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 CationsIon Systematic

NameOlder Name

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) Mercurou

s

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+