Question
Which equilibrium can be described as an acid-base reaction using the Lewis acid-base definition but not using the Bronsted-Lowry definition?
A.
\[2N{{H}_{3}}+{{H}_{2}}S{{O}_{4}}\rightleftharpoons 2NH_{4}^{+}+SO_{4}^{2-}\]
B.
\[N{{H}_{3}}+C{{H}_{3}}COOH\rightleftharpoons NH_{4}^{+}+C{{H}_{3}}CO{{O}^{-}}\]
C.
\[{{H}_{2}}O+C{{H}_{3}}COOH\rightleftharpoons {{H}_{3}}{{O}^{+}}+C{{H}_{3}}CO{{O}^{-}}\]
D.
\[{{\left[ Cu\left( {{H}_{2}}{{O}_{4}} \right) \right]}^{2+}}+4N{{H}_{3}}\rightleftharpoons {{\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]}^{2+}}+4{{H}_{2}}O\]
Answer :
\[{{\left[ Cu\left( {{H}_{2}}{{O}_{4}} \right) \right]}^{2+}}+4N{{H}_{3}}\rightleftharpoons {{\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]}^{2+}}+4{{H}_{2}}O\]
Solution :
$${\left[ {Cu{{\left( {{H_2}O} \right)}_4}} \right]^{2 + }} + 4N{H_3} \rightleftharpoons $$ $${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }} + 4{H_2}O$$ involves lose and gain of electrons. $${H_2}O$$ is coordinated to $$Cu$$ by donating electrons $$(LHS).$$ It is then removed by withdrawing electrons.