Magnetic moment of `[Cu(NH_(3))_(4)]^(2+)` ion is

Options

A. 1.414
B. 1.73
C. 2.23
D. 2.38

Explanation

To find the magnetic moment of the `[Cu(NH₃)₄]²⁺` ion, we will follow these steps:

### Step 1: Determine the oxidation state of copper in the complex
The complex `[Cu(NH₃)₄]²⁺` indicates that copper is in the +2 oxidation state because ammonia (NH₃) is a neutral ligand.

**Hint:** The charge of the complex ion gives the oxidation state of the central metal ion.

### Step 2: Write the electron configuration of copper
Copper (Cu) has an atomic number of 29. The electron configuration of neutral copper is:
- **Cu:** [Ar] 3d¹⁰ 4s¹

When copper loses two electrons to form Cu²⁺, the electron configuration becomes:
- **Cu²⁺:** [Ar] 3d⁹

**Hint:** Remember that when forming cations, electrons are removed from the outermost shell first.

### Step 3: Determine the number of unpaired electrons
In the 3d⁹ configuration, the distribution of electrons in the d-orbitals is as follows:
- 3d: ↑↓ ↑↓ ↑↓ ↑ ↑ (one unpaired electron)

Thus, there is **1 unpaired electron** in the 3d orbital of Cu²⁺.

**Hint:** Visualize the d-orbitals and fill them according to Hund's rule to identify unpaired electrons.

### Step 4: Use the formula for magnetic moment
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

Substituting \( n = 1 \):
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \]

### Step 5: Calculate the numerical value of the magnetic moment
Calculating the square root gives:
\[ \mu \approx 1.73 \, \text{BM (Bohr Magneton)} \]

### Final Answer
The magnetic moment of the `[Cu(NH₃)₄]²⁺` ion is approximately **1.73 BM**.

---