Chapter 15, Problem 15.123QP

Identify the Lewis acid and Lewis base that lead to the formation of the following species: (a) ${\text{AlCl}}_4^{-}$, (b) $\text{Cd}{(\text{CN})}_4^{2-}$, (c) ${\text{HCO}}_3^{-}$, (d) H₂SO₄.

Interpretation Introduction

Interpretation:

The Lewis acid and Lewis base that lead to the formation of $AlC{l}_4^{-}$, have to be identified.

Concept Information:

According to Lewis concept of acid and bases,

Lewis acid is an electron pair acceptor.

Lewis base is an electron pair donor.

Answer to Problem 15.123QP

The Lewis acid and Lewis base that lead to the formation of $AlC{l}_4^{-}$ is mentioned below,

$\underset{\begin{array}{l}Lewis\\ acid\end{array}}{AlC{l}_3}+\underset{\begin{array}{l}Lewis\\ base\end{array}}{C{l}^{-}}\to AlC{l}_4^{-}$

Explanation of Solution

Given species $AlC{l}_4^{-}$ can be formed by the reaction of $AlC{l}_3$ with $C{l}^{-}$. In this reaction $C{l}^{-}$ donates electron pairs, $AlC{l}_3$ accepts electron pairs and so according to the Lewis acid-base concepts, $C{l}^{-}$ acts as Lewis base, and $AlC{l}_3$ acts as Lewis acid.

The reaction of formation of given species $AlC{l}_4^{-}$ is written below,

$\underset{\begin{array}{l}Lewis\\ acid\end{array}}{AlC{l}_3}+\underset{\begin{array}{l}Lewis\\ base\end{array}}{C{l}^{-}}\to AlC{l}_4^{-}$

Interpretation Introduction

Interpretation:

The Lewis acid and Lewis base that lead to the formation of $Cd{\left(CN\right)}_4^{2-}$, have to be identified.

Concept Information:

According to Lewis concept of acid and bases,

Lewis acid is an electron pair acceptor.

Lewis base is an electron pair donor.

Answer to Problem 15.123QP

The Lewis acid and Lewis base that lead to the formation of $Cd{\left(CN\right)}_4^{2-}$ is mentioned below,

$\underset{\begin{array}{l}Lewis\\ acid\end{array}}{C{d}^{2+}}+\underset{\begin{array}{l}Lewis\\ base\end{array}}{4C{N}^{-}}\to Cd{\left(CN\right)}_4^{2-}$

Explanation of Solution

Given species $Cd{\left(CN\right)}_4^{2-}$ can be formed by the reaction of $C{d}^{2+}$ with $C{N}^{-}$. In this reaction $C{N}^{-}$ donates electron pairs, $C{d}^{2+}$ accepts electron pairs and so according to the Lewis acid-base concepts, $C{N}^{-}$ acts as Lewis base, and $C{d}^{2+}$ acts as Lewis acid.

The reaction of formation of given species $Cd{\left(CN\right)}_4^{2-}$ is written below,

$\underset{\begin{array}{l}Lewis\\ acid\end{array}}{C{d}^{2+}}+\underset{\begin{array}{l}Lewis\\ base\end{array}}{4C{N}^{-}}\to Cd{\left(CN\right)}_4^{2-}$

Interpretation Introduction

Interpretation:

The Lewis acid and Lewis base that lead to the formation of $HC{O}_3^{-}$, have to be identified.

Concept Information:

According to Lewis concept of acid and bases,

Lewis acid is an electron pair acceptor.

Lewis base is an electron pair donor.

Answer to Problem 15.123QP

The Lewis acid and Lewis base that lead to the formation of $HC{O}_3^{-}$ is mentioned below,

$\underset{\begin{array}{l}Lewis\\ acid\end{array}}{H^{+}}+\underset{\begin{array}{l}Lewis\\ base\end{array}}{C{O}_3^{-}}\to HC{O}_3^{-}$

Explanation of Solution

Given species $HC{O}_3^{-}$ can be formed by the reaction of $H^{+}$ with $C{O}_3^{-}$. In this reaction $C{O}_3^{-}$ donates electron pairs, $H^{+}$ accepts electron pairs and so according to the Lewis acid-base concepts, $C{O}_3^{-}$ acts as Lewis base, and $H^{+}$ acts as Lewis acid.

The reaction of formation of given species $HC{O}_3^{-}$ is written below,

$\underset{\begin{array}{l}Lewis\\ acid\end{array}}{H^{+}}+\underset{\begin{array}{l}Lewis\\ base\end{array}}{C{O}_3^{-}}\to HC{O}_3^{-}$

Interpretation Introduction

Interpretation:

The Lewis acid and Lewis base that lead to the formation of $H_{2}S{O}_4$, have to be identified.

Concept Information:

According to Lewis concept of acid and bases,

Lewis acid is an electron pair acceptor.

Lewis base is an electron pair donor.

Answer to Problem 15.123QP

The Lewis acid and Lewis base that lead to the formation of $H_{2}S{O}_4$ is mentioned below,

$\underset{\begin{array}{l}Lewis\\ acid\end{array}}{H^{+}}+\underset{\begin{array}{l}Lewis\\ base\end{array}}{HS{O}_4^{-}}\to H_{2}S{O}_4$

Explanation of Solution

Given species $H_{2}S{O}_4$ can be formed by the reaction of $H^{+}$ with $HS{O}_4^{-}$. In this reaction $HS{O}_4^{-}$ donates electron pairs, $H^{+}$ accepts electron pairs and so according to the Lewis acid-base concepts, $HS{O}_4^{-}$ acts as Lewis base, and $H^{+}$ acts as Lewis acid.

The reaction of formation of given species  $H_{2}S{O}_4$ is written below,

$\underset{\begin{array}{l}Lewis\\ acid\end{array}}{H^{+}}+\underset{\begin{array}{l}Lewis\\ base\end{array}}{HS{O}_4^{-}}\to H_{2}S{O}_4$