Final answer:
The complex [Mn(CN)6]^x is likely to have a valence electron configuration of d4 (low-spin) due to the strong-field ligand CN-, with an oxidation state of +3 for Mn, resulting in a charge of [Mn(CN)6]3-. The closest match to the experimental magnetic moment of 2.81 μB indicates that there are two unpaired electrons.
Step-by-step explanation:
To determine the valence electron configuration, spin state, and charge of the complex [Mn(CN)6]^x with an experimental magnetic moment of μeff = 2.81 μB, we must first note that the cyanide ion (CN-) is a strong-field ligand, which generally causes the metal ion to adopt a low-spin configuration.
Manganese commonly has an oxidation state of +2 to +7, with the +2 being d5, which would give five unpaired electrons in a high-spin complex. However, for a low-spin complex with a magnetic moment of approximately 2.81 μB, Mn would likely be in the +3 oxidation state, which is d4 configuration. Using the formula μeff = √n(n+2), where n is the number of unpaired electrons, and solving for n, we find that the closest value to 2.81 is 2 (μeff of a complex with two unpaired electrons is approximately 2.83 μB), which suggests a d4 low-spin configuration for Mn in this complex.
Consequently, the probable charge of the complex, considering that the cyano ligands are -1 each and there are six of them, would be [Mn(CN)6]3-. Hence, the [Mn(CN)6]^x complex is likely a d4 low-spin with a -3 charge, and the Mn ion has an oxidation state of +3.