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9-1 Dr. Wolf’s CHM 101 Chapter 9 Models of Chemical Bonding.

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Presentation on theme: "9-1 Dr. Wolf’s CHM 101 Chapter 9 Models of Chemical Bonding."— Presentation transcript:

1 9-1 Dr. Wolf’s CHM 101 Chapter 9 Models of Chemical Bonding

2 9-2 Dr. Wolf’s CHM 101 Models of Chemical Bonding 9.1 Atomic Properties and Chemical Bonds 9.2 The Ionic Bonding Model 9.3 The Covalent Bonding Model 9.4 Between the Extremes: Electronegativity and Bond Polarity 9.5 An Introduction to Metallic Bonding

3 9-3 Dr. Wolf’s CHM 101 A general comparison of metals and nonmetals

4 9-4 Dr. Wolf’s CHM 101 Types of Chemical Bonding 1. Metal with nonmetal: electron transfer and ionic bonding 2. Nonmetal with nonmetal: electron sharing and covalent bonding 3. Metal with metal: electron pooling and metallic bonding

5 9-5 Dr. Wolf’s CHM 101 The three models of chemical bonding

6 9-6 Dr. Wolf’s CHM 101 Lewis Electron-Dot Symbols For main group elements - Example: Nitrogen, N, is in Group 5A and therefore has 5 valence electrons. N:... The A group number gives the number of valence electrons. Place one dot per valence electron on each of the four sides of the element symbol. Pair the dots (electrons) until all of the valence electrons are used.

7 9-7 Dr. Wolf’s CHM 101 Lewis electron-dot symbols for elements in Periods 2 and 3

8 9-8 Dr. Wolf’s CHM 101 Ionic Bonding Ionic bonding results when there is a transfer of electrons between two atoms. Each atom achieves a full outer level of electrons. For many atoms in the 2nd and 3rd period, this would be 8 electrons, known as the octet rule.

9 9-9 Dr. Wolf’s CHM 101 SAMPLE PROBLEM 9.1Depicting Ion Formation PLAN: SOLUTION: PROBLEM:Use partial orbital diagrams and Lewis symbols to depict the formation of Na + and O 2- ions from the atoms, and determine the formula of the compound. Draw orbital diagrams for the atoms and then move electrons to make filled outer levels. It can be seen that 2 sodiums are needed for each oxygen. 3s3p Na 3s3p Na 2s2p O 2s2p O 2- 2 Na + : Na + O. :... 2Na + + O 2- : : :: Na 2 O

10 9-10 Dr. Wolf’s CHM 101 Electron configurations Li 1s 2 2s 1 Orbital diagrams Lewis electron-dot symbols +F 1s 2 2s 2 2p 5 Li + 1s 2 +F - 1s 2 2s 2 2p 6 Three ways to represent the formation of Li + and F - through electron transfer. Li 1s2s2p F 1s2s2p + Li + 1s2s2p F-F- 1s2s2p +. +F:: : Li. Li + +F - :: : :

11 9-11 Dr. Wolf’s CHM 101 Born-Haber Cycle Lattice Energy - Energy released when ions come together forming an ionic solid. Remember Hess’s law states that the enthalpy change between two states is the same as the sum of enthalpies in a multistep process that goes between the same two states.  H 0 f =   H 0 elements to atoms +   H 0 ions from atoms +  H 0 lattice So from this one equation if all of the  H 0 ‘s except one are known, e.g.  H 0 lattice, the value can be calculated.

12 9-12 Dr. Wolf’s CHM 101 The Born-Haber cycle for lithium fluoride

13 9-13 Dr. Wolf’s CHM 101 Trends in lattice energy

14 9-14 Dr. Wolf’s CHM 101 Covalent Bonding Covalent bonding results when two electrons are shared in an orbital between two atoms. Each atom achieves a full outer level of electrons resulting in a lower energy system. The pair of electrons used are called the shared or bonding pair. In terms of the octet rule, this pair of electrons counts for both atoms in completing the octet. The electron pairs that are not involved in bonding belong only to the atom with which they are associated. These are called lone pairs. BOND ORDER - When only one pair of electrons are shared between two atoms, it’s called a single bond. If two pairs of electrons are shared covalently between two atoms, it’s called a double bond; three pairs, triple bond.

