Question

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 \(\mathrm{PF}_{5}\) b. \(\mathrm{AsF}_{5}\) or \(\mathrm{AsI}_{5}\) c. \(\mathrm{NF}_{3}\) or \(\mathrm{NBr}_{3}\)

Short answer

In Pair A, \(\mathrm{PF}_{5}\) is stable due to the octet rule for phosphorus, while \(\mathrm{NF}_{5}\) is unstable due to the expanded octet of nitrogen. For Pair B, \(\mathrm{AsF}_{5}\) is stable because of strong and stable As-F bonds, but \(\mathrm{AsI}_{5}\) is unstable due to weak As-I bonding. In Pair C, \(\mathrm{NF}_{3}\) is stable due to strong N-F bonds and stable geometry, whereas \(\mathrm{NBr}_{3}\) is unstable because of weaker N-Br bonds and unstable molecular geometry.

Step-by-step solution

Understanding stability in terms of molecular structure and electron configuration

Stability of a compound can be determined by analyzing the molecular structure and electron configuration of the elements involved. A stable compound has a low energy configuration with complete octets for each atom, while an unstable compound has a high energy configuration with incomplete or expanded octets, making the molecule less stable. We will use this concept to analyze the given pairs of substances.

Analyzing Pair A - \(\mathrm{NF}_{5}\) and \(\mathrm{PF}_{5}\)

We first examine the electron configurations of nitrogen (N) and phosphorus (P), which are the central atoms for the two given molecules. Nitrogen (N) has 5 valence electrons, while phosphorus (P) has 5 valence electrons as well. In the case of \(\mathrm{NF}_{5}\), nitrogen would be sharing one electron with each of the five fluorine atoms, totaling 10 electrons in its valence shell which exceeds the octet rule. This expanded octet makes the nitrogen atom in \(\mathrm{NF}_{5}\) unstable. On the other hand, \(\mathrm{PF}_{5}\) meets the octet rule for phosphorus. Phosphorus will share one electron with each of the five fluorine atoms, totaling 10 electrons in its valence shell. Thus, \(\mathrm{PF}_{5}\) is the stable substance in Pair A, and \(\mathrm{NF}_{5}\) is unstable due to the expanded octet of nitrogen.

Analyzing Pair B - \(\mathrm{AsF}_{5}\) and \(\mathrm{AsI}_{5}\)

We first examine the electron configuration of arsenic (As), which is the central atom for both given molecules. Arsenic has 5 valence electrons. In both \(\mathrm{AsF}_{5}\) and \(\mathrm{AsI}_{5}\), arsenic shares one electron with each of the five attached atoms (fluorine in \(\mathrm{AsF}_{5}\) and iodine in \(\mathrm{AsI}_{5}\)), resulting in an expanded octet similar to \(\mathrm{PF}_{5}\). However, the difference between the two compounds lies in the nature of the bonding atoms. Fluorine (F) is more electronegative than iodine (I) and forms stronger and more stable bonds with arsenic (As). The large size and relatively weaker bonding of iodine (I) with arsenic (As) make it difficult to form stable compound \(\mathrm{AsI}_{5}\). Thus, \(\mathrm{AsF}_{5}\) is the stable substance in Pair B, and \(\mathrm{AsI}_{5}\) is unstable due to weak As-I bonding.

Analyzing Pair C - \(\mathrm{NF}_{3}\) and \(\mathrm{NBr}_{3}\)

We first examine the electron configuration of nitrogen (N), which is the central atom for both given molecules. Nitrogen has 5 valence electrons. In both \(\mathrm{NF}_{3}\) and \(\mathrm{NBr}_{3}\), nitrogen shares one electron with each of the three attached atoms (fluorine in \(\mathrm{NF}_{3}\) and bromine in \(\mathrm{NBr}_{3}\)), allowing nitrogen to have a complete octet. However, the difference between the two compounds lies in the polarity of the molecules. In \(\mathrm{NF}_{3}\), the N-F bonds are polar due to the electronegativity difference between nitrogen and fluorine. The \(\mathrm{NBr}_{3}\) has polar N-Br bonds as well. However, since the N-F bonds in \(\mathrm{NF}_{3}\) are stronger and more directional, the molecule has a stable geometry. In contrast, the N-Br bonds in \(\mathrm{NBr}_{3}\) are weaker, making the molecule more prone to decomposition and instability. Thus, \(\mathrm{NF}_{3}\) is the stable substance in Pair C, and \(\mathrm{NBr}_{3}\) is unstable due to its weaker bonding and unstable molecular geometry.

