Chapter 22: Problem 45
Determine the number of unpaired electrons in the following tetrahedral complexes. All tetrahedral complexes are high spin. (a) \(\left[\mathrm{FeCl}_{4}\right]^{2-}\) (c) \(\left[\mathrm{MnCl}_{4}\right]^{2}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CoCl}_{4}\right]\) \((\mathrm{d})\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{ZnCl}_{4}\right]\)
Short Answer
Step by step solution
Determine Metal Oxidation States
Write Electronic Configurations for Each Metal Ion
Apply High Spin Configurations in Tetrahedral Complexes
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
High Spin Complexes
- This occurs because of the relatively low energy difference between d-orbitals in tetrahedral complexes.
- Electrons fill each orbital singly as far as possible, following Hund's rule.
- This type of arrangement results in more unpaired electrons.
Oxidation States
- For example, chloride (\( ext{Cl}^-\)) is a common ligand with a -1 charge.
- In the complex \( ext{FeCl}_4^{2-}\), with each Cl carrying a -1 charge, iron must have an oxidation state of +2 to balance the -4 from the four chlorides plus the 2- overall charge.
- This principle applies similarly to the other complexes like \( ext{MnCl}_4^{2-}\) and \( ext{CoCl}_4^{2-}\).
Electronic Configurations
- For \( ext{Fe}^{2+}\), starting from neutral iron's configuration \([ ext{Ar}] 4s^2 3d^6\), 2 electrons are lost, resulting in \([ ext{Ar}] 3d^6\).
- Similarly, \( ext{Mn}^{2+}\) becomes \([ ext{Ar}] 3d^5\).
- This configuration directly ties into how electrons arrange in high or low spin states in complexes.
Hund's Rule
- In current examples, like \( ext{Mn}^{2+}\) with a \(d^5\) configuration, all five d-orbitals each have one electron, maximizing unpaired spins.
- Applying Hund's rule minimizes electron-electron repulsions, stabilizing the complex.
- This results in a clearer prediction of complexes' magnetic properties due to the number of unpaired electrons.