Question

Identify the Lewis acid and Lewis base in each of the following reactions: (a) \(\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)\) (c) \(\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\) (d) \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\)

Short answer

(a) Lewis acid: $\mathrm{HNO}_{2}$, Lewis base: $\mathrm{OH}^{-}$; (b) Lewis acid: $\mathrm{FeBr}_{3}$, Lewis base: $\mathrm{Br}^{-}$; (c) Lewis acid: $\mathrm{Zn}^{2+}$, Lewis base: $\mathrm{NH}_{3}$; (d) Lewis acid: $\mathrm{SO}_{2}$, Lewis base: $\mathrm{H}_{2} \mathrm{O}$.

Step-by-step solution

Identification of Lewis Acid and Lewis Base

In this reaction, the hydroxide ion \(\mathrm{OH}^{-}\) donates its electron pair to \(\mathrm{HNO}_{2}\). This makes \(\mathrm{OH}^{-}\) the Lewis base, while \(\mathrm{HNO}_{2}\) acts as the Lewis acid. L2) Reaction (b): \(\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)\)

Identification of Lewis Acid and Lewis Base

In this reaction, the bromide ion \(\mathrm{Br}^{-}\) donates its electron pair to \(\mathrm{FeBr}_{3}\). This makes \(\mathrm{Br}^{-}\) the Lewis base, while \(\mathrm{FeBr}_{3}\) acts as the Lewis acid. 3) Reaction (c): \(\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\)

Identification of Lewis Acid and Lewis Base

In this reaction, the ammonia molecules \(\mathrm{NH}_{3}\) donate their electron pairs to the \(\mathrm{Zn}^{2+}\) ion. This makes \(\mathrm{NH}_{3}\) the Lewis base, while \(\mathrm{Zn}^{2+}\) acts as the Lewis acid. 4) Reaction (d): \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\)

Identification of Lewis Acid and Lewis Base

In this reaction, the water molecule \(\mathrm{H}_{2} \mathrm{O}\) donates its electron pair to \(\mathrm{SO}_{2}\). This makes \(\mathrm{H}_{2} \mathrm{O}\) the Lewis base, while \(\mathrm{SO}_{2}\) acts as the Lewis acid.