The hybridization of central metal ion in the complex `[Mn(H_(2)O)_(6)]^(2+)` is

Options

A. `sp^(3)`
B. `sp^(3)d^(2)`
C. `d^(2)sp^(3)`
D. `dsp^(3)`

Answer

To determine the hybridization of the central metal ion in the complex \([Mn(H_2O)_6]^{2+}\), we can follow these steps:

### Step 1: Identify the oxidation state of manganese.
The complex is \([Mn(H_2O)_6]^{2+}\). Since water (H2O) is a neutral ligand, the oxidation state of manganese (Mn) in this complex is +2.

### Step 2: Determine the electronic configuration of the manganese ion.
The atomic number of manganese (Mn) is 25. The electronic configuration of neutral manganese is:
\[ [Ar] 4s^2 3d^5 \]
When manganese loses two electrons to form \(Mn^{2+}\), the configuration becomes:
\[ Mn^{2+}: [Ar] 3d^5 \]

### Step 3: Analyze the ligands and their effect on hybridization.
In this complex, we have six water molecules as ligands. Water is a weak field ligand, which means it does not cause pairing of electrons in the d-orbitals. Therefore, the 3d electrons will remain unpaired.

### Step 4: Determine the number of orbitals involved in hybridization.
Since we have six ligands, we need to accommodate six pairs of electrons. The hybridization that accommodates six pairs of electrons is \(sp^3d^2\).

### Step 5: Conclude the hybridization.
Thus, the hybridization of the central metal ion (Mn) in the complex \([Mn(H_2O)_6]^{2+}\) is \(sp^3d^2\).

### Final Answer:
The hybridization of the central metal ion in the complex \([Mn(H_2O)_6]^{2+}\) is \(sp^3d^2\).

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