1 / 54

Ch 8 Chemical Bonding and Molecular Structure

Ch 8 Chemical Bonding and Molecular Structure. AP Chemistry 2013-2014. Bonds are attractive forces that hold groups of atoms together and make them function as a unit .

elgin
Télécharger la présentation

Ch 8 Chemical Bonding and Molecular Structure

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. Ch 8 Chemical Bonding and Molecular Structure AP Chemistry 2013-2014

  2. Bonds are attractive forces that hold groups of atoms together and make them function as a unit. • Being bound requires less energy than existing in the elemental form. Energy is released when a bond is formed; energy is required to break a bond. The energy required to break a bond is called the bond energy. • Ionic bonds • Covalent bonds

  3. Coulomb’s Law is used to calculate the energy of an ionic bond (see equation on the right; k = 2.31 x 10-19Jnm, Q = charge on each ion) • The energy of an ionic bond will be negative; it indicates an attractive force so that the ion pair has lower energy than the separated ions.

  4. Chemical bond formation • When two hydrogen atoms approach each other, two repulsions & one attraction occur • Electron/electron repulsion • Proton/proton repulsion • Proton/electron attraction

  5. When the attractive forces offset the repulsive forces, the energy of the two atoms decreases and a bond is formed. Bond length is the distance between two nuclei where the energy is at a minimum.

  6. The electrons and nucleus of one atom strongly perturb or change the spatial distribution of the other atom’s valence electrons. A new orbital (wave function) is needed to describe the distribution of the bonding electrons bond orbital. The energy of the electrons in a bond orbital is lower that their energy in valence electron orbitals when they are in isolated atoms.

  7. In an ionic bond, the bonding orbital is strongly displaced toward one nucleus. In a covalent bond, the bond orbital is more or less evenly distributed and the electrons are shared by two nuclei. Most chemical bonds are somewhere between purely ionic and purely covalent.

  8. Electronegativity is the ability of an atom in a molecule to attract shared electrons to itself. • Fluorine is the most electronegative (4.0) due to highest effective nuclear charge (Zeff) and smallest radius—so that the nucleus is closest to the “action” • Francium is the least (0.7) due to lowest Zeff and largest radius so that the nucleus is farthest from the “action”

  9. This atomic trend is only used when atoms form molecules. • Ionic: ΔEN >1.67 • Covalent: ΔEN < 1.67 • Nonpolar covalent: ΔEN < 0.4

  10. Covalent bonding • Most compounds are covalently bonded, especially carbon compounds. • Localized electron (LE) bonding model: assumes that a molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms. Electron pairs are assumed to be localized on a particular atom [lone pairs] or in the space between two atoms [bonding pairs]. • Lewis structures describe the valence electron arrangement • Geometry of the molecule is predicted with VSEPR

  11. Number of bond pairs/Octet rule • Single and multiple bonds • Single bond: one pair of electrons shared (sigma σ bond) • Double bond: two pairs of electrons shared (one sigma bond, one pi bond) • Triple bond: three pairs of electrons shared (one sigma bond, two pi bonds)

  12. Double Bond (ethene)

  13. Double bond (Carbon dioxide)

  14. Triple Bond (Ethyne)

  15. Pi bonds are weaker than sigma bonds but never exist alone. Triple bonds are stronger than double bonds are stronger than single bonds. Multiple bonds are most often formed by C, N, O, and P.

  16. Multiple bonds increase the electron density between two nuclei and therefore decrease the nuclear repulsions while enhancing the attraction between nucleus/electrons; either way, the nuclei move closer together and the bond length is shorter for a double bond than a single bond.

  17. Exceptions to the Octet rule • Fewer than eight: H, Be, B • Expanded valence: can only happen if the central atom has d-orbitals and can thus be surround by more than four valence pairs in certain compounds. • Odd-electron compounds: ex. NO, NO2, ClO2

  18. Exercise 1 Relative Bond Polarities Order the following bonds according to polarity: H-H, O-H, Cl-H, S-H, and F-H

  19. Bond polarity and electronegativity: electronegativity determines polarity since it measures a nucleus’s attraction or “pull” on the bonded electron pair. • When two nuclei are the same, the sharing is equal and the bond is described as nonpolar. • When two nuclei are different, the electrons are not shared equally, setting up slight +/- poles, and the bond is described as polar. • When the electrons are shared very unequally, the bond is described as ionic.

  20. Ionic bonding • The final result of ionic bonding is a solid, regular array of cations and anions called a crystal lattice. • Enthalpy of dissociation: energy required to decompose an ion pair (from a lattice) into ions; a measure of the strength of the ionic bond (related to Coulomb’s law) • The energy of attraction depends directly on the magnitude of the charges and inversely on the distance between them (related to the size of the ion).

  21. Exercise 2 Comparing lattice energy Which compound in each pair will have the higher lattice energy? NaF or RbFMgOor LiCl

  22. Drawing Lewis structures • H is always a terminal atom • Atom with lowest EN goes in center • Find the total number of valence electrons by adding together the valence electrons of every atom in the compound • For ions, add for negative charges and subtract for positive charges

  23. Place one pair of electrons, a sigma bond, between each pair of bonded atoms. Complete the octets of all atoms with lone pairs. Leftover pairs are assigned to the central atom if it can accommodate them. Double/triple bonds may need to be used (pi bonds).

