Spin Only Magnetic Moment of Cr(CO)6

To calculate the spin-only magnetic moment of Cr(CO)6, we need to determine the number of unpaired electrons in the compound. The spin-only magnetic moment can be calculated using the formula:

μ = √(n(n+2)) BM

where μ is the spin-only magnetic moment in Bohr magneton units, n is the number of unpaired electrons, and BM is the Bohr magneton unit.

Step 1: Determine the number of valence electrons in Cr(CO)6

To determine the number of valence electrons in Cr(CO)6, we need to consider the electronic configuration of Cr.

The atomic number of Cr is 24, and its electronic configuration is [Ar]3d54s1. In Cr(CO)6, the oxidation state of Cr is zero, which means it has lost all its valence electrons.

Therefore, the number of valence electrons in Cr(CO)6 is 0.

Step 2: Determine the number of unpaired electrons

Since the number of valence electrons in Cr(CO)6 is 0, there are no unpaired electrons in the compound. This is because all the valence electrons have been lost by Cr.

Therefore, the number of unpaired electrons in Cr(CO)6 is 0.

Step 3: Calculate the spin-only magnetic moment

Using the formula μ = √(n(n+2)) BM, we can calculate the spin-only magnetic moment of Cr(CO)6.

Since the number of unpaired electrons is 0, the value of n is 0.

μ = √(0(0+2)) BM
= √(0) BM
= 0 BM

Therefore, the spin-only magnetic moment of Cr(CO)6 is 0 Bohr magneton units.

Conclusion

The correct answer is option A, 0. This is because there are no unpaired electrons in Cr(CO)6, resulting in a spin-only magnetic moment of 0 Bohr magneton units.