Question

Name the following species. (a) \(\mathrm{HIO}_{4}\) (b) \(\mathrm{BrO}_{2}^{-}\) (c) HIO (d) \(\mathrm{NaClO}_{3}\)

Short answer

Answer: The IUPAC names of the given chemical species are: (a) hydrogen periodate, (b) bromite, (c) hydrogen iodite, and (d) sodium chlorate.

Step-by-step solution

(a) Naming \(\mathrm{HIO}_{4}\)

To name this species, identify the elements present and their respective oxidation states. \(\mathrm{HIO}_{4}\) is comprised of a hydrogen cation (\(\mathrm{H}^{+}\)) and a periodate anion (\(\mathrm{IO}_{4}^{-}\)). Therefore, the IUPAC name for \(\mathrm{HIO}_{4}\) is hydrogen periodate.

(b) Naming \(\mathrm{BrO}_{2}^{-}\)

To name this species, identify the elements present and their respective oxidation states. \(\mathrm{BrO}_{2}^{-}\) is comprised of a bromine atom and two oxygen atoms, forming an anion with an overall charge of -1. The "-ate" ending is used for oxyanions with more oxygen atoms, while the "-ite" ending is used for those with fewer. In this case, the IUPAC name for \(\mathrm{BrO}_{2}^{-}\) is bromite.

(c) Naming HIO

To name this species, identify the elements present and their respective oxidation states. HIO is comprised of a hydrogen cation (\(\mathrm{H}^{+}\)) and an iodate anion (\(\mathrm{IO}^{-}\)). As with the bromite anion in part (b) above, the anion with fewer oxygen atoms has the "-ite" ending. Therefore, the IUPAC name for HIO is hydrogen iodite.

(d) Naming \(\mathrm{NaClO}_{3}\)

To name this species, identify the cation and anion present. \(\mathrm{NaClO}_{3}\) is comprised of a sodium cation (\(\mathrm{Na}^{+}\)) and a chlorate anion (\(\mathrm{ClO}_{3}^{-}\)). The sodium cation is named as "sodium," and the chlorate anion uses the "-ate" ending as it has more oxygen atoms. Therefore, the IUPAC name for \(\mathrm{NaClO}_{3}\) is sodium chlorate.