Problem 70 - Chemistry Matter and Change

Question

Using orbital notation, diagram the formation of an ionic bond between aluminum and fluorine.

Step-by-Step Solution

Answer

The ionic bond between aluminum (Al) and fluorine (F) is formed through the transfer of 3 valence electrons from aluminum to 3 fluorine atoms. When aluminum loses its 3 valence electrons, it forms a positively charged ion Al³⁺ with an electron configuration of \(1s^2 2s^2 2p^6\). Each fluorine gains 1 electron and forms a negatively charged ion F¹⁻ with an electron configuration of \(1s^2 2s^2 2p^6\). The strong electrostatic attraction between the Al³⁺ ion and the three F¹⁻ ions forms the ionic bond. The orbital notation for this process is shown as: Al(13): [↓↑]₁s [↓↑]₂s [↓↑]₂p (↓)₃s (↓)₃p + F(9): [↓↑]₁s [↓↑]₂s [↓↑]₂p (↓↓↓__) ↓ Al¹³⁺: [↓↑]₁s [↓↑]₂s [↓↑]₂p + 3 × F¹⁻: [↓↑]₁s [↓↑]₂s [↓↑]₂p (↓↓↓↓↑)

Step 1: Electron configuration for aluminum and fluorine

To understand the formation of an ionic bond between aluminum and fluorine, we need to first determine the electron configuration of these elements. Aluminum has 13 electrons with the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^1\). Fluorine has 9 electrons with the electron configuration: \(1s^2 2s^2 2p^5\).

Step 2: Determine the valence electrons

Valence electrons are the outermost electrons in an atom and are responsible for chemical bonding. Aluminum has 3 valence electrons in the 3s and 3p orbitals. Fluorine has 7 valence electrons in the 2s and 2p orbitals.

Step 3: Transfer of electrons and formation of ions

In order to achieve a stable electron configuration, aluminum will lose its 3 valence electrons, and fluorine will gain 1 electron to complete its octet. Aluminum will lose 3 electrons and form an ion with a +3 charge: Al^3+. Three fluorine atoms will each gain 1 electron to form ions with a -1 charge: F^-.

Step 4: Orbital notation for aluminum and fluorine ions

After the transfer of electrons, we need to represent the new electronic configuration using orbital notation. Aluminum ion (Al^3+): After losing 3 electrons, the electron configuration becomes \(1s^2 2s^2 2p^6\). The 3s and 3p orbitals will be empty. Fluorine ion (F^-): After gaining 1 electron, the electron configuration becomes \(1s^2 2s^2 2p^6\). The 2p orbital is now complete.

Step 5: Diagram the formation of the ionic bond using orbital notation

To diagram the formation of the ionic bond using orbital notation, we will show the transfer of electrons from aluminum's valence orbitals to fluorine's valence orbitals. Al(13): [↓↑]₁s [↓↑]₂s [↓↑]₂p (↓)₃s (↓)₃p + F(9): [↓↑]₁s [↓↑]₂s [↓↑]₂p (↓↓↓__) ↓ Al¹³⁺: [↓↑]₁s [↓↑]₂s [↓↑]₂p + 3 × F¹⁻: [↓↑]₁s [↓↑]₂s [↓↑]₂p (↓↓↓↓↑) In this diagram, the downward arrows represent electrons transferred from aluminum's valence orbitals to each fluorine atom's valence orbitals. After the transfer, we have one aluminum ion (Al³⁺) and three fluorine ions (F¹⁻). The strong electrostatic attraction between the positively charged aluminum ion and the negatively charged fluorine ions forms the ionic bond between them.

Key Concepts

Electron Configuration

Electron configuration refers to the way electrons are distributed in an atom's orbitals. Each atom follows a specific filling order known as the Aufbau principle to reach stability.
For example:

• Aluminum (Al): has 13 electrons. Its electron configuration is written as: \(1s^2 2s^2 2p^6 3s^2 3p^1\).
• Fluorine (F): with 9 electrons, its configuration is: \(1s^2 2s^2 2p^5\).

The electron configuration tells us which orbitals are filled and helps predict how an element will bond chemically.
In the case of ionic bonds, these configurations play a critical role in understanding how electrons are shared or transferred between atoms.

Valence Electrons

Valence electrons are the electrons located in the outermost shell of an atom. They are crucial because they determine how an atom can bond with others. These electrons can be gained, lost, or shared in chemical reactions.
Let's look at the valence electrons for both aluminum and fluorine:

• Aluminum (Al): Has 3 valence electrons found in the 3s and 3p orbitals.
• Fluorine (F): Possesses 7 valence electrons located in the 2s and 2p orbitals.

Aluminum tends to lose its valence electrons to form a stable ion, while fluorine tends to gain electrons to complete its octet.
This gain or loss of electrons is what leads to the formation of an ionic bond, as seen in aluminum fluoride.

Electrostatic Attraction

Electrostatic attraction is the force that holds ions together in an ionic compound. When an atom loses or gains electrons, it becomes charged, resulting in the creation of ions. These oppositely charged ions attract each other because of this electrostatic force.
In aluminum fluoride:

• Aluminum (Al): Loses three electrons, forming an Al³⁺ cation.
• Fluorine (F): Gains one electron to become an F⁻ anion. Three fluorine atoms are required to balance the charge of one Al³⁺.

The strong attraction between the positive aluminum ion and the negative fluoride ions forms the ionic bond.
This bond holds the compound together tightly, resulting in the solid state typical of many ionic compounds.

Orbital Notation

Orbital notation provides a visual representation of an atom's electron configuration using arrows and boxes to indicate electron placements in orbitals. This approach helps in understanding the distribution and pairing of electrons.
For aluminum and fluorine, the orbital notation illustrates the transfer of electrons that forms the ionic bond:

• Aluminum (Al) before bonding: [↓↑]₁s [↓↑]₂s [↓↑]₂p (↓)₃s (↓)₃p
• Fluorine (F) before bonding: [↓↑]₁s [↓↑]₂s [↓↑]₂p (↓↓↓__)

During the formation of an ionic bond, one aluminum atom loses electrons, filling the 2p orbital in three separate fluorine atoms. This results in:

• Al³⁺ ion: [↓↑]₁s [↓↑]₂s [↓↑]₂p
• F⁻ ion: [↓↑]₁s [↓↑]₂s [↓↑]₂p (↓↓↓↓↑) for each F⁻ ion

This notation clearly depicts how aluminum's electrons fill the empty spots in fluorine's valence shell, reinforcing their electrostatic attraction.