Problem 44 - Chemistry and Chemical Reactivity

Question

The following are high-spin complexes. Use the ligand field model to find the electron configuration of the central metal ion in each ion. Determine the number of unpaired electrons, if any, in each. (a) \(\mathrm{K}_{4}\left[\mathrm{FeF}_{6}\right]\) (c) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (b) \(\left|\mathrm{MnF}_{6}\right|^{4-}\) \(\left.(\mathrm{d})\left(\mathrm{NH}_{4}\right)_{3} | \mathrm{FeF}_{6}\right]\)

Step-by-Step Solution

Answer

(a) 4 unpaired, (b) 5 unpaired, (c) 4 unpaired, (d) 5 unpaired.

Step 1: Determine the Oxidation States

For each complex, identify the oxidation state of the metal ion. \(\text{(a) } \mathrm{K}_{4}[\mathrm{FeF}_{6}]: \text{potassium ion, } \text{K}^{+}, \text{so Fe must be } \text{Fe}^{4+}.\)\(\text{(b) } |\mathrm{MnF}_{6}|^{4-}: \text{charge is } 4-, \text{so Mn must be } \text{Mn}^{2+}.\)\(\text{(c) } [\mathrm{Cr(H_{2}O)}_{6}]^{2+}: \text{charge is } 2+, \text{so Cr must be } \text{Cr}^{2+}.\)\(\text{(d) } (\mathrm{NH}_{4})_{3}[\mathrm{FeF}_{6}]: \text{charge is } 3-, \text{so Fe must be } \text{Fe}^{3+}.\)

Step 2: Electron Configuration of Metal Ions

Determine the electron configuration for each metal ion in their given oxidation states.\(\text{(a) } \text{Fe}^{4+}: \text{starts as } 3d^6 \text{, so it becomes } 3d^4.\)\(\text{(b) } \text{Mn}^{2+}: \text{starts as } 3d^7 \text{, so it remains } 3d^5.\)\(\text{(c) } \text{Cr}^{2+}: \text{starts as } 3d^6 \text{, so it remains } 3d^4.\)\(\text{(d) } \text{Fe}^{3+}: \text{starts as } 3d^5 \text{, so it remains } 3d^5.\)

Step 3: Apply High-Spin Configuration

Use the ligand field model to apply high-spin configurations to each ion. - **High-spin complexes** have maximum unpaired electrons due to weak-field ligands like F- and \(\text{H}_{2}\text{O}\).- **(a)** For \(\text{Fe}^{4+}: 3d^4\), the electrons are distributed as \(\uparrow \uparrow \uparrow \uparrow\) (4 unpaired).- **(b)** For \(\text{Mn}^{2+}: 3d^5\), the electrons are distributed as \(\uparrow \uparrow \uparrow \uparrow \uparrow\) (5 unpaired).- **(c)** For \(\text{Cr}^{2+}: 3d^4\), the electrons are distributed as \(\uparrow \uparrow \uparrow \uparrow\) (4 unpaired).- **(d)** For \(\text{Fe}^{3+}: 3d^5\), the electrons are distributed as \(\uparrow \uparrow \uparrow \uparrow \uparrow\) (5 unpaired).

Step 4: Determine the Number of Unpaired Electrons

Count the unpaired electrons for each ion based on the high-spin configuration:- **(a)** \(\text{Fe}^{4+}: 4\) unpaired electrons.- **(b)** \(\text{Mn}^{2+}: 5\) unpaired electrons.- **(c)** \(\text{Cr}^{2+}: 4\) unpaired electrons.- **(d)** \(\text{Fe}^{3+}: 5\) unpaired electrons.

Key Concepts

High-Spin Complexes

High-spin complexes are fascinating forms of coordination compounds where ligands interact with central metal ions in a way that affects the electrons' spin state. In chemistry, a spin state refers to the orientation of the spin of electrons in a d orbital. When bonding occurs with weak-field ligands, such as \(\text{F}^-\) or \(\text{H}_2\text{O}\), the energy difference between d orbitals is small. This allows electrons to occupy higher energy orbitals without pairing up in lower ones. This tendency results in a configuration with a maximum number of unpaired electrons. Consequently, these complexes often appear colorful or have magnetic properties.

• Weak-field ligands create small splitting in energy levels.
• Electrons occupy high-energy orbitals for higher spins.
• Unpaired electrons result in unique physical properties.

In the case of the given complexes, you can see how high-spin configurations play a vital role in deciding the number of unpaired electrons in a compound.

Electron Configurations

The electron configuration of an element tells us how electrons are distributed in its atomic or molecular orbitals. Understanding electron configuration is crucial for predicting chemical properties and behaviors in high-spin complexes. In transition metals, the d orbitals are primarily involved, and each metal ion will have an electron arrangement altered by its oxidation state. For example, \({\text{Fe}^{4+}}\) becomes \({3d^4}\) from \({3d^6}\) when losing electrons.

• Electrons fill orbitals based on the Aufbau principle, Hund's rule, and the Pauli exclusion principle.
• High-spin positions prefer unpaired electrons in higher orbitals rather than pairing in lower ones.

For each complex, knowing the oxidation state helps determine the initial electron count, and adjustments can be made for the correct high-spin configuration. Thus, the electron configuration forms the foundation for understanding other dynamic processes in chemistry.

Oxidation States

Oxidation states reflect the charge of an ion after electrons have been transferred due to chemical bonding. They are pivotal in determining the electron configuration of a complex. Typically, oxidation states are indicated by the charges on the metal ions, and these provide a predictable framework for electron adjustments and calculations.
For example, in \(\text{K}_4\left[\text{FeF}_6\right]\), \(\text{Fe}\) is calculated to be \(\text{Fe}^{4+}\), and this reduction of four electrons helps determine its configuration as \(3d^4\). Similarly, each metal ion in this exercise was identified and recalibrated to its respective neutral atom before determining the number of unpaired electrons in a high-spin situation.

• Oxidation states help track electron movements in chemical reactions.
• Essential for configuring metal ions accurately in coordination chemistry.

The states provide clarity in redox reactions and are an indispensable part of understanding the chemical structure of high-spin complexes.

Unpaired Electrons

Unpaired electrons are a key feature in distinguishing between high-spin and low-spin complexes. They are significant in defining the magnetic properties of a compound. These electrons can cause paramagnetic behavior, meaning they become attracted by external magnetic fields.
In the step-by-step solution, counting unpaired electrons for various complexes provides insight into their chemical architecture. \(\text{Fe}^{4+}\) in \(\text{K}_4[\text{FeF}_6]\) displays four unpaired electrons, while \(\text{MnF}_6^{4-}\) shows five, influencing the extent of their magnetic properties.

• Paramagnetism arises from unpaired electron spins.
• Crucial for defining an element's reactivity and interaction with magnetic fields.

By knowing the number of unpaired electrons, chemists can predict molecular behavior in magnetic fields and various reactions, making this a vital aspect of coordination chemistry.