Problem 42 - Chemistry An Atoms First Approach

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

Step-by-Step Solution

Answer

The stable substances for each pair are: a. \(\mathrm{PF}_{5}\) because it can accommodate an expanded octet due to the presence of d orbitals, unlike nitrogen in \(\mathrm{NF}_{5}\). b. \(\mathrm{AsF}_{5}\) due to the stronger bonds formed with fluorine than those formed with iodine in \(\mathrm{AsI}_{5}\). c. \(\mathrm{NF}_{3}\) because it has more stable and stronger bonds than \(\mathrm{NBr}_{3}\), which has weaker bonds between nitrogen and bromine.

Step 1: Compare electron configurations

First, let's consider the electron configurations of nitrogen and phosphorus. Nitrogen has 7 electrons, with 5 valence electrons (configuration: 2s² 2p³). Phosphorus has 15 electrons, with 5 valence electrons (configuration: 3s² 3p³).

Step 2: Analyze the bonding

Now, let's look at the bonding in both compounds. In \(\mathrm{NF}_{5}\), nitrogen forms 5 bonds with 5 fluorine atoms, resulting in an expanded octet. But, nitrogen does not have any d orbitals available to accommodate the extra electrons required for an expanded octet, making \(\mathrm{NF}_{5}\) unstable. On the other hand, in \(\mathrm{PF}_{5}\), phosphorus forms 5 bonds with 5 fluorine atoms. Phosphorus has access to empty d orbitals, which allows it to accommodate the extra electrons required for the expanded octet, making \(\mathrm{PF}_{5}\) a stable compound. 2. Pair (b) - \(\mathrm{AsF}_{5}\) and \(\mathrm{AsI}_{5}\)

Step 1: Compare electronegativities

Analyzing electronegativities of fluorine (3.98) and iodine (2.66), we can see that fluorine is much more electronegative compared to iodine.

Step 2: Analyze bond strengths

Fluorine atoms have stronger bonds with arsenic (\(\mathrm{AsF}_{5}\)) because of high electronegativity difference. Therefore, \(\mathrm{AsF}_{5}\) is more stable compared to \(\mathrm{AsI}_{5}\). The large size of the iodine atom and weaker bonds in \(\mathrm{AsI}_{5}\) make it an unstable compound. 3. Pair (c) - \(\mathrm{NF}_{3}\) and \(\mathrm{NBr}_{3}\)

Step 1: Compare electronegativities

Comparing electronegativities of fluorine (3.98) and bromine (2.96), we can notice that fluorine is more electronegative.

Step 2: Analyze bond strengths

The higher electronegativity difference between nitrogen and fluorine results in more stable and stronger bonds in \(\mathrm{NF}_{3}\) compared to \(\mathrm{NBr}_{3}\). The weaker bonds between nitrogen and bromine in \(\mathrm{NBr}_{3}\) make it an unstable compound. So, for each pair of substances, the stable ones are: a. \(\mathrm{PF}_{5}\) b. \(\mathrm{AsF}_{5}\) c. \(\mathrm{NF}_{3}\)