Problem 71 - Chemistry The Molecular Science

Question

Draw the crystal-field splitting diagrams and put in the d electrons for these octahedral complexes. In those cases where they are possible, draw diagrams for both low-spin and high-spin cases. (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (c) \(\left[\mathrm{FeF}_{6}\right]^{\frac{3}{3}-}\) (d) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\)

Step-by-Step Solution

Answer

(a) High-spin: 4 electrons; (b) High-spin: 5 electrons; (c) High-spin: 5 electrons; (d) Low-spin: 3 electrons in t2g.

Step 1: Determine d electron count

For each complex, determine the number of d electrons for the metal ion. (a) Cr in \([\mathrm{Cr(OH}_2)]^{2+}\): Cr is in the +2 oxidation state; hence it has 4 d electrons (d^4).(b) Mn in \([\mathrm{Mn(OH}_2)]^{2+}\): Mn is in the +2 oxidation state; hence it has 5 d electrons (d^5).(c) Fe in \([\mathrm{FeF}_6]^{3-}\): Fe is in the +3 oxidation state; hence it has 5 d electrons (d^5).(d) Cr in \([\mathrm{Cr(en)}_3]^{3+}\): Cr is in the +3 oxidation state; hence it has 3 d electrons (d^3).

Step 2: Crystal Field Splitting in Octahedral Complexes

In octahedral complexes, the d orbitals split into two sets: the lower energy t2g orbitals (dxy, dxz, dyz) and the higher energy eg orbitals (dz^2, dx^2-y^2). The crystal field splitting energy is represented by \(\Delta\).

Step 3: Consider High-spin and Low-spin Cases

High-spin and low-spin arrangements refer to the distribution of electrons between these orbitals in response to the crystal field splitting. High-spin: Electrons fill the orbitals to maximize unpaired spins. Low-spin: Electrons fill in pairs in the lower energy orbitals first.

Step 4: Apply to Complexes

(a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is typically high-spin. Place 4 electrons with maximum unpaired spins in t2g and eg.(b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is typically high-spin. Place 5 electrons with maximum unpaired spins in t2g and eg.(c) \(\left[\mathrm{FeF}_{6}\right]^{3-}\): Fluoride is a weak field ligand, hence typically high-spin; place 5 electrons as unpaired as possible in t2g and eg.(d) \(\left[\mathrm{Cr(en)}_{3}\right]^{3+}\): Ethylenediamine (en) is a strong field ligand; in a low-spin case, place 3 d electrons in t2g orbitals.

Key Concepts

d electron count

In transition metal complexes, understanding the number of d electrons is crucial. This count determines how electrons arrange in different energy levels. Each metal ion has a defined number of d electrons, which depends on its oxidation state.

• Chromium (Cr) in \([\mathrm{Cr(OH}_2)]^{2+}\), with an oxidation state of +2, has 4 d electrons (d⁴).
• Manganese (Mn) in \([\mathrm{Mn(OH}_2)]^{2+}\), also in the +2 oxidation state, features 5 d electrons (d⁵).
• Iron (Fe) in \([\mathrm{FeF}_6]^{3-}\), at the +3 oxidation state, similarly has 5 d electrons (d⁵).
• Chromium (Cr) in \([\mathrm{Cr(en)}_3]^{3+}\), being in a +3 oxidation state, contains 3 d electrons (d³).

Identifying the d electron count helps predict how these electrons will be distributed in various ligand environments.

high-spin and low-spin cases

High-spin and low-spin configurations arise due to the arrangement of d electrons in a crystal field. This arrangement is influenced by the strength of the ligand field and the resulting crystal field splitting.

• High-spin complexes are typically formed with weak field ligands, where electrons occupy the higher energy levels before pairing up. This leads to maximum unpaired electrons.
• Low-spin complexes occur with strong field ligands, where electrons pair up in the lower energy \(t_{2g}\) orbitals before moving to higher levels. This results in fewer unpaired electrons.

The determination of high-spin or low-spin states depends on the balance between spin-pairing energy and crystal field splitting energy (\(\Delta\)). This energy impacts how electrons are distributed among the orbitals.
Consider \([\mathrm{Cr(en)}_3]^{3+}\) as an example. It typically forms a low-spin complex due to ethylenediamine's strong field nature, making it favorable for electrons to pair in lower energy orbitals.

octahedral complexes

Octahedral complexes are formed when six ligands surround a central metal ion, forming an octahedron. This geometry leads to a specific splitting of the d orbitals into two distinct energy levels.

• The lower energy set is the \(t_{2g}\) orbitals, which include \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\).
• The higher energy set is the \(e_g\) orbitals, comprising \(d_{z^2}\) and \(d_{x^2-y^2}\).

The energy difference between these sets is called the crystal field splitting energy, denoted by \(\Delta\).
Complexes like \([\mathrm{Cr(H_2O)_6}]^{2+}\) assume a typical high-spin configuration because water is a weak field ligand that results in a smaller \(\Delta\), allowing electrons to remain unpaired and spread across both \(t_{2g}\) and \(e_g\) orbitals. Understanding this splitting is key to predicting the magnetic and electronic properties of the complex.