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11/9/2011 Chapter 9 Bonding Ionic Compounds • Formed between metal and nonmetal: – Example: Mg (s) + Cl2 (g) ------> MgCl2 (s) • Ionic solids: ions are arranged in a regular lattice • Strong forces due to attraction of ions for each other – For example, melting points generally are high 1 11/9/2011 Lattice Energy • A measure of the stability of an ionic solid • Based on Coulomb’s Law – – – – q q F C 1 22 r F = force q1 and q2 = 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 369 - 371 Ionic Compounds • Stronger forces between ions for – Higher charges (most important) – Smaller ions • Strong forces => high MP • Example: Three compounds and three melting points are shown below. • Which MP goes with which compound? – Compounds: – Melting Points (oC): LiF, 610, MgO, LiCl 845, 2800 2 11/9/2011 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 Lewis Structures • Look at HF • # valence electrons = Group number – H: 1 valence electron – F: 7 valence electrons 3 11/9/2011 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 1. 4. Distribute remaining electrons to give all atoms an octet of electrons (Octet rule) 4 11/9/2011 Skeleton Structures • You should be able to write skeleton structures – From organic condensed structural formulas • CH2ClCH2Cl • CH3COOH – One central atom • NH3 Lewis Structures 1. 2. 3. 4. Draw a skeleton structure joining atoms by single bond Count the number of valence electrons, including the charge. Deduct 2 electrons for each bond from step 1. Distribute remaining electrons to give all atoms an octet of electrons • Examples: – OH– CO – CH2CH2 – CH3CHO – NH4+ 5 11/9/2011 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 11/9/2011 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 7 11/9/2011 Electronegativity Difference • Nature of a bond is related to the electronegativity difference between elements involved – O–H – H-F DEN = 3.5 – 2.1 = 1.4 DEN = 4.0 – 2.1 = 1.9 Electronegativity Difference • Rules of Thumb – – – – DEN = 0 DEN < 0.5 DEN >= 0.5 DEN >= 2.0 Totally covalent bond Essentially nonpolar bond (text doesn’t do this) Polar covalent bond 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 11/9/2011 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 • Earlier we did Lewis structure for formate ion, HCO2- • Would you predict that both CO bonds have the same length? 9 11/9/2011 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 11/9/2011 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 N3• Example: carbonate ion CO32- Benzene • Important organic chemical, C6H6 • Skeletal structure: • Frequent representation: 11 11/9/2011 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 Formal Charges • Example: Formaldehyde H2C=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: • O: • H: 4 assigned, 4 valence 6 assigned, 6 valence 1 assigned, 1 valence Formal charge = 0 Formal charge = 0 Formal charge = 0 – Sum of formal charges must = charge of species 12 11/9/2011 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 11/9/2011 Formal Charges • Look at two important examples from Organic Chem • What is formal charge of carbon on both structures? – What do these formal charges tell us? Formal Charges • Formal charges on carbon are the below • This implies that each species must be an ion – CH3+ (a carbocation) – CH3- (a carboanion) • Carbocations and carboanions are very important in Organic 14 11/9/2011 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 B2H6 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 5 valence 6 assigned 6 valence 6 assigned 5 assigned 5 valence 6 valence 15 11/9/2011 Exceptions to the Octet Rule • Expanded octets • Examples: SF6 • How is this possible? – 3rd, 4th, etc. period elements have empty d-orbitals available – S: 1s22s22p63s23p4 – 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 3rd period (and 4th, 5th, 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: PF5 • Example: XeF4 16 11/9/2011 Bond Enthalpies • Enthalpy required to break a bond • Apply only to gases • Average values Bond Enthalpies • ΔH = -[bonds formed – bonds broken] • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) • ΔH = -[2BE(C=O) + 4BE(OH) -4BE(C-H)-2BE(O=O)] • = -[2(803) + 4(467) – 4(416) – 2(498)] • = -814 kJ/mole 17