model: simplification of reality

Transcription

1 1 model: simplification of reality Covalent bonds are formed when atoms share electrons! 3 models: 1. Lewis electron dot structures & VSEPR 2. Valence Bond model 3. Molecular Orbital model 1st model: Lewis electron dot structures Remember examples: CH 4, NH 3, CO 2, BF 3, PF 5 VSEPR model for molecules above But what about shape of CH 2 =CH 2? Ch. 11 presents two more viewpoints, both of which emphasize orbitals: Valence bond method and molecular orbital method As background, consider two H atoms approaching one another. PHHM9 Figure 11-1 is similar to figure on next page

2 2

3 3 2nd model: Valence bond model -- form bond by overlap of atomic orbitals (AOs) Similar to Lewis e - dot structures but use AOs & optimize overlap of H 1s atomic orbital with H 1s atomic orbital Except for H-H, H-S-H (Figure 11-2), and PH 3 (Figure 11-3), often need Hybridization Play with orbitals on one atom. We must still be able to explain the shapes of CH 4, NH 3, etc. Formation of sp 3 hybrid orbitals p427 & Fig. 11-5, 11-6, & 11-7 slides 6-10, 11, 12,13 sp 2 hybridization p429 (a 4th e - may be in π bond); Figure 11-8 slides 14, 15

4 4 sp hybridization Figure 11-9 (slides 16, 17) Figure (slide 22): both VSEPR and hybridization begin with a Lewis electron dot structure CH 2 CH 2 Figure (slide 28-29) sigma vs pi bonds hybridization in H 2 C=C=CH 2 ) Valence bond model: form bond by overlap of AOs or hybrid orbitals Summary of hybridization VSEPR.pdf Why is O 2 paramagnetic? Figure 10-3 (Chapter 11, slide 50) Why is NO stable? Do we need resonance? We need another model! 3rd model: Molecular Orbital model Bonding electrons need not be localized between just 2 atoms. Formation of H 2 PHHM9 Figure 11-19(shows nuclei), 20, 21; Zumdahl 4th Fig14.25(bonding MO lower in energy),26,27,28,31,29 U:\_SC Student File Area\Pultz\Molecular_Orbitals.ppt slides 1-6 Guidelines for MO theory: 1. Orbitals from different atoms interact to form MOs of lower & higher energy. 2. # MOs = # AOs 3. E of bonding MO < E of parent AO E of antibonding MO > E of parent AO 4. Fill lowest E levels first remembering Pauli Principle & Hund's Rule 5. AOs combine most effectively with orbitals of the same type and similar energy. Zumdahl 4th Fig14.33, 34; Figure 11-23, 24 slides 8, 9, 10, 11 MO energy level diagrams for 2nd row diatomics Figure U:\_SC Student File Area\Pultz\MO for 2nd row diatomics_atkins Phys Chem 5 Fig14-27 O 2 is paramagnetic with a bond order of 2 I will always ask for MO energy level diagrams such as Figure and rather than the molecular orbital diagrams used at the bottom of p. 444 & on p. 445 Does the molecular orbital model or a Lewis electron dot structure account better for the bonding in NO? Explain.

5 NO MO energy level diagram & bond order On p. 446 PHHM9 says to use Figure 11-24a if one of the 2 elements is O or F. But use of Figure 11-24b gives the correct answer for bond order & magnetism. 5 MO theory allows us to avoid the use of resonance, a concept which arose from the limitations of Lewis electron dot structures. C 6 H 6 MO energy level diagram for π bonds: 3 bonding MOs, 3 antibonding MOs Figure 11-26, 27, 28, 29 (slide 59-62) The 6 2p e - s are best described as being in molecular orbitals which cover the entire molecule. O 3 MO energy level diagram for π bonds: bonding, nonbonding, antibonding Figure & 31; K&T4p468 (σ bonds formed from overlap of sp 2 hybrids) Some Chapter 11 questions: Concept Assessments on pp. 430, 431, 434, 438, 441, 445, 447, 456 Practice Examples on pp (Use MO energy level diagram when needed) Exercises # 5, 7, 9, 11, 13, 15, 17, 27, 31, 33, 35, 39, 43, 47, For each of the following molecules, write the Lewis electron dot structure, predict the molecular shape (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity. CH 4, CF 4, SiCl 4, NH 3, H 2 O, HCN, CO 2, PCl 3, PCl 5, SF 4, SF 6, BF 3, ClF 3, IF 5, XeF 2, XeF 4 2. Why must all six atoms in C 2 H 4 be in the same plane? Z What is the hybridization on each carbon atom in H 2 C=C=CH 2? Identify the number and type of unhybridized valence orbitals on each carbon atom. Make a drawing to show the orientation of all the valence orbitals. What is the spatial arrangement of the four hydrogen atoms? 4. Distinguish between bonding and antibonding molecular orbitals in terms of: a. charge density b. energy relative to the energy of the atomic orbitals that were combined. 5. Which of the following are predicted by the molecular orbital model to be stable diatomic species? Explain your answer. H + 2, H 2, H - 2, H 2-2, He + 2-2, He 2, F 2 Z Does the molecular orbital model or a Lewis electron dot structure account better for the bonding in NO? Explain. 7. Name K 2 O, K 2 O 2, and KO 2. Why can all three compounds exist? Explain their stability! 8. Which is the shortest bond in each group? C-C or C-N or C-O H-H or F-F or Cl-Cl C-F or C-Cl or C-Br C-C or C=C or C C C-H in C 2 H 2 or C-H in C 2 H 4 or C-H in CH 4 9. Distinguish between a localized and a delocalized model for bonding. Give example(s).

