Decide whether each pair of elements in the table below will form an ionic compound. If they will, write the empirical formula of the compound formed in the space provided.element #1 element #2 Forms ionic compound? empirical formula of ionic compoundmagnesium sulfur yesnosulfur bromine yesnochlorine rubidium yesnorubidium calcium yesno
Question
Decide whether each pair of elements in the table below will form an ionic compound. If they will, write the empirical formula of the compound formed in the space provided.element #1 element #2 Forms ionic compound? empirical formula of ionic compoundmagnesium sulfur yesnosulfur bromine yesnochlorine rubidium yesnorubidium calcium yesno
Solution
The table you provided is not fully visible. However, I can provide you with the general rules to determine if an ionic compound will form and how to write the empirical formula.
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Magnesium (Mg) and Sulfur (S): Yes, they will form an ionic compound. Magnesium is a metal and sulfur is a non-metal. Metals and non-metals usually form ionic compounds. The empirical formula is MgS.
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Sulfur (S) and Bromine (Br): No, they will not form an ionic compound. Both sulfur and bromine are non-metals and non-metals usually form covalent compounds, not ionic.
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Chlorine (Cl) and Rubidium (Rb): Yes, they will form an ionic compound. Chlorine is a non-metal and rubidium is a metal. The empirical formula is RbCl.
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Rubidium (Rb) and Calcium (Ca): No, they will not form an ionic compound. Both rubidium and calcium are metals and metals do not usually form ionic compounds with each other.
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To fill in the table, we need to balance the charges of the cations and anions to find the empirical formula for each ionic compound. Then we can name the compound accordingly. 1. For the cation Fe^3+ and the anion S^2-: - To balance the charges, we need two Fe^3+ ions for every three S^2- ions. - The empirical formula is Fe2S3. - The name of the compound is iron(III) sulfide. 2. For the cation Pb^4+ and the anion O^2-: - To balance the charges, we need two Pb^4+ ions for every four O^2- ions (which simplifies to one Pb^4+ for every two O^2-). - The empirical formula is PbO2. - The name of the compound is lead(IV) oxide. 3. For the cation Mn^2+ and the anion N^3-: - To balance the charges, we need three Mn^2+ ions for every two N^3- ions. - The empirical formula is Mn3N2. - The name of the compound is manganese(II) nitride. 4. For the cation Fe^2+ and the anion P^3-: - To balance the charges, we need three Fe^2+ ions for every two P^3- ions. - The empirical formula is Fe3P2. - The name of the compound is iron(II) phosphide. So the completed table should list the empirical formulas and names of the compounds as follows: - Fe2S3, iron(III) sulfide- PbO2, lead(IV) oxide- Mn3N2, manganese(II) nitride- Fe3P2, iron(II) phosphide
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