1. Type II Binary Ionic Compounds
Type II binary ionic compounds contain a metal that can form more than one type of cation.
(aka Transition Metals)

2. Some metals are predictable:
Group 1 alkali metals always form cations
Group 2 alkaline earth metals always form 2+ cations
Aluminum always form 3+ cations

3. Transition Metals can have more than one cation
Roman numerals are used to determine which cation is present.
We can determine the charge on the cation by looking at the anion whose charge doesn’t change.

4. FeCl2 Cl always has a 1- charge
So, if the compound has two Cl present the total negative charge is 2-
But, the compound must be neutral so the Fe must have a charge of 2+ to equal out the Cl
It is written as: Fe(II)Cl2

5. FeCl3 Here we have a 3- charge from the 3 Cl
So, the Fe must have a charge of 3+
Fe(III)Cl3

6. The Roman Numeral tells the charge on the ion, not the number of ions present

7. Common Type II Cations Ion Systematic Name Older Name Fe3+ Iron (III)
Ferric
Sn4+
Tin (IV)
Stannic
Fe2+
Iron (II)
Ferrous
Sn2+
Tin (II)
Stannous
Cu2+
Copper (II)
Cupric
Pb4+
Lead (IV)
Plumbic
Cu+
Copper (I)
Cuprous
Pb2+
Lead (II)
Plumbous
Co3+
Cobalt (III)
Cobaltic
Hg2+
Mercury (II)
Mercuric
Co2+
Cobalt (II)
Cobaltous
Hg22+
Mercury (I)
Mercurous

8. Practice CuCl Cu (I) because Cl is 1- Fe2O3 Fe(III) because O is 2-
PbCl4
Pb(IV) because Cl is 1-
MnO2
Mn(IV) because O is 2-

9. What is the formula for each of the following?
Sn(IV) and Cl
SnCl4
Pb(II) and I
PbI2
Co(III) and O
Co2O3

10. And
Cu(II) and SO4
CuSO4
Cu(I) and SO4
Cu2SO4
Fe(III) and NO3
Fe(NO3)3

11. Review of Type II Binary Ionic Compounds
The compound must be neutral
The anions will always be negative and they will always be the same
The cations will change – they are transition metals
We can determine the charge on the cation by finding the charge on the anion first

12. Let’s take ZnCl2 What is the charge on the Zn?
Since Cl is always 1- there is a 2- charge on the compound
So, that means Zn must be 2+

13. SnO2 In this case O is always 2- So, the overall negative charge is 4-
Therefore, Sn will have a charge of 4+

14. Fe2(SO4)3
Since SO4 is always 2- we have a total of 6- charge on the anion
Therefore, the Iron must have 6+ charge all together
Since there are 2 Fe the charge on each must be 3+

15. Ag2C8H4O2 The C8H4O2 always has a charge of 2-
Therefore the Ag must be 1+ each so that we have a total of 2+