Question

Properties of Transition Elements Give the electron configuration for each of the following ions, and tell whether each is paramagnetic or diamagnetic. (a) \(\mathrm{Cr}^{3+}\) (b) \(\mathrm{V}^{2+}\) (c) \(\mathrm{Ni}^{2+}\) (d) \(\mathrm{Cu}^{+}\)

Short answer

(a) \\mathrm{Cr}^{3+}\ is \\[\mathrm{[Ar] \ 3d^3}\], paramagnetic. (b) \\mathrm{V}^{2+}\ is \\[\mathrm{[Ar] \ 3d^3}\], paramagnetic. (c) \\mathrm{Ni}^{2+}\ is \\[\mathrm{[Ar] \ 3d^8}\], paramagnetic. (d) \\mathrm{Cu}^{+}\ is \\[\mathrm{[Ar] \ 3d^{10}}\], diamagnetic.

Step-by-step solution

Understand the Electron Configurations

Begin by finding the ground state electron configuration of the neutral atom. For transition metals, the 4s electrons are filled before the 3d electrons, but are the first to be removed when forming ions.

Chromium (Cr) and its Ion \\(\mathrm{Cr}^{3+}\\)

The electron configuration for neutral Chromium (Cr) is \[\mathrm{[Ar] \ 3d^5 \ 4s^1}\]. When it forms the \(\mathrm{Cr}^{3+}\) ion, three electrons are removed: first from 4s and then two from 3d, resulting in \(\mathrm{[Ar] \ 3d^3}\). This configuration has unpaired electrons, making \(\mathrm{Cr}^{3+}\) paramagnetic.

Vanadium (V) and its Ion \\(\mathrm{V}^{2+}\\)

The electron configuration for neutral Vanadium (V) is \[\mathrm{[Ar] \ 3d^3 \ 4s^2}\]. For \(\mathrm{V}^{2+}\), two electrons are removed from the 4s and 3d orbitals, resulting in \[\mathrm{[Ar] \ 3d^3}\]. With unpaired electrons in the 3d orbital, \(\mathrm{V}^{2+}\) is paramagnetic.

Nickel (Ni) and its Ion \\(\mathrm{Ni}^{2+}\\)

The electron configuration for neutral Nickel (Ni) is \[\mathrm{[Ar] \ 3d^8 \ 4s^2}\]. Removing two electrons for \(\mathrm{Ni}^{2+}\) leads to an electron configuration of \[\mathrm{[Ar] \ 3d^8}\]. The 3d orbital contains two unpaired electrons, making \(\mathrm{Ni}^{2+}\) paramagnetic.

Copper (Cu) and its Ion \\(\mathrm{Cu}^{+}\\)

Copper (Cu) has the electron configuration \[\mathrm{[Ar] \ 3d^{10} \ 4s^1}\]. Removing one electron to form \(\mathrm{Cu}^{+}\) results in \[\mathrm{[Ar] \ 3d^{10}}\]. The 3d orbital is fully filled, meaning all electrons are paired, so \(\mathrm{Cu}^{+}\) is diamagnetic.

Key concepts

Electron Configuration

Electron configuration describes the distribution of electrons around the nucleus of an atom. For transition metals, it's important to understand that their d orbitals play a critical role. Transition metals usually involve 3d subshells when describing electron configurations. Start with the ground state of the neutral atom and fill electrons according to the Aufbau principle, which states to fill orbitals from lowest to highest energy. Typically, for transition metals, the 4s orbital fills before the 3d. When these elements form ions, electrons are lost first from the 4s orbital, then 3d, as 4s is of higher energy once occupied.

• Remember: when ions are formed, remove electrons starting from the highest energy level
• Chromium's electron configuration: \([ ext{Ar}] \, 3d^5 \, 4s^1\)
• For \( ext{Cr}^{3+}\): remove from 4s first, ending with \([ ext{Ar}] \, 3d^3\)

This systematic approach eases the process of determining configurations.

Paramagnetic

A substance is considered paramagnetic if it has one or more unpaired electrons. These unpaired electrons result in a net magnetic moment, allowing the substance to be attracted by a magnetic field. This is a fundamental property of many transition metal ions.In the periodic table, several transition metal ions like \( ext{Cr}^{3+}\), \( ext{V}^{2+}\), and \( ext{Ni}^{2+}\) showcase paramagnetism. Their configurations leave some electrons unpaired, contributing to their magnetic properties.

• Check for unpaired electrons in the outermost d orbital.
• No complete pairing in 3d: substances align with a magnetic field.

Being aware of paramagnetism helps in understanding specific applications in material science and chemistry.

Diamagnetic

Diamagnetic energy levels are when all electrons are paired. Substances with this property are not attracted to a magnetic field, and in fact, may slightly repel a magnetic one. This is less common among transition metal ions but still relevant, as in certain configurations like \( ext{Cu}^+\).

• Complete electron pairing leads to magnetic neutrality.
• Common in filled d orbitals such as \( ext{Cu}^+\): \([ ext{Ar}] \, 3d^{10}\).

Knowing if an element is diamagnetic is useful in fields such as medicinal chemistry, where magnetic properties influence material behavior.

Transition Metal Ions

Transition metal ions form by the loss of electrons, primarily starting from the outermost s orbital and then the d orbitals. This ionization process affects the chemical behavior and properties of the metals. Typically found between groups 3 to 12 in the periodic table. For transition metals, they can exhibit multiple oxidation states due to the close energy levels of d and s orbitals. This flexibility results in their ability to form various compounds, each exhibiting unique properties.

• Focus on 3d and 4s orbitals for electron removal order.
• Chromium, Vanadium, Nickel, and Copper ions all showcase this; each revealing different magnetic properties based on their configuration.

Understanding how and why these ions are formed is crucial for applications in catalysts, electronics, and even in biochemistry.