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 problem 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: Identify the Coordination Number
The coordination number is the total number of ligand donor atoms that are bonded to the central metal atom.

- The complex has two oxalate ligands \((C_2O_4)^{2-}\). Each oxalate ligand is bidentate, meaning it can attach to the metal at two points. Therefore, two oxalate ligands contribute \(2 \times 2 = 4\) coordination sites.
- The complex also has two ammonia ligands \((NH_3)\). Ammonia is a monodentate ligand, meaning it attaches at one point. Therefore, two ammonia ligands contribute \(2 \times 1 = 2\) coordination sites.

Adding these together gives the total coordination number:
\[ \text{Coordination Number} = 4 \, (\text{from } C_2O_4) + 2 \, (\text{from } NH_3) = 6 \]

### Step 2: Determine the Oxidation State of Chromium
Next, we need to find the oxidation state of chromium in the complex.

- Let \(x\) be the oxidation state of chromium. The charge from the oxalate ligands is \(-4\) (since each oxalate is \(-2\) and there are two of them), and the ammonia ligands are neutral (0 charge). The overall charge of the complex is \(-1\).

Setting up the equation:
\[ x + (-4) + 0 = -1 \]
Solving for \(x\):
\[ x - 4 = -1 \implies x = +3 \]

Thus, the oxidation state of chromium is +3.

### Step 3: Determine the Electron Configuration
The electron configuration of chromium in its elemental state is:
\[ \text{Cr: } [Ar] \, 4s^1 \, 3d^5 \]
In the +3 oxidation state, the configuration becomes:
\[ \text{Cr}^{3+}: [Ar] \, 3d^3 \]
This means there are 3 electrons in the 3d subshell.

### Step 4: Determine the Magnetic Moment
To find the magnetic moment, we need to know the number of unpaired electrons. In the \(3d^3\) configuration, all three electrons are unpaired.

The formula for calculating the magnetic moment (\(\mu\)) in Bohr magnetons is:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.

Substituting \(n = 3\):
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]
Calculating \(\sqrt{15}\) gives approximately:
\[ \mu \approx 3.87 \, \text{Bohr magnetons} \]

### Final Answer
Thus, the coordination number of the complex \([Cr(C_2O_4)_2(NH_3)_2]^-\) is **6**, and the magnetic moment is approximately **3.87 Bohr magnetons**.

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