Key concepts

Electron Configuration

The concept of electron configuration is fundamental to understanding molecular stability. It refers to the arrangement of electrons in an atom's orbitals. Atoms seek stability by achieving a lower energy state, often by filling their valence shells, the outermost shell of electrons. For instance, nitrogen (N) has 5 valence electrons, while phosphorus (P) also has 5. When forming compounds such as \(\mathrm{NF}_{5}\) or \(\mathrm{PF}_{5}\), these electrons are shared with surrounding atoms. However, exceeding the typical octet can make certain configurations less stable. In molecules like \(\mathrm{PF}_{5}\), the expanded configuration is stable for phosphorus, while \(\mathrm{NF}_{5}\) is less stable due to nitrogen's difficulty in handling more than eight electrons. Understanding these configurations helps predict which molecular structures are energetically favorable.

Octet Rule

The octet rule is a key rule in chemistry that suggests atoms strive for eight electrons in their valence shell to reach stability, similar to the noble gases. When atoms form molecules, they tend to share or transfer electrons to satisfy this octet rule. For example, nitrogen typically forms compounds fulfilling the octet rule, like \(\mathrm{NF}_{3}\). However, in \(\mathrm{NF}_{5}\), nitrogen would need to accommodate ten electrons, which violates the octet rule, leading to instability. Phosphorus in \(\mathrm{PF}_{5}\), on the other hand, can handle an expanded octet, adhering to a rule for certain elements beyond the second period of the periodic table. This highlights how compliance with the octet rule supports the stability of molecular compounds.

Molecular Geometry

The shape of a molecule, known as its molecular geometry, greatly influences its stability. Molecular geometry results from the spatial arrangement of atoms around a central atom, influenced by electron pairs that repel each other to minimize energy. In stable molecules like \(\mathrm{NF}_{3}\), the geometry allows for a symmetrical, lower-energy structure due to the strong N-F bonds and the distribution of electrons. In contrast, \(\mathrm{NBr}_{3}\) has weaker N-Br bonds leading to possible asymmetry and less stability. Similarly, \(\mathrm{PF}_{5}\) and \(\mathrm{AsF}_{5}\) maintain stable geometries that accommodate their electron sharing without excess strain on the bonds, contributing to their overall stability.

Bond Strength

Bond strength refers to the energy required to break a bond between two atoms. Stronger bonds generally lead to more stable molecules. For example, the stability of \(\mathrm{AsF}_{5}\) over \(\mathrm{AsI}_{5}\) can be attributed to the stronger arsenic-fluorine bonds compared to the arsenic-iodine ones. Fluorine's high electronegativity creates strong bonds with arsenic, whereas iodine's larger atomic size and lower electronegativity result in weaker, less stable bonds. Similarly, \(\mathrm{NF}_{3}\) exhibits stronger N-F bonds compared to the N-Br bonds in \(\mathrm{NBr}_{3}\), leading to increased stability. Evaluating bond strength helps determine the durability and stability of molecular structures.

Electronegativity

Electronegativity refers to an atom's ability to attract shared electrons in a bond. It plays a crucial role in molecular stability. Atoms with high electronegativity like fluorine participate in strong, stable bonding. This is observed in the greater stability of \(\mathrm{AsF}_{5}\) compared to \(\mathrm{AsI}_{5}\). The high electronegativity of fluorine allows it to pull electrons closely, forming tight, stable bonds. In \(\mathrm{NF}_{3}\), the electronegativity difference between nitrogen and fluorine creates polar bonds, again contributing to geometrical and energetic stability. The comparison between \(\mathrm{NF}_{3}\) and \(\mathrm{NBr}_{3}\) again underscores this principle, where the more effective electron sharing due to fluorine’s electronegativity results in better orbital overlap and bond stability.