For each pair of molecules or ions, select the stronger base, and write its Lewis structure. (a) \(\mathrm{CH}_{3} \mathrm{~S}^{-}\)or \(\mathrm{CH}_{3} \mathrm{O}^{-}\) (b) \(\mathrm{CH}_{3} \mathrm{NH}^{-}\)or \(\mathrm{CH}_{3} \mathrm{O}^{-}\) (c) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\)or \(\mathrm{OH}^{-}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\)or \(\mathrm{H}^{-}\) (e) \(\mathrm{NH}_{3}\) or \(\mathrm{OH}^{-}\) (f) \(\mathrm{NH}_{3}\) or \(\mathrm{H}_{2} \mathrm{O}\) (g) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\)or \(\mathrm{HCO}_{3}^{-}\) (h) \(\mathrm{HSO}_{4}^{-}\)or \(\mathrm{OH}^{-}\) (i) \(\mathrm{OH}^{-}\)or \(\mathrm{Br}^{-}\)

Short Answer

Expert verified
a) \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) or \(\mathrm{CH}_{3} \mathrm{O}^{-}\): \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) is the stronger base. b) \(\mathrm{CH}_{3} \mathrm{NH}^{-}\) or \(\mathrm{CH}_{3} \mathrm{O}^{-}\): \(\mathrm{CH}_{3} \mathrm{O}^{-}\) is the stronger base. c) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) or \(\mathrm{OH}^{-}\): \(\mathrm{OH}^{-}\) is the stronger base. d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) or \(\mathrm{H}^{-}\): \(\mathrm{H}^{-}\) is the stronger base. e) \(\mathrm{NH}_{3}\) or \(\mathrm{OH}^{-}\): \(\mathrm{OH}^{-}\) is the stronger base. f) \(\mathrm{NH}_{3}\) or \(\mathrm{H}_{2} \mathrm{O}\): \(\mathrm{NH}_{3}\) is the stronger base. g) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) or \(\mathrm{HCO}_{3}^{-}\): \(\mathrm{HCO}_{3}^{-}\) is the stronger base. h) \(\mathrm{HSO}_{4}^{-}\) or \(\mathrm{OH}^{-}\): \(\mathrm{OH}^{-}\) is the stronger base. i) \(\mathrm{OH}^{-}\) or \(\mathrm{Br}^{-}\): \(\mathrm{OH}^{-}\) is the stronger base.

Step by step solution

01

(a) Comparing \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) and \(\mathrm{CH}_{3} \mathrm{O}^{-}\)

In this pair, the basicity depends on the relative ability of \(\mathrm{S}^{-}\) and \(\mathrm{O}^{-}\) to donate their lone electron pair. Since sulfur is larger and less electronegative than oxygen, it can more easily donate its electrons. Therefore, \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) is the stronger base. The Lewis structure for \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) is: ``` S- | S=C-H | H ```
02

(b) Comparing \(\mathrm{CH}_{3} \mathrm{NH}^{-}\) and \(\mathrm{CH}_{3} \mathrm{O}^{-}\)

In this pair, the basicity depends on the relative ability of \(\mathrm{NH}^{-}\) and \(\mathrm{O}^{-}\) to donate their lone electron pair. Oxygen has a higher electron density around it and is a better electron donor. Therefore, \(\mathrm{CH}_{3} \mathrm{O}^{-}\) is the stronger base. The Lewis structure for \(\mathrm{CH}_{3} \mathrm{O}^{-}\) is: ``` O- | R-C-H | H ```
03

(c) Comparing \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) and \(\mathrm{OH}^{-}\)

In this pair, the basicity depends on the relative ability of \(\mathrm{COO}^{-}\) and \(\mathrm{OH}^{-}\) to donate their lone electron pair. Since the electron density on the doubly bonded oxygen in \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) is delocalized across two oxygen atoms and the resonance stabilizes the ion, it is less basic than \(\mathrm{OH}^{-}\). Therefore, \(\mathrm{OH}^{-}\) is the stronger base. The Lewis structure for \(\mathrm{OH}^{-}\) is: ``` O-H | - ```
04

(d) Comparing \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) and \(\mathrm{H}^{-}\)

In this pair, \(\mathrm{H}^{-}\) has the most basic site which is the lone electron pair; it is most willing to donate its electrons due to low electronegativity. Therefore, \(\mathrm{H}^{-}\) is the stronger base. The Lewis structure for \(\mathrm{H}^{-}\) is simple: ``` H | - ```
05

