Supplemental packet page 73

1. Supplemental packet page 73

2. carbide ion phosphide ion phosphate ion borate ion nitride ion sulfide ion sulfate ion carbonate ion oxide ion chloride ion perchlorate ion nitrate ion fluoride ion X- charges

3. Acids named as a pure compound as a gas as a gas hydrogens hydrogens hydrogen perchlorate hydrogen perchlorate dihydrogen sulfate dihydrogen sulfate hydrogen chloride hydrogen chloride hydrogen nitrate hydrogen nitrate trihydrogen phosphate trihydrogen phosphate hydrogen acetate hydrogen acetate Acidsas a pure compoundhydrogen listed first

4. aqueous solution hydrogen perchlorate perchloric acid dihydrogen sulfate sulfuric acid hydrogen chloride hydrochloric acid hydrogen nitrate Acids named as a nitric acid trihydrogen phosphate pure compound phosphoric acid hydrogen acetate hydrogen perchlorate acetic acid dihydrogen sulfate hydrogen chloride hydrogen nitrate trihydrogen phosphate hydrogen acetate Acidsas an aqueous solution Acids named as an

5. Covalent Compounds carbon dioxide dihydrogen sulfide (g) as an acid in aqueous solution hydrosulfuric acid (aq) tetracarbon octahydride disulfur hexabromide

6. Covalent Compounds sulfur dioxide (smog) dinitrogen oxide better known as nitrous oxide laughing gas diphosphorus pentaoxide pentaphosphorus decaoxide

7. Covalent Compounds nitrogen triiodide dinitrogen tetraoxide phosphorus pentafluoride triarsenic heptaoxide

8. Ionic Compounds the sum of all oxidation numbers in a neutral species is zero, 0,

9. Ionic Compounds the sum of all oxidation numbers in a neutral species is zero, 0,

10. 1 1 TM+ roman numeral 1 2 2 1 1 2 A Roman numeral represents oxidation state: The larger the positive number, The higher the oxidation state of the ion. Only transition metals and heavier post-transition metals (e.g. Sn,Pb) use a Roman numeral in its name Ionic Compounds the sum of all oxidation numbers in a neutral species is zero, 0,

11. Ion charge calculation in ionic substances 1) Ion charge is called “oxidation state or number” 2) memorize the monatomic ions and their charge the “–ides” (C4- , N3- , O2- , F1- ). 3) memorize the polyatomic ions and their charge “–ates”( [PO4]3- , [SO4]2- , [ClO4]1- ) 4) All anions (-ides and -ates) seek out positively charged cations ( Na1+, Ca2+, Al3+ ) to achieve a balance of zero in overall substance charge.

12. SO4 Na Na SO4 2- Na Na SO4 2- Na Na 1+ 1+ 1 + (2-) + 1 = 0 Perhaps the easiest way to calculate an oxidation number for a metal in an ionic compound is to draw a visual. For example, Na2SO4 1) Separate the metal from the nonmetals in the formula, 2) Assign monatomics and polyatomics whose oxidation number was memorized, 3) Knowing the sum of all oxidation numbers in a neutral species is zero (0), solve for the oxidation number of the remaining element. sodium sulfate

13. PO4 Fe PO4 3- Fe Na PO4 3- Fe Na 1+ 3+ 3 + (3-) = 0 Calculate an oxidation number for a metal in an ionic compound of FePO4 1) Separate the metal from the nonmetals in the formula, 2) Assign monatomics and polyatomics whose oxidation number was memorized, 3) Knowing the sum of all oxidation numbers in a neutral species is zero (0), solve for the oxidation number of the remaining element. iron (III) phosphate

14. PO4 PO4 PO4 PO4 PO4 PO4 3- 3- 3- 3- Fe Fe Fe 2+ Fe Fe Fe 2+ Fe Fe Fe 2+ Calculate an oxidation number for a metal in an ionic compound of Fe3(PO4)2 1) separate the metal from the nonmetals in the formula, 2) Assign monatomics and polyatomics whose oxidation number was memorized, 3) Knowing the sum of all oxidation numbers in a neutral species is zero (0), solve for the oxidation number of the remaining element. iron (II) phosphate 2 + (3-) + 2 + (3-) + 2 = 0