Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Copyright©2000 by Houghton Mifflin Company. All rights reserved.
Bonds Forces that hold groups of atoms together and make them function as a unit. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Quartz grows in beautiful, regular crystals.
Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Bond Energies Bond breaking requires energy (endothermic).
Bond formation releases energy (exothermic). H = D(bonds broken)  D(bonds formed) energy required energy released Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 8.1: (a) The interaction of two hydrogen atoms. (b) Energy profile as a function of the distance between the nuclei of the hydrogen atoms. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Electronegativity The ability of an atom in a molecule to attract shared electrons to itself. Fluorine is the most electronegative element – 4.0 Francium is the least electronegative – 0.7 Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Three main types of bonds
Ionic Bonding Electrons are transferred from one atom to another Electronegativity difference: 1.7 and above Covalent Bonding Electrons are shared between atoms There are two types of covalent bonding: 1. Non-Polar Covalent Electrons are shared equally Electronegativity difference: 0.0 to 0.2 2. Polar Covalent Electrons are shared, but unequally Electronegativity difference: 0.3 to 1.6 Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Ionic Bonds Formed from electrostatic attractions of closely packed, oppositely charged ions. Formed when an atom that easily loses electrons reacts with one that has a high electron affinity. Hint: metal/non-metal!!! Ex. Na2O Na – 0.9; O – 3.5 Electronegativity difference: 2.6 – very ionic!!!! Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Ionic Bonds Q1 and Q2 = numerical ion charges r = distance between ion centers (in nm) Generally, the greater the charges on the ions, the stronger the bond!!!! Ex. Which bond is stronger – NaCl or MgO? MgO – has 2+ and 2- charge Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Polarity A molecule, such as HF, that has a center of positive charge and a center of negative charge is said to be polar, or to have a dipole moment. Means “slightly positive” Means “slightly negative” Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Question: Which of the following three bonds will be most polar  O-F, N-F, C-F? Answer: The most polar bond will be the bond in which the difference in electronegativity is the greatest. Therefore, the C-F bond will have the greatest electronegativity difference as C is the least electronegative among C, N, and O (It’s farther away from Fluorine – which is No. 1!!!) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 8.2: The effect of an electric field on hydrogen fluoride molecules. (a) When no electric field is present, the molecules are randomly oriented. (b) When the field is turned on, the molecules tend to line up with their negative ends toward the positive pole and their positive ends toward the negative pole. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 8.5: (a) The structure and charge distribution of the ammonia molecule. The polarity of the N—H bonds occurs because nitrogen has a greater electronegativity than hydrogen. (b) The dipole moment of the ammonia molecule oriented in an electric field. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 8.6: (a) The carbon dioxide molecule. (b) The opposed bond polarities cancel out, and the carbon dioxide has no dipole moment. When talking about molecules not bonds between atoms Non-polar molecule – no dipole moment Polar bond But…. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Isoelectronic Ions (O2, F, Na+, Mg2+, Al3+)
Ions containing the the same number of electrons (O2, F, Na+, Mg2+, Al3+) O2> F > Na+ > Mg2+ > Al3+ largest smallest Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Lattice Energy The change in energy when separated gaseous ions are packed together to form an ionic solid. M+(g) + X(g)  MX(s) Lattice energy is negative (exothermic) from the point of view of the system. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Formation of an Ionic Solid
1. Sublimation of the solid metal M(s)  M(g) [endothermic] 2. Ionization of the metal atoms M(g)  M+(g) + e [endothermic] 3. Dissociation of the nonmetal 1/2X2(g)  X(g) [endothermic] Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Formation of an Ionic Solid (continued)
4. Formation of X ions in the gas phase: X(g) + e  X(g) [exothermic] 5. Formation of the solid MX M+(g) + X(g)  MX(s) [quite exothermic] Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 8.8: The energy changes involved in the formation of solid lithium fluoride from its elements.

Figure 8.9: The structure of lithium fluoride.

Figure 8.10: Comparison of the energy changes involved in the formation of solid sodium fluoride and solid magnesium oxide. Mgo much greater exothermic reaction than Na2O Why?? 

