1 10 11 12 15 17 20 24 31 37 39 49 53 55 61 65 67 71 88 103

8.2 – Lattice Energy of Ionic Compounds The energy required to completely separate a mole of a solid ionic compound into its gaseous ions. It is always a + value The higher the Lattice Energy, the more stable the compound Cannot be measured directly, so use “Born-Haber Cycle” to calculate The energy associated with electrostatic interactions is governed by Coulomb’s law: The energy between two ions is directly proportional to the product of their charges and inversely proportional to the distance separating them. K = 8.99 x 109 Jm/C2 E =  Q1Q2 r

Some Lattice Energies:

Lattice Energy and the Formula of Ionic Compounds -Lattice Energy explains why multi-positive ions can form -Stability of “noble gas configuration” is not enough – Ionization energy is always + and increases with each subsequent IE -The lattice energy of the IONIC COMPOUND FORMED is high enough to compensate for the energy needed to remove the electrons from the cation

Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy to remove beyond that would be expended that cannot be overcome by lattice energies. Mg+2 is “worth” being formed when a compound like MgCl2 is formed Na+2 is not “worth” being formed because the lattice energy of NaCl2 is not high enough to compensate

Transition Metal Ions Lose valence s electrons first

Born – Haber Cycle Born-Haber Cycle – The Born-Haber cycle involves the formation of an ionic compound from the reaction of a metal with a non-metal (DHf). Born-Haber cycles are used primarily as a means of calculating lattice enthalpies, which cannot otherwise be measured directly. Relates Ionization Energy, Electron Affinity, and other properties in a Hess’ Law-type calculation

Example of Born-Haber Cycle Calculate the Lattice Energy of LiF: Goal: LiF(s) -> Li+1(g) + F-(g) Use: Enthalpy of sublimation Lithium = +155.2 kJ/mol Bond Dissociation Energy Fluorine Gas = +150.6 kJ/mol Ionization Energy of Lithium = +520 kJ/mol Electron Affinity of Fluorine = -328 kJ/mol Enthalpy of Formation of Lithium fluoride = -594.1 Reaction:

Example of Born-Haber Cycle Work:

The Born-Haber cycle can also be represented schematically DHf = IE + DHsub + 1/2DHdiss + EA + - LE

Another Example the of Born-Haber Cycle (not +1/-1) Calculate the Lattice Energy of CaCl2: Goal: CaCl2(s) -> Ca+2(g) + 2Cl-(g) Use: Enthalpy of sublimation Calcium = +121 kJ/mol Bond Dissociation Energy Chlorine Gas = +242.7 kJ/mol First Ionization Energy of Calcium= +589.5 kJ/mol Second Ionization Energy of Calcium = +1145 kJ/mol Electron Affinity of Chlorine = -349 kJ/mol Enthalpy of Formation of Calcium chloride= -795 kJ/mol Reaction:

Work:

8.3 – Covalent Bonding Electrons are shared Molecular Compounds have covalent bonds Lone Pairs = Pairs of electrons not in a covalent bond Lewis Structure – Representation of Covalent Bonding Octet Rule = Atoms bond to form a noble gas configuration (often 8 valence e-) Single Bond = 1 pair electrons Double Bond = 2 pairs electrons Triple Bond = 3 pairs electrons

8.3 Bond Length = Distance between nuclei of two covalently bonded atoms in a molecule. 1>2>3 Bond Energy = Energy needed to break a bond 3>2>1

8.4 - Electronegativity Ability to pull electrons; Pauli Scale (max 4.0) Polar Covalent = Unequal sharing of electrons – 0.3-1.7 difference Nonpolar Covalent = Equal sharing - <0.3 difference Ionic = Transfer of electrons - >1.7 difference (Some books give different ranges)

Comparison of Ionic and Molecular Compounds Ionic Bonds Covalent Bonds (Polar or Nonpolar) Very Strong Bonds Strong Bonds High mp/bp Low mp/bp Crystal Lattice Individual Molecule Conducts electricity when in water or melted Do not conduct electricity in water or when melted

8.4 – Bond Polarity and Electronegativity Although atoms often form compounds by sharing electrons, the electrons are not always shared equally. Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does. Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.

Polar Covalent Bonds When two atoms share electrons unequally, a bond dipole results. The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:  = Qr It is measured in debyes (D).

Polar Covalent Bonds H - F H - F The greater the difference in electronegativity, the more polar is the bond. d+ d- H - F H - F

8.5 – Drawing Lewis Structures Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.

