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 can follow these steps:

### Step 1: Determine the oxidation state of Chromium (Cr)
In the complex \([Cr(CO)_{6}]\), carbon monoxide (CO) is a neutral ligand, which means it does not contribute any charge. Since the overall charge of the complex is neutral, the oxidation state of chromium can be calculated as follows:

Let the oxidation state of Cr be \(x\).
\[ x + 6(0) = 0 \implies x = 0 \]
Thus, the oxidation state of chromium in \([Cr(CO)_{6}]\) is 0.

### Step 2: Write the electron configuration of Chromium
Chromium has an atomic number of 24. The electron configuration for chromium in its ground state is:
\[ [Ar] 4s^1 3d^5 \]
This indicates that chromium has one electron in the 4s orbital and five electrons in the 3d orbital.

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

### Step 4: Determine the number of unpaired electrons
In \([Cr(CO)_{6}]\), the five 3d electrons will pair up due to the strong field nature of CO. Therefore, all electrons will be paired, resulting in:
- Number of unpaired electrons = 0

### Step 5: Calculate the spin-only magnetic moment
The formula for calculating the spin-only magnetic moment (\(\mu\)) in Bohr magneton units is:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons. Since \(n = 0\):
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \text{ BM} \]

### Conclusion
The spin-only magnetic moment value of \([Cr(CO)_{6}]\) is **0 BM**.

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