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 each pair of substances, the stable compounds are \(\mathrm{PF}_{5}\), \(\mathrm{AsF}_{5}\), and \(\mathrm{NF}_{3}\). The unstable compounds are \(\mathrm{NF}_{5}\) (due to exceeding the octet rule), \(\mathrm{AsI}_{5}\) (due to steric hindrance and large size of iodine atoms), and \(\mathrm{NBr}_{3}\) (due to weaker N-Br bonds and steric hindrance).

Step-by-step solution

Pair a: \(\mathrm{NF}_{5}\) or \(\mathrm{PF}_{5}\)

For this pair, we will compare the octet rule and the sizes of the central atoms N and P. The octet rule states that atoms will seek to form bonds in a way that will give them a total of eight electrons in their valence shell. \(\mathrm{NF}_{5}\): The nitrogen atom has 5 valence electrons. If it forms five bonds with fluorine atoms, it will have a total of 10 electrons in its valence shell, which exceeds the octet rule. \(\mathrm{PF}_{5}\): The phosphorus atom has 5 valence electrons. If it forms five bonds with fluorine atoms, it will obey the octet rule as P is able to expand its valence shell due to the presence of d orbitals. Thus, \(\mathrm{PF}_{5}\) is the stable substance, while \(\mathrm{NF}_{5}\) is unstable.

Pair b: \(\mathrm{AsF}_{5}\) or \(\mathrm{AsI}_{5}\)

For this pair, we will consider the size and electronegativity of the halogen atoms (F and I) relative to the central atom As. \(\mathrm{AsF}_{5}\): Arsenic and fluorine have a significant electronegativity difference, which results in strong polar covalent bonds. Also, the As-F bond length is small enough for As to accommodate five F atoms. \(\mathrm{AsI}_{5}\): While arsenic and iodine have a smaller electronegativity difference, the iodine atoms are much larger than the arsenic atom, causing significant steric hindrance that destabilizes the compound. Thus, \(\mathrm{AsF}_{5}\) is the stable substance, while \(\mathrm{AsI}_{5}\) is unstable.

Pair c: \(\mathrm{NF}_{3}\) or \(\mathrm{NBr}_{3}\)

For this pair, we will consider the size of the central atom N relative to the size of the halogen atoms (F and Br). \(\mathrm{NF}_{3}\): Nitrogen and fluorine are both small atoms, and the N-F bond is strong and stable due to the electronegativity difference. Thus, \(\mathrm{NF}_{3}\) is a stable compound. \(\mathrm{NBr}_{3}\): Nitrogen is much smaller than bromine, which results in a weaker N-Br bond and significant steric hindrance due to the size difference. Additionally, there is less electronegativity difference between N and Br compared to N and F, making the bond weaker and less stable. Thus, \(\mathrm{NF}_{3}\) is the stable substance, while \(\mathrm{NBr}_{3}\) is unstable.