1. Formulas & Nomenclature
Transition Metals
Formulas & Nomenclature

2. Transition metals with ONE charge
Silver - always a “+1”
Cadmium - always a “+2”
Zinc - always a “+2”

3. Transition Metals with MORE than ONE charge
In order to write the formula of a compound that contains a transition metal (other than silver, cadmium or zinc), you need to know the oxidation number.

4. One way the oxidation number is expressed in the name is by the use of a ROMAN NUMERAL
The roman numeral indicates the oxidation number of the transition metal

5. Name of metal + (Roman Numeral) + Name of
(I) NONMETAL +
(II) “-ide” ending
(III)
(IV)

6. Write the formula for iron (III) oxide
Step 1 - determine the oxidation numbers for each element
Fe O-2
The roman numeral (III) shows that the charge on the iron is “+3”. The charge of the oxide ion is found on the periodic table.

7. Write the formula for iron (III) oxide
*Step 2 - “cross charges” if necessary
Fe+3 O-2
The iron ion will lose three electrons while the oxide ion gains two. You must have 2 iron ions for every 3 oxide ions for the charges to balance:
Fe2O3

8. Write the formula for iron (III) oxide
Step 3 - evaluate your answer
Check to be sure the overall charge on the compound is ZERO
Fe2O3
2(+3) + 3(-2) = 0

9. Practice Problems
Write the correct formula for the following ionic compounds:
Copper (I) nitride
Copper (II) nitride
Lead (II) bromide
Lead (IV) bromide
Chromium (II) sulfide
Chromium (III) sulfide

10. Naming Ionic Compounds with Transition Metals
Transition metals and metals on the right side of the periodic table often have more than one oxidation number. To tell which oxidation number is used the name of the formula must indicate the charge. The charge is written as a Roman Numeral in parentheses after the metal cation’s name.

11. CuO Oxide’s oxidation number is always -2
There is one oxide ion, therefore the overall negative charge is -2
There must be an overall +2 charge to equal the negative charge.
There is one copper ion, therefore its charge must be +2

12. CuO 1 Copper (x) + 1 Oxygen (-2) = 0 x -2 = 0 x = +2
Name : Copper (II) oxide

13. SnCl4 Chlorine’s charge is always -1
There are 4 chloride ions, so the overall negative charge is -4
There must be an overall charge of +4 to equal the negative charge.
There is one tin ion, therefore its charge must be +4

14. SnCl4
1(x) + 4(-1) = 0
x = 0
x = +4
Name: Tin (IV) chloride

15. Fe3N2 Nitride’s charge is always -3
There are 2 nitride ions, so the overall negative charge is -6
There must be an overall positive charge of +6 to equal the negative charge
There are 3 iron ions, so the charge of one iron ion must be +2

16. Fe3N2
3(x) + 2(-3) = 0
3x = 0
3x = 6
x = 2
Name: Iron (II) nitride

17. Practice Problems Write the correct names for the following ionic compounds
PbI2
PbI4
SnF2
SnF4
CrI3
CuCl
Fe2O3

18. Stock vs Classical Names
ELEMENT
CATION
STOCK NAME
CLASSICAL NAME
Copper
Cu+1
Cu+2
Copper (I)
Copper (II)
Cuprous
Cupric
Iron
Fe+2
Fe+3
Iron (II)
Iron (III)
Ferrous
Ferric
Lead
Pb+2
Pb+4
Lead (II)
Lead (IV)
Plumbous
Plumbic
Mercury
Hg+1
Hg+2
Mercury (I)
Mercury (II)
Mercurous
Mercuric
Tin
Sn+2
Sn+4
Tin (II)
Tin (IV)
Stannous
stannic

19. Practice Problems
Fe2O3
PbCl2
Sn3P4
Hg3N