Question

Write out the electron configuration for each of the following metal ions: (a) \(\mathrm{Sc}^{3+}\) (b) \(\mathrm{K}^{+}\) (c) \(\mathrm{Ti}^{4+}\) (d) \(\mathrm{Ba}^{2+}\)

Short answer

Sc^{3+}: [Ar]; K^{+}: [Ar]; Ti^{4+}: [Ar]; Ba^{2+}: [Xe].

Step-by-step solution

Understanding Neutral Atom Configuration

To write the electron configuration of ions, start by knowing the configuration of the neutral atom. This will help us know which electrons are removed when forming ions. For scandium (Sc), potassium (K), titanium (Ti), and barium (Ba), the number of electrons are 21, 19, 22, and 56, respectively.

Determine Neutral Electron Configuration

For each metal in its neutral state, determine the electron configuration: - Sc (atomic number 21): [Ar] 3d^1 4s^2 - K (atomic number 19): [Ar] 4s^1 - Ti (atomic number 22): [Ar] 3d^2 4s^2 - Ba (atomic number 56): [Xe] 6s^2

Removing Electrons for Cation Configuration: Scandium

Remove electrons from the outermost shell first. For Sc^{3+}, remove three electrons: two from 4s and one from 3d. The configuration becomes: Sc^{3+} : [Ar]

Removing Electrons for Cation Configuration: Potassium

For K^{+}, remove one electron from the 4s orbital: K^{+} : [Ar]

Removing Electrons for Cation Configuration: Titanium

For Ti^{4+}, remove four electrons: two from the 4s and two from the 3d orbitals. Ti^{4+} : [Ar]

Removing Electrons for Cation Configuration: Barium

For Ba^{2+}, remove two electrons from the 6s orbital: Ba^{2+} : [Xe]

Key concepts

Metal Ions

Metal ions are atoms that have lost electrons to become positively charged. When an atom of a metal loses electrons, it forms a cation—a positively charged ion. The number of electrons lost corresponds to the charge on the ion. For example, if a metal atom loses three electrons, it will have a charge of +3.

Understanding metal ions is crucial because their electron configurations can greatly affect their chemical behavior and interactions. In the process of losing electrons, these metal ions typically achieve a more stable electron configuration, often resembling the electron configuration of the nearest noble gas. This stability is why many metal ions frequently occur in biological systems and industrial applications.

Electron Configuration of Ions

When determining the electron configuration of ions, the first step is understanding the electron configuration of the neutral atom. From this base, electrons are removed to form the ion. 1. **Neutral Atom Configuration**: Begin with the configuration that represents the distribution of electrons in the neutral state.
2. **Removal/Addition of Electrons**: Adjust the electron count based on the ion's charge. A positive charge indicates loss of electrons, while a negative charge indicates gain.
3. **Identifying the Configuration**: The resulting electron configuration reflects the ionized state.

For example, the \(\mathrm{Sc}^{3+}\) ion is derived from scandium's neutral configuration of [Ar] 3d^1 4s^2 to simply [Ar], as three electrons are removed.

Removal of Electrons

Removing electrons from a neutral atom to form a cation generally follows a specific order. Electrons are first removed from the outermost shell, which often contains the highest energy electrons. Here's how it works:

• Electrons in the highest principal energy level (highest n value) are removed first.
• Within the same energy level, electrons are typically removed from the highest energy sublevel first (such as removing electrons from the 4s orbital before the 3d in transition metals).

For metal ions like \( \mathrm{Ti}^{4+} \), this process is critical. The neutral Ti atom configuration is [Ar] 3d^2 4s^2. To form \( \mathrm{Ti}^{4+} \), you remove two 4s and two 3d electrons, ending up with: [Ar].

This sequence ensures that the ion achieves a stable, lower energy state after the electrons are removed.

Neutral Atom Configuration

Neutral atom configuration refers to the original electron distribution in an atom before it forms an ion. The configuration is based on the atom's total number of electrons, which is equal to its atomic number.

• **Scandium (Sc)**: 21 electrons, configuration [Ar] 3d^1 4s^2.
• **Potassium (K)**: 19 electrons, configuration [Ar] 4s^1.
• **Titanium (Ti)**: 22 electrons, configuration [Ar] 3d^2 4s^2.
• **Barium (Ba)**: 56 electrons, configuration [Xe] 6s^2.

This configuration serves as the starting point for determining how many electrons need to be removed when the atom transitions into a cation form. Knowing the neutral configuration allows one to predict the resulting ion's electron configuration, which aligns with the goal of achieving the stable electron arrangement of noble gases.