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Chapter 20: Problem 48

In each of the following pairs of substances, one is stable and known, and the other is unstable. For each pair, choose the stable substance, and explain why the other is unstable. a. \(\mathrm{NF}_{5}\) or \(\mathrm{PF}_{5} \quad\) b. AsF_r $\mathrm{AsI}_{5} \quad\( c. \)\mathrm{NF}_{3}\( or \)\mathrm{NBr}_{3}$

Short Answer

Expert verified
In each pair, the stable molecules are: a. \(\mathrm{PF}_{5}\): \(\mathrm{NF}_{5}\) is unstable because nitrogen can only form a maximum of 4 bonds. b. \(\mathrm{AsF}_{5}\): \(\mathrm{AsI}_{5}\) is unstable due to weaker covalent bonds with arsenic and increased repulsion among larger iodine atoms. c. \(\mathrm{NF}_{3}\): \(\mathrm{NBr}_{3}\) is unstable because of weaker covalent bonds and increased electron repulsion with bromine atoms.

Step by step solution

01

Identify the stable molecule in pair a

In pair a, we have \(\mathrm{NF}_{5}\) and \(\mathrm{PF}_{5}\). Among these two, \(\mathrm{PF}_{5}\) is known to be stable. Therefore, \(\mathrm{NF}_{5}\) is the unstable molecule.
02

Explain why NF5 is unstable

To understand why \(\mathrm{NF}_{5}\) is unstable, we need to consider the electron configuration of nitrogen and fluorine. Nitrogen has 5 electrons in its valence shell, while fluorine has 7. A molecule of \(\mathrm{NF}_{5}\) would require nitrogen to share its 5 valence electrons with 5 fluorine atoms to form 5 covalent bonds. However, nitrogen can only form a maximum of 4 bonds, as it would require an extra electron pair to form another bond. This makes \(\mathrm{NF}_{5}\) unstable and unlikely to exist.
03

Identify the stable molecule in pair b

In pair b, we have \(\mathrm{AsF}_{5}\) and \(\mathrm{AsI}_{5}\). Among these two, \(\mathrm{AsF}_{5}\) is known to be stable. Therefore, \(\mathrm{AsI}_{5}\) is the unstable molecule.
04

Explain why AsI5 is unstable

To understand why \(\mathrm{AsI}_{5}\) is unstable, we need to consider the atom sizes and electron configurations of arsenic and iodine. The larger atomic size of iodine compared to fluorine results in weaker covalent bonds with arsenic. Additionally, there is increased repulsion among the larger iodine atoms, putting further strain on the molecular structure. As a result, \(\mathrm{AsI}_{5}\) is unstable and not commonly found.
05

Identify the stable molecule in pair c

In pair c, we have \(\mathrm{NF}_{3}\) and \(\mathrm{NBr}_{3}\). Among these two, \(\mathrm{NF}_{3}\) is known to be stable. Therefore, \(\mathrm{NBr}_{3}\) is the unstable molecule.
06

Explain why NBr3 is unstable

To understand why \(\mathrm{NBr}_{3}\) is unstable, we need to consider the atom size and electron configuration of nitrogen and bromine. Nitrogen has 5 valence electrons and can form three covalent bonds by sharing electrons with bromine atoms. However, the larger atomic size of bromine compared to fluorine results in weaker covalent bonds and increased electron repulsion. This makes \(\mathrm{NBr}_{3}\) prone to dissociation and, thus, unstable.

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Most popular questions from this chapter

Lead forms compounds in the \(+2\) and \(+4\) oxidation states. All lead(II) halides are known (and are known to be ionic). Only \(\mathrm{PbF}_{4}\) and \(\mathrm{PbCl}_{4}\) are known among the possible lead(IV) halides. Presumably lead(IV) oxidizes bromide and iodide ions, producing the lead(Il) halide and the free halogen: $$ \mathrm{PbX}_{4} \longrightarrow \mathrm{PbX}_{2}+\mathrm{X}_{2} $$ Suppose 25.00 g of a lead(IV) halide reacts to form 16.12 g of a lead(Il) halide and the free halogen. Identify the halogen.

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The resistivity (a measure of electrical resistance) of graphite is $(0.4 \text { to } 5.0) \times 10^{-4}\( ohm \)\cdot \mathrm{cm}$ in the basal plane. (The basal plane is the plane of the six-membered rings of carbon atoms.) The resistivity is 0.2 to 1.0 ohm \(\cdot \mathrm{cm}\) along the axis perpendicular to the plane. The resistivity of diamond is \(10^{14}\) to $10^{16} \mathrm{ohm} \cdot \mathrm{cm}$ and is independent of direction. How can you account for this behavior in terms of the structures of graphite and diamond?

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