(e) Comparing \(\mathrm{NH}_{3}\) and \(\mathrm{OH}^{-}\)

In this pair, the basicity depends on the relative ability of the lone electron pair on the nitrogen atom in \(\mathrm{NH}_{3}\) and the lone electron pair on the oxygen atom in \(\mathrm{OH}^{-}\). Oxygen is more basic due to higher electron density and greater willingness to donate its lone pair of electrons. Therefore, \(\mathrm{OH}^{-}\) is the stronger base, and its Lewis structure is the same as in (c).
06

(f) Comparing \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\)

In this pair, the basicity depends on the relative ability of the lone electron pair on the nitrogen atom in \(\mathrm{NH}_{3}\) and the lone electron pairs on the oxygen atom in \(\mathrm{H}_{2} \mathrm{O}\). The lone pairs on oxygen atoms in water are less likely to be donated due to hydrogen bonding. Therefore, \(\mathrm{NH}_{3}\) is the stronger base. The Lewis structure for \(\mathrm{NH}_{3}\) is: ``` N /|\ H H H ```
07

(g) Comparing \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) and \(\mathrm{HCO}_{3}^{-}\)

In this pair, both ions have resonance structures due to electron delocalization. However, the negative charge in \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) is more delocalized across the two O atoms, thus it is less likely to donate its electron pairs and is less basic than \(\mathrm{HCO}_{3}^{-}\). Therefore, \(\mathrm{HCO}_{3}^{-}\) is the stronger base. The Lewis structure for \(\mathrm{HCO}_{3}^{-}\) is: ``` O- | O=C-O-H ```
08

(h) Comparing \(\mathrm{HSO}_{4}^{-}\) and \(\mathrm{OH}^{-}\)

In this pair, the basicity depends on the relative ability of \(\mathrm{HSO}_{4}^{-}\) and \(\mathrm{OH}^{-}\) to donate their lone electron pairs. The \(\mathrm{OH}^{-}\) ion is more likely to donate its electrons since it has higher electron density, while the negative charge on \(\mathrm{HSO}_{4}^{-}\) is delocalized and stabilized. Consequently, \(\mathrm{OH}^{-}\) is the stronger base, and the Lewis structure is the same as in (c).
09

(i) Comparing \(\mathrm{OH}^{-}\) and \(\mathrm{Br}^{-}\)

In this pair, the basicity depends on the relative ability of the oxygen in \(\mathrm{OH}^{-}\) and the bromine in \(\mathrm{Br}^{-}\) to donate their lone electron pairs. Since oxygen is smaller and has a higher affinity for electrons, it is a better electron donor than the larger bromine ion. Therefore, \(\mathrm{OH}^{-}\) is the stronger base, and its Lewis structure is the same as in (c).

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Key Concepts

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

Comparative Basicity
Understanding the comparative basicity of various molecules and ions is crucial when predicting their chemical behavior. Basicity refers to the ability of a species to donate an electron pair. In general, a stronger base has a greater tendency to donate its electron pair to a proton (H+).

When comparing the basicity of two species, such as \textbf{CH}\(_3\)S- and \textbf{CH}\(_3\)O-, we look at factors like atom size and electronegativity. Since sulfur is larger than oxygen, \textbf{CH}\(_3\)S- can better accommodate the negative charge and is thus a stronger base compared to \textbf{CH}\(_3\)O-. This concept is similarly applied to other comparisons in the exercise.

For example, in comparing hydride (H-) with an ethoxide ion (\textbf{CH}\(_3\)\textbf{CH}\(_2\)O-), we consider that the hydride ion, being smaller and less electronegative, is more willing to donate its electron pair and thus acts as a stronger base.

In each case, drawing the Lewis structure can visually highlight the presence and disposition of lone electron pairs, making the comparison of basicities more understandable.
Resonance Stabilization
Resonance stabilization significantly influences the basicity of molecules. A molecule or ion is resonance-stabilized when it has a structure that allows for the delocalization of electrons across multiple atoms. This delocalization leads to a stabilization of the molecule or ion, making it less likely to donate its electron pair.

For instance, the carboxylate ion (\textbf{CH}\(_3\)\textbf{COO}-) has resonance structures that delocalize the negative charge between two oxygen atoms. This dispersal of charge over more than one atom makes the ion less reactive as a base, compared to \textbf{OH}-, which doesn't have such delocalization.

