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)

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

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.

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

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

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

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

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-

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

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

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

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+

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

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+

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