Hybridization and magnetic moment of `[Cu(NH_(3))_(4)]^(2+)` and `[Mn(CN)_(6)^(3-)` ions respectively are

Options

A. `dsp^(2), 1.73BM: d^(2)sp^(3), 2.83 BM`
B. `sp^(3),zero, sp^(3)d^(2), 4.9BM`
C. `sp^(3), 1.73 BM, sp^(3)d^(2), 4.9BM`
D. `dsp^(2), 1.73 BM, d^(2)sp^(3), zero`

Explanation

To determine the hybridization and magnetic moment of the complexes \([Cu(NH_3)_4]^{2+}\) and \([Mn(CN)_6]^{3-}\), we can follow these steps:

### Step 1: Analyze the first complex \([Cu(NH_3)_4]^{2+}\)

1. **Identify the oxidation state of Copper (Cu)**:
- Ammonia (NH₃) is a neutral ligand, so the oxidation state of Cu in \([Cu(NH_3)_4]^{2+}\) is +2.

2. **Determine the electron configuration of Cu**:
- The atomic number of Cu is 29. The electron configuration is \([Ar] 3d^{10} 4s^1\).
- For Cu²⁺, we remove two electrons (one from 4s and one from 3d), resulting in \(3d^9\).

3. **Determine the hybridization**:
- The complex has 4 ligands (NH₃), which will lead to \(dsp^2\) hybridization.
- In this case, we can consider the pairing of electrons in the presence of a strong field ligand (NH₃), which can promote one electron from 3d to 4p, resulting in \(dsp^2\) hybridization.

4. **Count unpaired electrons**:
- In \(3d^9\), there is 1 unpaired electron.

5. **Calculate the magnetic moment**:
- The formula for magnetic moment (\(\mu\)) is given by \(\mu = \sqrt{n(n + 2)}\), where \(n\) is the number of unpaired electrons.
- Here, \(n = 1\), so \(\mu = \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.73\) Bohr magnetons.

### Step 2: Analyze the second complex \([Mn(CN)_6]^{3-}\)

1. **Identify the oxidation state of Manganese (Mn)**:
- Cyanide (CN⁻) is a negatively charged ligand. The overall charge of the complex is -3, which means Mn must be in the +3 oxidation state.

2. **Determine the electron configuration of Mn**:
- The atomic number of Mn is 25. The electron configuration is \([Ar] 3d^5 4s^2\).
- For Mn³⁺, we remove three electrons (two from 4s and one from 3d), resulting in \(3d^4\).

3. **Determine the hybridization**:
- The complex has 6 ligands (CN⁻), which will lead to \(d^2sp^3\) hybridization.
- CN⁻ is a strong field ligand, which means it will cause pairing of electrons in the 3d subshell.

4. **Count unpaired electrons**:
- In \(3d^4\) with all electrons paired due to the strong field nature of CN⁻, there are 4 unpaired electrons.

5. **Calculate the magnetic moment**:
- Using the same formula, \(n = 4\), so \(\mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.89\) Bohr magnetons.

### Final Answers:
- For \([Cu(NH_3)_4]^{2+}\): Hybridization is \(dsp^2\) and magnetic moment is approximately \(1.73\) Bohr magnetons.
- For \([Mn(CN)_6]^{3-}\): Hybridization is \(d^2sp^3\) and magnetic moment is approximately \(4.89\) Bohr magnetons.