Question

Place the species in each of the following groups in order of increasing base strength. Give your reasoning in each case. a. \(\mathrm{IO}_{3}^{-}, \mathrm{BrO}_{3}^{-}\) b. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}\) c. \(\mathrm{OCl}^{-}, \mathrm{OI}^{-}\)

Short answer

The order of increasing base strength for each pair is: a. \(\mathrm{BrO}_{3}^{-} < \mathrm{IO}_{3}^{-}\) based on the size and electronegativity of the central atom (Iodine is larger and less electronegative than Bromine). b. \(\mathrm{NO}_{3}^{-} < \mathrm{NO}_{2}^{-}\) because the negative charge on the Nitrogen atom is less spread out in \(\mathrm{NO}_{2}^{-}\), making it easier to donate a lone pair of electrons. c. \(\mathrm{OCl}^{-} < \mathrm{OI}^{-}\) as the lone pair of electrons on Oxygen in \(\mathrm{OI}^{-}\) will be less tightly held due to Iodine being larger and less electronegative than Chlorine.

Step-by-step solution

Recall the role of electronegativity and central atom size in basicity

Electronegativity of an atom measures its ability to attract electrons. On the periodic table, electronegativity increases as we move from left to right and decreases as we move from top to bottom. Basicity (base strength) generally increases with the decrease in electronegativity of the central atom and the increase in size of the central atom. This is because larger atoms and less electronegative atoms hold on to their electrons less tightly, making it easier for them to donate a lone pair of electrons and act as bases.

Comparing base strengths in Set a (\(\mathrm{IO}_{3}^{-}, \mathrm{BrO}_{3}^{-}\))

In both of these species, the central atoms are iodine (I) and bromine (Br) respectively. They belong to the same group in the periodic table and are located directly above and below each other. Iodine is larger and less electronegative than bromine, which means the \(\mathrm{IO}_{3}^{-}\) ion is likely to be a stronger base compared to \(\mathrm{BrO}_{3}^{-}\). Therefore, the order of increasing base strength for this pair is: \(\mathrm{BrO}_{3}^{-} < \mathrm{IO}_{3}^{-}\).

Comparing base strengths in Set b (\(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}\))

In both of these species, the central atom is nitrogen (N). The difference between the two species lies in the number of oxygen atoms bonded to the nitrogen atom and the overall charge of the species. \(\mathrm{NO}_{2}^{-}\) has a -1 charge and \(\mathrm{NO}_{3}^{-}\) has a -1 charge as well. Considering the charge is equal, the next factor to look at is the number of oxygen atoms: \(\mathrm{NO}_{2}^{-}\) has less oxygen atoms than \(\mathrm{NO}_{3}^{-}\). This means that in \(\mathrm{NO}_{2}^{-}\), the negative charge on the nitrogen atom is less spread out and therefore, it can more easily donate a lone pair of electrons, making it a stronger base. Thus, the order of increasing base strength for this pair is: \(\mathrm{NO}_{3}^{-} < \mathrm{NO}_{2}^{-}\).

Comparing base strengths in Set c (\(\mathrm{OCl}^{-}, \mathrm{OI}^{-}\))

In both of these species, the central atom is oxygen (O), but the difference lies in the atoms bonded to oxygen: chlorine (Cl) in \(\mathrm{OCl}^{-}\) and iodine (I) in \(\mathrm{OI}^{-}\). Iodine is larger and less electronegative than chlorine, which means the lone pair of electrons on oxygen in the \(\mathrm{OI}^{-}\) ion will be less tightly held and more available to be donated, making it a stronger base. Therefore, the order of increasing base strength for this pair is: \(\mathrm{OCl}^{-} < \mathrm{OI}^{-}\). In conclusion, the order of increasing base strength for each pair is: a. \(\mathrm{BrO}_{3}^{-} < \mathrm{IO}_{3}^{-}\) b. \(\mathrm{NO}_{3}^{-} < \mathrm{NO}_{2}^{-}\) c. \(\mathrm{OCl}^{-} < \mathrm{OI}^{-}\)