`[Cu(NH_3)_4]^(2+)` has hybridisation and magnetic moment

Options

A. `sp^3,1.73 B.M`
B. `sp^3d`, 1.73 B.M
C. `dsp^2`,2.83 B.M
D. `dsp^2` 1.73 B.M

Answer

To determine the hybridization and magnetic moment of the complex ion \([Cu(NH_3)_4]^{2+}\), we can follow these steps:

### Step 1: Determine the oxidation state of copper
The complex ion \([Cu(NH_3)_4]^{2+}\) has a charge of \(2+\). Since ammonia (\(NH_3\)) is a neutral ligand, the oxidation state of copper in this complex is \(+2\).

### Step 2: Write the electronic configuration of copper
The atomic number of copper (Cu) is 29. The electronic configuration of neutral copper is:
\[ \text{Cu: } [Ar] 4s^1 3d^{10} \]
For \(Cu^{2+}\), we remove two electrons. The first electron is removed from the \(4s\) orbital and the second from the \(3d\) orbital:
\[ \text{Cu}^{2+}: [Ar] 3d^9 \]

### Step 3: Analyze the ligands and their effect
Ammonia (\(NH_3\)) is a strong field ligand and will cause pairing of electrons in the \(d\) orbitals. In the case of \(Cu^{2+}\) with a \(3d^9\) configuration, one of the \(3d\) electrons will be promoted to the \(4p\) orbital due to the strong field nature of \(NH_3\).

### Step 4: Determine the hybridization
With four ligands (ammonia) and one \(3d\) electron promoted to the \(4p\) orbital, the hybridization can be determined:
- The \(3d\) orbitals are involved along with \(4s\) and \(4p\) orbitals.
- This leads to \(dsp^2\) hybridization.

### Step 5: Count the number of unpaired electrons
In the \(3d^9\) configuration, after pairing and promoting one electron to the \(4p\) orbital, we have:
- \(8\) paired electrons in the \(3d\) orbitals.
- \(1\) unpaired electron in the \(4p\) orbital.

Thus, the number of unpaired electrons \(N = 1\).

### Step 6: Calculate the magnetic moment
The formula for magnetic moment (\(\mu\)) is given by:
\[ \mu = \sqrt{N(N + 2)} \text{ BM} \]
Substituting \(N = 1\):
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \text{ BM} \]
Calculating the value:
\[ \sqrt{3} \approx 1.73 \text{ BM} \]

### Final Answer
- **Hybridization**: \(dsp^2\)
- **Magnetic Moment**: \(1.73 \text{ BM}\)

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