The type of magnetism exhibited by `[Mo(H_(2)O)_(6)]^(2+)` ion is _______.

Explanation

To determine the type of magnetism exhibited by the complex ion \([Mo(H_2O)_6]^{2+}\), we will follow these steps:

### Step 1: Determine the Oxidation State of Molybdenum
The oxidation state of molybdenum (Mo) in the complex can be calculated as follows:

- Let the oxidation state of Mo be \(x\).
- The water molecule (H₂O) is a neutral ligand, contributing 0 to the overall charge.
- The overall charge of the complex is +2.

Thus, we have:
\[ x + 0 = +2 \implies x = +2 \]

### Step 2: Identify the Electronic Configuration of Molybdenum
The ground state electronic configuration of molybdenum (Mo) is:
\[ [Kr] 4d^5 5s^1 \]

When it is in the +2 oxidation state, it loses two electrons. The electrons are removed first from the 5s orbital and then from the 4d orbital. Therefore, the electronic configuration for \([Mo]^{2+}\) is:
\[ [Kr] 4d^4 \]

### Step 3: Analyze the Ligand Field Strength
The ligand in this complex is H₂O, which is considered a weak field ligand. However, in the context of 4d and 5d transition metals, H₂O can behave as a stronger field ligand compared to first-row transition metals.

### Step 4: Determine the Spin State
Since H₂O is a strong field ligand, it will cause the electrons to pair up in the lower energy orbitals (t2g) before occupying the higher energy orbitals (eg).

For the configuration \(4d^4\):
- The electrons will pair up in the lower energy orbitals, leading to a low-spin configuration.

### Step 5: Count the Number of Unpaired Electrons
In the low-spin state of \(4d^4\):
- The electrons will be arranged as follows:
- Pairing occurs in the t2g orbitals.

This results in:
- 4 electrons paired (2 pairs) and no unpaired electrons.

### Step 6: Determine the Type of Magnetism
Since there are no unpaired electrons in the complex, it is classified as:
- **Diamagnetic**.

### Final Answer
The type of magnetism exhibited by \([Mo(H_2O)_6]^{2+}\) ion is **diamagnetic**.

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