Spin - only magnetic moment of `[Co(NH_(3))_(3)(H_(2)O)_(3)]Cl_(3)(` in Bohr Magnetons ) is `:`

Options

A. `3.5`
B. `4.0`
C. `4.5`
D. `3.3`

Explanation

To find the spin-only magnetic moment of the cobalt complex \([Co(NH_3)_3(H_2O)_3]Cl_3\) in Bohr magnetons, we will follow these steps:

### Step 1: Determine the oxidation state of cobalt
The complex contains three chloride ions (Cl\(^-\)), which contribute a total charge of -3. Since the complex is neutral overall, the oxidation state of cobalt must be +3 to balance the -3 charge from the chlorides.

**Hint:** Remember that the overall charge of a neutral complex is zero, so the sum of the oxidation states must equal zero.

### Step 2: Write the electronic configuration of cobalt in the +3 oxidation state
Cobalt (Co) has an atomic number of 27, and its ground state electronic configuration is \([Ar] 3d^7 4s^2\). In the +3 oxidation state, cobalt loses three electrons (two from the 4s and one from the 3d), resulting in the configuration \([Ar] 3d^6\).

**Hint:** For transition metals, the 4s electrons are removed before the 3d electrons when determining oxidation states.

### Step 3: Identify the type of ligands and their effect on the electronic configuration
In this complex, ammonia (NH\(_3\)) is a strong field ligand, while water (H\(_2\)O) is a weak field ligand. However, since ammonia is present in a greater number (3 NH\(_3\) vs. 3 H\(_2\)O), the overall effect will be that the complex behaves as a low-spin complex due to the strong field ligands.

**Hint:** Strong field ligands tend to cause pairing of electrons in the lower energy d-orbitals, leading to a low-spin configuration.

### Step 4: Draw the d-orbital splitting diagram
For a low-spin complex with a \(d^6\) configuration, the electrons will fill the lower energy orbitals first, resulting in all six electrons being paired in the \(t_{2g}\) orbitals (the lower energy set of d-orbitals).

**Hint:** Remember that in octahedral complexes, the d-orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy).

### Step 5: Count the number of unpaired electrons
In this case, since all six electrons are paired in the \(t_{2g}\) orbitals, there are 0 unpaired electrons.

**Hint:** Unpaired electrons contribute to the magnetic moment; paired electrons do not.

### Step 6: Calculate the spin-only magnetic moment
The formula for the spin-only magnetic moment (\(\mu\)) in Bohr magnetons is given by:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons. Since \(n = 0\):
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \text{ Bohr magnetons} \]

**Hint:** The formula shows that if there are no unpaired electrons, the magnetic moment will be zero.

### Final Answer
The spin-only magnetic moment of \([Co(NH_3)_3(H_2O)_3]Cl_3\) is **0 Bohr magnetons**.

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