complete the table below by writing the symbols for the cation and anion that make up each ionic compound…

Explanation:

Step1: Determine the cation for each ionic compound

For (MnBr_4), the cation is (Mn^{4 +}) (since bromide (Br^-) has a charge of (- 1) and there are 4 (Br^-) ions, the manganese ion must have a charge of (+4) to balance the charges). For (FeCl_2), the cation is (Fe^{2+}) (as chloride (Cl^-) has a charge of (-1) and there are 2 (Cl^-) ions, the iron ion must have a charge of (+2)). For (CrF_3), the cation is (Cr^{3+}) (because fluoride (F^-) has a charge of (-1) and there are 3 (F^-) ions, the chromium ion must have a charge of (+3)). For (Mn_2O_3), the cation is (Mn^{3+}) (since oxide (O^{2 -}) has a charge of (-2) and there are 3 (O^{2-}) ions with a total negative charge of (-6), and there are 2 manganese ions, so each manganese ion has a charge of (+3)).

Step2: Determine the anion for each ionic compound

For (MnBr_4), the anion is (Br^-). For (FeCl_2), the anion is (Cl^-). For (CrF_3), the anion is (F^-). For (Mn_2O_3), the anion is (O^{2-}).

Answer:

ionic compound	(MnBr_4)	(FeCl_2)	(CrF_3)	(Mn_2O_3)	(NaCl)
cation	(Mn^{4+})	(Fe^{2+})	(Cr^{3+})	(Mn^{3+})	(Na^{+})
anion	(Br^-)	(Cl^-)	(F^-)	(O^{2 -})	(Cl^-)