Problem 24 - 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) \(\left[\mathrm{MnF}_{6}\right]^{4-}\) (d) \(\left(\mathrm{NH}_{4}\right)_{3}\left[\mathrm{FeF}_{6}\right]\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Metal and Oxidation State

(a) In \( \mathrm{K}_{4}\left[\mathrm{FeF}_{6}\right] \), the complex anion is \( \mathrm{FeF}_{6}^{4-} \). Potassium provides a +1 charge each, giving a total charge of -4 for \( \left[\mathrm{FeF}_{6}\right] \). The oxidation state of Fe is +2. (b) In \( \left[\mathrm{MnF}_{6}\right]^{4-} \), given that the anion charge is -4, Mn is in the +2 oxidation state. (c) In \( \left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+} \), Cr is in the +2 oxidation state. (d) The complex anion is the same as in (a): \( \left[\mathrm{FeF}_{6}\right]^{3-} \), giving Fe a +3 oxidation state.

Step 2: Determine the Electron Configuration for Each Metal

(a) Fe in +2 oxidation state has an electron configuration of \([\mathrm{Ar}]\,3d^{6}\). (b) Mn in +2 oxidation state has an electron configuration of \([\mathrm{Ar}]\,3d^{5}\). (c) Cr in +2 oxidation state gives \([\mathrm{Ar}]\,3d^{4}\). (d) Fe in +3 oxidation state has \([\mathrm{Ar}]\,3d^{5}\).

Step 3: Apply the Ligand Field Model

The complexes are high-spin, meaning the electrons are maximally unpaired. This results in: (a) \([\mathrm{Ar}]\,3d^{6}\): \(\uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow\) - 4 unpaired electrons. (b) \([\mathrm{Ar}]\,3d^{5}\): \(\uparrow, \uparrow, \uparrow, \uparrow, \uparrow\) - 5 unpaired electrons. (c) \([\mathrm{Ar}]\,3d^{4}\): \(\uparrow, \uparrow, \uparrow, \uparrow\) - 4 unpaired electrons. (d) \([\mathrm{Ar}]\,3d^{5}\): \(\uparrow, \uparrow, \uparrow, \uparrow, \uparrow\) - 5 unpaired electrons.

Key Concepts

Understanding High-Spin Complexes

High-spin complexes are a fascinating feature in chemistry, particularly in transition metal complexes. These complexes allow electrons to spread out, filling each available orbital before pairing up.
This results in a maximum number of unpaired electrons. - In high-spin complexes, ligands create a weaker field. - The weak field doesn't cause a large energy gap between orbitals.
- Electrons prefer to remain unpaired in different orbitals to minimize repulsion. This behavior is the opposite of low-spin complexes, where a stronger field causes electrons to pair up to avoid higher-energy orbitals. High-spin complexes are more common when weak field ligands like fluoride (F⁻) or water (H₂O) are present.

Exploring Electron Configuration

Electron configuration is the distribution of electrons in an atom's or ion's orbitals.
It describes where each electron resides, typically starting from lower to higher energy levels.- For a neutral atom, electrons fill orbitals in order of increasing energy.- When considering ions, adjust the configuration based on added or removed electrons.For example, an iron (Fe) atom normally has \(\text{[Ar]}\, 3d^6 \) electron configuration. However, in an ionized state, such as Fe²⁺, two electrons are removed, leading to the same configuration but reflecting the charge. Understanding this helps predict chemical reactivity and bonding.

Determining Oxidation States

Oxidation states indicate the charge on an atom in a compound or complex, representing electron loss or gain.
Knowing the oxidation state helps predict the electron configuration.- The oxidation state is calculated based on the charges of surrounding atoms or ions.- In transition metals, knowing the oxidation state is crucial for determining the number of d-electrons.For example, in \(\text{K}_4[\text{FeF}_6]\), the complex has potassium providing a charge balance.\
In this case, Fe carries a +2 oxidation state leading to an electron configuration of \(\text{[Ar]}\, 3d^6 \). Calculating this correctly is key to understanding the complex's properties.

Counting Unpaired Electrons

Unpaired electrons significantly influence the magnetic properties of a substance.
In high-spin complexes, the number of unpaired electrons is maximum due to weak field ligands.- More unpaired electrons mean higher magnetic moments (paramagnetism).- High-spin complexes result in more unpaired electrons compared to low-spin counterparts.Consider \(\text{[MnF}_6]^{4-}\).
The manganese here is in a +2 oxidation state, resulting in a configuration of \(\text{[Ar]}\, 3d^5 \). Since it's a high-spin complex, all five d-orbitals have one unpaired electron each, providing substantial paramagnetic properties.