6 11-7 Metals, Semiconductors, and Insulators as seen by Band Theory PHHM9 Figure 11-34; Kotz&Treichel 4th U:\_SC Student File Area\Pultz\MO_*_K&T4* metal crystal is molecule with MOs extending over whole crystal metallic bonding (nondirectional covalent bonding) band of energy levels in metal is only partially filled ==> high electrical conductivity metal: partially filled valence band PHHM9 Figure electrical insulator: completely filled valence band & empty conduction band at much higher energy; large band gap intrinsic semiconductors: small band gap; electrical conductivity increases with temperature and decreases with size of band gap; e.g., Si or Ge with 4 valence electrons extrinsic semiconductors: dope Si or Ge with group 13 or 15 element to give deficit or excess of electrons; decreases band gap relative to pure element and hence increases electrical conductivity Figure from Zumdahl 4 th, Figure other semiconductors: GaAs (small band gap ==> all visible light absorbed ==> black color), InSb, HgCdTe Average number of valence electrons must be 4 to facilitate formation of octets in the bulk. A Light-Emitting Diode (LED) is a semiconductor designed to emit photons when electron changes energy levels. LEDs are monochromatic, but combining light from different color LEDs is one way to create "white" light. AlGaAs is used to make red LEDs AlGaP is used to make green LEDs InGaN is used to make blue LEDs C&EN, 3 Dec 2007, p50

7 7 ANSWER KEY to Some Chapter 11 questions 1. For each of the following molecules, write the Lewis electron dot structure, predict the molecular shape (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity. Species Lewis structure Molecular shape Hybrid Polarity H CH 4 H-C-H tetrahedral sp 3 nonpolar H THIS ELECTRONIC VERSION MISSES SOME LONE PAIRS CF 4 :F-C-F: tetrahedral sp 3 nonpolar :Cl: SiCl 4 :Cl-Si-Cl: tetrahedral sp 3 nonpolar :Cl: NH 3 H-N-H trigonal pyramidal sp 3 polar ~109 H H 2 O H-O-H bent sp 3 polar ~109 HCN H-C N: linear, 180 sp polar CO 2 :O::C::O: linear, 180 sp nonpolar PCl 3 :Cl-P-Cl: trigonal pyramidal sp 3 polar ~109 :Cl: :Cl: PCl 5 :Cl-P-Cl: trigonal bipyramidal sp 3 d nonpolar :Cl: :Cl: 120 and 90 SF 4 :F-S-F: seesaw sp 3 d polar ~120 and ~90

8 8 SF 6 :F-S-F: octahedral sp 3 d 2 nonpolar 90 BF 3 :F-B-F: trigonal planar sp 2 nonpolar 120 ClF 3 Cl T-shaped sp 3 d polar ~90 and ~180 IF 5 :F-I-F: square pyramidal sp 3 d 2 polar ~90 XeF 2 Xe linear sp 3 d nonpolar 180 XeF 4 :F-Xe-F: square planar sp 3 d 2 nonpolar Why must all six atoms in C 2 H 4 be in the same plane? Z3-9.7 Each C is sp 2 hybridized with one unhybridized p orbital. The sp 2 hybrid orbital from one C overlaps endwise the sp 2 hybrid orbital from the second C, forming a sigma bond. In order to maximize sidewise overlap of the unhybridized p orbitals on the two carbon atoms, they must be parallel. Since the p orbital on a particular C is perpendicular to the sp 2 hybrid orbitals on that C, all six sp 2 hybrid orbitals must be in the same plane. Finally, the H 1s atomic orbitals overlap the ends of the other four sp 2 hybrid orbitals. So all six atoms are in the same plane. It may be helpful to make a drawing. 3. What is the hybridization on each carbon atom in H 2 C=C=CH 2? sp 2 hybridization on the end carbons sp hybridization on the central C Identify the number and type of unhybridized valence orbitals on each carbon atom. one unhybridized p orbital on the end carbons two unhybridized p orbitals on the central carbon Make a drawing to show the orientation of all the valence orbitals. What is the spatial arrangement of the four hydrogen atoms? They are in mutually perpendicular planes.

