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Consider the base hydroxylamine, \(\mathrm{NH}_{2} \mathrm{OH}\) . (a) What is the conjugate acid of hydroxylamine? (b) When it acts as a base, which atom in hydroxylamine accepts a proton? (c) There are two atoms in hydroxylamine that have nonbonding electron pairs that could act as proton acceptors. Use Lewis structures and formal charges (Section 8.5) to rationalize why one of these two atoms is a much better proton acceptor than the other.

Short Answer

Expert verified
The conjugate acid of hydroxylamine, \(\mathrm{NH}_2\mathrm{OH}\), is \(\mathrm{NH}_3\mathrm{OH}^+\). When hydroxylamine acts as a base, the oxygen atom, with higher electronegativity and two lone pairs, accepts the proton. Using Lewis structures and formal charges, the oxygen atom has a formal charge of -1 and is more likely to attract a proton than the nitrogen atom, which has a formal charge of +1. Hence, oxygen is the better proton acceptor in hydroxylamine.

Step by step solution

01

(a) Conjugate Acid of Hydroxylamine

To find the conjugate acid of hydroxylamine, we need to add a proton (H+) to hydroxylamine's formula. The formula of hydroxylamine is \(\mathrm{NH}_2\mathrm{OH}\). Adding a proton to it will result in the formation of its conjugate acid: \[\mathrm{NH}_2\mathrm{OH} + \mathrm{H}^+ \rightarrow \mathrm{NH}_3\mathrm{OH}^+\]Thus, the conjugate acid of hydroxylamine is \(\mathrm{NH}_3\mathrm{OH}^+\).
02

(b) Proton Acceptor in Hydroxylamine

When hydroxylamine acts as a base, it accepts a proton (H+) from an acid. The oxygen atom in hydroxylamine has two lone pairs of electrons, and the nitrogen atom has one lone pair of electrons. Oxygen with its higher electronegativity and having two lone pairs is more likely to attract the proton and form a bond. Therefore, the oxygen atom in hydroxylamine accepts the proton.
03

(c) Better Proton Acceptor: Nitrogen or Oxygen

There are two atoms in hydroxylamine that have nonbonding electron pairs that could act as proton acceptors: nitrogen (N) and oxygen (O). We need to use Lewis structures and formal charges to rationalize which atom is a better proton acceptor. According to the Lewis structure, the nitrogen atom in hydroxylamine has a partial positive charge, and the oxygen atom has a partial negative charge. This can also be confirmed using the formal charge formula: \[Formal\;Charge = Valence\;electrons - (Nonbonding\;electrons + \frac{1}{2}Bonding\;electrons)\] For Nitrogen: Formal Charge = 5 - (2 + (8/2)) = +1 For Oxygen: Formal Charge = 6 - (4 + (4/2)) = -1 Since the oxygen atom has a partial negative charge, it is more likely to attract the proton and accept it, leading to the formation of a new bond. Therefore, oxygen is a much better proton acceptor than nitrogen in hydroxylamine.

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Key Concepts

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

Conjugate Acid-Base Pairs
Understanding how conjugate acid-base pairs work is crucial in acid-base chemistry. Take hydroxylamine, \(\mathrm{NH}_{2} \mathrm{OH}\), for instance; when it gains a proton (\(\mathrm{H}^+\)), it becomes its conjugate acid \(\mathrm{NH}_3\mathrm{OH}^+\). This transformation illustrates the concept that bases turn into their conjugate acids after accepting a proton—an essential process in neutralization reactions. Similarly, when acids donate a proton, they form their conjugate bases.

Seeing the pair as two sides of the same coin helps clarify their roles in reversible reactions. They demonstrate how substances can switch between being an acid or base depending on the reaction circumstances. This fluid identity is at the heart of the Brønsted–Lowry acid-base theory, which characterizes acids and bases by their ability to donate and accept protons, respectively.
Lewis Structures
Lewis structures are like blueprints for molecules, mapping out their skeletons. They show how atoms are bonded and where the electrons are located, which is pivotal when considering how these molecules will interact. In the case of hydroxylamine, Lewis structures illuminate the molecule's shape and indicate the electron pairs that can participate in bonding.

Let's get hands-on with dots and lines representing electrons and bonds, respectively. In hydroxylamine, you'd see nitrogen and oxygen atoms surrounded by dots (electrons) and linked by lines (bonds). These structures aren't just for show—they guide us in predicting how a molecule like hydroxylamine might behave as a proton acceptor or form its conjugate acid.
Formal Charges
The formal charge is like a molecule's bank statement, showing the 'credit' or 'debit' of electrons for an atom. Calculating it involves a simple formula but provides profound insight into the stability and reactivity of molecules. In hydroxylamine, for instance, the formal charges reveal that the nitrogen atom has a positive charge and the oxygen atom carries a negative charge.

Why does this matter? Molecular charge distribution can predict where reactions are likely to occur. It's a clincher in our detective work to determine the favored proton acceptor in a molecule. Armed with this knowledge, we can deduce why oxygen, with its negative formal charge, is eager to grab that incoming proton, leaving nitrogen, the positively charged wallflower, in the background.
Proton Acceptors
The term 'proton acceptor' makes you think of a dance floor where atoms wait to catch a dance partner—that partner being a proton. In our acid-base dance, an effective proton acceptor should have a high electronegativity and available electron pairs. Hydroxylamine's oxygen fits the bill perfectly, with a pair of nonbonding electrons ready to tango.

What makes oxygen the better dancer than nitrogen in this case? It's all down to the electron date outfit: oxygen's electronegativity and formal charge make it more attractive to an incoming proton. So, when acids are looking for a base to react with, it's the oxygen in hydroxylamine that's most likely to catch the proton and waltz away as part of a new molecule—the conjugate acid.

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Most popular questions from this chapter

How many milliliters of concentrated hydrochloric acid solution \((36.0 \%\) HCl by mass, density \(=1.18 \mathrm{g} / \mathrm{mL})\) are required to produce 10.0 \(\mathrm{L}\) of a solution that has a pH of 2.05\(?\)

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\) Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

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) \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{3}(a q)\)

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) Acid strength in a series of \(\mathrm{H}-\) A molecules increases with increasing size of \(\mathrm{A} .\) (b) For acids of the same general structure but differing electronegativities of the central atoms, acid strength decreases with increasing electronegativity of the central atom.(c) The strongest acid known is HF because fluorine is the most electronegative element.

Calculate the \(\mathrm{pH}\) of each of the following solutions \((K_{a}\) and \(K_{b}\) values are given in Appendix \(\mathrm{D} ) :\) (a) 0.095\(M\) propionicacid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right),(\mathbf{b}) 0.100 M\) hydrogen chromate ion \(\left(\mathrm{HCrO}_{4}^{-}\right),(\mathbf{c}) 0.120 M\) pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right) .\)

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