Problem 46 - Chemistry and Chemical Reactivity

Question

Determine the number of unpaired electrons in the following tetrahedral complexes. All tetrahedral complexes are high spin. (a) \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) (c) \(\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) (b) VOCI_ \((d)\left[C u(e n)_{2}\right]^{2+}\)

Step-by-Step Solution

Answer

(a) 0 unpaired electrons; (b) 5 unpaired electrons; (c) 2 unpaired electrons; (d) 1 unpaired electron.

Step 1: Determine the oxidation state and electron configuration for Zn

For the complex \( \left[\mathrm{Zn}\left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\right]^{2+} \), zinc typically has an oxidation state of +2. The electronic configuration for \( \text{Zn}^{2+} \) is \( [\text{Ar}] 3d^{10} \) because the 4s electrons are lost first. Since \( 3d^{10} \) is a fully filled d-orbital, there are no unpaired electrons.

Step 2: Determine the electron configuration for Mn in the complex

For the complex \( \mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} \), manganese can be assumed to be in the +2 oxidation state for simplicity, given it commonly forms these ions. \( \text{Mn}^{2+} \) has an electronic configuration of \( [\text{Ar}] 3d^5 \). As a high spin complex, all five electrons will occupy different orbitals, resulting in five unpaired electrons.

Step 3: Determine the oxidation state and electron configuration for V in VOCl_

In VOCl, vanadium is usually in the +3 oxidation state due to the VO component, leaving it as \( \text{V}^{3+} \) which has an electronic configuration of \( [\text{Ar}] 3d^2 \). This means there are two unpaired electrons in the d-orbitals for this high spin tetrahedral complex.

Step 4: Determine the electron configuration for Cu in the complex

For the complex \( \left[\text{Cu(en)}_2\right]^{2+} \), copper typically has a +2 oxidation state, resulting in \( \text{Cu}^{2+} \) with an electronic configuration of \( [\text{Ar}] 3d^9 \). With only one electron missing from a filled d shell, there is a single unpaired electron in the \( 3d \) orbitals.

Key Concepts

Tetrahedral Complexes

In coordination chemistry, tetrahedral complexes are a common arrangement where a central metal ion is surrounded by four ligands placed at the corners of a tetrahedron. This geometric configuration is significant because it lacks the symmetry found in octahedral complexes, resulting in different electron interactions.
Tetrahedral complexes are typically found with metal ions that are less prone to strong field splitting. This means that they exhibit different electronic properties compared to other geometries.

• The arrangement causes the d-orbitals to split into two levels with an energy difference less than that in octahedral complexes.
• High spin states are often found in tetrahedral complexes due to smaller ligand field splitting energy.

Tetrahedral complexes are generally high spin due to this smaller splitting, which means electrons are more likely to occupy higher energy orbitals to minimize electron repulsion.

Electron Configuration

Electron configuration refers to the distribution of electrons in an atom or ion across different orbitals. This concept is crucial for understanding the chemical behavior of atoms and their complexes.
When determining the electron configuration, it involves filling the atomic orbitals in a given sequence, generally starting from the lowest energy level. The Aufbau principle guides this filling process.
For coordination complexes, like those mentioned in the exercise, the electron configuration of the metal ion is affected by its oxidation state:

• For instance, Zn in \( \left[\mathrm{Zn} \left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\right]^{2+} \) is in a +2 oxidation state resulting in electron configuration of \[ [\text{Ar}] 3d^{10} \].
• Manganese in \( \mathrm{Mn} \left(\mathrm{NH}_{3} \right)_{2} \mathrm{Cl}_{2} \) as a +2 oxidation state shows \[ [\text{Ar}] 3d^5 \].

Understanding these configurations helps in predicting the number of unpaired electrons and the magnetic properties of the complex.

High Spin Complexes

In coordination chemistry, high spin complexes are those in which the electrons fill the higher energy d-orbitals in the presence of a weak field ligand. This occurs because the energy required to pair electrons in lower energy orbitals is greater than placing them in the higher energy orbitals.
High spin complexes are typical in cases where the splitting energy (abla) between orbitals is less than the pairing energy required:

• Tetrahedral complexes, as mentioned before, generally form high spin states due to the smaller splitting of d-orbitals.
• In \( \mathrm{Mn} \left(\mathrm{NH}_{3} \right)_{2} \mathrm{Cl}_{2} \), manganese experiences a high spin situation where it has five unpaired electrons due to this minimal splitting.

The implications of high spin on physical properties include increased magnetic characteristics, such as being paramagnetic.

Unpaired Electrons

Electrons that do not have a pairing partner within an orbital are termed unpaired electrons. The presence of these unpaired electrons is what contributes to the magnetic properties of a complex.
The number of unpaired electrons in a complex can influence its behavior in a magnetic field:

• For zinc in the complex \( \left[\mathrm{Zn} \left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\right]^{2+} \), all electrons are paired, resulting in a diamagnetic nature.
• The manganese complex \( \mathrm{Mn} \left(\mathrm{NH}_{3} \right)_{2} \mathrm{Cl}_{2} \) with high spin has five unpaired electrons, showing strong paramagnetic properties.
• Similarly, copper in \( \left[\text{Cu(en)}_2\right]^{2+} \) has one unpaired electron.

Understanding and determining the number of unpaired electrons is crucial for predicting and explaining the magnetic properties of complexes.