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) \(\left[\mathrm{MnCl}_{4}\right]^{2}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CoCl}_{4}\right]\) \((\mathrm{d})\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{ZnCl}_{4}\right]\)

Short answer

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

Step-by-step solution

Determine Metal Oxidation States

Identify the oxidation state of each central metal. (a) In \([\text{FeCl}_{4}]^{2-}\), iron has oxidation state +2. (b) In \(\text{Na}_{2}[\text{CoCl}_{4}]\), cobalt has oxidation state +2. (c) In \([\text{MnCl}_{4}]^{2-}\), manganese has oxidation state +2. (d) In \((\text{NH}_{4})_{2}[\text{ZnCl}_{4}]\), zinc has oxidation state +2.

Write Electronic Configurations for Each Metal Ion

Determine the electron configuration for each metal ion based on their oxidation states. Iron(II) \([\text{Fe}^{2+}]\) is \([\text{Ar}] 3d^6\). Cobalt(II) \([\text{Co}^{2+}]\) is \([\text{Ar}] 3d^7\). Manganese(II) \([\text{Mn}^{2+}]\) is \([\text{Ar}] 3d^5\). Zinc(II) \([\text{Zn}^{2+}]\) is \([\text{Ar}] 3d^{10}\).

Apply High Spin Configurations in Tetrahedral Complexes

For high spin complexes, electrons fill orbitals to maximize spin, following Hund's rule. (a) \([\text{FeCl}_{4}]^{2-}\): 4 unpaired electrons \((3d^4)\). (b) \(\text{Na}_{2}[\text{CoCl}_{4}]\): 3 unpaired electrons \((3d^7)\). (c) \([\text{MnCl}_{4}]^{2-}\): 5 unpaired electrons \((3d^5)\). (d) \((\text{NH}_{4})_{2}[\text{ZnCl}_{4}]\): 0 unpaired electrons \((3d^{10})\).

Key concepts

High Spin Complexes

In chemistry, high spin complexes are a type of coordination compound where the metal center interacts with ligands, resulting in a specific arrangement of electrons in the metal's d-orbitals. These complexes are particularly common in tetrahedral coordination, where transition metals are surrounded by four ligands. The term "high spin" refers to the way electrons fill the d-orbitals, prioritizing maximum unpaired spins. This means that instead of pairing up electrons in lower energy orbitals, electrons will occupy orbitals of the same energy with parallel spins.

• This occurs because of the relatively low energy difference between d-orbitals in tetrahedral complexes.
• Electrons fill each orbital singly as far as possible, following Hund's rule.
• This type of arrangement results in more unpaired electrons.

High spin complexes are important for understanding the magnetic properties of compounds. More unpaired electrons typically mean stronger magnetic properties, making these complexes paramagnetic.

Oxidation States

Oxidation states are vital in determining how many electrons a metal ion has lost or gained compared to its elemental form. For transition metals in complexes, the oxidation state influences the number of valence electrons available for bonding and interaction with ligands. To determine the oxidation state, consider the overall charge of the complex and the known charges of the surrounding ligands.

• For example, chloride (\( ext{Cl}^-\)) is a common ligand with a -1 charge.
• In the complex \( ext{FeCl}_4^{2-}\), with each Cl carrying a -1 charge, iron must have an oxidation state of +2 to balance the -4 from the four chlorides plus the 2- overall charge.
• This principle applies similarly to the other complexes like \( ext{MnCl}_4^{2-}\) and \( ext{CoCl}_4^{2-}\).

The oxidation state helps elucidate the electronic configuration of the metal ion, crucial for predicting the number of unpaired electrons.

Electronic Configurations

Electronic configurations describe the distribution of electrons in an atom or ion. These configurations are particularly significant when evaluating unpaired electron numbers in transition metals. If you know the oxidation state, you can deduce the number of electrons removed from the neutral atom's electronic configuration. This tells you the ion's electronic configuration.

• For \( ext{Fe}^{2+}\), starting from neutral iron's configuration \([ ext{Ar}] 4s^2 3d^6\), 2 electrons are lost, resulting in \([ ext{Ar}] 3d^6\).
• Similarly, \( ext{Mn}^{2+}\) becomes \([ ext{Ar}] 3d^5\).
• This configuration directly ties into how electrons arrange in high or low spin states in complexes.

Knowing electronic configurations allows us to apply Hund's rule to find out how many unpaired electrons are present in these complexes.

Hund's Rule

Hund's rule is an essential principle in chemistry that dictates how electrons will fill available orbitals. It states that electrons must occupy every orbital singly before any orbital receives a second electron. In orbitals of the same energy level, electrons will go into each one singly and parallel, maximizing the number of unpaired electrons. This is especially relevant for high spin complexes.

• In current examples, like \( ext{Mn}^{2+}\) with a \(d^5\) configuration, all five d-orbitals each have one electron, maximizing unpaired spins.
• Applying Hund's rule minimizes electron-electron repulsions, stabilizing the complex.
• This results in a clearer prediction of complexes' magnetic properties due to the number of unpaired electrons.

Hund's rule, together with understanding oxidation states and electronic configurations, helps explain the formation and properties of high spin complexes.