Chapter 9. Ionic Compounds
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1 Chapter 9 Bonding Ionic Compounds Formed between metal and nonmetal Ionic solids: ions are arranged in a regular lattice Strong forces: attraction of ions for each other 1
2 Lattice Energy A measure of the stability of an ionic solid Based on Coulomb s Law F = force q 1 and q 2 = charges r = distance between charges C = constant Forces between ions are stronger when there are Higher charges (most important) Smaller ions We will not cover the Born-Haber Cycle pp F q1q C r 2 2 Ionic Compounds Stronger forces between ions for Higher charges (most important) Smaller ions Strong forces => high MP Example Which MP goes with which compound? Compounds: LiF, MgO, LiCl Melting Points ( o C): 610, 845,
3 Covalent Compound Formed from two nonmetals Bonding is due to shared electrons Covalent bond: a bond in which two electrons are shared by two atoms A covalent compound contains only covalent bonds We represent covalent compounds with Lewis structures Look at HF Lewis Structures # valence electrons = Group number H: 1 valence electron F: 7 valence electrons 3
4 Lewis Structures Represent valence electrons as dots H: 1 valence electron F: 7 valence electrons Share electrons to give both elements full shell configuration (H = 2; F = 8) Two shared electrons are a covalent bond Bond represented by a line Lewis Structures Need orderly process for complex molecules Follow the rules below: 1. Draw a skeleton structure joining atoms by single bond. 2. Count the number of valence electrons, including the charge. 3. Deduct 2 electrons for each bond from step Distribute remaining electrons to give all atoms an octet of electrons (Octet rule) 4
5 Skeleton Structures You should be able to write skeleton structures From organic condensed structural formulas CH 2 ClCH 2 Cl CH 3 COOH One central atom NH 3 Lewis Structures 1. Draw a skeleton structure joining atoms by single bond 2. Count the number of valence electrons, including the charge. 3. Deduct 2 electrons for each bond from step Distribute remaining electrons to give all atoms an octet of electrons Examples: OH - CH 2 CH 2 CO CH 3 CHO NH 4 + 5
6 Bond Lengths Distance between two bonded atoms Bond lengths vary from compound to compound Average values below for carbon-carbon bonds C-C 154 pm C=C 134 pm C C 120 pm In general, Single bond > double bond > triple bond Polar and Nonpolar Bonds Electrons in bonds not necessarily shared equally The F atom attracts electrons more than H atom Get a polar bond e - shared unequally 6
7 Polar and Nonpolar Bonds We can picture the formation of a ions as a situation in which the bond is so polar that electrons have been completely transferred Nonpolar Polar Ionic Electronegativity Electronegativity: the ability of an atom to attract electrons when in chemical bond High EN = strong attraction for e - Low EN = weak attraction for e - Useful for making predictions of bond polarity Useful for understanding reactivity (or unreactivity) of elements 7
8 Electronegativity Difference Nature of a bond is related to the electronegativity difference between elements involved O H DEN = = 1.4 H-F DEN = = 1.9 Electronegativity Difference Rules of Thumb DEN = 0 Totally covalent bond DEN < 0.5 Essentially nonpolar bond (text doesn t do this) DEN >= 0.5 Polar covalent bond DEN >= 2.0 Ionic bond Questions Which is more polar: HF or HCl? Which bond has greater ionic character: BO or CO? Is the C-Se bond polar? 8
9 Bond Polarity Don t always have electronegativity tables available You should recognize that F, N, O, and Cl all have high EN. Almost all their bonds are polar C and H have about the same EN. The C-H bond is essentially nonpolar Resonance Lewis structure for formate ion, HCO 2 - Would you predict that both CO bonds have the same length? 9
10 Resonance The C-O bond lengths in the formate ion are equal Bond length: 126 pm Midway between C-O (135 pm) and C=O (120 pm) Actual structure is what we call a resonance hybrid of two separate resonance structures Resonance A resonance hybrid basically is a weakness of the Lewis Structure model Cannot use one structure to describe a molecule Use more than one Actual structure is intermediate between the two resonance structures Mule = horse donkey 10
11 Resonance If you have a choice of ways to distribute multiple bonds in a structure, then actual structure is resonance hybrid of the choices Example: azide ion N 3 - Example: carbonate ion CO 3 2- Benzene Important organic chemical, C 6 H 6 Skeletal structure: Frequent representation: 11
12 Formal Charges Book-keeping method to help describe distribution of electrons in a molecule Not real charges Frequently useful in organic chemistry Starting point for predicting chemical and physical properties Example: Formaldehyde Formal Charges H 2 C=O Draw circle around each atom, dividing bonds Two electrons per bond Each bonded atom is assigned one of them Both electrons in a lone pair are assigned to one atom Call these the assigned electrons Formal charge = - (# assigned e - - # valence e - ) C: 4 assigned, 4 valence Formal charge = 0 O: 6 assigned, 6 valence Formal charge = 0 H: 1 assigned, 1 valence Formal charge = 0 Sum of formal charges must = charge of species 12
13 Formal Charges Consider one of carbonate resonance structures Oxygen has 6 valence e - and carbon 4 Formal charge of oxygen atoms with 3 lone pairs is -1 This is as it should be Sum of formal charges = -2 Charge of ion = -2 Formal Charges Formal charges help gauge importance of resonance hybrid structures. If possible, don t want Positive charge on highly electronegative element Large separation of charge 13
14 Exceptions to Octet Rule The octet rule works best for second period elements (C, N, O) There are exceptions Compounds of B and Be Example: Diborane B 2 H 6 Exceptions to Octet Rule Odd-electron molecules Impossible to satisfy octet rule Example: NO Can write two Lewis structures with incomplete octets Use formal charges to choose the best one 5 assigned 6 assigned 5 valence 6 valence 6 assigned 5 assigned 5 valence 6 valence 14
15 Exceptions to the Octet Rule Expanded octets Examples: SF 6 How is this possible? 3 rd, 4 th, etc. period elements have empty d-orbitals available S: 1s 2 2s 2 2p 6 3s 2 3p 4 n quantum number = 3 for valence electrons If n = 3, l can be 0, 1, or 2. 3d orbitals are available for bonding Exceptions to Octet Rule For 3 rd period (and 4 th, 5 th, etc) elements the central atom can have more than four bonds or pairs of electrons if necessary Usually easy to identify Central atom bonded to more than four atoms Example: PF 5 Example: XeF 4 15
16 Bond Enthalpies Enthalpy required to break a bond Apply only to gases Average values Used to estimate enthalpies of compounds Bond Enthalpies ΔH = -[bonds formed bonds broken] ΔH = -[2BE(C=O) + 4BE(OH) -4BE(C-H)-2BE(O=O)] = -[2(803) + 4(467) 4(416) 2(498)] = -814 kj/mole 16
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