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Chapter 8 Chemical Bonds

Chapter 8 Chemical Bonds.

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Chapter 8 Chemical Bonds

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  1. Chapter 8 Chemical Bonds The block being supported by soap bubbles is a piece of SEAgel, a solid foam made from agar. Aerogels like SEAgel have a porous structure that encapsulates gases. It is claimed that the density of an aerogel can be equal to or less than that of air. This material, which was developed for the space program, is a good insulator and can be used in packaging. The properties of SEAgel and other materials are related to the types of bonds between their atoms. In this chapter we examine the formation of bonds.

  2. Assignment for Chapter 8 8.46, 8.50,8.51, 8.56, 8.60, 8.65, 8.68, 8.71

  3. How atoms are linked together? A+BAB EA+EB>EAB Generally, An+Bm+Ck+…AnBmCk… Ionic bonds Valence bonds

  4. Figure 8.1 Gilbert Newton Lewis (1875–1946).

  5. Lewis Symbols for Atoms and Ions H He N O K Mg K + Cl K [ Cl ] duplet - octet

  6. N Nitrogen atomic orbitals Lewis symbol

  7. Figure 8.2 When a main-group metal atom forms a cation, it loses its valence s- and p-electrons and acquires the electron configuration of the preceding noble-gas atom. The heavier atoms in Groups 13 and 14 behave similarly, but the resulting core consists of the noble-gas configuration and an additional complete subshell of d-electrons. 1s2 Li+ Be2+ B3+ +2s2 2p6 Na+ Mg2+ +3s2 3p6 K+ Ga3+ +4s2 4p6 3d10 Rb+ In3+/5+ +5s2 5p6 4d10 Cs+ Tl3+/5+ Fr+ Ra2+ +6s2 6p6 4f14 5d10

  8. O2- P3- Se2- Te2- Bi3- H- Figure 8.3 When p-block elements acquire electrons and form anions, they do so until they have reached the electron configuration of the following noble gas. 1s2 +2s2 2p6 +3s2 3p6 +4s2 4p6 3d10 +5s2 5p6 4d10 +6s2 6p6 4f14 5d10

  9. - + Ionic Bonds • An ionic bond is the electric attraction between the opposite charges of cations and anions. The formation of ionic bonds is represented in terms of Lewis symbols by the loss or gain of electrons until both species have Reached an octet or duplet of electrons.

  10. 2+ 2- 4+ 2- Figure 8.4 When tin(II) oxide is heated in air, it becomes incandescent as it reacts to form tin(IV) oxide. Even without being heated, it smolders and can ignite. SnO + O2 SnO2 2- Sn [ O ] + O[ O ]Sn[ O ] Lewis symbols and formulae are not really written for transition metals.

  11. Variable Valence Inert-pair effect The tendency to form cations two units lower in charge than expected from the group number, i.e., the p-block elements to form cations that have octet configuration. • 4p,5p,6p • 4s,5s,6s More than one type of cation can be formed for some elements:

  12. Figure 8.5 The typical ions formed by the heavy elements in Groups 13–15 show the influence of the inert-pair effect. These elements have the tendency to form compounds in which the oxidation numbers differ by2.

  13. Figure 8.6 Considerable energy is needed to produce cations and anions from neutral gas-phase atoms: the ionization energy of the metal atoms must be supplied, and it is only partly recovered from the electron affinity of the nonmetal atoms. The overall lowering of energy that drags the ionic solid into existence is due to the strong attraction between cations and anions in the solid. It takes 145 kJ/mol to produce the ions from the elements, and the solid compound is 787 kJ/mol lower in energy than the separated ions.

  14. Figure 8.7 A two-dimensional slice of an ionic crystal. The greater the distance between two ions, the weaker their attractions. Therefore, the strongest attractions to an ion are those of the adjacent ions of opposite charge. The blue circles denote the distances over which the two closest repulsions occur, and the yellow circles denote the two closest attractions.

  15. Figure 8.8 In a Born-Haber cycle, we select a sequence of steps that starts and ends at the same point (the elements, for instance). The lattice enthalpy is the enthalpy change of the step in which the solid is formed from a gas of ions (the dark red arrow). The sum of enthalpy changes for the complete cycle (yellow line) is 0 because enthalpy is a state property. The “Enthalpy of ionization” step includes the energies required to produce both cations and anions from their respective atoms.

  16. Figure 8.9 The Born-Haber cycle used to determine the lattice enthalpy of potassium chloride. Enthalpy of ionization (Cl) Enthalpy of ionization (K) Enthalpy of atomizaiton (Cl) Enthalpy of atomizaiton (K) -Enthalpy of formation Lattice enthalpy:

  17. Figure 8.10 An ionic solid—here we see a fragment of sodium chloride, with the sodium ions represented by pink spheres and the chloride ions by green spheres—consists of an almost infinite array of cations and anions stacked together to give the lowest energy arrangement. The pattern shown here is repeated throughout the crystal.

  18. Figure 8.11 This sequence of images illustrates why ionic solids are brittle. (a) The original solid consists of an orderly array of cations and anions. (b) A hammer blow can push the ions into positions where cations are next to cations and anions are next to anions; there are now strong repulsive forces acting (as depicted by the double-headed arrows). (c) As a result of these repulsive forces, the solid springs apart in fragments. (d) This chunk of calcite consists of several large crystals joined together. (e) The blow of a hammer has shattered the crystal, leaving flat, regular surfaces consisting of planes of ions.