Q1, Q2 = charges on the ions r = shortest distance between centers of the cations and anions Generally, the greater the charges on the ions, the greater the lattice energy… Ex. The lattice energies of FeCl3, FeCl2, and Fe2O3 are (in no particular order) -2631, -5359, and -14,774 kJ/mol. Match the appropriate formula to each lattice energy. And, explain…. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Q1, Q2 = charges on the ions r = shortest distance between centers of the cations and anions Generally, the greater the charges on the ions, the greater the lattice energy… Answer: Lattice energy is proportional to the charge of the cation x charge of the anion… Fe2O3  (3+)(2-) = -6 ( kJ/mol) FeCl3  (3+)(1-) = -3 (-5359 kJ/mol) FeCl2  (2+)(1-) = -2 (-2631 kJ/mol) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Example 2: Which has the most exothermic lattice energy, NaCl or KCl? NaCl has the most exothermic lattice energy since all of the ions in both compounds have either -1 or +1 charges (the distance between the charges needs to be considered). Na+ is smaller than K+. So, the distance between the centers of the sodium and chlorine ions is less than the distance between the potassium and chloride ions So…. Answer is: NaCl Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Localized Electron Model
A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Localized Electron Model
1. Description of valence electron arrangement (Lewis structure). 2. Prediction of geometry (VSEPR model). 3. Description of atomic orbital types used to share electrons or hold long pairs. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Lewis Structure Shows how valence electrons are arranged among atoms in a molecule. Reflects central idea that stability of a compound relates to noble gas electron configuration. The Octet Rule states: “chemical compounds tend to form so that each atom, by gaining, losing, or sharing electrons, has an OCTET, which is 8 electrons in its highest energy level!” Most main group elements form bonds according to the Octet Rule. Hydrogen is an exception as it only requires 2_ electrons to completely fill its outer shell. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Comments About the Octet Rule
2nd row elements C, N, O, F observe the octet rule. 2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive. 3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals. When writing Lewis structures, satisfy octets first, then place electrons around elements having available d orbitals. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Determining Lewis Structures Ex. AsF3
Count the valence electrons of each atom and then add them together. As – 1 atom x 5 valence electrons F - 3 atoms x 7 valence electrons Total valence electrons: 26 Copyright©2000 by Houghton Mifflin Company. All rights reserved.

2. Draw the skeletel structure of the molecule and place two electrons (a dash) between each pair of bonded atoms. If the molecule contains three or more atoms, the least electronegative will usually occupy the central position. (If carbon is present, it is always in the middle; if hydrogen is present, it is never in the middle!) F AS F F Copyright©2000 by Houghton Mifflin Company. All rights reserved.

2. Add electrons to the surrounding atoms first until each has a complete outer shell (remember, hydrogen needs 2 electrons, all others need 8… F AS F F Copyright©2000 by Houghton Mifflin Company. All rights reserved.

4. Add the remaining electrons to the central atom, then look at the central atom: If the central atom has a complete octet, you are finished!! If the central atom has too few electrons, remove an electron pair from an outer atom and add another bond between the outer atom and the central atom. Do this until the central atom has a complete octet. If the central atom has more than 8 electrons, that is okay under certain circumstances. Leave them on the central atom. F AS F F Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Resonance Occurs when more than one valid Lewis structure can be written for a particular molecule. These are resonance structures. The actual structure is an average of the resonance structures. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Bond Energy Calculations
Example: Using Table 8.4, estimate ∆H (bond energy) for the following reaction: 2C2H O2  4CO H2O H = D(bonds broken)  D(bonds formed) {2[1(C-C) + 6(C-H)] + 7(O=O)} – {4(2)(C=O) + 6(2)(O-H)} {2[1(347 kJ) + 6(413 kJ)] + 7(495 kJ)} – {4(2)(799 kJ) + 6(2)(467 kJ)} 6637 kJ – kJ = kJ Exothermic!!! Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Formal Charge The difference between the number of valence electrons (VE) on the free atom and the number assigned to the atom in the molecule. We need: 1. # VE on free neutral atom 2. # VE “belonging” to the atom in the molecule Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Formal Charge To calculate the formal charge we need to: 1. Take the sum of the loan pair electrons and ½ of the shared electrons. This is the number of valence electrons assigned to the atom in the molecule 2. Subtract the number of assigned electrons from the number of valence electrons on the free, neutral atom to obtain the formal charge Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Formal Charge Atoms in molecules try to achieve formal charges as close to zero as possible Any negative formal charges are expected to reside on the most electronegative atoms Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Formal Charge Single-bonded O Triple-bonded O Free O atom = 6 v.e. Free O atom = 6 v.e. Single-bond O: 6 +1/2(2) = 7v.e. Triple-bond O: 2 + ½(6) = 5 v.e. Oxygen’s single-bond atom = -1 f.c. Oxygen’s triple-bond atom +1 f.c Free C atom Double-bonded O Free C atom = 4 v.e. Free O atom = 6 v.e. Carbon: ½(8) = 4v.e. double-bond O: 4 + ½(2) = 6 v.e. Carbon atom = O f.c. Oxygen’s double-bond atom O f.c Not the best Copyright©2000 by Houghton Mifflin Company. All rights reserved. Good!!

