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 \(\mathrm{NBr}_{3}\)

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.

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.

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.

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.