Yes, a low-spin complex can indeed be paramagnetic. Paramagnetism arises from the presence of one or more unpaired electrons in a chemical species. Low-spin complexes are characterized by strong-field ligands that cause a large crystal field splitting, leading to the pairing of electrons in the lower energy orbitals (t2g in octahedral complexes) before occupying the higher energy orbitals (eg). Even with this pairing, if the total number of d-electrons is such that some electrons must remain unpaired, the complex will exhibit paramagnetism.
Understanding Paramagnetism and Low Spin
To clarify how a low-spin complex can be paramagnetic, let's break down the key concepts:
- Paramagnetism: A substance is paramagnetic if it contains unpaired electrons. These unpaired electrons create tiny magnetic moments that align with an external magnetic field, leading to a weak attraction. The more unpaired electrons, the stronger the paramagnetism.
- Low-Spin Complexes: These complexes form when ligands exert a strong crystal field, causing a significant energy gap between the t2g and eg orbitals (Δo). Electrons prefer to pair up in the lower energy t2g orbitals first, even if it requires overcoming electron-electron repulsion, rather than jumping to the higher energy eg orbitals. This pairing results in fewer unpaired electrons compared to their high-spin counterparts.
The crucial point is that "fewer unpaired electrons" does not necessarily mean "zero unpaired electrons."
How Low-Spin Complexes Retain Paramagnetism
For a low-spin complex to be paramagnetic, it simply needs to have at least one unpaired electron after the electron configuration follows the low-spin rule. This typically occurs for certain d-electron counts:
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d1, d2, d3 electron configurations: In octahedral complexes, these configurations will always have unpaired electrons, regardless of whether they are high-spin or low-spin (as the t2g orbitals are filled before pairing is necessary). For example, d1 has 1 unpaired electron (t2g¹), d2 has 2 unpaired electrons (t2g²), and d3 has 3 unpaired electrons (t2g³). All these will be paramagnetic.
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d4, d5, d6, d7 electron configurations: These are the key configurations where the distinction between high-spin and low-spin significantly impacts the number of unpaired electrons. For low-spin complexes with these electron counts, some unpaired electrons can still exist:
- d4 (low-spin): Electrons fill t2g orbitals: (↑↓, ↑, ↑). This leaves 2 unpaired electrons, making the complex paramagnetic.
- d5 (low-spin): Electrons fill t2g orbitals: (↑↓, ↑↓, ↑). This leaves 1 unpaired electron, making the complex paramagnetic.
- d6 (low-spin): Electrons fill t2g orbitals: (↑↓, ↑↓, ↑↓). This results in 0 unpaired electrons, making the complex diamagnetic.
- d7 (low-spin): Electrons fill t2g orbitals: (↑↓, ↑↓, ↑↓, ↑). This leaves 1 unpaired electron in the eg orbitals, making the complex paramagnetic.
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d8, d9, d10 electron configurations: These typically have specific paramagnetic or diamagnetic properties regardless of strong/weak field ligands in octahedral geometry. For instance, d10 is always diamagnetic.
Example: (Cr(CN)6)4-
A prime example of a complex that is both low-spin and paramagnetic is the hexacyanochromate(II) ion, (Cr(CN)6)4-.
Let's analyze its properties:
- Oxidation State of Chromium: Let Cr be x. x + 6(-1) = -4. So, x = +2. Chromium is in the +2 oxidation state (Cr(II)).
- d-Electron Count: Chromium (Cr) has an electron configuration of [Ar] 3d5 4s1. As Cr(II), it loses the 4s electron and one 3d electron, resulting in a 3d4 configuration.
- Ligand Type: Cyanide (CN-) is a very strong-field ligand. This means it will induce a large crystal field splitting, leading to a low-spin electron configuration.
- Electron Configuration (Low-Spin d4): In an octahedral environment, the four d-electrons will occupy the t2g orbitals as follows:
- t2g: ↑↓ ↑ ↑
- eg: ---
- Paramagnetism: As shown, there are two unpaired electrons in the t2g orbitals. Therefore, despite being a low-spin complex, (Cr(CN)6)4- is paramagnetic.
This example clearly illustrates that low-spin does not equate to diamagnetic; it simply implies a specific electron filling order due to strong field ligands, which can still leave unpaired electrons.
Summary of Unpaired Electrons in Octahedral Complexes
The table below summarizes the number of unpaired electrons for various d-electron configurations in both high-spin and low-spin octahedral complexes, highlighting when low-spin complexes are paramagnetic.
| d-Electron Count | High Spin (Unpaired Electrons) | Low Spin (Unpaired Electrons) | Paramagnetic (Low Spin)? |
|---|---|---|---|
| d1 | 1 | 1 | Yes |
| d2 | 2 | 2 | Yes |
| d3 | 3 | 3 | Yes |
| d4 | 4 | 2 | Yes |
| d5 | 5 | 1 | Yes |
| d6 | 4 | 0 | No (Diamagnetic) |
| d7 | 3 | 1 | Yes |
| d8 | 2 | 2 | Yes |
| d9 | 1 | 1 | Yes |
| d10 | 0 | 0 | No (Diamagnetic) |
As the table demonstrates, several low-spin configurations (d1, d2, d3, d4, d5, d7, d8, d9) result in paramagnetic complexes. Only d6 and d10 low-spin complexes are diamagnetic.
For further reading on Crystal Field Theory and magnetism in coordination compounds, you can explore resources like LibreTexts Chemistry.
