Step-by-step Derivation

1. Determine the oxidation state of the central metal atom (Cr). The complex given is Cr(CO)₆. The ligand is carbonyl (CO), which is a neutral molecule. Therefore, its charge is 0. Let the oxidation state of Chromium (Cr) be 'x'. The overall charge on the complex is 0. $x + 6 \times (0) = 0$ $x = 0$ So, the oxidation state of Cr in Cr(CO)₆ is 0.

2. Write the electronic configuration of the central metal atom. The atomic number of Chromium (Cr) is 24. Its ground state electronic configuration is an exception to the Aufbau principle: $\text{Cr (Z=24): [Ar] } 3d^5 4s^1$ Since the oxidation state is 0, we consider the configuration of the neutral Cr atom. It has a total of 6 valence electrons (5 in 3d and 1 in 4s).

3. Analyze the ligand and the complex geometry. The ligand is carbonyl (CO). CO is a very strong field ligand. Strong field ligands cause a large crystal field splitting ($\\%Delta_o$). This large energy gap between the $t_{2g}$ and $e_g$ orbitals forces the electrons to pair up in the lower energy $t_{2g}$ orbitals before occupying the higher energy $e_g$ orbitals. Such complexes are called low-spin complexes. The coordination number is 6, which corresponds to an octahedral geometry. In an octahedral field, the d-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).

4. Determine the distribution of electrons in the d-orbitals. We have 6 valence electrons from Cr(0) to fill into the d-orbitals. Since CO is a strong field ligand, these electrons will first fill the $t_{2g}$ orbitals and pair up completely before any electron enters the $e_g$ orbitals. The filling proceeds as follows:

   • The first 3 electrons singly occupy the three $t_{2g}$ orbitals ($d_{xy}, d_{yz}, d_{zx}$).
   • The next 3 electrons pair up with the electrons already in the $t_{2g}$ orbitals. This results in the electronic configuration: $t_{2g}^6 e_g^0$.

5. Calculate the number of unpaired electrons (n). From the configuration $t_{2g}^6 e_g^0$, we can see that all 6 electrons are in pairs in the $t_{2g}$ orbitals. There are no unpaired electrons. $n = 0$

6. Calculate the spin-only magnetic moment (μ_s). The formula for the spin-only magnetic moment is: $\mu_s = \sqrt{n(n+2)} \text{ Bohr Magneton (BM)}$ Substituting the value of n = 0 into the formula: $\mu_s = \sqrt{0(0+2)} = \sqrt{0} = 0 \text{ BM}$ Thus, the complex Cr(CO)₆ is diamagnetic.

Conclusion

The spin-only magnetic moment of Cr(CO)₆ is 0 BM. This corresponds to option A.