  24. Exercise 3 Writing Lewis Structures Give the Lewis structure for each of the following. a) HF

  25. Exercise 3 Writing Lewis Structures Give the Lewis structure for each of the following. b) N2

  26. Exercise 3 Writing Lewis Structures Give the Lewis structure for each of the following. c) NH3

  27. Exercise 3 Writing Lewis Structures Give the Lewis structure for each of the following. d) CH4

  28. Exercise 3 Writing Lewis Structures Give the Lewis structure for each of the following. e) CF4

  29. Exercise 3 Writing Lewis Structures Give the Lewis structure for each of the following. f) NO+

  30. Exercise 4 Lewis Structures for Molecules that Violate the Octet Rule Give the Lewis structure for each of the following. a) PCl5

  31. Exercise 4 Lewis Structures for Molecules that Violate the Octet Rule Give the Lewis structure for each of the following. b) ClF3

  32. Exercise 4 Lewis Structures for Molecules that Violate the Octet Rule Give the Lewis structure for each of the following. c) XeO3

  33. Exercise 4 Lewis Structures for Molecules that Violate the Octet Rule Give the Lewis structure for each of the following. d) RnCl2

  34. Exercise 4 Lewis Structures for Molecules that Violate the Octet Rule Give the Lewis structure for each of the following. e) BeCl2

  35. Exercise 4 Lewis Structures for Molecules that Violate the Octet Rule Give the Lewis structure for each of the following. f) ICl4-

  36. Resonance structures • Ex. ozone has equal bond lengths and equal bond strengths, implying that there are an equal number of bond pairs on each side of the central oxygen atom. The Lewis structure does not agree with this; instead, we have to use a composite to describe the reality. This composite depicts the blending of resonance structures for ozone. Instead of truly having a single bond and a double bond, both of its C-O bondsO-O bonds could be thought of as “a bond and a half”.

  37. Resonance structures differ only in the assignment of electron pair positions, never atom positions. They differ in the number of bond pairs between a given pair of atoms. Note that the resonance structures and composite are drawn with brackets (required for full credit on AP exam).

  38. Exercise 5Resonance Structures Draw every resonance structure for the carbonate ion. Also draw the composite structure.

  39. Bond properties • Bond order: simply the number of bonding electron pairs shared by two atoms in a molecule. • 1 = one shared pair; sigma bond between two atoms • 2 = two shared pairs; sigma bond and pi bond • 3 = three shared pairs; sigma bond and two pi bonds • Fractional for resonance structures (3/2 for ozone, 2/3 for carbonate) Bond order = number of shared pairs linking X and Y number of X-Y links

  40. Bond length: the distance between the nuclei of two bonded atoms • Higher bond order = shorter length • Bond energy: the greater the number of electron pairs between a pair of atoms, the shorter the bond. This implies that atoms are held together more tightly when there are multiple bonds, so there is a relationship between bond order and the energy required to break a bond. • Bond dissociation energy (D): the energy supplied to break a chemical bond • Endothermic; D is positive

  41. Bonds in reactants are broken while bonds in products are formed. Energy released is greater than energy absorbed in exothermic reactions. The converse is also true. ΔH°rxn = ΣmD(bonds broken) - ΣnD(bonds made) ΔH°rxn = reactants(E cost) - products(E payoff) • Note that this is “backwards” from thermodynamics. First we must break the bonds of the reactants (costs energy) then subtract the energy gained by forming new bonds in the products.

  42. Exercise 6 ΔH from Bond Energies Using the bond energies from the table on the right, calculate ΔH° for the reaction of methane with chlorine and fluorine to give Freon-12 (CF2Cl2). CH4 + 2Cl2 + 2 F2 CF2Cl2 + 2 HF + 2 HCl

  43. CH4 + 2Cl2 + 2 F2 CF2Cl2 + 2 HF + 2 HCl

  44. Formal charge • Formal charge is the difference between the number of valence electrons on a free element, and the number of electrons assigned to the atom once it is in a molecule. • Formal charge = group number – [# of lone electrons – 2(# of bonding electrons)]

  45. The idea of formal charge allows us to determine the most favored structure out of a set of nonequivalent Lewis structures. • Oxidation states of more than +/- are questionable, while formal charges are more realistic. • The sum of the formal charges on an ion must equal the ion’s overall charge. • Use formal charges along with the following to determine resonance structure • Atoms in molecules (or ions) should have formal charges that are as small (close to zero) as possible • A molecule (or ion)is most stable when any negative formal charge resides on the most electronegative atom.

  46. Ex. There are three possible structures for the sulfate ion shown below (note that these are not resonance structures). The third is the most valid of the three; it results in the fewest (and smallest) formal charges.

  47. Exercise 7Formal Charges Give possible Lewis structures for XeO3, an explosive compound of Xenon. Which Lewis structure or structures are most appropriate according to the formal charges?

  48. Valence shell electron pair repulsion theory (VSEPR) • Molecular shape changes with the number of sigma bonds plus lone pairs about the central atom • Molecular geometry is the arrangement in space of the atoms bonded to a central atom • lone pairs take up more space around an atom than bonds • Each lone pair or bond pair repels all other electron pairs; they try to avoid each other making as wide an angle as possible. • Ex. Water: the two lone pairs on oxygen “warp” the normal 109.5 angle through repulsion, resulting in a bond angle of 104.5.

  49. To determine molecular geometry • Sketch the Lewis dot structure • Describe the structural pair or electronic geometry (the shape of the molecule considering both its bonds and lone pairs) • Focus on the bond locations (ignore lone pairs) and assign a molecular geometry based on their locations • Molecular geometry and electronic geometry are only the same in the absence of lone pairs on the central atom. • Works well for elements of the s and p-blocks; does not apply to transition element compounds (exceptions)

More Related