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Use Lewis structures to diagram the following reaction in the manner of reaction (16.19) $$2 \mathrm{NH}_{3}+\mathrm{Ag}^{+} \longrightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$$ Identify the Lewis acid and Lewis base.

Short Answer

Expert verified
The Lewis acid is the Silver ion (Ag+) and the Lewis base is the ammonia (NH3). The Lewis acid (Ag+) accepts an electron pair from the Lewis base (NH3) to form a complex ion [Ag(NH3)2]+

Step by step solution

01

Determine the Lewis Structures

Firstly, Lewis structures should be drawn for the reactants and products. The nitrogen atom in ammonia (NH3) has a lone pair of electrons that can be donated, and Ag+ is a positively charged ion that can accept a pair of electrons to achieve a stable electron configuration.
02

Identify the Lewis Acid

A Lewis acid is defined as an atom, ion or molecule that accepts an electron pair to form a covalent bond. In this case, the silver ion \(Ag^{+}\) acts as a Lewis acid because it accepts a pair of electrons from ammonia.
03

Identify the Lewis Base

A Lewis base is an atom, ion or molecule that donates an electron pair to form a covalent bond. In this case, the nitrogen in the ammonia (NH3) is the Lewis base, because it donates its lone pair of electrons.
04

Visualize the Reaction

The reaction is visualized as two ammonia (NH3) molecules donates their lone pair of electrons to the silver ion (Ag+) to form a coordination compound. The Lewis base is the electron-donating species (NH3), and the Lewis acid is the electron-accepting species (Ag+).

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Key Concepts

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

Lewis acid
In the world of chemistry, Lewis acids are central players when it comes to forming bonds.
They are substances that can accept a pair of electrons. It's like they act as electron "receivers" in a bond. The prime example given in our exercise involves the silver ion, denoted as \( \text{Ag}^+ \). This ion is positively charged, meaning it has a shortage of electrons, making it eager to accept additional electrons from a donor.

When a Lewis acid accepts an electron pair, it forms a stable chemical bond. This makes the substance involved more stable and lowers its energy, which is often the driving force behind chemical reactions.
Lewis acids are not limited to just ions like silver here, but can include molecules that exhibit similar electron-pair accepting qualities. So, whenever you spot a positively charged or electron-deficient molecule in a reaction, there's a good chance it's acting as a Lewis acid.
Lewis base
If Lewis acids are the "receivers" of electrons, then Lewis bases are the generous "givers."
A Lewis base is any molecule or ion that donates an electron pair to form a covalent bond. In the context of our exercise, ammonia (\( \text{NH}_3 \)) takes on this role.

Take a closer look at ammonia: the nitrogen atom at its center has a lone pair of electrons. This pair is key because it can be easily shared or donated. Ammonia, like all Lewis bases, works on forming bonds by giving this lone pair to another molecule that needs it, aka the Lewis acid.
This electron donation is not only crucial for bond formation but also for stabilizing the involved atoms or ions, which results in a more favorable chemical reaction. Identifying Lewis bases in reactions can often be as simple as spotting molecules with lone pairs that aren’t tightly held, ready to make connections.
Coordination compound
Coordination compounds bring an interesting twist into chemical reactions.
These compounds form when a Lewis base donates a lone pair of electrons to a Lewis acid, resulting in a more complex structure. In other words, they are the products of coordinate covalent bonds where both electrons in the bond originate from the base or donor.

In our exercise, the coordination compound this reaction produces is \( \left[\text{Ag}(\text{NH}_3)_2\right]^+ \). The formation happens once the \( \text{Ag}^+ \) ion accepts electron pairs from two \( \text{NH}_3 \) molecules.
This new compound encapsulates both the qualities of its individual parts—namely the electron-deficient silver and the electron-rich ammonia—and creates a unique, stable molecular entity. Coordination compounds are fundamental not only in chemistry but also in real-world applications, such as catalysis, dye formation, and even medicine.

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Most popular questions from this chapter

The molecular solid \(\mathrm{I}_{2}(\mathrm{s})\) is only slightly soluble in water but will dissolve to a much greater extent in an aqueous solution of \(\mathrm{KI}\), because the \(\mathrm{I}_{3}^{-}\) anion forms. Write an equation for the formation of the \(I_{3}^{-}\) anion, and indicate the Lewis acid and Lewis base.

Write the formula of the conjugate base in the reaction of each acid with water. (a) \(\mathrm{HIO}_{3} ;\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (c) \(\mathrm{HPO}_{4}^{2-} ;\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)

For the ionization of phenylacetic acid, $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \\\ K_{\mathrm{a}}=4.9 \times 10^{-5} \end{array}$$ (a) What is \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\right]\) in \(0.186 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) (b) What is the \(\mathrm{pH}\) of \(0.121 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\)

The solubility of 1 -naphthylamine, \(\mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{2}, \mathrm{a}\) substance used in the manufacture of dyes, is given in a handbook as 1 g per \(590 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). What is the approximate \(\mathrm{pH}\) of a saturated aqueous solution of 1-naphthylamine? $$\begin{array}{r} \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \\ \mathrm{p} K_{\mathrm{b}}=3.92 \end{array}$$

Each of the following is a Lewis acid-base reaction. Which reactant is the acid, and which is the base? Explain. (a) \(\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\) (b) \(\operatorname{Zn}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})\)

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