15 9-15 Dr. Wolf’s CHM 101 Covalent bond formation in H 2.

16 9-16 Dr. Wolf’s CHM 101 The attractive and repulsive forces in covalent bonding.

17 9-17 Dr. Wolf’s CHM 101 Bond Energy - The amount of energy required to break a bond. The greater the energy, the stronger the bond. Bond breaking is an endothermic process, so bond breaking enthalpies are positive.

18 9-18 Dr. Wolf’s CHM 101 Bond Length - In general, the closer the electrons are held by the atoms, the shorter the bond length and the higher the bond energy. Multiple bonds result in stronger, shorter bonds.

19 9-19 Dr. Wolf’s CHM 101 SAMPLE PROBLEM 9.2Comparing Bond Length and Bond Strength PROBLEM: PLAN: SOLUTION: Using the periodic table, but not Tables 9.2 and 9.3, rank the bonds in each set in order of decreasing bond length and bond strength: (a) S - F, S - Br, S - Cl(b) C = O, C - O, C O (a) The bond order is one for all and sulfur is bonded to halogens; bond length should increase and bond strength should decrease with increasing atomic radius. (b) The same two atoms are bonded but the bond order changes; bond length decreases as bond order increases while bond strength increases as bond order increases. (a) Atomic size increases going down a group. Bond length: S - Br > S - Cl > S - F Bond strength: S - F > S - Cl > S - Br (b) Using bond orders we get Bond length: C - O > C = O > C O Bond strength: C O > C = O > C - O

20 9-20 Dr. Wolf’s CHM 101 Electronegativity and Bond Polarity Electronegativity, (EN), is the ability of an atom to attract electron density of shared electrons. To the extent that an atom attracts extra electron density away from the other atom, it has a partial negative charge. The other atom has a corresponding positive charge. This creates a polar bond. The greater the difference in EN between the two atoms, the more polar the bond. The extreme “polar bond” is where the electrons are completely with one atom from the other. What has already been identified as an ionic bond. On the Pauling scale of EN, fluorine is the most electronegative with a value of 4.0. All other elements are less electronegative with Cesium being the least with a 0.7 value.

21 9-21 Dr. Wolf’s CHM 101 The Pauling electronegativity (EN) scale.

22 9-22 Dr. Wolf’s CHM 101 Electronegativity and atomic size.

23 9-23 Dr. Wolf’s CHM 101 SAMPLE PROBLEM 9.3Determining Bond Polarity from EN Values PROBLEM: PLAN: SOLUTION: (a) Use a polar arrow to indicate the polarity of each bond: N-H, F-N, I -Cl. (b) Rank the following bonds in order of increasing polarity: H-N, H-O, H-C. (a) Use Figure 9.16(button at right) to find EN values; the arrow should point toward the negative end. (b) Polarity increases across a period. (a) The EN of N = 3.0, H = 2.1; F = 4.0; I = 2.5, Cl = 3.0 N - HF - N I - Cl (b) The order of increasing EN is C < N < O; all have an EN larger than that of H. H-C < H-N < H-O

24 9-24 Dr. Wolf’s CHM 101  EN 3.0 2.0 0.0 Boundary ranges for classifying ionic character of chemical bonds.

25 9-25 Dr. Wolf’s CHM 101 Percent ionic character of electronegativity difference (  EN).

26 9-26 Dr. Wolf’s CHM 101 Properties of the Period 3 chlorides.

27 9-27 Dr. Wolf’s CHM 101 End of Chapter 9

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