2. Writing Formulas for Simple Binary Compunds If the compound’s name ends in “ide” chances are it’s a binary compound. (There are a few exceptions which will be covered later).

3. Find the 1st element on the periodic table and determine its oxidation number. If there is more than 1 oxidation state then go on to the section entitled Writing formulas of Complex Binary Compounds.

4. 1 23 2 4 -4 3 5 -3 2 4 6 -2 3 5 7 Oxidation numbers can be determined from position in the periodic table below.

5. Writing Formulas for Simple Binary Compunds Find the oxidation state of the 2nd element remembering the first element is always positive and the second is always negative. Once both oxidation numbers are determined the goal is to combine the positives and the negatives in ratios where the total charge adds up to zero.

6. Example 1 - magnesium nitride. Mg 2+ N 3- Since 3 Mg 2+ particles have a charge of 6+ and 2 N 3- particles have a charge of 6- the total charge adds up to zero. The correct formula must be: Mg 2+ N 3- Mg 3 N 2 2 N 3- particles 3 Mg 2+ particles

7. Writing Formulas for Simple Binary Compunds Example 2 - sodium phosphide. Sodium is in group 1 so its oxidation number is 1+. Phosphide is short for phosphorus. Its oxidation number must be negative since it’s the 2nd name so phosphide has a charge of 3-. Na 1+ P 3- If the total charge is going to add up to zero 3 Na 1+ and 1 P 3- must be used. Na 1+ The final answer is: Na 3 P

8. Writing Formulas for Complex Binary Compounds - “ous,ic ” If the 1st name ends in “ous” or “ic” find the element on the periodic table. Latin names are sometimes used. Below is a list of the latin names most frequently used: ferrous, ferric - iron- 26 Fe cuprous, cupric - copper- 29 Cu stannous, stannic - tin- 50 Sn plumbous, plumbic - lead- 82 Pb 2+ 3+ 1+ 2+ 4+ 2+ 4+

9. Generally “ous” is used for the lowest positive oxidation state and “ic” is the next highest. There are exceptions to this rule. Example - Stannous chloride Sn 2+ Cl 1- SnCl 2 Example - plumbic sulfide Pb 4+ S 2- - PbS 2 Cl 1- S 2- -

10. Writing Formulas for Complex Binary Compounds Roman Numeral System - IUPAC When names for compounds have Roman Numerals present, this number represents the charge of the positive ion. This charge can be used to determine the number of negative particles needed to create a combination of particles with an overall charge of zero.

11. Example - sulfur(VI) oxide The Roman Numeral VI stands for 6. (V is 5 and I after it means add 1 to 5. The Roman Numeral for 4 is IV) S 6+ If sulfur particles have a charge of 6+, 3 oxygen particles, each with a charge of 2- are needed to create a collection of particles where the total charge adds up to zero. O 2- SO 3

12. Writing Formulas for Complex Binary Compounds - Prefix System Prefix No. of Atoms mono- 1 di- 2 tri- 3 tetra- 4 penta- 5 hexa- 6 hepta- 7 octa- 8 When a compound ending in “ide” also contains prefixes like mono, di, tri, etc. the formula can be written using these prefixes instead of using charges. Example - dicarbon tetrahydride 2 carbons4 hydrogens C2H4C2H4

13. Writing Formulas of Oxyacids The “ic” acids HO HO HO HO HO N C Cl S P 1 2 1 2 3 3 3 3 4 4 nitric acid carbonic acid chloric acid sulfuric acid phosphoric acid

14. Writing Formulas of Oxyacids - “ous” and “ic” HNO 3 H 2 CO 3 HClO 3 H 2 SO 4 H 3 PO 4 remove 1 O, HNO 2 H 2 CO 2 HClO 2 H 2 SO 3 H 3 PO 3 remove 2 O’s HNO H 2 CO HClO H 2 SO 2 H 3 PO 2 add 1 O HNO 4 H 2 CO 4 HClO 4 H 2 SO 5 H 3 PO 5 Per_____ic _____ ic ______ ous hypo ____ ous

15. Family Substitutions Since elements in the same family have the same number of valence shell electrons they can sometimes be freely substituted for one another. Some examples of family substitutions are:

16. F Cl Br I S Se Te Po P As phosphate arsenate PO 4 3- AsO 4 3- sulfite selenite SO 3 2- SeO 3 2- hypochlorite hypofluorite hypobromite ClO 1- FO 1- BrO 1- perchoric acid perbromic acid HClO 4 HBrO 4

17. HNO 3 H 2 CO 3 HClO 3 H 2 SO 4 H 3 PO 4 Deriving Polyanions From the Oxyacids Each of the oxyacids below is made up of positive H ions and negative polyanions H 1+ NO 3 1- H 1+ CO 3 2- H 1+ ClO 3 1- H 1+ SO 4 2- H 1+ PO 4 3- nitrate carbonate chlorate sulfate phosphate

18. Deriving “ites” from “ates” NO 3 1- CO 3 2- ClO 3 1- SO 4 2- PO 4 3- remove 1 O, NO 2 1- CO 2 2- ClO 2 1- SO 3 2- PO 3 3- remove 2 O’s NO 1- CO 2- ClO 1- SO 2 2- PO 2 3- add 1 O NO 4 1- CO 4 2- ClO 4 1- SO 5 2- PO 5 3- per_____ate _____ ates ______ ites hypo ____ite