  19. Figure 8.12 These micrographsshow the porous structure of bone. The calcium in bone is extracted by the body if the level of calcium in the diet is low. (a) Healthy bone tissue. (b) Bone that has suffered calcium loss through osteoporosis. The overlay shows the regular arrangement of the calcium and phosphate ions in healthy bone. (a) (b)

  20. Covalent Bonds • In covalent bond formation, atoms go as far as possible toward completing their octets (duplets) by sharing electron pairs.

  21. Figure 8.13 The formation of a covalent bond between two hydrogen atoms. (a) Two separate hydrogen atoms, each with one electron. (b) The electron cloud that forms when the spins pair and the orbitals merge is most dense between the two nuclei. (c) The boundary surface that we shall use to depict a covalent bond.

  22. Covalent Bonds • In covalent bond formation, atoms go as far as possible toward completing their octets (duplets) by sharing electron pairs. That’s why F2 is so reactive. Lone pairs The Lewis structure gives not only bond positions (shared electron pairs) But also reactivity-how reactive is the molecules and which part(s) is(are) reactive (lone pairs).

  23. The Structures of Ployatomic Species • Single, double, triple bonds may be formed.

  24. How to write Lewis structure of a polyatomic species • Count the total number of valence electrons on each atom and divide by 2 to obtain the number of electron pairs. If the species is polyatomic ion, add one more electron for each negative charge or subtract one electron for each positive charge. • Write the chemical symbols of the atoms to show their layout in the molecule. One can predict the most likely arrangements of atoms by using common patterns and the rules of thumb (largely correct but exceptions possible) given earlier. • Place one electron pair between each pair of bonded atoms. • Complete the octet (duplet for H) of each atom by placing any remaining electron pairs as lone pairs around the atoms. If there are not enough electrons to form octets, form multiple bonds.

  25. More Examples NH3: 3+5=8 valence electrons = 4 pairs. 3 pairs to form three bonds, leaving one lone pair Formic acid HCOOH 2+12+4=18 valence electrons = 9 pairs 5 pairs to form five bonds, leaving 4 lone pairs. Hypobromous acid HBrO: 1+6+7=14 valence electrons = 7 pairs Two pairs to form two bonds, Leaving 5 lone pairs.

  26. Classroom Exercise • Write the Lewis structure of CH3COOH

  27. Resonance: Blending of Structures All valid. We cannot find two bond lengths (hypothetical N-O vs N=O)

  28. Resonance: Blending of Structures

  29. Resonance: Blending of Structures

  30. Quiz Write the Lewis structure of the following compounds: (1) Methanol (2) Nitrate (3) Carbon Monoxide Answer:

  31. Formal Charge • The real number of valence electrons each atom in a molecule “owns” may be different from the number of normal valence electrons of that atom, rendering it positively or negatively charged. Formal charge (FC) = number of valence electrons on the free atom (V) - number of electrons present as lone pairs (L) - ½ number of electrons shared in bonds (S/2) Formal charge can be understood as surplus or deficit of valence electrons after deducting lone pairs and shared pairs. Typically, the most stable Lewis structures are those in which the formal charge of the individual nonmetal atoms are close to zero

  32. Plausibility of a structure

  33. Classroom Exercise • Suggest a plausible structure for the poisonous gas phosgene, COCl2. Resonance states 0 0 0 0 0 0 0 0 Formal charges are all zero, too. But why is this one not favored? Experiment shows the oxygen has double bond rather than single bond.

  34. Exceptions to the Octet Rule • B, C, N, O, F observe the octet rule. • P, S, Cl can have more than 8 valence electrons. • Radicals: odd-electron species, highly reactive. • Biradicals: with two unpaired electrons.

  35. .. .. .. . . .. .. O2 is a biradical! Wrong!

  36. Classroom Exercise Write a Lewis structure for the hydrogenperoxyl radical HOO., which plays an important role in atmospheric chemistry and which in the body has been implicated in the degeneration of neurons.

  37. Case Study 8 (a) The equipment in the illustration monitors air quality from a rooftop in Los Angeles. The concentration of NO2 in the air due to automobile traffic increases during the day, contributing to the typical brown color of the afternoon sky.

  38. Case Study 8 (b) The gas NO (left) is colorless. When it is exposed to air (right), it is rapidly oxidized to brown NO2. Oxygen molecules are actually biradicals.

  39. Expanded valence shells • Nonmetal atoms in Period 3 or higher can accommodate 10, 12 or more electrons (expanded valence shell). e.g., P, S, Cl.

  40. Figure 8.16 (a) A model using small spheres to represent atoms and (b) a space-filling model of PCl5, showing how closely the chlorine atoms must pack around the central phosphorus atom. (c) A nitrogen atom is significantly smaller than a phosphorus atom, and five chlorine atoms cannot pack around it.

  41. Figure 8.17 Phosphorus trichloride is a colorless liquid. When it reacts with chlorine (the pale yellow-green gas in the flask), it forms the very pale yellow solid phosphorus pentachloride (at the bottom of the flask).

  42. +

  43. Figure 8.18 Two circumstances in which a central atom assumes an expanded octet. First, there are too many electrons to be accommodated in octets because either (a) there are too many atoms attached, or (b) the central atom must accommodate additional lone pairs. (c) Second,a resonance structure with multiple bonds has a favorable energy.

  44. S accommodates 10 electrons

  45. How many electrons does Xe accommodate?

  46. Which is most plausible?

  47. 12 valence electrons!

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