VSEPR Model (3D geometry)
The structure around a given atom is determined principally by minimizing electron pair repulsions. It states: “ repulsion between the sets of valence-level electrons surrounding an atom causes these sets to be oriented as FAR APART AS POSSIBLE!!! Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Predicting a VSEPR Structure
1. Draw Lewis structure. 2. Put pairs as far apart as possible. 3. Determine positions of atoms from the way electron pairs are shared. 4. Determine the name of molecular structure from positions of the atoms. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

VSEPR is tetrahedral (4 “sides” of electron pairs)
Figure 8.14: The molecular structure of methane. The tetrahedral arrangement of electron pairs produces a tetrahedral arrangement of hydrogen atoms. VSEPR is tetrahedral (4 “sides” of electron pairs) Molecular geometry (shape) is also tetrahedral

Figure 8.15:(a) The tetrahedral arrangement of electron pairs around the nitrogen atom in the ammonia molecule. (b) Three of the electron pairs around nitrogen are shared with hydrogen atoms as shown and one is a lone pair. Although the arrangement of electron pairs is tetrahedral, as in the methane molecule, the hydrogen atoms in the ammonia molecule occupy only three corners of the tetrahedron. A lone pair occupies the fourth corner. (c) Note that molecular geometry is trigonal pyramidal, not tetrahedral. VSEPR is tetrahedral Molecular geometry (shape) is trigonal pyramidal Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 8.16: (a) The tetrahedral arrangement of the four electron pairs around oxygen in the water molecule. (b) Two of the electron pairs are shared between oxygen and the hydrogen atoms and two are lone pairs. (c) The V-shaped molecular structure of the water molecule. Molecular geometry (shape) is BENT! VSEPR is TETRAHEDRAL H -- O -- H Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 8.17: The bond angles in the CH4, NH3, and H2O molecules.
VSEPR – All Tetrahedral Different shapes cause slight difference in bond angle (nuclear pull from only one or both atoms) Copyright©2000 by Houghton Mifflin Company. All rights reserved. Tetrahedral Trigonal Pyramidal Bent

Figure 8.20: Three possible arrangements of the electron pairs in the I3- ion. 3 loan pairs will orient as far apart as possible, so are in the 120 ° equatorial domain, not 90 ° axial I3 – 22 v.e. I – I – I Molecular Geometry (shape we see) is LINEAR VSEPR is TRIGONAL BIPYRAMIDAL Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Octahedral electron arrangement for Xe
Eight faces, six vertices. Each face is an equilateral triangle!!

Figure 8.19: Possible electron-pair arrangements for XeF4.
Lone pairs repulse so orient 180 degrees apart! Molecular geometry (shape) is SQUARE PLANAR VSEPR is OCTAHEDRAL Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Example: SF4 Draw the Lewis structure Identify the VSEPR electron-domain geometry Identify the molecular geometry (shape) F Valence = 34 F – S – F F b) Trigonal bipyramidal c) See-Saw Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Example: BrF5 Draw the Lewis structure Identify the VSEPR electron-domain geometry Identify the molecular geometry (shape) F F Valence electrons: 42 F – Br – F F b) Octahedral c) Square pyramidal Copyright©2000 by Houghton Mifflin Company. All rights reserved.

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