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: Determine the oxidation state of manganese (Mn).
In the complex \([Mn(H_2O)_6]^{2+}\), the overall charge is \(2+\). Since water (H₂O) is a neutral ligand, the oxidation state of Mn can be calculated as follows:

\[ \text{Oxidation state of Mn} = +2 \]

### Step 2: Write the electron configuration of manganese.
The atomic number of manganese (Mn) is 25. The electron configuration of Mn in its elemental form is:

\[ \text{Mn: } [Ar] 4s^2 3d^5 \]

Since Mn is in the +2 oxidation state, it loses two electrons (the two 4s electrons):

\[ \text{Mn}^{2+}: [Ar] 3d^5 \]

### Step 3: Identify the number of d-electrons.
In the \(Mn^{2+}\) state, there are 5 d-electrons present in the 3d subshell.

### Step 4: Determine the coordination number and geometry.
The complex \([Mn(H_2O)_6]^{2+}\) has six water molecules coordinated to the manganese ion. This gives a coordination number of 6, which typically leads to an octahedral geometry.

### Step 5: Determine the hybridization.
In an octahedral complex, the hybridization involves the mixing of one s orbital, three p orbitals, and two d orbitals. Therefore, the hybridization can be determined as follows:

\[ \text{Hybridization} = sp^3d^2 \]

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

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