Binary Ionic Compounds

Binary ionic compounds are made up of only two elements. The positive ion, or cation, is an ion consisting of only one atom. The name of this ion is the same as the name of the element. The positive ion, or cation, is an ion consisting of only one atom. The name of this ion is the same as the name of the element. e.g. K + is a potassium ion; Mg 2+ is a magnesium ion. e.g. K + is a potassium ion; Mg 2+ is a magnesium ion. The negative ion, or anion, is an ion consisting of only one atom. The name of this ion is formed by changing the ending of the element to –ide. The negative ion, or anion, is an ion consisting of only one atom. The name of this ion is formed by changing the ending of the element to –ide. e.g. F - is a fluoride ion; O 2- is an oxide ion. e.g. F - is a fluoride ion; O 2- is an oxide ion.

For atoms in the s and p blocks, the charges of the ions may be determined from their position in the periodic table.

Transition elements can sometimes form cations with more than one charge. These cations are named by including their charge as a Roman numeral in parentheses. These cations are named by including their charge as a Roman numeral in parentheses. e.g. Cu + is the Copper (I) ion; e.g. Cu + is the Copper (I) ion; Cu 2+ is the Copper (II) ion. Cu 2+ is the Copper (II) ion.

Cations and anions must be combined such that the total positive charges equal the total negative charges. Example: Potassium bromide Example: Potassium bromide K is found in the first column. Its charge is therefore +1. K is found in the first column. Its charge is therefore +1. Bromine is found in Group 17. Bromide’s charge is therefore -1. Bromine is found in Group 17. Bromide’s charge is therefore -1. To produce a neutral compound, one K + is needed for every one Br -. To produce a neutral compound, one K + is needed for every one Br -. Therefore the formula is KBr. Therefore the formula is KBr.

Cations and anions must be combined such that the total positive charges equal the total negative charges. Example: Magnesium chloride Example: Magnesium chloride Mg is found in the second column. Its charge is therefore +2. Mg is found in the second column. Its charge is therefore +2. Chlorine is found in Group 17. Chloride’s charge is therefore -1. Chlorine is found in Group 17. Chloride’s charge is therefore -1. To produce a neutral compound, one Mg 2+ is needed for every two Cl -. To produce a neutral compound, one Mg 2+ is needed for every two Cl -. Therefore the formula is MgCl 2. Therefore the formula is MgCl 2.

Cations and anions must be combined such that the total positive charges equal the total negative charges. Example: Aluminum oxide Example: Aluminum oxide Al is found in the Group 13. Its charge is therefore +3. Al is found in the Group 13. Its charge is therefore +3. Oxygen is found in Group 16. Oxide’s charge is therefore -2. Oxygen is found in Group 16. Oxide’s charge is therefore -2. To produce a neutral compound, two Al 3+ are needed for every three O 2-. To produce a neutral compound, two Al 3+ are needed for every three O 2-. Therefore the formula is Al 2 O 3. Therefore the formula is Al 2 O 3.

Transition elements can often have more than one charge. This charge will be the same as the Roman numeral. Example: Iron (II) chloride Example: Iron (II) chloride The iron ion has a charge of +2. The iron ion has a charge of +2. Chlorine is found in Group 17. Chloride’s charge is therefore -1. Chlorine is found in Group 17. Chloride’s charge is therefore -1. To produce a neutral compound, one Fe 2+ is needed for every two Cl -. To produce a neutral compound, one Fe 2+ is needed for every two Cl -. Therefore the formula is FeCl 2. Therefore the formula is FeCl 2.

Naming binary compounds Name the ion with the cation first, the anion second. Examples: Name the ion with the cation first, the anion second. Examples: NaCl: Sodium chloride NaCl: Sodium chloride MgCl 2 : Magnesium chloride MgCl 2 : Magnesium chloride CuCl 2 : Copper (II) chloride CuCl 2 : Copper (II) chloride Fe 2 O 3 : Iron (III) oxide Fe 2 O 3 : Iron (III) oxide Note that the first word may or may not be capitalized, but the second word (the name of the anion) is never capitalized. Note that the first word may or may not be capitalized, but the second word (the name of the anion) is never capitalized.

For atoms that form more than one ion, the Roman numeral indicating the charge must appear as part of the name. Example: CuCl 2 Example: CuCl 2 Cu can have more than one charge. Assume that this is true for any transition element. Cu can have more than one charge. Assume that this is true for any transition element. Chlorine is in group 17, so the chloride ion must have a charge of -1. Chlorine is in group 17, so the chloride ion must have a charge of -1. Since there are two Cl - for every one copper ion, the copper ion must have a charge of +2. Since there are two Cl - for every one copper ion, the copper ion must have a charge of +2. Therefore, the name is Copper (II) chloride. Therefore, the name is Copper (II) chloride.

For atoms that form more than one ion, the Roman numeral indicating the charge must appear as part of the name. Example: Fe 2 O 3 Example: Fe 2 O 3 Fe can have more than one charge. Assume that this is true for any transition element. Fe can have more than one charge. Assume that this is true for any transition element. Oxygen is in group 16, so the oxide ion must have a charge of -2. Oxygen is in group 16, so the oxide ion must have a charge of -2. Since there are three 0 -2 for every two iron ions, the iron ions must have a charge of +3. Since there are three 0 -2 for every two iron ions, the iron ions must have a charge of +3. Therefore, the name is Iron (III) oxide. Therefore, the name is Iron (III) oxide.