \(-1\). Since there are 4 cyanide ions, their total contribution is \(4 \times (-1) = -4\).
3. The overall charge of the complex is \(-2\).
4. Therefore, we can set up the equation:
\[ x + (-4) = -2 \]
5. Solving for \( x \):
\[ x - 4 = -2 \implies x = +2 \]
6. Thus, the oxidation state of nickel in \([Ni(CN)_4]^{2-}\) is \( +2 \).

### Step 2: Determine the Hybridization
1. The electronic configuration of nickel (atomic number 28) in its ground state is:
\[ [Ar] 4s^2 3d^8 \]
2. In the +2 oxidation state, nickel loses the two 4s electrons:
\[ [Ar] 3d^8 \]
3. The \(3d\) orbitals will be filled as follows:
\[ 3d: \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow \uparrow \]
4. Since cyanide is a strong field ligand, it will cause the pairing of the electrons in the \(3d\) orbitals, leaving one \(3d\) orbital vacant.
5. To accommodate the four cyanide ligands, we need four hybrid orbitals. We can use one \(d\), one \(s\), and two \(p\) orbitals for hybridization:
\[ \text{Hybridization: } d^2sp^2 \]

### Step 3: Determine the Geometry
1. The hybridization \(d^2sp^2\) corresponds to a square planar geometry.
2. Therefore, the geometry of the complex \([Ni(CN)_4]^{2-}\) is square planar.

### Summary of Results
- **Oxidation State of Nickel**: \( +2 \)
- **Type of Hybridization**: \( d^2sp^2 \)
- **Geometry**: Square planar