Writing Lewis Structures Old method of connecting dots does not always work. Here is the other method: Find the sum of valence electrons of all atoms in the polyatomic ion or molecule. If it is an anion, add one electron for each negative charge. If it is a cation, subtract one electron for each positive charge. PCl3 5 + 3(7) = 26

Writing Lewis Structures The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds. Keep track of the electrons: 26  6 = 20

Writing Lewis Structures Fill the octets of the outer atoms. Keep track of the electrons: 26  6 = 20  18 = 2

Writing Lewis Structures Fill the octet of the central atom. Keep track of the electrons: 26  6 = 20  18 = 2  2 = 0

Writing Lewis Structures HCN If you run out of electrons before the central atom has an octet… …form multiple bonds until it does.

Lewis Structures HNO3 CO3-2

8.5 – Formal Charges Then assign formal charges. For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms. Subtract that from the number of valence electrons for that atom: The difference is its formal charge. Total formal charge on a neutral molecule = 0 Total formal charge on a polyatomic cation /anion= charge

Writing Lewis Structures The best Lewis structure… 1. Most times FOLLOW THE OCTET RULE 2. When more than one structure is possible: …is the one with the fewest formal charges. …puts a negative charge on the most electronegative atom.

8.6 - Resonance -Structure has more than one equal placement of electrons. This is the Lewis structure we would draw for ozone, O3. + -

Resonance But this is at odds with the true, observed structure of ozone, in which… …both O—O bonds are the same length. …both outer oxygens have a charge of 1/2.

Resonance One Lewis structure cannot accurately depict a molecule such as ozone. We use multiple structures, resonance structures, to describe the molecule.

Resonance Just as green is a synthesis of blue and yellow… …ozone is a synthesis of these two resonance structures. Ozone resonance structure =

Resonance Formate Ion: The electrons that form the second C—O bond in the double bonds below do not always sit between that C and that O, but rather can move among the two oxygens and the carbon. They are not localized, but rather are delocalized.

Resonance The organic compound benzene, C6H6, has two resonance structures. It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring.

8.7 - Exceptions to the Octet Rule There are three types of ions or molecules that do not follow the octet rule: Ions or molecules with an odd number of electrons. Ions or molecules with less than an octet. Ions or molecules with more than eight valence electrons (an expanded octet).

Odd Number of Electrons Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons. Ex – NO; NO2; ClO2

Incomplete Octet Consider BF3: OR Consider BF3: Giving boron a filled octet places a negative charge on the boron and a positive charge on fluorine. This would not be an accurate picture of the distribution of electrons in BF3. B stays at less than an octet!!

Incomplete Octet Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.

Incomplete Octet The lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom. (Group 13 is stable with incomplete octet)

Expanded Octet The only way PCl5 can exist is if phosphorus has 10 electrons around it. It is allowed to expand the octet of atoms on the 3rd row or below. Presumably d orbitals in these atoms participate in bonding.

Expanded Octet Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, the better structure puts a double bond between the phosphorus and one of the oxygens.

Expanded Octet This eliminates the charge on the phosphorus and the charge on one of the oxygens. The lesson is: When the central atom is on the 3rd row or below and expanding its octet eliminates some formal charges, do so.

8.8 – Strengths of Covalent Bonds Most simply, the strength of a bond is measured by determining how much energy is required to break the bond. This is the bond enthalpy = Bond Dissociation Energy The bond enthalpy for a Cl—Cl bond, D(Cl—Cl), is measured to be 242 kJ/mol.

Average Bond Enthalpies This table lists the average bond enthalpies for many different types of bonds. Average bond enthalpies are positive, because bond breaking is an endothermic process.

Average Bond Enthalpies NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than the C—H bond in chloroform, CHCl3.

Enthalpies of Reaction Yet another way to estimate H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed. In other words, Hrxn = (bond enthalpies of bonds broken) + -(bond enthalpies of bonds formed)

Enthalpies of Reaction CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) In this example: Broken: Four C—H bonds One Cl—Cl bond Made: One C—Cl One H—Cl bond Three C-H bonds

Enthalpies of Reaction Hrxn = [4D(C—H) + D(Cl—Cl) + -[3D(C—H) +D(C—Cl) + D(H—Cl)] = [4(413 kJ) + 242 kJ] + -[3(413 kJ) + 431 kJ + 328 kJ] = (1894kJ) + (-1998 kJ) = 104 kJ

Bond Enthalpy and Bond Length We can also measure an average bond length for different bond types. As the number of bonds between two atoms increases, the bond length decreases.