Find the oxidation state, type of hybridization and geometry of the following complexes.
`[MnCl_4]^(2-)`

Explanation

To solve the problem of finding the oxidation state, type of hybridization, and geometry of the complex \([MnCl_4]^{2-}\), we can follow these steps:

### Step 1: Determine the Oxidation State of Manganese (Mn)

1. **Identify the charge on the complex**: The overall charge of the complex \([MnCl_4]^{2-}\) is -2.
2. **Identify the charge of the ligands**: Each chloride ion (Cl) has a charge of -1. Since there are 4 chloride ions, their total contribution is \(4 \times (-1) = -4\).
3. **Set up the equation**: Let the oxidation state of manganese be \(x\). The equation can be set up as follows:
\[ x + (-4) = -2 \]
4. **Solve for \(x\)**:
\[ x - 4 = -2 \\ x = -2 + 4 \\ x = +2 \]
Thus, the oxidation state of manganese in \([MnCl_4]^{2-}\) is +2.

### Step 2: Determine the Type of Hybridization

1. **Write the electron configuration of manganese**: Manganese (Mn) has an atomic number of 25. Its electron configuration is:
\[ [Ar] 4s^2 3d^5 \]
2. **Consider the oxidation state**: In the +2 oxidation state, manganese loses the two 4s electrons, leaving it with:
\[ [Ar] 3d^5 \]
3. **Determine the hybridization**: Since there are 4 ligands (Cl) surrounding the manganese, we need to consider the hybridization that accommodates 4 electron pairs. Chloride is a weak field ligand and does not cause pairing of the 3d electrons. Therefore, the hybridization will involve one 3s and three 3p orbitals:
\[ \text{Hybridization} = sp^3 \]

### Step 3: Determine the Geometry of the Complex

1. **Identify the geometry based on hybridization**: The \(sp^3\) hybridization corresponds to a tetrahedral geometry.
2. **Visualize the arrangement**: In a tetrahedral geometry, the four chloride ligands will be positioned at the corners of a tetrahedron around the manganese ion.

### Summary of Results

- **Oxidation State**: +2
- **Type of Hybridization**: \(sp^3\)
- **Geometry**: Tetrahedral