Which equilibrium can be described as an acid-base reaction using the Lewis acid-base definition, but not using Bronsted-Lowry definition?

Question

Which equilibrium can be described as an acid-base reaction using the Lewis acid-base definition, but not using Bronsted-Lowry definition? (A) NH3+CH3COOHleftharpoons CH3COO-+NH4+ (B) H2O+CH3COOHleftharpoons H3O++CH3COO- (C) 4 NH 3+[ Cu ( H 2 O )4]2+ leftharpoons[ Cu ( NH 3)4]2++4 H 2 O (D) 2NH3+H2SO4leftharpoons2NH4++SO42 -

Tardigrade Admin · Accepted Answer

According to Lewis acid-base concept, bases are electron pair donor and acids are electron pair acceptor. In reaction

4 N H_{3} + [C u (H_{2} O)_{4}]^{2 +} ⇌ [C u (N H_{3})_{4}]^{2 +} + 4 H_{2} O

N H_{3}

molecules donates electron pairs, so it is a Lewis base and

C u^{2 +}

accepts the electron pairs donated by the

N H_{3}

molecules, so it acts as Lewis acid.

Q. Which equilibrium can be described as an acid-base reaction using the Lewis acid-base definition, but not using Bronsted-Lowry definition?

$N H_{3} + C H_{3} COO H ⇌ C H_{3} CO O^{-} + N H_{4}^{+}$

$H_{2} O + C H_{3} COO H ⇌ H_{3} O^{+} + C H_{3} CO O^{-}$

$4 N H_{3} + [C u (H_{2} O)_{4}]^{2 +} ⇌ [C u (N H_{3})_{4}]^{2 +} + 4 H_{2} O$

$2 N H_{3} + H_{2} S O_{4} ⇌ 2 N H_{4}^{+} + S O_{4}^{2 -}$

Q. Which equilibrium can be described as an acid-base reaction using the Lewis acid-base definition, but not using Bronsted-Lowry definition?

NH3​+CH3​COOH⇌CH3​COO−+NH4+​

H2​O+CH3​COOH⇌H3​O++CH3​COO−

4NH3​+[Cu(H2​O)4​]2+⇌[Cu(NH3​)4​]2++4H2​O

2NH3​+H2​SO4​⇌2NH4+​+SO42−​

$N H_{3} + C H_{3} COO H ⇌ C H_{3} CO O^{-} + N H_{4}^{+}$

$H_{2} O + C H_{3} COO H ⇌ H_{3} O^{+} + C H_{3} CO O^{-}$

$4 N H_{3} + [C u (H_{2} O)_{4}]^{2 +} ⇌ [C u (N H_{3})_{4}]^{2 +} + 4 H_{2} O$

$2 N H_{3} + H_{2} S O_{4} ⇌ 2 N H_{4}^{+} + S O_{4}^{2 -}$