04CHM114_0305 - Cite-M m(p.231 3,009 m Irma 1—6 cru RE A...

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Unformatted text preview: Cite-M m; (p.231 3,009, m: Irma; 1—6 cru RE A: 0. {q 3. We may find non-Lewis structures, that are still the best representation (a) electron deficient (< 8e-) 22:3 .‘ F3 J-h‘u ‘BI-F: .. . l a. B’ r- .1:- u “55“ ' 2F... ‘6‘“ Owl?! ‘0? F 3": a I; tug-‘1 Pan-0h“. H§:/‘:PB¢! _9 NfigiJfi ‘5. M n . . (b) electron surplus ( > 39-) (mostlyuh' . fluorine-containing... F breaks any rule to get “Sling-m; hands on electronsl!) PF5 and PE; ( for lithium batteries, until LiBOB came along) IF7 14 e' around iodine!!! no... Bond energies and chemical reaction enthalpies: We did this section early in our treatment of heats of reaction and thermodynamics Revise this section later part of section*8.8 (see notes at end of our Chapter 5 treatment), because I would like to ask a question using it, now that you know more about chemical bonds) ‘ Concept in Fig. 8.13 N B ’ Do Problems 8.83, 8.85 . . ' fl # Chapter 9. Molecular geometry and molecular orbitals CHAPTER 9 SKIP ps 314-42 Section 9.7 (molecular orbitals) We will do a little study of this subject so (i) you can see the basics of chemical bonding in modern terms (ii) you can get an idea about metallic bonds (important for engineers) *F‘lrst, let us look at the relation between the atomic orbital "skeleton" (hydrogen atom energy level scheme) we used for building multielectron atoms, and the molecular orbital "skeleton" we will use for understanding molecular bonding: Atomic orbitals were obtained by solution of Schrodinger's wave equation for system with m electron and one center of positive charge (at nucleus) (the H atom problem, or the He+ ion problem) Molecular orbitals are obtained by solution of Schrofidinger's wave equation for system of m electron and tWO 01‘ more centers of _- ositive charge. H atom problem H; ion problem $76 6" UG-tl.‘ is“. 9 .52?“ ‘1 l "ht-+5. €- 6 who «John... M Miflhfium Out {poise-l. “Nb“. 4.!“ boa-a Ian-5w " 6.5 ® Key finding: the number of orbitals with quantum no 1 increases linearly with the number of centers of positive charge. Ic_ om- ® OBI..- (13’9er {139 Q +qu tar-L4...” ‘; [Bug @ fidu— orb {1436' Flut‘fi . \_ , *1» I rem-+8. h‘h:+:: {II-Ii“ “it '0‘?- \ I ”bi-‘5‘: 5“?“I‘ll‘v ‘Mdt‘un no '6 N n ghoul-J [hull-I- l How do we represent them? boui-Lg "5:4“ ”—3 6—K "- flw‘t’hl‘v‘t—a “ 6—“ "(3" is the Greek symbol for "s" bonding 0', and antibonding 5*, orbitals 10 ‘ 5'. «$5 “ H 0 Lu» ' ' ' " H5 H2 (1 electron) (2 electrons) bond order 1/2 bond order 1 (new term) _ . 4:! " 9% anr Affir‘oml/ECcuf-u: ”Ike... bard-m1 my Where are e electrons? (electron distributions are non symmetrical in molecular orbitals because there are two or more nuclei involved.) L... 6'? 9'33 11 “'6: Can.» Lust. ‘ ' bud-1 0;“ Bonding of helium species " according to M0 theory. H930. 1‘._*—SI: Lair-:4.” m (5‘51.th tit-“:1; The multinuclear case and .. band theory of metals. 9““- Want to explain: *high electronic conductivity *difference between metals and semiconductors Remember The number of HS orbitals with a given quantum no. 11 is equal to the number of positive Charge centers (number of atoms). .4 3‘ . Lt Points to recognize. '3 J: ‘3 H I F 'L —-""" ‘J - - . unn— ‘th # =35 5 I \ Mg... 5 12 _ ' H “‘3‘“ *Formation of bands as 11 increases *filling up of whole region between bonding and antibonding orbitals of diatomic molecule. Let‘s do it 0 sep ation of individual orbital energies becomes vanishingly small, so electrons always "free". 0 filled vs half-filled bands 0 3 band and p band overlap (later) 13 ...
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