Considering another example, bicarbonate (HCO3-) versus carboxylate (\textbf{CH}\(_3\)\textbf{COO}-), the negative charge on bicarbonate is not as delocalized as in the carboxylate ion, which has two oxygen atoms to share the negative charge, making bicarbonate a slightly stronger base due to less resonance stabilization.
Electronegativity and Basicity
Electronegativity plays a pivotal role in determining basicity. It is the measure of an atom's ability to attract electrons towards itself. In a basicity comparison, atoms with lower electronegativity are typically better electron pair donors, hence stronger bases.

When comparing oxygen and nitrogen, such as in the case of \textbf{CH}\(_3\)\textbf{NH}- and \textbf{CH}\(_3\)\textbf{O}-, oxygen, being more electronegative, holds onto its electrons more tightly than nitrogen. This makes \textbf{CH}\(_3\)\textbf{O}- a stronger base than \textbf{CH}\(_3\)\textbf{NH}- because it better utilizes its lone pair for bonding with a proton.

Additionally, the size of the atom is also a factor; larger atoms with the same charge as smaller atoms are typically less electronegative. For example, this is why bromide (Br-) is a weaker base than hydroxide (\textbf{OH}-). Despite having a full negative charge, the larger size and lower electronegativity of bromine compared to oxygen means that bromide is less efficient in donating its lone pair.
Lone Electron Pair Donation
The tendency of an atom within a molecule or ion to donate a lone electron pair is the cornerstone of basicity. This property is influenced by various factors, including the atom's electronegativity, the size of the atom, its resonance effects, and the overall molecular structure.

Base strength increases with the availability of the lone pair for donation. For example, in ammonia (\textbf{NH}\(_3\)), the lone pair on the nitrogen is readily available, making it a stronger base than water (\textbf{H}\(_2\)\textbf{O}), where the lone pairs are more involved in hydrogen bonding, thus less available for donation.

Furthermore, in a species like hydroxide (\textbf{OH}-), the negatively charged oxygen atom has a high electron density which makes it more inclined to donate its lone pair compared to a neutral atom like in water. Properly visualizing the Lewis structures of these species can help students understand the location and availability of lone pairs which translates to the molecule's or ion's basic potential.

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Most popular questions from this chapter

Answer true or false to the following statements about the mechanism of acid- base reactions. (a) The acid and base must encounter each other by a collision in order for the proton to transfer. (b) All collisions between acids and bases result in proton transfer. (c) During an acid-base reaction the lone pair on the base fills the A-H antibonding sigma orbital.

Methyl isocyanate, \(\mathrm{CH}_{3}-\mathrm{N}=\mathrm{C}=\mathrm{O}\), is used in the industrial synthesis of a type of pesticide and herbicide known as a carbamate. As a historical note, an industrial accident in Bhopal, India, in 1984 resulted in leakage of an unknown quantity of this chemical into the air. An estimated 200,000 persons were exposed to its vapors, and over 2000 of these people died. (a) Write a Lewis structure for methyl isocyanate, and predict its bond angles. What is the hybridization of its carbonyl carbon? Of its nitrogen atom? (b) Methyl isocyanate reacts with strong acids, such as sulfuric acid, to form a cation. Will this molecule undergo protonation more readily on its oxygen or nitrogen atom? In considering contributing structures to each hybrid, do not consider structures in which more than one atom has an incomplete octet.

In acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), the \(\mathrm{OH}\) hydrogen is more acidic than the \(\mathrm{CH}_{3}\) hydrogens. Explain.

An ester is a derivative of a carboxylic acid in which the hydrogen of the carboxyl group is replaced by an alkyl group (Section 1.3E). Draw a structural formula of methyl acetate, which is derived from acetic acid by replacement of the \(\mathrm{H}\) of its \(-\mathrm{OH}\) group by a methyl group. Determine if proton transfer to this compound from \(\mathrm{HCl}\) occurs preferentially on the oxygen of the \(\mathrm{C}=\mathrm{O}\) group or the oxygen of the \(\mathrm{OCH}_{3}\) group.

Following is a structural formula for guanidine, the compound by which migratory birds excrete excess metabolic nitrogen. The hydrochloride salt of this compound is a white crystalline powder, freely soluble in water and ethanol. (a) Write a Lewis structure for guanidine showing all valence electrons. (b) Does proton transfer to guanidine occur preferentially to one of its \(-\mathrm{NH}_{2}\) groups \((\) cation A) or to its = NH group (cation B)? Explain.

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