The spin only magnetic moment value of `Cr(CO)_(6)` is

Options

A. 2.84 B.M
B. 4.90 B.M.
C. 5.92 B.M
D. 0 B.M.

Answer

To find the spin-only magnetic moment value of `Cr(CO)_(6)`, we can follow these steps:

### Step 1: Determine the oxidation state of chromium in `Cr(CO)_(6)`.
- In `Cr(CO)_(6)`, carbon monoxide (CO) is a neutral ligand. Therefore, the oxidation state of chromium (Cr) is 0.

### Step 2: Write the electronic configuration of chromium.
- The atomic number of chromium (Cr) is 24. Its electronic configuration is:
\[ \text{Cr: } [Ar] 3d^5 4s^1 \]

### Step 3: Analyze the ligand field.
- Carbon monoxide (CO) is a strong field ligand. Strong field ligands cause pairing of electrons in the d-orbitals.

### Step 4: Determine the hybridization of `Cr(CO)_(6)`.
- In an octahedral complex like `Cr(CO)_(6)`, the hybridization is typically `d^2sp^3`. This means that two d-orbitals, one s-orbital, and three p-orbitals are used for bonding.

### Step 5: Count the number of unpaired electrons.
- Since CO is a strong field ligand, it will cause the electrons in the 3d orbitals to pair up. In the case of Cr in the 0 oxidation state, the 3d orbitals will be filled as follows:
- 3d: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ (all electrons are paired)
- Therefore, there are no unpaired electrons in `Cr(CO)_(6)`.

### Step 6: Calculate the spin-only magnetic moment.
- The formula for calculating the spin-only magnetic moment (μ) is given by:
\[ μ = \sqrt{n(n + 2)} \text{ BM} \]
where \( n \) is the number of unpaired electrons. Since \( n = 0 \) for `Cr(CO)_(6)`, we have:
\[ μ = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \text{ BM} \]

### Conclusion:
- The spin-only magnetic moment value of `Cr(CO)_(6)` is **0 BM**.

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