9 9 4. Distinguish between bonding and antibonding molecular orbitals in terms of: a. charge density Bonding MOs have enhancement of electron density in the region of space between the two nuclei. Antibonding MOs have a decrease in electron density in the region of space between the two nuclei. b. energy relative to the energy of the atomic orbitals that were combined. Bonding MOs are lower in energy than the atomic orbitals from which they were formed. Antibonding MOs are higher in energy than the atomic orbitals from which they were formed. 5. Which of the following are predicted by the molecular orbital model to be stable diatomic species? Explain your answer. one valence electron ==> bond order = ½ ==> stable H 2 two valence electrons ==> bond order = 1 ==> stable - H 2 three valence electrons ==> bond order = ½ ==> stable 2- H 2 four valence electrons ==> bond order = 0 ==> unstable + He 2 three valence electrons ==> bond order = ½ ==> stable He 2 four valence electrons ==> bond order = 0 ==> unstable H 2 + E σ* 1s for H 2 2- and He 2 σ 1s F valence e - s in diagram below ==> 0 bond order ==> unstable 6. Does the molecular orbital model or a Lewis electron dot structure account better for the bonding in NO? Explain. The MO model explains the bonding in NO best. NO has 11 valence electrons, an odd number, so it is impossible to have octets on both atoms. But normally we equate octets with stability, so we would predict NO to be unstable. According to the MO diagram, NO has a bond order of 2.5 and hence is predicted to be stable. E σ* 2p π* 2p σ 2p π 2p σ* 2s σ 2s

10 7. Name K 2 O, K 2 O 2, and KO 2. Why can all three compounds exist? Explain their stability! 10 K 2 O potassium oxide K + and O 2- ions have noble gas electron configurations K 2 O 2 potassium peroxide KO 2 potassium superoxide K 2 O 2 and KO 2 also have a K + ion. O 2-2 and O - 2 are seen to be stable if you use the MO diagram on p444. Draw this molecular orbital energy level diagram. O 2-2 has a total of (2x8)+2=18 electrons (or 14 valence electrons), so the bond order is 1. O - 2 has a total of (2x8)+1=17 electrons (or 13 valence electrons), so the bond order is 1½. 8. Which is the shortest bond in each group? C-C or C-N or C-O since O has smallest atomic radius H-H or F-F or Cl-Cl C-F or C-Cl or C-Br C-C or C=C or C C since bond length as # bonds C-H in C 2 H 2 or C-H in C 2 H 4 or C-H in CH 4 The carbons have sp, sp 2, and sp 3 hybridization respectively. The hybrid orbitals with the most s character (and least p character) will be closest to the carbon nucleus, forcing the 1s atomic orbital on H to come closest for effective overlap. 9. Distinguish between a localized and a delocalized model for bonding. Give example(s). A localized bonding model always places a pair of bonding electrons between just two nuclei. For example, the Lewis electron dot structure for CH 4 assumes that a particular pair of electrons is located between a particular hydrogen and the central carbon atom. The valence bond / hybridization model says the same thing, emphasizing that the bonding electrons correspond to enhanced electron density (due to overlap of bonding orbitals) directly between the 2 nuclei with their positive charges. The Lewis electron dot structure for O 3 puts 2 electron pairs between one end oxygen and the central oxygen, and one electron pair between the other end oxygen and the central oxygen. Or we can view O 3 as having sp 2 hybrid orbitals on the central atom and one end atom, with two orbitals overlapping to form a sigma bond. The unhybridized p orbitals on those two atoms would overlap to form a pi bond where the enhancement of electron density would be above and below a line connecting the two nuclei. The other end atom would have sp 3 hybridization, with one hybrid orbital overlapping an sp 2 hybrid orbital on the central atom, forming a sigma bond. For both models I have described only one resonance structure. The molecular orbital model allows for delocalized electrons which eliminates the need for resonance. In the case of O 3 it is simplest to use sp 2 hybrid orbitals on all three atoms and form essentially localized sigma bonds between the central atom and each end atom. The remaining two electron pairs are then put into molecular orbitals formed from the three unhybridized 2p atomic orbitals. These molecular orbitals extend over all three atoms, and hence are said to be delocalized.