19. Writing Formulas - “ates” & “ites” Example 1 - aluminum carbonite Determine the charges of each particle Al 3+ CO 2 2- Compile groups of particles where the sum of positive charges equals the sum of negative particles. This way the total charge is zero. 2 Al 3+ equals 6+ and 3 CO 2 2- equals 6-. Al 3+ CO 2 2- 6+ 6- zero charge The answer is Al 2 (CO 2 ) 3

20. Naming “ates” & “ites” Example 1 - Fe(NO 3 ) 2 Since the positive ion has more than one oxidation state, its specific oxidation state must be determined. This is done by figuring out the charge on the negative particle. Since the total negative charge is 2- the positive charge on the single particle of Fe must be 2+. NO 3 1- Remember oxyacid is HNO 3 Fe 2+ 2+ 2- zero charge Using the “ous” “ic” method, the name is ferrous. Using the IUPAC method the name is iron(II). NO 3 1- is called nitrate. The answer is ferrous nitrate or iron(II) nitrate

21. Polyatomic Anions Containing H H 1+ ions can be added to any of the “ates”or “ites” with a 2- or 3- charges, reducing the overall charge of the newly formed polyatomic negative ion by one. This is shown below: CO 2 2- carbonite H 1+ HCO 2 1- hydrogen carbonite or bicarbonite If 2 H 1+ ions are added to any of the “ates”or “ites” with a 3- charge the resulting particle is named: PO 4 3- H 1+ phosphate dihydrogen phosphate H 2 PO 4 1-

22. Polyatomic Anions Containing H Write the formula for manganese (III) dihydrogen phosphite All formulae in this course consists of a positive particle (cation) and a negative particle (anion). The positive particle is Mn 3+. This is determined from the Roman Numeral in the name. Mn 3+ The negative particle is a dihydrogen phosphite. Phosphite is: PO 3 3- After adding 2 hydrogens to the phosphite the charge decreases to 1- and the anion becomes: H 2 PO 3 1- To balance charges 3 of these dihydrogen phosphites are needed. H 2 PO 3 1- The resulting answer is: Mn(H 2 PO 3 ) 3

23. Special Polyatomic Ions Water is H 1+ O 2- In some instances one of these H 1+ particles is lost. The anion OH 1- is formed. This particle is called hydroxide. OH 1- hydroxide Vinegar is a 5% solution of acetic acid. Its stucture is shown below. Sometimes it loses a H 1+ particle to water forming: C H H HC O O H HC2H3O2HC2H3O2 C H H HC O O C 2 H 3 O 2 1- acetate

24. Special Polyatomic Ions Sulfur is in group VIB and sulfate has the formula SO 4 2-. Chromium is in group VIA and chromate has the formula CrO 4 2- O CrOO O chromate If two of these chromate particles are combined one of the oxygen atoms is lost forming: O CrOO O O OO O O OO O O OO O dichromate Cr 2 O 7 2-

25. Special Polyatomic Ions Cyanide salts used in execution chambers in the U.S. are combined with sulfuric acid producing poisonous hydrogen cyanide gas. The anion cyanide has the formula: CN 1- cyanide cyanate salts contain the anion CNO 1- cyanate The prefix “thio” is used whenever an O particle is replaced by a sulfur particle since both O and S are in the same family. The anion thiocyanate has the formula: CNO 1- S thiocyanate Thiosulfate has the formula: SO 4 2- S 2 O 3 2-

26. Special Polyatomic Ions Manganese is in group VIIA and chlorine is in group VIIB. This allows some creative substitution. Chlorate has the formula ClO 3 1- chlorate perchlorate is ClO 4 1- perchlorate permanganate is MnO 4 1- permanganate A poisonous substance found in the leaves of rhubarb, potatoes, tomatoes and countless other plants is called oxalic acid. Its formula is H2C2O4H2C2O4 If this acid loses two H 1+ particles it creates the anion oxalate. C 2 O 4 2- oxalate

27. Special Polyatomic Ions The poisonous gas, nitrogen trihydride, NH 3, has the common name ammonia. H N H H ammonia When ammonia molecules dissolve in water and collide with water molecules they sometimes form H N H H H O H H O H HO H Notice the water molecule left a H 1+ particle behind forming ammonium - NH 4 1+

28. hydroxide OH 1- acetate C 2 H 3 O 2 1- chromate CrO 4 1- dichromate Cr 2 O 7 2- cyanide CN 1- cyanate CNO 1- thiosulfate S 2 O 3 2- permanganate MnO 4 1- oxalate C 2 O 4 2- oxalate C 2 O 4 2- ammonium NH 4 1+

29. Naming Chemical Compounds Are there more than 2 elements? YesNo It is a binary compound so the name must end in ide. It has polyatomic ions. Does the negative ion come from an oxyacid? YesNo Use the HO, HO table to determine its name. It must be a special polyatomic ion. Does the 1st element have more than 1 oxidation state? YesNo Name it. Determine the oxidation state of the 1st element and use it to name the compound.

30. Writing Formulas for Chemical Compounds Does the name end in “ide”? Yes No It is probably a binary compound.It’s made up of polyatomic ions. Does the negative ion come from an oxyacid? YesNo Use the HO, HO table to determine its formula. It must be a special polyatomic ion. Does the 1st element have more than 1 oxidation state? YesNo Determine the formula. Determine the oxidation state of the 1st element and use it to determine the formula.

31. Binary Acids Hydrochloric acidHCl Hydrobromic acid HBr Hydroiodic acidHI Hydrofluoric acidHF Hydrosulfuric acidH 2 S Hydroselenic acid H 2 Se