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:
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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
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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
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Cl always has a 1- charge
So, if the compound has two Cl present the
total negative charge is 2But, 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
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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
Ion
Systematic
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
Older
Name
Practice
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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?
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Sn(IV) and Cl
SnCl4
Pb(II) and I
 PbI2
Co(III) and O
 Co2O3
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And
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Cu(II) and SO4
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Cu(I) and SO4
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CuSO4
Cu2SO4
Fe(III) and NO3

Fe(NO3)3
Review of Type II Binary Ionic
Compounds
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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
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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
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In this case O is always 2So, the overall negative charge is 4Therefore, Sn will have a charge of 4+
Fe2(SO4)3
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Since SO4 is always 2- we have a total of 6charge 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
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The C8H4O2 always has a charge of 2Therefore the Ag must be 1+ each so that
we have a total of 2+