Question

How many unpaired electrons are in the high-spin and low-spin octahedral complexes of \(\mathrm{Cr}^{2+} ?\)

Short answer

High-spin: 4 unpaired electrons; Low-spin: 1 unpaired electron.

Step-by-step solution

Determine Electron Configuration

First, we need to determine the electron configuration of the chromium ion, \(\mathrm{Cr}^{2+}\). Chromium, \(\mathrm{Cr}\), has an atomic number of 24, meaning it has 24 electrons in its neutral state. In its \(\mathrm{Cr}^{2+}\) state, it has lost 2 electrons, so we must remove them from the neutral configuration. The ground state electronic configuration of \(\mathrm{Cr}\) is \([\mathrm{Ar}] \, 3d^5 \, 4s^1\). After losing two electrons, the configuration of \(\mathrm{Cr}^{2+}\) is \([\mathrm{Ar}] \, 3d^4\).

Determine High-Spin Configuration

In an octahedral complex, the \(\mathrm{d}\)-orbitals split into two energy levels. In a high-spin complex, electrons avoid pairing in the lower energy \(t_{2g}\) orbitals and instead fill the \(e_g\) orbitals. For \(\mathrm{Cr}^{2+}\), which is \(d^4\), we distribute the electrons in the split \(d\)-orbitals as: \(t_{2g}^3 e_g^1\). This means three electrons are unpaired in \(t_{2g}\), and one electron is unpaired in \(e_g\). Thus, in a high-spin configuration, there are 4 unpaired electrons.

Determine Low-Spin Configuration

In a low-spin complex, electrons prefer to pair up in the lower energy \(t_{2g}\) orbitals before occupying the higher \(e_g\) orbitals. For \(\mathrm{Cr}^{2+}\), \(d^4\), these are distributed as \(t_{2g}^4 e_g^0\). Each of the three \(t_{2g}\) orbitals will have paired electrons, and one will have an unpaired electron. Therefore, in a low-spin configuration, there is 1 unpaired electron.

Key concepts

High-spin complexes

High-spin complexes occur when the ligands in a chemical compound create a weaker field. This is due to the ligand's ability to split the d-orbitals (also known as crystal field splitting) by a smaller amount of energy. When this occurs, the electrons prefer to remain unpaired and occupy the higher energy orbitals. Instead of forming pairs in the lower, more stable energy level, electrons will jump to the higher level if the energy difference is small.

• Maximize unpaired electrons to reduce electron-electron repulsion.
• Higher magnetic properties due to unpaired electrons.
• Common in ligands that produce a weak field such as halides.

In the context of a chromium ion with an \(\mathrm{Cr}^{2+}\) state, the electronic configuration of \(3d^4\) in a high-spin complex means having four unpaired electrons distributed as \(t_{2g}^3 \, e_g^1\).
High-spin complexes generally have higher magnetic moments, a consequence of the number of unpaired electrons.

Low-spin complexes

Low-spin complexes arise when the surrounding ligands are strong field ligands. They cause a significant energy split between lower and higher d-orbitals in an octahedral coordination.
In such complexes, electrons tend to pair up in the lower energy set of orbitals, known as the \(t_{2g}\) set, before moving into the higher energy \(e_g\) orbitals. This minimizes the number of unpaired electrons but increases stability.

• Reduced magnetic properties due to electron pairing.
• Typically formed by strong field ligands like CN⁻ and CO.
• Less tendency for unpaired electrons, resulting in low-spin configurations.

For a \(\mathrm{Cr}^{2+}\) state with a \(d^4\) electronic configuration, a low-spin complex would have only one unpaired electron. These electrons are distributed as \(t_{2g}^4 \, e_g^0\).

Unpaired electrons

Unpaired electrons play a crucial role in determining the magnetic and chemical properties of a coordination complex. When electrons are not paired, they contribute to a paramagnetic nature, making the compound interact with magnetic fields.
Such interactions are based on the number of unpaired electrons available, which form the basis for various experimental techniques such as Electron Spin Resonance.

• Unpaired electrons lead to paramagnetism.
• An odd number of electrons typically means at least one unpaired electron.
• They result in higher energy and less stability compared to paired electrons.

In the \(\mathrm{Cr}^{2+}\) ion's high-spin and low-spin complexes, you can observe differing properties based on the varied number of unpaired electrons, highlighting the importance of electron distribution.

Octahedral complexes

Octahedral complexes are a type of coordination geometry where six ligands surround the central metal atom or ion. In such complexes, the d-orbitals of the central atom split into two groups due to the ligand fields: \(t_{2g}\) and \(e_g\).
The energy difference between these split orbitals greatly influences whether the complex is high-spin or low-spin.

• The number of ligands is typically six, forming an octahedral shape.
• The \(t_{2g}\) orbitals have lower energy compared to the \(e_g\) orbitals.
• Both high-spin and low-spin complexes can occur depending on the ligands and their interaction with metal ions.

In octahedral complexes of \(\mathrm{Cr}^{2+}\), electron distribution in the split orbitals exemplifies both electronic configurations, emphasizing the effect of ligand strength and arrangement.

Chromium ion

The chromium ion, specifically \(\mathrm{Cr}^{2+}\), serves as an excellent example when examining electronic configurations in coordination chemistry. The neutral chromium has an electron configuration of \([\mathrm{Ar}] \, 3d^5 \, 4s^1\).
Upon losing two electrons to form \(\mathrm{Cr}^{2+}\), the most stable configuration becomes \[ \mathrm{Cr}^{2+} = [\mathrm{Ar}] \, 3d^4 \]. This shedding of electrons highlights the transition of chromium into a positively charged ion.

• The \(3d\) electrons are often involved in complex formations.
• The chromium ion can form both high-spin and low-spin octahedral complexes.
• Influences a variety of properties including color, magnetism, and reactivity.

Understanding the electron loss and rearrangement in \(\mathrm{Cr}^{2+}\) allows us to predict complex behavior across various chemical interactions.