Question

Draw the crystal-field energy-level diagrams and show the placement of \(d\) electrons for each of the following: (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (four unpaired electrons), \((\mathbf{b})\left[\operatorname{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (a high-spin complex), (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+}\) (a low-spin complex) \((\mathbf{d})\left[\operatorname{Ir} \mathrm{Cl}_{6}\right]^{2-}\) (a low-spin complex) \((\mathbf{e})\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\) \((\mathbf{f})\left[\mathrm{NiF}_{6}\right]^{4-}.\)

Short answer

(a) For [Cr(H2O)6]2+ (high-spin): t2g: _↑_ _↑_ _↑_ eg: _↑_ _↓_ Four unpaired electrons. (b) For [Mn(H2O)6]2+ (high-spin): t2g: _↑_ _↑_ _↑_ eg: _↑_ _↑_ Five unpaired electrons. (c) For [Ru(NH3)5(H2O)]2+ (low-spin): t2g: _↑↓_ _↑↓_ _↑↓_ eg: _↓_ _↓_ No unpaired electrons. (d) For [IrCl6]2- (low-spin): t2g: _↑↓_ _↑↓_ _↑↓_ eg: _↑_ _↑↓_ One unpaired electron. (e) For [Cr(en)3]3+ (low-spin): t2g: _↑↓_ _↑↓_ _↑_ eg: _↓_ _↓_ One unpaired electron. (f) For [NiF6]4- (high-spin): t2g: _↑_ _↑_ _↑_ eg: _↑_ _↑_ Four unpaired electrons.

Step-by-step solution

(a) Cr(H2O)6 2+

: For the complex [Cr(H2O)6]2+, the central metal ion is Cr2+. The electronic configuration of Cr2+ is [Ar] 3d4. As the ligand is H2O, a weak field ligand, the complex will be high-spin, hence the filling of d electrons will be according to the increasing energy levels. The crystal field energy level diagram would look like this: t2g: _↑_ _↑_ _↑_ eg: _↑_ _↓_ Thus, we can see there are four unpaired electrons.

(b) Mn(H2O)6 2+

: Similarly, for the complex [Mn(H2O)6]2+, the central metal ion is Mn2+. The electronic configuration of Mn2+ is [Ar] 3d5. As H2O is a weak field ligand, the complex is high-spin. The crystal field energy level diagram would look like this: t2g: _↑_ _↑_ _↑_ eg: _↑_ _↑_ There are five unpaired electrons.

(c) Ru(NH3)5(H2O) 2+

: For the complex [Ru(NH3)5(H2O)]2+, the central metal ion is Ru2+. The electronic configuration of Ru2+ is [Kr] 4d6. As NH3 is a strong field ligand, the complex is low-spin. The crystal field energy level diagram would look like this: t2g: _↑↓_ _↑↓_ _↑↓_ eg: _↓_ _↓_ There are no unpaired electrons.

(d) IrCl6 2-

: For the complex [IrCl6]2-, the central metal ion is Ir2-. The electronic configuration of Ir2- is [Xe] 5d9. Since Cl- is a weak field ligand, the complex is low-spin. The crystal field energy level diagram would look like this: t2g: _↑↓_ _↑↓_ _↑↓_ eg: _↑_ _↑↓_ There is just one unpaired electron.

(e) Cr(en)3 3+

: For the complex [Cr(en)3]3+, the central metal ion is Cr3+. The electronic configuration of Cr3+ is [Ar] 3d3. Since en is a strong field ligand, the complex is low-spin. The crystal field energy level diagram would look like this: t2g: _↑↓_ _↑↓_ _↑_ eg: _↓_ _↓_ There is only one unpaired electron.

(f) NiF6 4-

: For the complex [NiF6]4-, the central metal ion is Ni4-. The electronic configuration of Ni4- is [Ar] 3d6. Since F- is a weak field ligand, the complex is high-spin. The crystal field energy level diagram would look like this: t2g: _↑_ _↑_ _↑_ eg: _↑_ _↑_ There are four unpaired electrons.

Key concepts

High-spin complex

High-spin complexes occur when electrons in the d-orbitals of a transition metal are arranged in such a way that they occupy higher energy orbitals to minimize electron pairing. This happens when the ligands surrounding the metal are weak field ligands, which do not sufficiently increase the energy gap between the lower t2g orbitals and the higher eg orbitals.

As a result, a small crystal field splitting leads to higher overall energy but more unpaired electrons. High-spin complexes are often found with ligands like water (H2O) and fluoride (F-), which do not generate significant crystal field splitting.

For example, in the case of \(\left[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6\right]^{2+} \), the Mn2+ ion has an electron configuration that places five unpaired electrons in the five 3d orbitals, leading to high spin. This arrangement is also evident in \(\left[\mathrm{NiF}_6\right]^{4-}\), where Ni4- ion also forms a high-spin complex resulting in four unpaired electrons.

Low-spin complex

Low-spin complexes form when the ligands attached to the transition metal are strong field ligands. This means the energy gap between the lower t2g and the higher eg orbitals is sufficiently large to discourage electron occupation of the higher energy orbitals.

Strong field ligands, such as ammonia (NH3) or ethylenediamine (en), encourage electrons to pair up in the lower energy t2g set before filling the eg set. As a result, low-spin complexes have fewer unpaired electrons and tend to have lower magnetic moments compared to high-spin complexes.

For instance, \(\left[\mathrm{Ru}(\mathrm{NH}_3)_5(\mathrm{H}_2\mathrm{O})\right]^{2+} \) exhibits a low-spin configuration due to the presence of NH3, resulting in all electrons being paired in the 4d orbitals. Similarly, \(\left[\mathrm{Cr}(\mathrm{en})_3\right]^{3+}\), due to the strength of the en ligand, also forms a low-spin complex with only one unpaired electron.

Ligands

Ligands are molecules or ions that bind to a central metal atom or ion to form a coordination complex. They donate pairs of electrons to fill the metal's orbitals and facilitate bonding. The nature of the ligands — whether they are strong or weak field — significantly influences the electronic configuration of the metal center and determines whether the resulting complex is high-spin or low-spin.

• Weak Field Ligands (e.g., H2O, F-, Cl-): These ligands cause small splitting in d-orbitals, resulting in high-spin complexes with more unpaired electrons.
• Strong Field Ligands (e.g., NH3, en): These cause larger splitting, leading to low-spin complexes with fewer unpaired electrons.

The compound \(\left[\mathrm{Ir}\mathrm{Cl}_6\right]^{2-}\) illustrates the slight exception where Cl- is a weak field ligand but still forms a low-spin complex. It highlights that while ligand strength is critical, other factors like metal ion and overall charge may also play a role.

Transition metal complexes

Transition metal complexes consist of a central metal atom or ion surrounded by molecules or ions known as ligands. The metal is typically a transition metal, which means it has partially filled d-orbitals. These orbitals can interact with the lone pair electrons from the ligands, forming a range of coordination compounds with diverse properties.

• Central Metal: Determines the complex’s electronic configuration and coordination number.
• Ligands: Influence the color, magnetism, solubility, and stability of the complex.
• Geometry: Affects properties like color and reactivity; common shapes include octahedral, tetrahedral, and square planar.

The interplay between the central metal's electronic configuration and the ligands determines whether the complex is high-spin or low-spin, impacting its magnetic and spectral characteristics. Examples from the exercise, such as \(\left[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6\right]^{2+} \) and \(\left[\mathrm{Ir}\mathrm{Cl}_6\right]^{2-}\), show how changes in ligands and metal ions alter the crystal field splitting and therefore the final arrangement of d-electrons.