Spin Only Magnetic Moment of Chromium Hexacarbonyl

To determine the spin-only magnetic moment of $\text{Cr(CO)}_6$:

1. Oxidation State of Chromium:
   Carbon monoxide ($\text{CO}$) is a neutral ligand with an oxidation number of $0$. Therefore, the oxidation state of chromium in $\text{Cr(CO)}_6$ is zero:
   $\text{Oxidation state of Cr} = 0$

2. Electronic Configuration:
   The atomic number of chromium ($\text{Cr}$) is $24$. The ground-state electronic configuration of a neutral chromium atom is:
   $\text{Cr} : [\text{Ar}]\, 3d^5 4s^1$
   Thus, chromium has $6$ valence electrons.

3. Ligand Field and Electron Pairing:
   $\text{CO}$ is a strong field ligand (located at the high end of the spectrochemical series) and forms an octahedral complex. The strong crystal field causes large orbital splitting ($\Delta_o > P$, where $P$ is the pairing energy) and induces rearrangement/pairing of all valence electrons into the lower-energy $t_{2g}$ set of $d$-orbitals:
   $\text{Electronic configuration in octahedral field}: t_{2g}^6 e_g^0$

4. Calculation of Spin-Only Magnetic Moment:
   Since all electrons are completely paired, the number of unpaired electrons is:
   $n = 0$

   The spin-only magnetic moment ($\mu_s$) in Bohr Magnetons ($\text{BM}$) is given by:
   $\mu_s = \sqrt{n(n + 2)}\, \text{BM} = \sqrt{0(0 + 2)}\, \text{BM} = 0\, \text{BM}$

Therefore, the spin-only magnetic moment value is $0$.

Correct Option: A