Question

Determine the number of unpaired electrons in the following tetrahedral complexes. All tetrahedral complexes are high spin. (a) \(\left[\mathrm{FeCl}_{4}\right]^{2-}\) (c) \(\left[\mathrm{Mn} \mathrm{Cl}_{4}\right]^{2-}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CoCl}_{4}\right]\) (d) \(\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{ZnCl}_{4}\right]\)

Short answer

(a) 4 unpaired electrons, (b) 3 unpaired electrons, (c) 5 unpaired electrons, (d) 0 unpaired electrons.

Step-by-step solution

Determine the oxidation state of the metal

First, find the oxidation state of the metal in each complex. For (1) \([\mathrm{FeCl}_{4}]^{2-}\), iron has an oxidation state of +2 because each chloride carries a -1 charge.(2) \([\mathrm{MnCl}_{4}]^{2-}\), manganese is +2.(3) \([\mathrm{CoCl}_{4}]^{2-}\) in \(\mathrm{Na}_{2}[\mathrm{CoCl}_{4}]\), cobalt is +2.(4) \([\mathrm{ZnCl}_{4}]^{2-}\) in \((\mathrm{NH}_{4})_{2}[\mathrm{ZnCl}_{4}]\), zinc is +2.

Determine the electron configuration of the metal

Next, calculate the electron configuration of each metal ion: - Fe²⁺: 3d⁶ - Mn²⁺: 3d⁵ - Co²⁺: 3d⁷ - Zn²⁺: 3d¹⁰

Determine high-spin electron configuration in tetrahedral complexes

Tetrahedral complexes with weak field ligands like Cl⁻ typically result in a high-spin configuration. Based on their electron configuration: - For Fe²⁺: 3d⁶ configuration, use all orbitals singly before pairing. This gives us 4 unpaired electrons. - For Mn²⁺: 3d⁵ configuration, this gives 5 unpaired electrons. - For Co²⁺: 3d⁷ configuration, this gives 3 unpaired electrons. - For Zn²⁺: 3d¹⁰ configuration, all electrons are paired, so 0 unpaired electrons.

Key concepts

Tetrahedral Complexes

Tetrahedral complexes are a fascinating arrangement of coordination compounds where a metal ion is surrounded by four ligands. The geometry of these complexes forms a pyramid with a triangular base, much akin to the shape of a pyramid in 3D. This structure arises because the ligands position themselves to minimize repulsion in a spatial arrangement around the central metal ion. Generally, tetrahedral complexes are less symmetrical compared to other geometric forms, such as octahedral complexes.

This geometry affects the splitting of d-orbitals. In an octahedral complex, five d-orbitals split into two groups of different energies under the influence of the ligands. In tetrahedral complexes, a similar phenomenon occurs, but the energy gap is smaller due to fewer ligands interacting with the central metal ion. The smaller splitting typically results in high-spin configurations, as we'll explore next, since it's less energetically favorable for electrons to pair up in d-orbitals.

High Spin

High-spin configurations occur in coordination complexes where ligands are weak field, such as chloride ions. In a high-spin complex, the energy required to pair up electrons in the same orbital is greater than the energy difference between the split d-orbitals. Thus, electrons prefer to occupy additional empty orbitals rather than pairing up with others already present.

For example, in a high-spin tetrahedral complex with a metal such as iron, whose oxidation state is +2, the electrons fill the d-orbitals singly before any pairing occurs, resulting in unpaired electrons. High-spin complexes are common in tetrahedral geometry because the ligands are usually weakly interacting, causing less splitting of the d-orbitals as compared to strong field ligands in different geometries.

Electron Configuration

Electron configuration is key to understanding how electrons occupy orbitals in an atom or ion. For transition metals, the valence electrons are found in the d and sometimes s orbitals. Knowing the electron configuration helps us determine unpaired electrons in a complex. The general rule is that when filling these orbitals, electrons fill lower energy levels first, and once a level is filled, they move to the next.

In the example of \([\text{FeCl}_4]^{2-}\), the electron configuration for Fe²⁺ is 3d⁶, meaning there are six electrons occupying the five d-orbitals. In a high-spin arrangement, each of the d-orbitals gets filled with one electron before any orbital gets an additional electron, resulting in four unpaired electrons. Recognizing the electron configurations such as for Mn²⁺ (3d⁵), Co²⁺ (3d⁷), and Zn²⁺ (3d¹⁰) helps in identifying how many electrons remain unpaired in different tetrahedral complexes.

Oxidation States

Oxidation states represent the degree of oxidation of an atom in a compound. Determining the oxidation state of a metal in a complex helps us understand how its electrons will arrange themselves. This is crucial for calculating electron configuration.

The oxidation state can be determined by considering the charges of the ligands and the overall charge of the complex. In \([\text{FeCl}_4]^{2-}\), each chloride ion has a charge of -1. With four chloride ions, that gives a total charge of -4 from the ligands. The entire complex has a -2 charge, so the oxidation state of iron must be +2 to balance it. Similarly, the process applies to all metals in other given complexes, including manganese, cobalt, and zinc, helping us to systematically determine how electrons are distributed and how many remain unpaired.