Question

Write the electron configurations of the following ions: \(\mathrm{V}^{5+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Cu}^{2+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}\)

Short answer

Remove electrons to form ions: - \( \mathrm{V}^{5+} \): [Ar] - \( \mathrm{Cr}^{3+} \): [Ar] 3d³ - \( \mathrm{Mn}^{2+} \): [Ar] 3d⁵ - \( \mathrm{Fe}^{3+} \): [Ar] 3d⁵ - \( \mathrm{Cu}^{2+} \): [Ar] 3d⁹ - \( \mathrm{Sc}^{3+} \): [Ar] - \( \mathrm{Ti}^{4+} \): [Ar].

Step-by-step solution

Understand Neutral Atom Configurations

Start with the neutral atoms for each of the elements. Recognize that we need the electron configuration for the neutral atoms: Vanadium (V), Chromium (Cr), Manganese (Mn), Iron (Fe), Copper (Cu), Scandium (Sc), and Titanium (Ti).

Write Neutral Atom Configurations

Write the electron configuration for each neutral atom: - V: [Ar] 3d³ 4s² - Cr: [Ar] 3d⁵ 4s¹ - Mn: [Ar] 3d⁵ 4s² - Fe: [Ar] 3d⁶ 4s² - Cu: [Ar] 3d¹⁰ 4s¹ - Sc: [Ar] 3d¹ 4s² - Ti: [Ar] 3d² 4s²

Consider Lost Electrons for Ions

Determine that ions lose electrons from the highest energy level first (typically 4s before 3d in these examples).

Calculate Electron Configuration for Each Ion

Now, calculate electron configurations by removing electrons (from 4s first, then 3d if needed) according to the charge:- \( \mathrm{V}^{5+} \): Remove 5 electrons from [Ar] 3d³ 4s², resulting in [Ar].- \( \mathrm{Cr}^{3+} \): Remove 3 electrons from [Ar] 3d⁵ 4s¹, resulting in [Ar] 3d³.- \( \mathrm{Mn}^{2+} \): Remove 2 electrons from [Ar] 3d⁵ 4s², resulting in [Ar] 3d⁵.- \( \mathrm{Fe}^{3+} \): Remove 3 electrons from [Ar] 3d⁶ 4s², resulting in [Ar] 3d⁵.- \( \mathrm{Cu}^{2+} \): Remove 2 electrons from [Ar] 3d¹⁰ 4s¹, resulting in [Ar] 3d⁹.- \( \mathrm{Sc}^{3+} \): Remove 3 electrons from [Ar] 3d¹ 4s², resulting in [Ar].- \( \mathrm{Ti}^{4+} \): Remove 4 electrons from [Ar] 3d² 4s², resulting in [Ar].

Key concepts

Transition Metal Ions

Transition metal ions are fascinating entities because they exhibit unique properties due to their electron configurations.
Transition metals occupy the d-block of the periodic table and are known for their colorful compounds, variable oxidation states, and catalytic abilities.
In their elemental form, they contain electrons in the s and d subshells of their outermost electron shell. When these metals form ions, they lose some of their outermost electrons to achieve a more stable electron configuration.
The ions of transition metals are essential in understanding chemistry, as they often act as catalysts in various chemical reactions.
They can form complexes with other ions or molecules, leading to diverse chemical behavior and applications.

Electron Removal

Electron removal is the process by which atoms lose electrons to form ions.
For transition metals, understanding which electrons are removed first is vital due to the involvement of both 4s and 3d orbitals.
When forming ions, electrons are typically removed from the outermost energy level or shell. This means removing from the 4s orbital before the 3d.
This pattern may seem counterintuitive because 4s is filled before 3d in a neutral atom.
However, when it comes to electron removal in transition metal ions, the energy level of the electrons plays a role in their loss, making the 4s electrons more accessible to be removed first.

Electron Energy Levels

Electron energy levels depict the arrangement of electrons around a nucleus in a structured manner.
These levels, defined by quantum mechanics, are critical in dictating an atom's electron configuration.
Understanding these levels allows scientists to predict the chemical behavior of an element. For transition metals, the interaction between 4s and 3d orbitals is crucial.
The energy levels are not fixed but differ based on the atomic number and the surrounding environment of electrons.
Lower energy levels fill first, but during ionization, we see deviations in electron removal strategies.

• The 4s orbital fills and loses electrons before the 3d orbital.
• As electron energy increases, electrons can jump between levels, influencing the properties of the element and the ion forms.

This dynamic behavior explains the versatile chemistry of transition metals.

Neutral Atom Configuration

Neutral atom configuration refers to the number and arrangement of electrons in an atom when it is not charged.
This configuration provides a stable baseline from which ions are derived.
For the transition metals in question—vanadium, chromium, manganese, iron, copper, scandium, and titanium—understanding their neutral electron configuration is the first step in determining their ionic forms.
Here’s what we remember about their neutral configurations:

• Vanadium (V): [ ext{Ar}] 3d³ 4s²
• Chromium (Cr): [ ext{Ar}] 3d⁵ 4s¹
• Manganese (Mn): [ ext{Ar}] 3d⁵ 4s²
• Iron (Fe): [ ext{Ar}] 3d⁶ 4s²
• Copper (Cu): [ ext{Ar}] 3d¹⁰ 4s¹
• Scandium (Sc): [ ext{Ar}] 3d¹ 4s²
• Titanium (Ti): [ ext{Ar}] 3d² 4s²

Understanding these configurations enables the calculation of their ionic states by appropriately removing electrons for each ion’s specific charge.