Question

Which equilibrium can be described as an acid-base reaction using the Lewis acid-base definition but bot using the Bronsted-Lowry definition

• A. $2{NH}_3+H_2{SO}_4\leftrightharpoons 2{NH}_4^{+}+{SO}_4^{2-}$
• B. ${NH}_3+{CH}_{3}COOH\leftrightharpoons {NH}_4^{+}+{CH}_{3}COO{O}^{-}$
• C. $H_{2}O+{CH}_{3}COOH\leftrightharpoons H_3{O}^{+}+{CH}_{3}CO{O}^{-}$
• D. ${\left[Cu\right(H_{2}O{)}_4]}^{2-}+4{NH}_3\leftrightharpoons {\left[Cu{\left({NH}_3\right)}_4\right]}^{2+}+4H_{2}O$

Answer 1

The correct option is D ${\left[Cu\right(H_{2}O{)}_4]}^{2-}+4{NH}_3\leftrightharpoons {\left[Cu{\left({NH}_3\right)}_4\right]}^{2+}+4H_{2}O$

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.

${\left[Cu{\left(H_{2}O\right)}_4\right]}^{2+}+4{NH}_3\rightleftharpoons {\left[Cu{\left({NH}_3\right)}_4\right]}^{2+}+4H_{2}O$ involves loss and gain of electrons.

$H_{2}O$ is coordinated to $Cu$ by donating electrons (LHS). It is then removed by withdrawing electrons.