Chapter 16: Problem 95

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

Expert verified

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 $\mathrm{NH}_{3}$.