Problem 41 - Chemistry and Chemical Reactivity

Question

Halides of the Group 3 A elements are excellent Lewis acids. When a Lewis base such as \(\mathrm{Cl}^{-}\) interacts with \(\mathrm{AlCl}_{3}\). the ion \(\mathrm{AlCl}_{4}^{-}\) is formed. Draw a Lewis electron dot structure for this ion. What structure is predicted for \(\mathrm{AlCl}_{4}^{-} ?\) What hybridization is assigned to the aluminum atom in \(\mathrm{AlCl}_{4}^{-} ?\)

Step-by-Step Solution

Answer

The Lewis structure of \( \mathrm{AlCl}_{4}^{-} \) is tetrahedral, with \( \mathrm{sp^3} \) hybridization on Al.

Step 1: Determine the Total Valence Electrons

First, calculate the total number of valence electrons available in the \( \mathrm{AlCl}_{4}^{-} \) ion. Aluminum (\( \mathrm{Al} \)) has 3 valence electrons, each chloride ion (\( \mathrm{Cl} \)) has 7 valence electrons, and there are four chloride ions. Additionally, the negative charge contributes 1 extra electron. Thus, the total number of valence electrons is\[ 3 + 4 \times 7 + 1 = 32.\]

Step 2: Draw the Skeletal Structure

In the skeletal structure of \( \mathrm{AlCl}_{4}^{-} \), place the aluminum atom at the center and distribute the chloride ions around it. Each chlorine is single-bonded to aluminum, resulting in four Al-Cl bonds.

Step 3: Distribute the Electrons

Start by placing pairs of electrons between each Al and Cl to represent the bonds. Then distribute the remaining electrons around the chlorine atoms to satisfy their octet. Since chlorine needs 8 electrons to complete its octet and already has 2 from the bonding, add 6 more electrons (3 lone pairs) to each chlorine atom.

Step 4: Check the Formal Charges

For each chlorine, the formal charge is 0 because they have full octets, sharing 1 electron with Al and having 6 non-bonding, for a total of 7 electrons, matching their valence electrons from their isolation state. Aluminum starts with 3, has contributions due to sharing across 4 bonds (1 from each bond), so has a formal charge of -1 due to one excess electron provided as part of the complete 32 system and the charge of the ion.

Step 5: Determine the Molecular Geometry and Hybridization

The molecular geometry of \( \mathrm{AlCl}_{4}^{-} \) is determined by considering the arrangement of bonds around the aluminum. With 4 bonds and no lone pairs, it adopts a tetrahedral shape. This means aluminum undergoes \( \mathrm{sp^3} \) hybridization to form four equivalent bond orbitals.

Key Concepts

Lewis electron dot structure

A Lewis electron dot structure is a way to visually represent the valence electrons around atoms within a molecule or ion. It's a simple method that helps us understand how atoms bond and the electron distribution in a compound. For the ion \( \mathrm{AlCl}_4^- \), the steps are as follows:

• Calculate the total number of valence electrons. For our ion, this is 32 electrons.
• Draw the skeletal structure by placing the aluminum (\( \mathrm{Al} \)) atom in the center, with the four chlorine (\( \mathrm{Cl} \)) atoms arranged around it.
• Distribute electron pairs to form bonds between the aluminum and chlorine atoms, and then complete the octet around each chlorine atom.
• Check formal charges to confirm the structure makes sense: each chlorine should have a formal charge of 0, while the aluminum carries a -1 to match the ion's charge.

This structure is fundamental to predicting further properties like geometry and hybridization.

molecular geometry

Molecular geometry refers to the 3D arrangement of atoms in a molecule. It is crucial because it influences the molecule's properties and reactivity.
The exercise asks us to determine the geometry of \( \mathrm{AlCl}_4^- \). Here, aluminum forms four bonds with chlorine, and with no lone pairs on aluminum, the geometry is tetrahedral.

• A tetrahedral shape forms when there are four regions of electron density around the central atom, maximizing the distance between electron pairs to minimize repulsion.
• This arrangement gives each \( \angle \text{Cl-Al-Cl} \) an angle of 109.5°.

Understanding this helps in visualizing how this ion might interact with other substances in a reaction.

hybridization

Hybridization is the concept of mixing atomic orbitals of similar energies to form new 'hybrid' orbitals. This helps explain the observed shapes and bonding patterns in molecules.
For \( \mathrm{AlCl}_4^- \):

• Aluminum starts in the ground state with an electron configuration of \( [{\rm Ne}]3s^23p^1 \).
• To form four equivalent bonds, the \( 3s \) and four \( 3p \) orbitals mix to form four \( \, sp^3 \) hybrid orbitals.
• This allows aluminum to form four equal-spaced bonds, facilitating the tetrahedral geometry.

Learning about hybridization provides insight into why molecules like \( \mathrm{AlCl}_4^- \) adopt certain geometries and enhances understanding of bond formation.

valence electrons

Valence electrons are the electrons in the outermost shell of an atom available for bonding. They are crucial for determining how atoms will interact chemically.
For atoms involved in \( \mathrm{AlCl}_4^- \):

• Aluminum (\( \mathrm{Al} \)) brings three valence electrons to the structure.
• Each chlorine (\( \mathrm{Cl} \)) contributes seven valence electrons, contributing to satisfy its desire for a full octet.
• The extra negative charge implies one additional electron, making a total of 32 valence electrons available.

By understanding how these electrons are distributed and shared, we can predict and explain the bonding behavior and stability of compounds such as \( \mathrm{AlCl}_4^- \). This forms the basis of many concepts in chemistry.