Question

Identify the Lewis acid and Lewis base among the reactants in each of the following reactions: (a) \(\operatorname{Fe}\left(\mathrm{ClO}_{4}\right)_{3}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(a q)+3 \mathrm{ClO}_{4}^{-}(a q)\) (b) \(\mathrm{CN}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q)\) (c)\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(g)+\mathrm{BF}_{3}(g) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NBF}_{3}(s)\) (d)\(\operatorname{HIO}(l q)+\mathrm{NH}_{2}^{-}(l q) \Longrightarrow \mathrm{NH}_{3}(l q)+\mathrm{IO}^{-}(l q)\) (\(lq\) denotes liquid ammonia as solvent)

Short answer

In the given reactions: (a) \(\mathrm{Fe}^{3+}\) is a Lewis acid and \(\mathrm{H}_{2}\mathrm{O}(l)\) is a Lewis base; (b) \(\mathrm{CN}^{-}(a q)\) is a Lewis base and \(\mathrm{H}_{2} \mathrm{O}(l)\) is a Lewis acid; (c) \(\mathrm{BF}_{3}(g)\) is a Lewis acid and \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(g)\) is a Lewis base; (d) \(\mathrm{HIO}(l q)\) is a Lewis acid and \(\mathrm{NH}_{2}^{-}(l q)\) is a Lewis base.

Step-by-step solution

In this reaction, the metal ion \(\mathrm{Fe}^{3+}\) is a Lewis acid, as it is electron-pair acceptor and can form a complex with 6 water molecules. Water is a Lewis base which acts as an electron-pair donor. (b) $\mathrm{CN}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q)$ #Step 2: Identify the Lewis acid and Lewis base

In this reaction, the \(\mathrm{CN}^{-}\) ion is a Lewis base, as it donates the electron pair to form \(\mathrm{HCN}\). Water acts as a Lewis acid by accepting an electron pair from the \(\mathrm{CN}^{-}\) ion. (c)$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(g)+\mathrm{BF}_{3}(g) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NBF}_{3}(s)$ #Step 3: Identify the Lewis acid and Lewis base

In this reaction, the \(\mathrm{BF}_{3}\) molecule is a Lewis acid, as it can accept an electron pair from the nitrogen atom of \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{N}\). \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{N}\) acts as a Lewis base by donating its lone pair of electrons on the nitrogen atom. (d)$\operatorname{HIO}(l q)+\mathrm{NH}_{2}^{-}(l q) \Longrightarrow \mathrm{NH}_{3}(l q)+\mathrm{IO}^{-}(l q)$ #Step 4: Identify the Lewis acid and Lewis base

In this reaction, \(\mathrm{HIO}\) acts as a Lewis acid, as it accepts an electron pair from the \(\mathrm{NH}_{2}^{-}\) ion. The \(\mathrm{NH}_{2}^{-}\) ion is a Lewis base, as it donates its electron pair to form \(\mathrm{NH}_{3}\).

Key concepts

Electron-Pair Acceptor

In the world of chemistry, a Lewis acid is known for its ability to accept an electron pair. This defining trait sets it apart from other chemical species. Electron-pair acceptors don't donate electrons; they have vacancies or positively charged centers that draw in electrons from a donor. For example, in the reaction involving the formation of the complex \[\text{Fe}\left(\mathrm{ClO}_{4}\right)_{3}(s)+6 \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \left[\mathrm{Fe}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{3+}(aq)+3 \mathrm{ClO}_{4}^{-}(aq) \], the iron ion \( \mathrm{Fe}^{3+} \) is the Lewis acid. It has a positive charge and seeks electron pairs to fill its empty orbitals. These types of ions or molecules play a crucial role in reactions as they form bonds by accepting electrons from Lewis bases. By understanding electron-pair acceptors, you'll better grasp how chemical reactions proceed and how different species interact to form stable compounds.

Electron-Pair Donor

Lewis bases set themselves apart by their willingness and ability to donate electron pairs. They offer these electrons to form bonds with electron-pair acceptors. Typically, Lewis bases possess lone electron pairs that aren't involved in bonding, which they can donate to form new chemical interactions. In the reaction \( \mathrm{CN}^{-}(aq)+\mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{HCN}(aq)+\mathrm{OH}^{-}(aq) \), the \( \mathrm{CN}^{-} \) ion acts as a Lewis base because it donates its electron pair to the \( \mathrm{H}_{2} \mathrm{O} \), forming \( \mathrm{HCN} \). This ability to donate electron pairs is crucial in various chemical reactions. Furthermore, not only ions but neutral molecules with lone pairs, such as the ammonia molecule \( \mathrm{NH}_3 \), can also act as electron-pair donors. Recognizing these donors is key to predicting how they will interact in different chemical environments.

Complex Formation

The formation of a complex involves a Lewis acid and a Lewis base coming together through electron pair sharing. This creates a new structure known as a "complex". Complexes are integral in chemistry, often leading to colorful and interesting compounds. One classic example of complex formation is the reaction between boron trifluoride \( \mathrm{BF}_3 \) and trimethylamine \( \left(\mathrm{CH}_3\right)_3\mathrm{N} \), \[ \left( \mathrm{CH}_3 \right)_3 \mathrm{N}(g)+ \mathrm{BF}_3(g)\rightleftharpoons \left( \mathrm{CH}_3 \right)_3 \mathrm{NBF}_3(s) \], where the Lewis base \( \left(\mathrm{CH}_3\right)_3\mathrm{N} \) donates an electron pair to form a bond with the Lewis acid \( \mathrm{BF}_3 \). The resulting \( \left(\mathrm{CH}_3\right)_3 \mathrm{NBF}_3 \) is a stable complex. Understanding complex formation is not only pivotal in inorganic chemistry but also in areas such as biochemistry where metal ions frequently bind with biomolecules to form complexes with special functions.