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1 10 11 12 15 17 20 24 31 37 39 49 53 55 61 65 67 71 88 103. Q 1 Q 2 r. E = . 8.2 – Lattice Energy of Ionic Compounds. Lattice Energy: The energy required to completely separate a mole of a solid ionic compound into its gaseous ions. It is always a + value
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Q1Q2 r E = 8.2 – Lattice Energy of Ionic Compounds • Lattice Energy: 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 109Jm/C2
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:
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:
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)
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 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.
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 Writing Lewis Structures Old method of connecting dots does not always work. Here is the other method: 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 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.