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Identify the Lewis acid and Lewis base in each of the following reactions: (a) \(\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)\) (c) \(\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\) (d) \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{3}(a q)\)

Short Answer

Expert verified
In the given reactions: (a) Lewis base: OH鈦; Lewis acid: HNO鈧 (b) Lewis base: Br鈦; Lewis acid: FeBr鈧 (c) Lewis base: NH鈧; Lewis acid: Zn虏鈦 (d) Lewis base: H鈧侽; Lewis acid: SO鈧

Step by step solution

01

Reaction (a)

In this reaction, the OH鈦 ion donates a pair of electrons to the hydrogen in HNO鈧, forming a bond and creating H鈧侽. The HNO鈧 molecule accepts the electron pair. Thus, OH鈦 is the Lewis base, and HNO鈧 is the Lewis acid: Lewis base: OH鈦 Lewis acid: HNO鈧
02

Reaction (b)

In this reaction, the Br鈦 ion donates a pair of electrons to the FeBr鈧 molecule, creating a bond and forming FeBr鈧勨伝. The FeBr鈧 molecule accepts the electron pair. Thus, Br鈦 is the Lewis base, and FeBr鈧 is the Lewis acid: Lewis base: Br鈦 Lewis acid: FeBr鈧
03

Reaction (c)

In this reaction, the NH鈧 molecules each donate a pair of electrons to the Zn虏鈦 ion, creating bonds and forming Zn(NH鈧)鈧劼测伜. The Zn虏鈦 ion accepts the electron pairs. Thus, NH鈧 is the Lewis base, and Zn虏鈦 is the Lewis acid: Lewis base: NH鈧 Lewis acid: Zn虏鈦
04

Reaction (d)

In this reaction, the H鈧侽 molecule donates a pair of electrons to the sulfur atom in the SO鈧 molecule, creating a bond and forming H鈧侽鈧. The SO鈧 molecule accepts the electron pair. Thus, H鈧侽 is the Lewis base, and SO鈧 is the Lewis acid: Lewis base: H鈧侽 Lewis acid: SO鈧

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Pair Donor
In Lewis acid-base theory, one important player is the electron pair donor, known as the Lewis base. A Lewis base is any species that donates an electron pair to form a chemical bond. This donation occurs during chemical reactions and helps create new compounds.
In reaction (a), the hydroxide ion (\(\text{OH}^-\)) donates an electron pair to \(\text{HNO}_2\), forming water (\(\text{H}_2\text{O}\)).
This makes \(\text{OH}^-\) the Lewis base here. In reaction (b), bromide ions (\(\text{Br}^-\)) act as the electron pair donors when they interact with \(\text{FeBr}_3\) to form \(\text{FeBr}_4^-\), establishing \(\text{Br}^-\) as the Lewis base.
Similarly, in reaction (c), ammonia molecules (\(\text{NH}_3\)) provide electron pairs to the zinc ion (\(\text{Zn}^{2+}\)) to form \(\text{Zn(NH}_3)_4^{2+}\), making \(\text{NH}_3\) the electron pair donor.
Finally, in reaction (d), water (\(\text{H}_2\text{O}\)) provides electron pairs to \(\text{SO}_2\), becoming the Lewis base as it transforms into \(\text{H}_2\text{O}_3\).
In essence, whenever you identify a Lewis base, look for the species that gives away electron pairs to form new bonds.
Electron Pair Acceptor
The counterpart to the electron pair donor in Lewis acid-base theory is the electron pair acceptor, or the Lewis acid. A Lewis acid is a species that accepts an electron pair during a chemical reaction. This acceptance allows the formation of new chemical bonds and compounds.
In a close examination of these reactions, you will notice the trend where the Lewis acid generally becomes more stable through the acceptance of electron pairs.
For instance:
  • In reaction (a), \(\text{HNO}_2\) accepts an electron pair from \(\text{OH}^-\). Thus, \(\text{HNO}_2\) is the Lewis acid.
  • In reaction (b), \(\text{FeBr}_3\) is the one accepting the electron pair from \(\text{Br}^-\), making \(\text{FeBr}_3\) the Lewis acid.
  • For reaction (c), \(\text{Zn}^{2+}\) ion is the electron pair acceptor from \(\text{NH}_3\), classifying it as the Lewis acid.
  • Lastly, in reaction (d), \(\text{SO}_2\) accepts electron pairs from water (\(\text{H}_2\text{O}\)), reinforcing \(\text{SO}_2\) as the Lewis acid in this scenario.
The general role of Lewis acids is crucial because they facilitate the formation of chemical bonds by accepting electron pairs.
Chemical Reactions Analysis
To truly understand Lewis acid-base interactions, one must analyze the chemical reactions involved. This involves recognizing which species in a reaction donates or accepts electron pairs, thus identifying the Lewis base and acid.
Each reaction symbolizes a dynamic exchange of electrons, forming new chemical bonds. In a detailed analysis, spotting the changes in how electrons are shared or transferred is key.
  • For instance, in reaction (a) between \(\text{HNO}_2\) and \(\text{OH}^-\), tracking electron movement reveals the transformation into water, emphasizing the \(\text{OH}^-\) as a donor, and \(\text{HNO}_2\) as an acceptor.
  • In reaction (b) involving \(\text{FeBr}_3\) and \(\text{Br}^-\), the shift to form \(\text{FeBr}_4^-\) showcases a clear pairing of donator and acceptor roles.
  • Similarly, in reaction (c), the electron exchange between \(\text{Zn}^{2+}\) and \(\text{NH}_3\) enlarges the molecular understanding by forming a stable complex ion, \(\text{Zn(NH}_3)_4^{2+}\).
  • Lastly, reaction (d) displays how molecular interactions transform \(\text{SO}_2\) and water into a new compound, \(\text{H}_2\text{O}_3\), marking another classic example of electron pair movement.
Studying these reactions helps break down complex interactions into more manageable steps, solidifying your grasp of how compounds engage during reactions.

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Most popular questions from this chapter

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} .\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature.

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=0.00045 M ;\) (b) \(\left[\mathrm{OH}^{-}\right]=8.8 \times 10^{-9} \mathrm{M} ;\) (c) a solution in which \(\left[\mathrm{OH}^{-}\right]\) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\) .

The acid-dissociation constant for chlorous acid \(\left(\mathrm{HClO}_{2}\right)\) is \(1.1 \times 10^{-2} .\) Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{ClO}_{2}^{-}\) and \(\mathrm{HClO}_{2}\) at equilibrium if the initial concentration of \(\mathrm{HClO}_{2}\) is 0.0125 \(\mathrm{M} .\)

Arrange the following 0.10\(M\) solutions in order of increasing acidity: (i) \(\mathrm{NH}_{4} \mathrm{NO}_{3},(\mathrm{ii}) \mathrm{NaNO}_{3},(\mathrm{iii}) \mathrm{CH}_{3} \mathrm{COONH}_{4},(\mathrm{iv})\) \(\mathrm{NaF},(\mathrm{v}) \mathrm{CH}_{3} \mathrm{COONa}.\)

Ammonia, \(\mathrm{NH}_{3},\) acts as an Arrhenius base, a Bronsted-Lowry base, and a Lewis base, in aqueous solution. Write out the reaction \(\mathrm{NH}_{3}\) undergoes with water and explain what properties of ammonia correspond to each of the three definitions of "base."

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