How can you tell if [ MnCl6]4- or [ Mn(CN)6 ] 4- has the longer wavelength of light? and then how do you figure out how many unpaired electrons each has?
To determine which complex ion, [MnCl6]4- or [Mn(CN)6]4-, has the longer wavelength of light, you need to consider the ligands surrounding the central manganese atom.
First, let's understand the relationship between ligands and the wavelength of light absorbed by the complex ion. Ligands are molecules or ions that donate electron pairs to the central metal atom, forming coordination complexes. Different ligands have different electronic properties, which affect the energy difference between the metal's d orbitals.
In general, ligands can be classified into two types based on their field strength: high-field ligands and low-field ligands. High-field ligands are strongly donating and cause a large splitting of the d orbitals, resulting in a higher energy difference between them. Low-field ligands, on the other hand, are weakly donating and cause a smaller energy difference between the d orbitals.
To determine the wavelength of light absorbed by a complex ion, we need to analyze the ligands present in the complex and their field strength. In this case, the ligands in [MnCl6]4- are chloride ions (Cl-), whereas the ligands in [Mn(CN)6]4- are cyanide ions (CN-).
Chloride ions (Cl-) are considered a weak-field ligand, while cyanide ions (CN-) are strong-field ligands. Therefore, [MnCl6]4- has a smaller energy difference between its d orbitals, resulting in the absorption of light with a longer wavelength. In contrast, [Mn(CN)6]4- has a larger energy difference between its d orbitals, leading to the absorption of light with a shorter wavelength.
Now, regarding the number of unpaired electrons in each complex ion, we can determine it based on the central metal atom's oxidation state and the overall charge of the complex ion.
In both complex ions, the manganese (Mn) atom has a +4 oxidation state, consistent with its position in the periodic table (Group 7A, so it can lose four electrons to reach a stable configuration).
To calculate the number of unpaired electrons, we can apply the "spin-only" formula:
Number of unpaired electrons = √(n(n+2))
Where n represents the total number of electrons in the d orbitals. For Mn2+ (d5 configuration), n = 5.
For [MnCl6]4-, the overall charge is 4-. Therefore, the Mn atom must lose four additional electrons from its d orbitals. Subtracting these four electrons from the d5 configuration, we get one unpaired electron remaining.
For [Mn(CN)6]4-, the cyanide ligand (CN-) has a charge of -1. Since the overall charge of the complex is 4-, the Mn atom needs to lose four additional electrons from its d orbitals. Subtracting these four electrons from the d5 configuration, we find that all the electrons are paired (zero unpaired electrons).
To summarize:
- [MnCl6]4- absorbs light with a longer wavelength due to weak-field ligands (Cl-) and has one unpaired electron.
- [Mn(CN)6]4- absorbs light with a shorter wavelength due to strong-field ligands (CN-) and has zero unpaired electrons.