Naming Ions and Naming and Writing Ionic Compounds

Monoatomic ions cations lose valence electrons and therefore are positively charged –transition metals form more than one cation with different ionic charges –charge is determined from the # of electrons lost Fe 2+ lost two electrons Fe 3+ lost three electrons –naming Stock system –a Roman numeral is used to show the charge of the transition metal »Fe 2+ is iron(II) and Fe +3 is iron(III) Naming Ions

Classical system –the older name of the element is used to form the name of the cation »ferrum is Latin for iron the suffix -ous is used to name the cation with the lower charge Fe 2+ is the ferrous ion the suffix -ic is used to name the cation with the higher charge Fe +3 is the ferric ion

TermDefinition Ag 1+ silver Cd 2+ cadmium Zn 2+ zinc Au 1+ gold(I)/aurous Au 3+ gold(III)/auric Co 2+ cobalt(II)/cobaltous Co 3+ cobalt(III)/cobaltic Cr 2+ chromium(II)/chromous Cr 3+ chromium(III)/chromic Cu 1+ copper(I)/cuprous Cu 2+ copper(II)/cupric Fe 2+ iron(II)/ferrous Fe 3+ iron(III)/ferric Hg 2 2+ mercury(I)/mercurous Hg 2+ mercury(II)/mercuric Mn 2+ manganese(II) Mn 3+ manganese(III) Mn 4+ manganese(IV) Ni 2+ nickel(II)/nickelous Ni 3+ nickel(III)/nickelic Pb 2+ lead(II)/plumbous Pb 4+ lead(IV)/plumbic H 1+ hydrogen H 1- hydride

– anions gain valence electrons and therefore are negatively charged change ending to -ide

Polyatomic ions composed of more than one atom that form a single unit/ion with a charge –most end with -ite or –ate the –ite ending indicates one less oxygen than the –ate ending –ionic charge is the same for both

Naming ionic compounds Binary Compounds –cation is written first, followed by the anion with and –ide ending Cs 2 O  cesium oxide SrF 2  strontium fluoride CuO  copper(II) oxide –oxygen is always 2- and therefore copper will be 2+ Cu 2 O  copper(I) oxide –oxygen is 2- and therefore needed two copper atoms with 1- charge

–SnF 2  ? fluorine is always 1- and therefore tin will be 2+ –Stock system  tin(II) fluoride –Traditional system  stannous fluoride –SnS 2  ? sulfur is always 2- and therefore tin will be 4+ –Stock system  tin(IV) sulfide –Traditional system  stannic sulfide

Writing formulas for ionic compounds write symbol of cation and then anion add subscripts to balance the charges –calcium bromide Ca 2+ and Br 1- is CaBr 2 –potassium sulfide K +1 and S 2- is K 2 S –iron(III) oxide Fe +3 and O 2- is Fe 2 O 3

crisscross method the ionic charge number of each ion is crossed over and becomes the subscript for the other ion

Compounds with polyatomic ions –do same as binary compounds use the name of the polyatomic ion with the mono atomic ion –calcium nitrate  ? »Ca is 2+ and nitrate is 1- Ca(NO 3 ) 2 –Li 2 CO 3  ? »lithium carbonate

Names to Formulas aluminum chloride Al Cl 3 2. Determine number of ions 1. Write symbols of elements Final Formula x(+3) + y(-1) = 0 1 For aluminum, a 1 is multiplied times the +3 to give a +3 charge If there is only one atom the “1” is not shown If there is no Roman Numeral, you must know the oxidation number of the positive ion. +3 If there is no Roman Numeral, you must know the oxidation number of the positive ion. What is the Lowest Common Multiple “LCM” of 3 and 1. Do not worry about the sign(+/-) (Al ) x (Cl -1 ) y What is the Lowest Common Multiple “LCM” of 3 and 1. Do not worry about the sign(+/-) For chloride, a 3 is multiplied times the -1 to give a -3 charge 1(+3) + 3(-1) = 0 Remember all compounds are neutral, thus the total positive charge must equal the total negative charge Next This formula says that the +3 charge of one Al atom will cancel the-3 charge from 3 Cl atoms

Formula to Name Fe(NO 3 ) 3 Choose the correct name for the compound 1. Iron trinitrate 2. iron(I) nitrate 3. iron(III) nitrite 4. iron(III) nitrate 5. none of the above next problem Polyatomic IonsPeriodic Chart

Name to Formula sodium chlorite Choose the correct formula for the compound 1. NaCl 2. NaClO 3. NaClO 2 4. Na(ClO) 2 5. none of the above next problem Prefixes Periodic Chart