Type II Binary Ionic Compounds Type II binary
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: n n n 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 n n 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.
Fe. Cl 2 n n 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)Cl 2
Fe. Cl 3 n n n Here we have a 3 - charge from the 3 Cl So, the Fe must have a charge of 3+ Fe(III)Cl 3
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 Fe 3+ Older Name Iron (III) Ferric Sn 4+ Tin (IV) Stannic Fe 2+ Iron (II) Ferrous Sn 2+ Tin (II) Stannous Cu 2+ Copper (II) Cupric Pb 4+ Lead (IV) Plumbic Cu+ Copper (I) Cuprous Pb 2+ Lead (II) Plumbous Co 3+ Cobalt (III) Cobaltic Hg 2+ Mercury (II) Mercuric Co 2+ Cobalt (II) Cobaltous Hg 22+ Mercury (I) Mercurous
Practice n n Cu. Cl n Cu (I) because Cl is 1 Fe 2 O 3 n Fe(III) because O is 2 Pb. Cl 4 n Pb(IV) because Cl is 1 Mn. O 2 n Mn(IV) because O is 2
What is the formula for each of the following? n Sn(IV) and Cl Sn. Cl 4 n Pb(II) and I n Pb. I 2 n Co(III) and O n Co 2 O 3 n
And n Cu(II) and SO 4 n n Cu(I) and SO 4 n n Cu. SO 4 Cu 2 SO 4 Fe(III) and NO 3 n Fe(NO 3)3
Review of Type II Binary Ionic Compounds n n 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 Zn. Cl 2 n n n 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+
Sn. O 2 n n n In this case O is always 2 So, the overall negative charge is 4 Therefore, Sn will have a charge of 4+
Fe 2(SO 4)3 n n n Since SO 4 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+
Ag 2 C 8 H 4 O 2 n n The C 8 H 4 O 2 always has a charge of 2 Therefore the Ag must be 1+ each so that we have a total of 2+
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