1 / 57

Bonding

Bonding. 8 out of 75 M/C Questions Free Response—Every year. In General:. All bonds occur because of electrostatic attractions. Formation of molecules and the state of matter of a substance depends on the attractions between electron clouds of one atom and nucleus of another atom.

shadi
Télécharger la présentation

Bonding

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Bonding 8 out of 75 M/C Questions Free Response—Every year

  2. In General: • All bonds occur because of electrostatic attractions. • Formation of molecules and the state of matter of a substance depends on the attractions between electron clouds of one atom and nucleus of another atom.

  3. Bonding—General Rules • A metal and a nonmetal bond so that charges on the ions cancel. • When two nonmetals react to form a covalent bond, they share electrons in a way that gives both atoms a Noble gas configuration.

  4. Bonding—General Rules • nonmetal + representative metal binary ionic compound: ions form to give the nonmetal the valence electron configuration of the next noble gas atom, and valence orbitals of the metal are emptied

  5. Bonding—General Rules • Most bonds are combinations of ionic/covalent character • The more different the atoms bonding are, the more ionic character of the bond. • More similar—more covalent

  6. Covalent Sharing Molecules Structural formula Ionic Transfer / charged Compounds Formula Unit Word Association

  7. Coulomb’s Law • Describes energy of interaction between ions • E = k J*nm(Q1Q2 / r) • Q1 & Q2numerical ion charges • r distance between ion centers • Negative ans. means ion pair is more stable than individual ions.

  8. Coulomb’s Law Simplified • As charges on ions increase, the energy of the bond increases. • As the radius of the atoms increases, the energy of the bond decreases.

  9. Which bond has greater energy? • NaF or MgO? • NaCl or CsI? • MgCl2 or NaCl? • MgCl2 or BaBr2?

  10. Bond Length • Distance between bonding atoms at which energy is minimized • Atoms position themselves to minimize repulsions and maximize attraction & thus achieve lowest possible energy.

  11. Predicting Formulas for Ionic Compounds • Metal—positive charge equal to # of valence electrons • Nonmetal—negative charge equal to # of electrons away from next Noble gas • Compound—charges must cancel

  12. Predict Formulas for: • Potassium sulfide • Barium chloride • Aluminum oxide • Magnesium phosphide

  13. Exceptions to the Rule: • Sn—forms both +2 and +4 ions • Pb—forms both +2 and +4 ions • Bi—forms +3 and +5 ions • Tl—forms +1 and +3 ions  “no simple explanation for this behavior”

  14. Lattice Energy • Change in energy taking place when separated gaseous ions are packed together to form an ionic solid • Energy released when an ionic solid forms from its ions

  15. Calculating Energy Change • The sum of energy changes in each step • Formation of lithium fluoride: • Sublimation of solid lithium: 161 kJ per mole • Ionization of lithium: 520 kJ per mole

  16. Calculating Energy Change • Dissociation of F2: 77 kJ per mole of F atoms • Ionization of F: -328 kJ per mole (Electron affinity of F) • Formation of LiF: -1047 kJ per mole

  17. Calculating Energy Change • Sum: 161 + 520 + 77 – 328 – 1047 = -617 kJ per mole LiF (negative sign means process is exothermic & product has lower energy)

  18. Calculating Lattice Energy • Lattice energy can be calculated using a form of Coulomb’s Law: Lattice energy = k(Q1Q2 / r) k = proportionality constant dependent on solid’s structure & electron configuration of the solid

  19. P. 365—energy changes in forming LiF and MgO

  20. Lattice Energy • Much larger changes in Mg and O because of greater Q1 and Q2 • Even though Mg and O must ionize twice (which requires lots of energy) the overall change is very exothermic

  21. DH for Covalent Compunds • Breaking bonds requires energy (+). • Forming bonds releases energy (-). • Determine which bonds break and which bonds form. • Add values for broken bonds. • Subtract values for formed bonds.

  22. Find DH CH4 + Cl2 + F2 CF2Cl2 + 2HF + 2HCl

  23. Polar Covalent Bonds • Most bonds have both ionic and covalent character • Polar covalent bonds—unequally shared electrons • Not different enough to be purely ionic nor similar enough to be purely covalent

  24. Polar Covalent Bonds • Strength of polarity depends on electronegativity (ability of an atom to attract electrons to itself). • Higher electronegativity results in higher negative charge. • Partial charge results in a stronger bond & stronger inter-molecular attraction.

