Formulas & Nomenclature Transition Metals Formulas & Nomenclature

Transition metals with ONE charge Silver - always a “+1” Cadmium - always a “+2” Zinc - always a “+2”

Transition Metals with MORE than ONE charge In order to write the formula of a compound that contains a transition metal (other than silver, cadmium or zinc), you need to know the oxidation number.

One way the oxidation number is expressed in the name is by the use of a ROMAN NUMERAL The roman numeral indicates the oxidation number of the transition metal

Name of metal + (Roman Numeral) + Name of (I) NONMETAL + (II) “-ide” ending (III) (IV)

Write the formula for iron (III) oxide Step 1 - determine the oxidation numbers for each element Fe+3 O-2 The roman numeral (III) shows that the charge on the iron is “+3”. The charge of the oxide ion is found on the periodic table.

Write the formula for iron (III) oxide *Step 2 - “cross charges” if necessary Fe+3 O-2 The iron ion will lose three electrons while the oxide ion gains two. You must have 2 iron ions for every 3 oxide ions for the charges to balance: Fe2O3

Write the formula for iron (III) oxide Step 3 - evaluate your answer Check to be sure the overall charge on the compound is ZERO Fe2O3 2(+3) + 3(-2) = 0

Practice Problems Write the correct formula for the following ionic compounds: Copper (I) nitride Copper (II) nitride Lead (II) bromide Lead (IV) bromide Chromium (II) sulfide Chromium (III) sulfide

Naming Ionic Compounds with Transition Metals Transition metals and metals on the right side of the periodic table often have more than one oxidation number. To tell which oxidation number is used the name of the formula must indicate the charge. The charge is written as a Roman Numeral in parentheses after the metal cation’s name.

CuO Oxide’s oxidation number is always -2 There is one oxide ion, therefore the overall negative charge is -2 There must be an overall +2 charge to equal the negative charge. There is one copper ion, therefore its charge must be +2

CuO 1 Copper (x) + 1 Oxygen (-2) = 0 x -2 = 0 x = +2 Name : Copper (II) oxide

SnCl4 Chlorine’s charge is always -1 There are 4 chloride ions, so the overall negative charge is -4 There must be an overall charge of +4 to equal the negative charge. There is one tin ion, therefore its charge must be +4

SnCl4 1(x) + 4(-1) = 0 x - 4 = 0 x = +4 Name: Tin (IV) chloride

Fe3N2 Nitride’s charge is always -3 There are 2 nitride ions, so the overall negative charge is -6 There must be an overall positive charge of +6 to equal the negative charge There are 3 iron ions, so the charge of one iron ion must be +2

Fe3N2 3(x) + 2(-3) = 0 3x - 6 = 0 3x = 6 x = 2 Name: Iron (II) nitride

Practice Problems Write the correct names for the following ionic compounds PbI2 PbI4 SnF2 SnF4 CrI3 CuCl Fe2O3

Stock vs Classical Names ELEMENT CATION STOCK NAME CLASSICAL NAME Copper Cu+1 Cu+2 Copper (I) Copper (II) Cuprous Cupric Iron Fe+2 Fe+3 Iron (II) Iron (III) Ferrous Ferric Lead Pb+2 Pb+4 Lead (II) Lead (IV) Plumbous Plumbic Mercury Hg+1 Hg+2 Mercury (I) Mercury (II) Mercurous Mercuric Tin Sn+2 Sn+4 Tin (II) Tin (IV) Stannous stannic

Practice Problems Fe2O3 PbCl2 Sn3P4 Hg3N