Problem 134 - Chemistry

Question

Identify the Lewis acid and the Lewis base in each of the following reactions. a. \(\mathrm{Fe}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)\) b. \(\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q)\) c. \(\mathrm{HgI}_{2}(s)+2 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}{\underline{\phantom{xx}}}^{2-}(a q)\)

Step-by-Step Solution

Answer

In the given reactions: a. Lewis acid: \(\mathrm{Fe}^{3+}\) and Lewis base: \(\mathrm{H}_{2} \mathrm{O}\) b. Lewis acid: \(\mathrm{H}_{2} \mathrm{O}\) and Lewis base: \(\mathrm{CN}^{-}\) c. Lewis acid: \(\mathrm{HgI}_{2}\) and Lewis base: \(\mathrm{I}^{-}\)

Step 1: a. Identify the Lewis acid and Lewis base

In the reaction \(\mathrm{Fe}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)\), the \(\mathrm{Fe}^{3+}\) ion has a positive charge and can accept electron pairs to form a coordinate covalent bond with the negatively charged oxygen atom in \(\mathrm{H}_{2} \mathrm{O}\). Therefore, \(\mathrm{Fe}^{3+}\) is the Lewis acid, and \(\mathrm{H}_{2} \mathrm{O}\) is the Lewis base.

Step 2: b. Identify the Lewis acid and Lewis base

In the reaction \(\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q)\), the \(\mathrm{CN}^{-}\) ion can donate an electron pair to the hydrogen atom of the water molecule, forming the HCN molecule. The water molecule acts as an electron pair acceptor in this case. Therefore, \(\mathrm{H}_{2} \mathrm{O}\) is the Lewis acid, and \(\mathrm{CN}^{-}\) is the Lewis base.

Step 3: c. Identify the Lewis acid and Lewis base

In the reaction \(\mathrm{HgI}_{2}(s)+2 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}{\underline{\phantom{xx}}}^{2-}(a q)\), HgI\(_2\), which is a solid, can accept electron pairs from the \(\mathrm{I}^{-}\) ions. As a result, HgI\(_4$$^{2-}\) is formed. In this case, \(\mathrm{HgI}_{2}\) is the Lewis acid, and \(\mathrm{I}^{-}\) is the Lewis base.