The coordination number and magnetic moment of the complex `[Cr(C_(2)O_(4))_(2)(NH_(3))_(2)]^(-)`Jet respectively is

Options

A. 6,3.87 BM
B. 4,3.87 BM
C. 6, 3.47 BM
D. 4,1.73 BM

Answer

To solve the question regarding the coordination number and magnetic moment of the complex \([Cr(C_2O_4)_2(NH_3)_2]^-\), we will follow these steps:

### Step 1: Determine the Coordination Number
1. **Identify the ligands**: The complex contains oxalate \((C_2O_4)^{2-}\) and ammonia \((NH_3)\).
2. **Count the binding sites**:
- Each oxalate ligand is a bidentate ligand, meaning it has 2 binding sites. Since there are 2 oxalate ligands, they contribute \(2 \times 2 = 4\) binding sites.
- Ammonia is a monodentate ligand, contributing 1 binding site each. There are 2 ammonia ligands, contributing \(2 \times 1 = 2\) binding sites.
3. **Calculate the total coordination number**:
\[ \text{Total Coordination Number} = \text{Binding sites from oxalate} + \text{Binding sites from ammonia} = 4 + 2 = 6 \]

### Step 2: Determine the Charge on Chromium
1. **Identify the charge of the ligands**:
- Each oxalate ligand has a charge of \(-2\). Therefore, for 2 oxalate ligands, the total charge is \(-4\).
- Ammonia is neutral, contributing 0 charge.
2. **Set up the equation for the overall charge**:
\[ \text{Let the charge on Chromium be } X. \]
The overall charge of the complex is \(-1\):
\[ X - 4 + 0 = -1 \implies X - 4 = -1 \implies X = +3 \]

### Step 3: Determine the Electron Configuration of Chromium
1. **Find the electron configuration of neutral chromium**:
- The atomic number of chromium (Cr) is 24, so its electron configuration is:
\[ [Ar] 4s^2 3d^4 \]
2. **Adjust for the +3 oxidation state**:
- When chromium loses 3 electrons (2 from \(4s\) and 1 from \(3d\)), the configuration becomes:
\[ [Ar] 3d^3 \]

### Step 4: Determine the Number of Unpaired Electrons
1. **Fill the \(3d\) orbitals**:
- The \(3d\) subshell has 5 orbitals. With 3 electrons, they will occupy separate orbitals before pairing occurs:
\[ \uparrow \quad \uparrow \quad \uparrow \]
- Thus, there are 3 unpaired electrons.

### Step 5: Calculate the Magnetic Moment
1. **Use the formula for magnetic moment**:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.
2. **Substitute \(n = 3\)**:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \, \text{BM} \]

### Final Answer
- The coordination number of the complex \([Cr(C_2O_4)_2(NH_3)_2]^-\) is **6**.
- The magnetic moment is approximately **3.87 BM**.