Total number of d-orbitals involved in the hybridization of central metal ion in complex ion `[MnO_(4)]^(-)` is

Options

A. 6
B. 4
C. 2
D. 3

Answer

To determine the total number of d-orbitals involved in the hybridization of the central metal ion in the complex ion \([MnO_4]^-\), we can follow these steps:

### Step 1: Identify the Central Metal Ion
The central metal ion in the complex \([MnO_4]^-\) is manganese (Mn).

### Step 2: Determine the Oxidation State of Manganese
In \([MnO_4]^-\), oxygen typically has an oxidation state of -2. Since there are four oxygen atoms, the total contribution from oxygen is \(4 \times (-2) = -8\). The overall charge of the complex is -1. Therefore, the oxidation state of manganese can be calculated as follows:

\[ x + (-8) = -1 \implies x = +7 \]

So, manganese is in the +7 oxidation state.

### Step 3: Determine the Electron Configuration of Manganese
The electron configuration of manganese (Mn) in its elemental form (atomic number 25) is:

\[ [Ar] 4s^2 3d^5 \]

In the +7 oxidation state, manganese loses all its 4s and 3d electrons:

\[ Mn^{7+} : [Ar] \]

### Step 4: Determine the Hybridization
The complex \([MnO_4]^-\) has a tetrahedral geometry. In tetrahedral complexes, the hybridization is typically \(sp^3\). However, for transition metals, we consider the involvement of d-orbitals in hybridization.

For tetrahedral complexes, the hybridization can be represented as \(s\) and \(p\) orbitals hybridizing with some \(d\) orbitals. The hybridization for \([MnO_4]^-\) is \(sd^3\), which means one \(s\) orbital and three \(d\) orbitals are involved.

### Step 5: Identify the d-Orbitals Involved
In this case, the d-orbitals involved in the hybridization are:

- \(d_{xy}\)
- \(d_{yz}\)
- \(d_{zx}\)

Thus, a total of 3 d-orbitals are involved in the hybridization.

### Final Answer
The total number of d-orbitals involved in the hybridization of the central metal ion in the complex ion \([MnO_4]^-\) is **3**.

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