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

To draw the Lewis structure of the \(\mathrm{AsCl}_{4}^{+}\) ion, first determine the total number of valence electrons in the molecule. Arsenic (As) has 5 valence electrons and each chlorine (Cl) atom has 7 valence electrons. The ion has a positive charge, so we must subtract 1 electron. Therefore, there are a total of 5 + 4(7) - 1 = 33 valence electrons. Place the As atom in the center and the four Cl atoms around it, forming single bonds between each Cl and the As atom. Distribute the remaining 29 valence electrons as lone pairs around the Cl atoms. Each Cl atom will have 3 lone pairs. The Lewis structure for the \(\mathrm{AsCl}_{4}^{+}\) ion is: \[ \hspace{4cm}\text{Cl}\overset{\ominus}{\text{As}}\left(\underset{\ominus}{\text{Cl}}\right)_3^{+1} \]

To draw the Lewis structure of the \(\mathrm{AsCl}_{6}^{-}\) ion, first determine the total number of valence electrons in the molecule. Arsenic (As) has 5 valence electrons and each chlorine (Cl) atom has 7 valence electrons. The ion has a negative charge, so we must add 1 electron. Therefore, there are a total of 5 + 6(7) + 1 = 47 valence electrons. Place the As atom in the center and the six Cl atoms around it, forming single bonds between each Cl and the As atom. Distribute the remaining 41 valence electrons as lone pairs around the Cl atoms. Each Cl atom will have 3 lone pairs. The Lewis structure for the \(\mathrm{AsCl}_{6}^{-}\) ion is: \[ \hspace{4cm}\text{Cl}\overset{\ominus}{\text{As}}\left(\underset{\ominus}{\text{Cl}}\right)_5^{-1} \]

To determine the type of reaction, analyze the changes occurring in the reaction. We have: $$ 2 \mathrm{AsCl}_5(g) \longrightarrow \mathrm{AsCl}_{4}^{+}(s) + \mathrm{AsCl}_{6}^{-}(s) $$ One \(\mathrm{AsCl}_5\) molecule is donating an electron to another \(\mathrm{AsCl}_5\) molecule. This leads to the formation of \(\mathrm{AsCl}_{4}^{+}\), which has one more Chlorine atom but is missing one of electron, and \(\mathrm{AsCl}_{6}^{-}\), which has one extra Chlorine atom and one extra electron. Since we observe an electron transfer from one molecule to another, this reaction is an oxidation-reduction (redox) reaction.