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Chapter #9 - 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. Sodium Chloride.
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Chapter #9 - 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
Depicting Ion Formation with Orbital Diagrams and Electron Dot Symbols - I Problem: Use orbital diagrams and Lewis structures to show the formation of magnesium and chloride ions from the atoms, and determine the formula of the compound. Plan: Draw the orbital diagrams for Mg and Cl. To reach filled outer levels Mg loses 2 electrons, and Cl will gain 1 electron. Therefore we need two Cl atoms for every Mg atom. Solution: Mg + Mg+2 + 2 Cl- .. 2 Cl . .. .. .. . Cl Cl . .. .. .. .. Mg + Mg+2 + 2 Cl . .. ..
Depicting Ion Formation from Orbital Diagrams and Electron Dot Symbols - II Problem: Use Lewis structures and orbital diagrams to show the formation of potassium and sulfide ions from the atoms, and determine the formula of the compound. Plan: Draw orbital diagrams for K and S. To reach filled outer orbitals, sulfur must gain two electrons, and potassium must lose one electron. Solution: 2 K + 2 K+ + S - 2 S . .. . 2 - . .. .. .. .. .. K . + S 2 K+ + S K
Three Ways of Showing the Formation ofLi+ and F - through Electron Transfer
The Reaction between Na and Br to Form NaBr The Elements The Reaction!
Melting and Boiling Points of Some Ionic Compounds Compound mp( oC) bp( oC) CsBr 636 1300 NaI 661 1304 MgCl2 714 1412 KBr 734 1435 CaCl2 782 >1600 NaCl 801 1413 LiF 845 1676 KF 858 1505 MgO 2852 3600
Figure 9.10: The electron probability distribution for the H2 molecule.
Covalent bonds http://wine1.sb.fsu.edu/chm1045/notes/Bonding/Covalent/Bond04.htm animation http://www.chem.ox.ac.uk/vrchemistry/electronsandbonds/intro1.htm
For elements larger than Boron, atoms usually react to develop octets by sharing electrons. H, Li and Be strive to “look” like He. B is an exception to the noble gas paradigm. It’s happy surrounded by 6 electrons so the compound BH3 is stable. Try drawing a Lewis structure for methane.
Draw Lewis dot structures for the halogens. Notice that these all follow the octet rule! Try oxygen and nitrogen. These also follow the octet rule!
The Charge Density of LiF Fig. 9.20
Figure 9.12: Molecular model of nitro-glycerin. What is the formula for this compound?
Rules for drawing Lewis structures 1. Count up all the valence electrons 2. Arrange the atoms in a skeleton 3. Have all atoms develop octets (except those around He)
Make some Lewis Dot Structures with other elements: SiH4 H2O NH3 CH2O C2H6 C2H6O
CH3I Figure 9.9: Model of CHI3Courtesy of Frank Cox.
Make some Lewis Dot Structures with other elements: CH4 H2O NH3 CH2O C2H6 C2H6O
Rules for drawing Lewis structures 1. Count up all the valence electrons 2. Arrange the atoms in a skeleton 3. Have all atoms develop octets (except those around He) 4. Satisfy bonding preferences!
The Relation of Bond Order,Bond Length and Bond Energy Bond Bond Order Average Bond Average Bond Length (pm) Energy (kJ/mol) C O 1 143 358 C O 2 123 745 C O 3 113 1070 C C 1 154 347 C C 2 134 614 C C 3 121 839 N N 1 146 160 N N 2 122 418 N N 3 110 945 Table 9.4
The Periodic Table of the Elements 2.1 He 0.9 1.5 2.0 2.5 3.0 3.5 4.0 Ne Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Ar 0.8 1.0 1.3 1.5 1.6 1.6 1.5 1.8 1.8 1.8 1.9 1.6 1.6 1.8 2.0 2.4 2.8 Kr 0.8 1.0 1.2 1.4 1.6 1.8 1.9 2.2 2.2 2.2 1.9 1.7 1.7 1.8 1.9 2.1 2.5 Xe 0.7 0.9 1.1 1.3 1.5 1.7 1.9 2.2 2.2 2.2 2.4 1.9 1.8 1.8 1.9 2.0 2.2 Rn 0.7 0.9 1.1 Ce Pr Nd Pm Yb Lu 1.1 1.1 1.1 1.2 1.2 1.1 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.3 1.3 1.5 1.7 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.5 Th Pa U Np No Lr
Determining Bond Polarity from Electronegativity Values Problem: (a)Indicate the polarity of the following bonds with a polarity arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S (b) rank those bonds in order of increasing polarity. Plan: (a) We use Fig. 9.16 to find the EN values, and point the arrow toward the negative end. (b) Use the EN values. Solution: a) the EN of O = 3.5 and of H = 2.1: O - H the EN of O = 3.5 and of Cl = 3.0: O - Cl the EN of C = 2.5 and of P = 2.1: C - P the EN of P = 2.1 and of N = 3.0: P - N the EN of N = 3.0 and of S = 2.1: N - S the EN of C = 2.5 and of Br = 2.8: C - Br the EN of As = 2.0 and of O = 3.5: As - O b) C - Br < C - P < O - Cl < P - N < N - S < O - H < As - O 0.3 < 0.4 < 0.5 < 0.9 < 0.9 < 1.4 < 1.5
Percent Ionic Character as a Function ofElectronegativity Difference (En) Fig. 9.19
The Charge Density of LiF Fig. 9.20