Ionic Bonding Ch 8 CVHS Chemistry EQ: How does the electron configuration of elements impact the Ionic Bonding?

Ionic Compounds Ionic Bonds Metallic bond Chemical compounds Held together by chemical bonds Ionic Compounds Electrons from a metal are transferred to a nonmetal Resulting ions held by electrostatic force Chemical Bonds Ionic Bonds: Metal & Nonmetal Metallic Bonds: Metal & Metal Covalent (molecular) Bonds: Nonmetal & Nonmetal Ionic Bonds Metallic bond

The Octet Rule & Why Ionic Bonds form Noble gases are extremely stable Each noble gas has eight valence electrons, except for helium, which has two The octet rule: atoms can become stable by having eight electrons in their outer energy level Ex: Neon, (or two electrons in the case of some of the smallest atoms). Elements become stable by attaining the same configuration as the noble gases.

The Octet Rule

Properties of ionic compounds Practice Problem: What bond would form between Ca and Br? Answer: CaBr2 Ca (4s2) + Br (4s2 4p5) 2 Br atoms take the 2 electrons from Ca and then both atoms have a noble gas configuration Well organized, tightly bound ions w/ a crystal lattice structure. Crystalline solids @ room temp. so it takes a lot of energy to melt them to break the bonds Electrolytes: dissolve in water or melt and then conduct electricity

Formation of an Ionic Compound Ionic Bonds are held together by the force of the oppositely charged particles attracting each other. Explain the formation of an ionic compound from the elements aluminum and oxygen: Aluminum: 3 Valence electrons Oxygen: 6 valence electrons To acquire noble gas configuration each oxygen takes 2 electrons and each aluminum gives 3 electrons To make the charges equal zero: Al2O3

Names and formulas for ionic compounds Formula Unit: Simplest ratio of atoms in an ionic compound Overall charge is 0 Monatomic Ions: one atom Charge can be determined by location on table Group 1A: 1+ Group 2A: 2+ Group 3A: 3+ Group 7A: 1-

Oxidation numbers Mg + Br Mg2+ (3s2) Br1- (4s2 3d10 4p6) Cation is written first, Charge is written as an exponent MgBr2 Subscripts represent # of atoms Multiply subscripts by exponents charge will equal 0 The charge on the ion # of electrons transferred Used to determine chemical formulas Multiply the oxidation # by the subscript, they must = 0.

Ionic Formulas & Oxidation Numbers Ionic Formula Rules: Metals always form “+” ions by losing one or more valence electrons from the “s” and/or “p” sublevels. Na = (Ne) 3s1 Lose the 3s e- Mg = (Ne) 3s2 Lose both 3s e- Al = (Ne) 3s2 3p1 Lose all 3 e- Nonmetals will gain one or more electrons to form a full outer valence by the Octet Rule. S = (Ne) 3s2 3p4 Gains 2e- in 3p Br = (Ar) 4s2 3d10 4p5 Gains e- in 4p Can form Co2+ (Ar) 4s2 3d5 Or Co3+ (Ar) 4s1 3d5

Ionic Formulas & Oxidation Numbers Transition metals will lose electrons to create a stable “d” configuration and lose the outermost “s” electrons. Note that a half-full sublevel is also stable, but less so than an empty or full one. Co = (Ar) 4s2 3d7 Can form Co2+ (Ar) 4s2 3d5 Or Co3+ (Ar) 4s1 3d5

Practice Now that we know the charges, let’s write Ionic Formulas (Be sure charges sum to zero!): Na++ Cl- NaCl Mg2+ + Br-  MgBr2 K++ O2-  K2O Al3++F- AlF3

Polyatomic ions A group of atoms that act as an ion. Formulas containing Polyatomic Ions: Na+ + OH- NaOH Na+ +SO42-Na2(SO4) Mg2++NO3-Mg(NO3)2 NH4++CO32-(NH4)2CO3