Considering `H_2O` as a weak field ligand, the number of unpaired electrons in `[Mn(H_2O)_6]^(2+)` will be (At. no. of `Mn=25`)

Options

A. 3
B. 5
C. 2
D. 4

Explanation

To determine the number of unpaired electrons in the complex \([Mn(H_2O)_6]^{2+}\), we can follow these steps:

### Step 1: Determine the oxidation state of manganese in the complex.
The overall charge of the complex is \(+2\). Since water (\(H_2O\)) is a neutral ligand, the oxidation state of manganese can be calculated as follows:

Let the oxidation state of manganese be \(x\). The equation can be set up as:
\[ x + 6(0) = +2 \]
This simplifies to:
\[ x = +2 \]
Thus, the oxidation state of manganese in \([Mn(H_2O)_6]^{2+}\) is \(+2\).

**Hint:** Remember that the charge of neutral ligands like water is zero when calculating oxidation states.

### Step 2: Write the electron configuration of manganese.
The atomic number of manganese (Mn) is 25. The electron configuration of neutral manganese is:
\[ \text{Mn: } [Ar] 4s^2 3d^5 \]
When manganese is in the \(+2\) oxidation state, it loses two electrons, which are removed from the \(4s\) orbital first. Therefore, the electron configuration for \([Mn]^{2+}\) is:
\[ \text{Mn}^{2+}: [Ar] 3d^5 \]

**Hint:** For transition metals, always remove electrons from the \(s\) orbital before the \(d\) orbital when determining the oxidation state.

### Step 3: Analyze the \(3d\) orbital for unpaired electrons.
In the \(3d^5\) configuration, the five electrons will occupy the five \(3d\) orbitals singly due to Hund's rule, which states that electrons will fill degenerate orbitals singly before pairing up. Thus, the \(3d\) orbitals will look like this:
\[ \begin{align*} 3d: & \uparrow \quad \uparrow \quad \uparrow \quad \uparrow \quad \uparrow \end{align*} \]
This indicates that all five \(3d\) electrons are unpaired.

**Hint:** Recall Hund's rule when filling orbitals; it helps predict the number of unpaired electrons.

### Step 4: Determine the hybridization and geometry of the complex.
Given that \(H_2O\) is a weak field ligand, it does not cause pairing of the \(d\) electrons. The coordination number of the complex is 6 (due to six water ligands), which leads to an octahedral geometry. The hybridization of the complex can be represented as \(sp^3d^2\).

**Hint:** The coordination number helps determine the geometry and hybridization of the complex.

### Step 5: Count the number of unpaired electrons.
Since all five \(3d\) electrons are unpaired, the number of unpaired electrons in \([Mn(H_2O)_6]^{2+}\) is:
\[ \text{Number of unpaired electrons} = 5 \]

**Final Answer:** The number of unpaired electrons in \([Mn(H_2O)_6]^{2+}\) is **5**.