  25. Dipoles • + indicates direction of dipole—points to negative end • Dipolar molecules orient themselves one way in an electric field. • Dipoles can cancel.

  26. Canceling Dipoles • If evenly spacedidentical bonds exist, dipoles cancel each other. • Linear—2 identical bonds separated by 180* (CO2) • Trigonal planar—3 identical bonds separated by 120* (SO3) • Tetrahedral—4 identical bonds separated by 109.5* (CH4)

  27. Ionic Bonds with Covalent Character • No totally ionic bonds exist. • Percent ionic character can be calculated: (measured dipole moment of X-Y) (calculated dipole moment of X+Y-) X 100%

  28. Ionic Bonds with Covalent Character • X-Y represents the molecule • X+Y- represents the ions • Graph—p. 367—Bonds with more than 50% ionic character are classified as ionic

  29. New Definition • Ionic compound—any substance that conducts electricity in its liquid state (melted, not dissolved)

  30. Beyond the Bond—Intermolecular Forces & States of Matter

  31. Intermolecular Forces • Weaker than chemical bond • Affect structure and state of matter

  32. Dipole-Dipole Forces • Positive and negative ends of polar molecules attract each other. • About 1% as strong as covalent or ionic bonds • Weaken as distance between molecules increases

  33. Hydrogen Bonding • Especially strong dipole-dipole force • Occurs when H bonds to a strongly electronegative atom—O, N, or F • Very strong because 1) molecule is very polar & 2) small size of H

  34. H Bonding • Example—water • More pronounced in molecules formed from small atoms (dipoles can come closer) • High boiling point

  35. London Dispersion Forces • Forces that exist in all substances but are important only in Noble gases and nonpolar molecules • Result from temporary dipoles formed when electrons distribute themselves unevenly—can induce a dipole in a neighboring atom • VERY WEAK

  36. Bonding models

  37. Problems with Current Model • If s & p orbitals are different, bonds formed from them should be different. • Since p orbitals are perpendicular to one another, we would expect 90o bond angles. • Neither of these things is true.

  38. Hybridization • Explains discrepancies • The mixing of native atomic orbitals to form special orbitals for bonding • Hybrid orbitals have shapes and energies that are between those of the native orbitals.

  39. Example: CH4 • Four hydrogen atoms combine with carbon in each of four orbitals. The one s and three p orbitals hybridize to form four hybrid sp3 orbitals.

  40. Hybridization and Geometry • The number of areas of charge density (bonded atoms + unshared pairs) relates to both hybridization and geometry. • sp3—4 areas/orbitals—tetrahedral • sp2—3 areas/orbitals—trigonal planar • sp—2 areas/orbitals—linear

  41. sp2 A molecule such as BH3 or an ion such as NO3- exhibit sp2 hybridization. This results in a trigonal planar geometry.

  42. Multiple Bonds • First bond in a hybrid orbital (sigma) s • Second or third in an unhybridized orbital (pi) p

  43. Bonding in Ethene

  44. sp • One s and one p orbital combine to form sp hybrids that will be at 90o to one another as in BeF2or CO2.

  45. dsp3 • Five electron pairs require the use of a d orbital. Only elements in level 3 or above can do this as in PF5 which has trigonal bipyramidal geometry.

  46. d2sp3 • The six electron pairs in a compound such as SF6 result in the use of two d orbitals and octahedral geometry.

  47. Localized Electron Model • Assumes electrons are located between two atoms • Includes several parts: • Description of valence electron arrangement (Lewis Structure) • Prediction of Geometry (VSEPR) • Description of types of orbitals (Hybridization)

  48. Shortcomings • Localized electron model • Assumes electrons are localized so resonance must be added. • Does not adequately address unpaired electrons. • Gives no information about bond energies.

  49. Molecular Orbital Model • Uses molecular orbitals (instead of atomic orbitals) to address shortcomings of localized electron model.

  50. Molecular Orbitals • Combining atoms forms two molecular orbitals—not one atomic orbital.

More Related