Ionic compounds with a transition metal: Almost all transition metals (in the centre of the periodic table) are able to form more than one cation. i.e.: Copper can form the cations Cu 1+, and Cu 2+ Iron can be Fe 2+ or Fe 3+
Roman Numerals I = 1 II = 2 III = 3 IV = 4 V = 5 VI = 6
Ionic compounds with transition metals – Nomenclature (naming) Name to formula: i.e. :copper (I) chloride 1.Write the elements’ symbols with their charges (the charge of the cation (the transition metal) is given in the brackets) Cu 1+ Cl - 2.If the charges cancel (add to zero), you are done Rewrite the formula CuCl.
Ionic compounds with transition metals – Nomenclature (naming) If the charges do not add to zero: Use the crossover method to determine the number of each atom required to have a neutral compound i.e. :iron (II) chloride Fe 2+ Cl -1 FeCl 2 See if you can reduce the subscripts. If not, you are done
Give it a try! What is the formula for: osmium (II) sulfide Os 2+ S 2- The charges add to zero so the answer is: OsS
Give it a try! What is the formula for: nickel (III) fluoride Don’t forget that the III (3) in the brackets tells us the charge of nickel Ni 3+ F 1- The charges do not add to zero, so crossover method: NiF 3
Ionic compounds with transition metals – Nomenclature (naming) Formula to name: The charge of the transition metal is written in brackets using Roman numerals 1.Write the name of the cation (the transition metal) 2.Use the inverse crossover method to find the charge of the transition metal 3.Write the charge next to the name of the cation in brackets (use Roman numerals) 4.Write the name of the anion with the ending –ide.
Ionic compounds with transition metals – Nomenclature (naming) i.e.: Fe 2 O 3 iron… (II or III?) Inverse crossover to find the charge of iron ( 2+ or 3+ ) This time the subscripts go up to become the charges Fe 2 O 3 So we have Fe 3+ = iron (III) and O 2- Make sure the charge found for the anion (the non- metal) is its actual charge according to the periodic table. If it is, you are done. The complete name is: iron (III) oxide
Give it a try! NiF 3 nickel …. (II or III?) Inverse crossover Ni 1 F 3 Ni 3+ F 1- 1- is the actual charge of fluoride so this works We have nickel (III) here The name is: nickel (III) fluoride
Ionic compounds with transition metals – Nomenclature (naming) PbO 2 lead… (II or IV?) Inverse crossover: Pb 1 O 2 Pb 2+ O 1- Is that the actual charge of the non-metal (oxygen)? No, so we have to multiply the charges of both atoms to have the actual (correct) charge on O Multiply both charges by 2: Pb 2+ x 2 O 1- x 2 Pb 4+ O 2- so it’s lead (IV) Complete name: lead (IV) oxide
Give it a try! FeO iron…(II) or (III)? Inverse crossover Fe 1 O 1 Fe 1+ O 1- *not the charge of O Multiply both charges by 2 to have the actual (correct) charge on O: Fe 1+x 2 O 1+ x 2 = Fe 2+ O 2- It is iron (II) Complete name: iron (II) oxide
Careful! We NEVER use Roman numerals for metals with one charge!
To do: Notes 6 Practice– Ionic Compounds with a Transition Metal