Problem 137 - Chemistry An Atoms First Approach

Question

Identify the Lewis acid and the Lewis base in each of the following reactions. a. \(\mathrm{B}(\mathrm{OH})_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{B}(\mathrm{OH})_{4}^{-}(a q)+\mathrm{H}^{+}(a q)\) b. \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)\) c. \(\mathrm{BF}_{3}(g)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BF}_{4}^{-}(a q)\)

Step-by-Step Solution

Answer

In the given reactions: a. Lewis base: \(\mathrm{H}_{2} \mathrm{O}\) (water), Lewis acid: \(\mathrm{B}(\mathrm{OH})_{3}\) b. Lewis base: \(\mathrm{NH}_{3}\) (ammonia), Lewis acid: \(\mathrm{Ag}^{+}\) (silver ion) c. Lewis base: \(\mathrm{F}^{-}\) (fluoride ion), Lewis acid: \(\mathrm{BF}_{3}\) (boron trifluoride)

Step 1: a. Identifying Lewis acid and Lewis base

For the reaction: \(\mathrm{B}(\mathrm{OH})_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{B}(\mathrm{OH})_{4}^{-}(a q)+\mathrm{H}^{+}(a q)\) In this reaction, water \(\mathrm{H}_{2}\mathrm{O}\) is donating a pair of electrons to \(\mathrm{B}(\mathrm{OH})_{3}\) to form \(\mathrm{B}(\mathrm{OH})_{4}^{-}\). So the Lewis base is: \(\mathrm{H}_{2} \mathrm{O}\) (water) And the Lewis acid is: \(\mathrm{B}(\mathrm{OH})_{3}\)

Step 2: b. Identifying Lewis acid and Lewis base

For the reaction: \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)\) Here, ammonia \(\mathrm{NH}_{3}\) is donating a pair of electrons to the silver ion \(\mathrm{Ag}^{+}\) to form the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\). So the Lewis base is: \(\mathrm{NH}_{3}\) (ammonia) And the Lewis acid is: \(\mathrm{Ag}^{+}\) (silver ion)

Step 3: c. Identifying Lewis acid and Lewis base

For the reaction: \(\mathrm{BF}_{3}(g)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BF}_{4}^{-}(a q)\) In this reaction, the fluoride ion \(\mathrm{F}^{-}\) is donating a pair of electrons to boron trifluoride \(\mathrm{BF}_{3}\) to form the tetrafluoroborate ion \(\mathrm{BF}_{4}^{-}\). So the Lewis base is: \(\mathrm{F}^{-}\) (fluoride ion) And the Lewis acid is: \(\mathrm{BF}_{3}\) (boron trifluoride)