Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 2 - Atoms, Molecules, and Ions - Exercises: 72

Answer

a.) Mercury (I) Oxide b.) Iron (III) Bromide c.) Cobalt (II) Sulfide d.) Titanium (IV) Chloride e.) $Sn_3N_2$ f.) $CoI_3$ g.) $HgO$ h.) $CrS_3$

Work Step by Step

All of these compounds are ionic compounds (composed of a cation and an anion (usually a non-metal and a metal)) so no Greek prefixes need to be added to the beginning of any element names. Since all of these compounds contain transition metals, a roman numeral must be included when naming that certain transition metal. These roman numerals denote the magnitude of the positive charge of the ion. It will always be a positive charge. a.) $Hg_2O$ can be split up into $2Hg^{+} + O^{2-}$ = Mercury (I) Oxide b.) $FeBr_3$ can be split up into $Fe^{3+} + 3Br^{-}$ = Iron (III) Bromide c.) $CoS$ can be split up into $Co^{2+} + S^{2-}$ = Cobalt (II) Sulfide d.) $TiCl_4$ can be split up into $Ti^{4+} + 4Cl^{-}$ = Titanium (IV) Chloride e.) Tin (II) is $Sn^{2+}$ and nitride corresponds to nitrogen which is $N^{3-}$ = $Sn_3N_2$ f.) Cobalt (III) is $Co^{3+}$ and iodide corresponds to iodine which is $I^{-}$ = $CoI_3$ g.) Mercury (II) is $Hg^{2+}$ and oxide corresponds to oxygen which is $O^{2-}$ = $HgO$ h.) Chromium (VI) is $Cr^{6+}$ and sulfide corresponds to sulfur which is $S^{2-}$ = $CrS_3$
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