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) \(\mathrm{VOCl}_{3}\) (d) \(\left[\mathrm{Cu}(\mathrm{en})_{2}\right]^{2+}\)

Short answer

(a) 0, (b) 0, (c) 5, (d) 1.

Step-by-step solution

- Determining the Oxidation State and d-electrons for Zn

For the complex \(Zn(H_2O)_4^{2+}\), Zn is in the state \(2+\). Zinc is located in the 12th group, meaning its pure metal form has a configuration ending in \(3d^{10}4s^2\). Note that all d electrons are retained even after ionization as Zn, hence the electron configuration becomes \(3d^{10}\) in the \(2+\) state.

Conclusion

Since all d-electrons in Zn are paired (\(3d^{10}\)), there are no unpaired electrons for \([Zn(H_2O)_4]^{2+}\).

- Determining Oxidation State and d-electrons for Mn

For \(Mn(NH_3)_2Cl_2\), since this tetrahedral complex is overall neutral and ammonia is a neutral ligand, we analyze the oxidation state of Mn:\(-2\) from \(2Cl^-\) yields the oxidation state \(2+\) for Mn. Mn's electron configuration starts as \([Ar]3d^54s^2\). In the \(2+\) state, Mn transitions to \(3d^{5}\).

Conclusion

With \(3d^5\), Mn has 5 unpaired electrons in a high-spin configuration.

- Determining Oxidation State and d-electrons for V

Considering \(VOCl_3\): with \(Cl^- = -1\) each, they total \(-3\). Oxygen also contributes \(-2\), suggesting \( V\) is in the \( 5+\) state. Starting with \([Ar]3d^34s^2\), oxidized to \(3d^0\).

Conclusion

There are no unpaired electrons as vanadium in \(VOCl_3\) becomes \(3d^0\) in this highest oxidation state.

- Determining the Oxidation State and d-electrons for Cu

In \( [Cu(en)_2]^{2+}\), copper has an oxidation state of \( 2+ \), hence configuration \( [Ar]3d^{9}\).

Conclusion

Due to its configuration (\( 3d^9\)), copper has one unpaired electron.

Key concepts

Tetrahedral Complexes

Tetrahedral complexes are a fascinating aspect of coordination chemistry. In these structures, the central metal atom is surrounded by four ligands positioned at the corners of a tetrahedron. This arrangement gives them a distinct geometric shape and influences their electronic and magnetic properties.

Due to the spatial configuration in tetrahedral complexes, there is less repulsion experienced by the d-electrons compared to octahedral complexes. This reduced repulsion results in a smaller crystal field splitting energy, meaning there is less difference in energy between the split d-orbitals.

• Lower crystal field splitting energy.
• Sp3 hybridization is common in tetrahedral complexes.
• Often leads to high-spin configurations.

Understanding the geometry and electronic structure of tetrahedral complexes is crucial in predicting their magnetic behavior and reactivity.

High-Spin Configuration

In coordination chemistry, a high-spin configuration refers to how electrons are distributed in the d-orbitals of a metal complex. In a high-spin state, electrons are spread out across the available d-orbitals. This results in a maximized number of unpaired electrons. The term high-spin indicates a configuration where less pairing among electrons occurs.

High-spin systems typically occur when the crystal field splitting energy, the energy difference between the split d-orbitals, is small. This encourages electrons to occupy higher energy orbitals instead of pairing up in lower ones. Consequently, tetrahedral complexes, due to their smaller splitting energy, often favor high-spin states.

• Common in tetrahedral complexes.
• Typically results in more unpaired electrons.
• Lower pairing energy compared to octahedral complexes.

The magnetic properties of a compound can be deduced by its spin state, with high-spin complexes often being paramagnetic due to the presence of unpaired electrons.

Oxidation States

Oxidation states are a fundamental concept in chemistry, referring to the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. For transition metals in complexes, determining the oxidation state helps in predicting the electron distribution and ligand interactions.

To determine the oxidation state, one examines the charges contributed by ligands and the overall charge of the complex. By subtracting the total negative charge of the ligands from the known charge of the complex, the oxidation state of the central metal ion is deduced.

• Important for determining electron configuration.
• Influences the magnetic and chemical behavior of the complex.
• Helps in predicting reactivity and thermal stability.

Having a clear understanding of oxidation states is essential for analyzing metalloproteins and catalysts, as well as for predicting the reactivity of metal complexes in various environments.

d-Electron Configuration

The d-electron configuration of a metal ion describes the distribution of electrons within the d-orbitals. It plays a vital role in the chemical and physical characteristics of a complex. For transition metals, the configuration arises from the filling of the d-orbitals as one progresses across the periodic table.

When forming complexes, electrons fill or vacate these d-orbitals based on the oxidation state of the metal and the strength of the interacting ligands. The arrangement of d-electrons determines key properties such as color, magnetism, and reactivity. In a metal complex, knowing the d-electron configuration can give insight into the presence of unpaired electrons.

• Directly related to the magnetic properties of the complex.
• Influences the color of the complex due to d-d transitions.
• Helps in understanding the nature of metal-ligand bonding.

An accurate determination of the d-electron configuration helps not only in determining the number of unpaired electrons but also in predicting many other properties of metal complexes.