In many reactions, the addition of \(\mathrm{AlCl}_{3}\) produces the same effect as the addition of \(\mathrm{H}^{+} .\) (a) Draw a Lewis structure for \(\mathrm{AlCl}_{3}\) in which no atoms carry formal charges, and determine its structure using the VSEPR method. (b) What characteristic is notable about the structure in part (a) that helps us understand the acidic character of AlCl \(_{3} ?\) (c) Predict the result of the reaction between \(\mathrm{AlCl}_{3}\) and \(\mathrm{NH}_{3}\) in a solvent that does not participate as a reactant. (d) Which acid-base theory is most suitable for discussing the similarities between \(\mathrm{AlCl}_{3}\) and \(\mathrm{H}^{+}\) ?

Short Answer

Expert verified
In summary, the Lewis structure of AlCl3 is Trigonal Planar, and its acidic character stems from the Aluminum atom's ability to accept electron pairs, making it a Lewis acid. When reacted with NH3, a Lewis base, it forms a complex: AlCl3NH3. The most suitable acid-base theory for discussing the similarities between AlCl3 and H+ is the Lewis Acid-Base Theory, as both can accept electron pairs from other species.

Step by step solution

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(a) Lewis structure and VSEPR structure of AlCl3

First, we draw the Lewis structure of AlCl3: 1. Count the total number of valence electrons: Al has 3 valence electrons and Cl has 7 valence electrons. AlCl3 has 3 chlorine atoms, so we have \(3 + 3 \times 7 = 24\) valence electrons. 2. Place the least electronegative atom in the center (Al) and surround it with the more electronegative atoms (Cl). 3. Distribute the 24 valence electrons as bonding electron pairs between the central Al atom and Cl atoms, and as lone pairs on the Cl atoms. Now we use the VSEPR method to determine its structure: 1. Count the electron groups around the central atom: There are 3 bonding electron pairs around the Al atom (no lone pairs), so the electron group geometry is Trigonal Planar. Therefore, the Lewis structure of AlCl3 is Trigonal Planar.
02

(b) Acidic character of AlCl3

Observing the Lewis structure of AlCl3, we notice that the Aluminum atom has only 3 valence electrons forming 3 covalent bonds with Cl atoms. The Aluminum atom has a vacant p-orbital, which can accept electron pairs from another species. The ability to accept electron pairs makes AlCl3 a Lewis acid. This characteristic is responsible for the acidic character of AlCl3.
03

(c) Reaction between AlCl3 and NH3

We are asked to predict the result of the reaction between AlCl3 and NH3 in a solvent that does not participate as a reactant. Keeping in mind that AlCl3 is a Lewis acid and NH3 is a Lewis base (due to the lone pair of electrons on the Nitrogen atom), we can expect a Lewis acid-base reaction to form a complex: \[AlCl3 + NH3 \rightarrow AlCl3NH3\] This reaction involves the transfer of a lone pair of electrons from the Nitrogen atom of NH3 to the vacant p-orbital of the Aluminum atom of AlCl3.
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(d) Suitable acid-base theory

The most suitable acid-base theory for discussing the similarities between AlCl3 and H+ is the Lewis Acid-Base Theory. This is because both AlCl3 and H+ can accept electron pairs from other species, which is the defining characteristic of a Lewis acid. This theory allows comparing the actions of both AlCl3 and H+ in reactions without explicitly referring to the proton transfer, which is the main focus of other acid-base theories such as the Arrhenius and Brønsted-Lowry theories.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Structure
Understanding the Lewis structure is fundamental to grasping the nature of chemical bonds and reactions. For compounds like aluminum chloride (AlCl3), the Lewis structure provides a visual representation of electron arrangement among atoms.

To create the Lewis structure, one should first count the total valence electrons available for bonding. Aluminum has three valence electrons, and each chlorine atom has seven. With three chlorine atoms, AlCl3 has a total of 24 valence electrons to be distributed. In the structure, aluminum is centered, with chlorine atoms surrounding it. Electrons are then paired to form bonds between aluminum and chlorine, while additional electrons fill in as lone pairs on chlorine.

This simplified diagram eschews complex conventions for clarity: The dots around Cl represent its seven valence electrons, and the lines or dashes between Al and Cl show the bonds — a clear, approachable method for visualizing molecular structure.
VSEPR Theory
VSEPR theory, which stands for Valence Shell Electron Pair Repulsion, is a model used to predict the shape of individual molecules. Based on the electron pairs surrounding the central atom, VSEPR theory postulates that these electron groups will arrange themselves as far apart as possible to minimize repulsion. For AlCl3, which has three bonding pairs and no lone pairs around the central aluminum atom, the electron groups form a trigonal planar shape.

An intuitive way to think about VSEPR theory is to imagine each electron group is like a balloon filled with repulsion. Balloons will naturally push away from each other, finding a configuration that minimizes their mutual repulsion. Apply this to electron groups, and you can predict shapes of molecules ranging from linear to octahedral.
Lewis Acid
In chemical reactions, a Lewis acid is defined as an electron pair acceptor. Let's consider aluminum chloride, AlCl3. The absence of a full octet on the central aluminum atom gives it a hunger for additional electrons, making it a prime candidate for accepting electron pairs from other species — hence its classification as a Lewis acid.

The Lewis acid concept broadens the scope of acid-base reactions to include reactions without the transfer of protons (H+ ions), which is otherwise a constraint in other acid-base theories like Arrhenius and Brønsted-Lowry. In context, AlCl3 can be likened to a magnet for electrons — it's not about donating or receiving a proton; it's about the affinity for pairs of electrons.
Electron Pair Acceptance
Electron pair acceptance is a core feature of Lewis acid-base chemistry. In the reaction between AlCl3 and a species like ammonia (NH3), AlCl3 acts as a Lewis acid by accepting an electron pair from the lone pair on the nitrogen atom of NH3.

To visualize this, imagine AlCl3 as having an open door (the vacant p-orbital). When NH3 comes along with its lone pair, it essentially 'walks through the open door', forming an AlCl3NH3 compound. This interaction is crucial in many chemical processes, including catalysts in organic synthesis and in the formation of complex ions. It's a dance of give-and-take at the atomic level, with electron pairs as the currency of chemical change.

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Most popular questions from this chapter

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} .\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature.

For each of these reactions, identify the acid and base among the reactants, and state if the acids and bases are Lewis, Arrhenius, and/or Bronsted-Lowry: (a) \(\mathrm{PCl}_{4}^{+}+\mathrm{Cl}^{-} \longrightarrow \mathrm{PCl}_{5}\) (b) \(\mathrm{NH}_{3}+\mathrm{BF}_{3} \longrightarrow \mathrm{H}_{3} \mathrm{NBF}_{3}\) (c) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}\right]^{2+}+\mathrm{H}_{3} \mathrm{O}^{+}\)

Phenol, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH},\) has a \(K_{a}\) of \(1.3 \times 10^{-10}.\) (a) Write out the \(K_{a}\) reaction for phenol. (b) Calculate \(K_{b}\) for phenol's conjugate base. (c) Is phenol a stronger or weaker acid than water?

Pyridinium bromide \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\right)\) is a strong electrolyte that dissociates completely into \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{Br}^{-} .\) An aqueous solution of pyridinium bromide has a pH of \(2.95 .\) (a) Write out the reaction that leads to this acidic pH. (b) Using Appendix D, calculate the \(K_{a}\) for pyridinium bromide. (c) A solution of pyridinium bromide has a pH of 2.95 . What is the concentration of the pyridinium cation atequilibrium, in units of molarity?

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), (B) \(\mathrm{HCO}_{3}^{-}\)

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