Question

Draw Lewis structures for the AsCl \(_{4}^{+}\) and \(\mathrm{AsCl}_{6}-\) ions. What type of reaction (acid-base, oxidation-reduction, or the like) is the following? $$ 2 \mathrm{AsCl}_{5}(g) \longrightarrow \mathrm{AsCl}_{4} \mathrm{As} \mathrm{Cl}_{6}(s) $$

Short answer

The Lewis structures for the AsCl\(_{4}^{+}\) and AsCl\(_{6}^-\) ions are: AsCl\(_{4}^{+}\): $$ \begin{array}{c} \displaystyle \text{Cl } \overset{\ddots}{\text{As}}- \text{Cl} \\ \ddots \\ \text{Cl} \end{array} $$ AsCl\(_{6}^-\): $$ \begin{array}{c} \displaystyle \text{Cl } \overset{\ddots}{\text{As}}- \text{Cl} \\ \ddots \qquad \ddots \\ \text{Cl} \qquad \text{Cl} \end{array} $$ The given reaction: $$ 2 \mathrm{AsCl}_{5}(g) \longrightarrow \mathrm{AsCl}_{4}^{+} \mathrm{As} \mathrm{Cl}_{6}^{-}(s) $$ is a redox (reduction-oxidation) reaction.

Step-by-step solution

Draw Lewis structure for AsCl\(_{4}^{+}\) ion

1. Identify the central atom: As (Arsenic) is the central atom, as it has the lowest electronegativity in the AsCl\(_{4}^{+}\) ion. 2. Calculate the total number of valence electrons in the ion: As has 5 valence electrons, and Cl has 7 valence electrons. As there are 4 Cl atoms, the total valence electrons in the structure would be: \(5 + 4 \times 7 = 33\) Since it is a positively charged ion, we subtract 1 electron. \(33 - 1 = 32\) 3. Place a single bond (2 electrons) between the central atom (As) and each surrounding atom (Cl), and complete the octet rule for each surrounding atom (Cl):+$$ \begin{array}{c} \displaystyle \text{Cl } \overset{\ddots}{\text{As}}- \text{Cl} \\ \ddots \\ \text{Cl} \end{array} $$ 4. Now, the remaining electrons are given to the central atom (As), and since we've used 32 electrons, all electrons have been used up. The Lewis structure for the AsCl\(_{4}^{+}\) ion is complete.

Draw Lewis structure for AsCl\(_{6}^-\) ion

1. Identify the central atom: As (Arsenic) is the central atom in the AsCl\(_{6}^-\) ion. 2. Calculate the total number of valence electrons in the ion: As has 5 valence electrons, and Cl has 7 valence electrons. As there are 6 Cl atoms, the total valence electrons in the structure would be: \(5 + 6 \times 7 = 47\) Since it is a negatively charged ion, we add 1 electron. \(47 + 1 = 48\) 3. Place a single bond (2 electrons) between the central atom (As) and each surrounding atom (Cl), and complete the octet rule for each surrounding atom (Cl):+$$ \begin{array}{c} \displaystyle \text{Cl } \overset{\ddots}{\text{As}}- \text{Cl} \\ \ddots \qquad \ddots \\ \text{Cl} \qquad \text{Cl} \end{array} $$ 4. Now, the remaining electrons are given to the central atom (As), and since we've used 48 electrons, all electrons have been used up. The Lewis structure for the AsCl\(_{6}^-\) ion is complete.

Identify the type of reaction

We are given the following reaction: $$ 2 \mathrm{AsCl}_{5}(g) \longrightarrow \mathrm{AsCl}_{4}^{+} \mathrm{As} \mathrm{Cl}_{6}^{-}(s) $$ The reaction is a disproportionation reaction, which is a specific type of redox (reduction-oxidation) reaction. In this reaction, As is both reduced and oxidized. In the AsCl\(_{5}\) molecule, As has an oxidation state of +5. In the AsCl\(_{4}^{+}\) ion, As has an oxidation state of +4 (it is reduced). In the AsCl\(_{6}^-\) ion, As has an oxidation state of +6 (it is oxidized). Therefore, the given reaction is a redox (reduction-oxidation) reaction.

Key concepts

Valence Electrons

Understanding valence electrons is key when dealing with Lewis Structures as these electrons are the ones involved in bonding. Valence electrons are located in the outer shell of an atom and determine how the atom will bond with others. For example, in the element arsenic (As), there are 5 valence electrons. This helps us predict it will form bonds to fulfill the octet rule.
When constructing a Lewis structure, start by counting the total valence electrons from all atoms involved. For instance, in the ion AsCl\(_4^{+}\), arsenic contributes 5 while each chlorine contributes 7, resulting in a collective total of 33 valence electrons before adjusting for the charge.
It's important to account for any charges. For a positive ion, remove electrons corresponding to the charge, whereas for a negative ion, add electrons. This adjusts the total, making AsCl\(_4^{+}\) have 32 electrons to distribute among the atoms in the structure.

Disproportionation Reaction

Disproportionation reactions are fascinating because they involve a single substance being both oxidized and reduced. This type of redox reaction results in different products, each showcasing distinct oxidation states of the same element.
For example, consider the reaction given: 2 AsCl\(_5\)(g) \rightarrow AsCl\(_4^{+} \) + AsCl\(_6^{-}\). In this reaction, arsenic in AsCl\(_5\) begins with an oxidation state of +5. As the reaction progresses, it is both oxidized to AsCl\(_6^{-}\) with an oxidation state of +6, and reduced to AsCl\(_4^+\) with an oxidation state of +4.
This double change in oxidation state characterizes a disproportionation reaction. It's particularly intriguing because it showcases the versatility of elements like arsenic, which can stabilize different electron configurations simultaneously.

Redox Reaction

Redox reactions, short for reduction-oxidation reactions, are vital in chemical processes where electrons are transferred between substances. Understanding the movement of electrons and change in oxidation states can elucidate these reactions.
A substance undergoes reduction when it gains electrons, leading to a decrease in its oxidation state. Conversely, oxidation involves the loss of electrons, increasing its oxidation state. Observing a reaction like: 2 AsCl\(_5\)(g) \rightarrow AsCl\(_4^{+}\) + AsCl\(_6^{-}\), we identify it as a redox reaction due to arsenic's change in oxidation states.
In the context of this exercise, arsenic in AsCl\(_5\) initially has an oxidation state of +5, but is reduced to +4 in AsCl\(_4^{+}\), and oxidized to +6 in AsCl\(_6^{-}\). Recognizing these changes is crucial for identifying and understanding redox reactions in chemical equations.