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(Bonding)

Nodal Plane. -. -. -. -. +. +. +. +. A. B. A. B. *. s. 2p. -2p. 2p. 2p. 2p. 2p. z. z. z. z. z. z. -. -. -. -. +. +. +. A. B. A. B. (Bonding). s. z Axis is taken to be line through A, B. z axis. (Antibonding). 2p z (A) + 2p z (B). z axis.

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(Bonding)

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  1. Nodal Plane - - - - + + + + A B A B * s 2p -2p 2p 2p 2p 2p z z z z z z - - - - + + + A B A B (Bonding) s z Axis is taken to be line through A, B z axis (Antibonding) 2pz(A) + 2pz(B) z axis 2pz(A) - 2pz(B) Molecular Orbitals Atomic Orbitals of Atoms A,B Bonding with 2p Orbitals

  2. Nodal Plane - + - + A B A B - - + + p* p -2px 2p 2p 2px (antibonding) x x 2p + + + A B A B - - - (bonding) 2p Note: orbital is  to bond (z) axis x z 2px(A) - 2px(B) x A,B Nodal Plane z 2px(A) + 2px(B) Molecular Orbitals Atomic Orbitals of Atoms A,B Bonding with 2p Orbitals (cont)

  3. Notational Detail Oxtoby uses two different notations for orbitals in the 4th and 5th editions of the class text: 2pz in the 4th edition becomes g2pz in the 5th edition 2pz* in the 4th edition becomes u2pz* in the 5th edition 2px in the 4th edition becomes u2px in the 5th edition 2px* in the 4th edition becomes g2px* in the 5th edition

  4. CORRELATION DIAGRAM for second period diatomic molecules s* 2p z p* p* 2p 2p y x 2p 2p 2p 2p s 2p z p p 2p 2p x y p* p* p p 2p 2p 2p 2p y y x x 2s 2s 2s 2s s* s* s* s s s 2s 2s 2s 2s 2p 2p z z E B2 Z ≤ 7 O2 Z ≥ 8 Molecular Orbitals Molecular Orbitals Note unpaired electrons (3 Valence electrons for each atom) (6 Valence electrons for each atom) Note order of filling 2pz Degenerate  orbitals

  5. Molecular Orbitals of Homonuclear Diatomic Molecules # Valence Electrons Bond Order Bond Length Bond Energy (kJ/mole) Mole- cule Valence Electron Configuration (Å) 0.74 431 2 1 H2 (1s)2 (1s)2(1s*)2 52 .000008 He2 4 0 2.67 105 (2s)2 Li2 2 1 2.45 9 (2s)2(2s*)2 Be2 4 0 1.59 289   (2s)2(2s*)2(2p)2 B2 6 1 1.24 599 C2 8 2 (2s)2(2s*)2(2p)4 1.10 942 (2s)2(2s*)2(2p)4 (2pz)2 N2 10 3 (2s)2(2s*)2(2pz)2(2p)4(2p*)2   1.21 494 O2 12 2 1.41 154 (2s)2(2s*)2(2pz)2(2p)4(2p*)4 F2 14 1 (2s)2(2s*)2(2pz)2(2p)4(2p*)4(2pz*)2 - - Ne2 16 0

  6. Bonding in Polyatomic Molecules Basically two ways to approach polyatomics. First is to use delocalized M.O.’s where e- are not confined to a single bond (region between 2 atoms) but can wander over 3 or more atoms. We will use this approach later for C bonding. Second is to use hybridization of atomic orbitals and then use these to form localized (usually) bonds.

  7. Localized BeH2 orbitals: sp+ } Add sp+ + 1s to get localized Be-H bond . . + – + + H H Be } Add sp- + 1s to get localized Be-H bond . + . + – + sp-

  8. H 1s + (sp-) H 1s + (sp+) and . . . . + + – – H Be Be H

  9. BH3 Fragment: 120˚ sp2 hybrid Third sp2 points along this direction Second sp2 points along this direction + • - Boron Nucleus

  10. 3 sp2 hybrids 120˚

  11. + + + H H H Overlap with H 1s to give 3 B-H bonds in a plane pointing at 120º with respect to each other: BH3 120˚ + + B Note B has 3 valence electrons + 1s22s22p

  12. 2s + 2px + 2py + 2pz sp3 gives 4 hybrid orbitals which point to the corners of a tetrahedron. Angle between is 109º28’ sp3 [1/4 s, 3/4 p]. Tetrahedral hybrids. 4 H atoms have 4 x 1s  4 valence e -

  13. E  2p 2s2 sp3 sp3 sp3 sp3 1s2 1s2

  14. Geometry of carbon sp3/H 1s Bonds in methane (CH4): sp3 hybridization on C leads to 4 bonds. CH4 is a good example. H C H H H

  15. Summary of Hybridization Results Example Groups Attached Hybrid Geometry to Center Atom BeH2 2 sp linear H-Be-H BH3 3 sp2 trig. plane (120º H-B-H angle) CH4 4 sp3 tetrahedral (109º28’ H-C-H angles)

  16. 2pz 2py Localized Bonds and Lone Pair Electrons NH3 3H, 1s (no choice) N: 1s22s22p3 Nitrogen nucleus 1s 1s This predicts 90º geometry 2px H atom 1s orbital 1s

  17. Geometry of NH3 found to be Trigonal Pyramidal N H H H 107˚

  18. 4 N sp3 orbitals combine with 3 1s H orbitals to give 3 sp3, 1s M.O.’s leaving one sp3 hybrid left Of 5 valence e - in N, 2 go into one sp3 orbital, 3 go into other 3 sp3,s. (combined with H (1s)) One of the driving forces for the tetrahedral configuration is that it puts bonding and lone pair electron groups as far away from each other as possible.

  19. •• Lone Pair N H H H

  20. N Electron pair repulsion effect is largest for small central atoms like B, N, O. As go to larger central atoms (e.g. S or metals) frequently find this effect not so large and start to get things closer to pure p orbital bonds (90º structures) For example, H2S has H-S-H angle of 92º.

  21. • • • • H2O: 2H 1s (2e-) O 1s22s22p4 (6 valence e-) 4 sp3 hybrids. Use 2 to make MO bonds with H 1s. Lone pairs stick into and out of paper Predicts H-O-H bond of 109º28” - Find 105º O H H

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