The spin only magnetic moment value (in Bohr magneton unit) of `[Cr(CO)_(6)]` is _____________ BM.

Explanation

To find the spin-only magnetic moment value of \([Cr(CO)_6]\) in Bohr magneton units, we will follow these steps:

### Step 1: Determine the oxidation state of chromium in \([Cr(CO)_6]\).
- Since CO is a neutral ligand, the oxidation state of chromium (Cr) in this complex is 0.

### Step 2: Write the electronic configuration of chromium.
- The electronic configuration of chromium (Cr) in its elemental state (oxidation state 0) is:
\[ [Ar] 3d^5 4s^1 \]

### Step 3: Analyze the ligand field.
- CO is a strong field ligand, which means it will cause pairing of electrons in the d-orbitals.

### Step 4: Determine the arrangement of electrons in the d-orbitals.
- In the presence of a strong field ligand like CO, the 3d electrons will pair up. The configuration will be:
\[ 3d^6 \quad (all \, paired) \]
- This results in all the 3d electrons being paired, leading to no unpaired electrons.

### Step 5: Count the number of unpaired electrons.
- Since all the electrons are paired, the number of unpaired electrons (n) is:
\[ n = 0 \]

### Step 6: Use the formula for spin-only magnetic moment.
- The formula for the spin-only magnetic moment (\(\mu\)) in Bohr magnetons is given by:
\[ \mu = \sqrt{n(n + 2)} \]
- Substituting \(n = 0\):
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \, \text{BM} \]

### Conclusion
- The spin-only magnetic moment value of \([Cr(CO)_6]\) is **0 BM**.

---