Which equilibrium can be described as an acidbase reaction using the Lewis acid-base definition but not…
- $2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} ightleftharpoons 2 \mathrm{NH}_{4}^{+}+\mathrm{SO}_{4}^{2-}$
- $\mathrm{NH}_{3}+\mathrm{CH}_{3} \mathrm{COOH} ightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}$
- $\mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3} \mathrm{COOH} ightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}$
- $\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}ight)_{4}ight]^{2+}+4 \mathrm{NH}_{3} ightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}ight)_{4}ight]^{2+}+4 \mathrm{H}_{2} \mathrm{O}$
Solution
According to Bronsted-Lowry concept of acids and bases, an acid is a substance which can give a proton and a base is a substance which accepts a proton.
According to Lewis concept of acids and bases, an acid is a substance which can accept a lone pair of electrons whereas a base is a substance which can donate a lone pair of electrons.
$\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}ight)_{4}ight]^{2+}+4 \mathrm{NH}_{3} ightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}ight)_{4}ight]^{2+}+4 \mathrm{H}_{2} \mathrm{O}$ involves loss and gain of electrons.
$\mathrm{H}_{2} \mathrm{O}$ is coordinated to $\mathrm{Cu}$ by donating electrons (LHS). It is then removed by withdrawing electrons.
Asked in: JEE-TOPICTESTS-CHEMISTRY