Hw9K-G1-MolStruct-PDF

Hw9K-G1-MolStruct-PDF - HW SET 9 - KEY MOLECULAR STRUCTURE...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW SET 9 - KEY MOLECULAR STRUCTURE CHM 2045 MGR-G1 VOL Valence Bond Theory! Hybridization of Atomic Orbitals ‘l. a) :5—5l‘. b) Two 0 bonding pairs are present, one between each chlorine atom " and the central oxygen atom. Two a lone pairs or nonbonding pairs are located on the oxygen atom. 0) Oxygen has a ground state valence electron configuration of 2522p”: leaving two 2p orbitals (with one electron each) that are available for the formation of two covalent bonds. Each chlorine atom has a 3523p5 configuration. with one half filled 3p orbital available to form one bond. The central atom determines the positioning of O the peripheral atoms, and this would be determined by the orientation of the two 2p ©0®CIO Orbitals of oxygen which are 90° apart from each g other. 40» 5' d) l Ji L _* spa sp‘a sp‘3 sp3 - 'i Energy Hybridization Bonding c7C” ',O\_ I CI' 61' ' .Ll .L _L 1095"" _u 2p 2D 2p 25 it u..— _1_ sp‘ 893 in sin Cl Cl 1 t_l ll. 5 3 3 If the 5p3 hybridization of oxygen were the only consideration, the Cl—O—Cl bond angle would be 109.5“. and the molecule would have a net angular or bent shape. Remember the net shape of the molecule is determined from the geometric placement of the atgmic nuclei within the molecule (These dense regions of matter are fairly easily located by experiment). it is also important to rememberthat where nuclei are placed is determined by the locations of both the bonding and nonbonding (lone pairs) electrons in their reSpective orbitals. e) The observed Cl-O-Cl bond angle of 110.8" is very close to the 109.5° bond angle expected fortetrahedral sp3 orbital hybridization versus the 90° angle anticipated for bonding with two unhybridized 2p orbitals on oxygen. The slight increase (1.3”) observed in the angle relative to a pure tetrahedron is likely to come from the steric repulsion between the two large chlorine atoms which tends to widen the angle. 2. Usually there is a distinct advantage to bonding with hybrid orbitals over unhybridized orbitals. The n bonds formed with hybrid orbitals are generally capable of greater volume of orbital overlap, forming stronger bonds. Hybrid orbitals also allow the placement of both bonding and nonbonding electron pairs as far apart from one another as possible around the central atom, minimizing electron pair repulsions. Finally hybridization m permits the formation of a These three factors all contribute to a lower energy greater number of covalent bonds. conditign and greater stability. 3) One 5 and two p yields three so2 hybrid orbitals with trigonal planar placement and 120° orbital angles. b) One 5, three p, and one d orbitals form five sp3d hybrid orbitals with trigonal bipyramidal placement. Two axial and three equatorial positions are produced. The angle between any two equatorial orbitals is 120° while the angle between an equatorial and an axial orbital is 90° c) One 5, three p, and two d orbitals derive six sp3d2 hybrid orbitals of octahedral orientation. The angle between adjacent spE‘d2 orbitals is 90". 51) Linear: - Two Sp hybrid orbitals 180° apart b) Tetrahedral: - Four spa hybrids 109.5" apart c) Trigonal planar: - Three sp2 hybrids 120' apart d) Octahedral: Six spad2 hybrids with 90° angles between adjacent orbitals e) Trigonalbipyramidal: - orbitals and 90' between axial and equatorial orbitals. a) Nitrogen's five valence electrons occupy the n = 2 principal energy level within which there are one 25 orbital and three 2p orbitals for a total of four orbitals. The maximum state of hybridization that nitrogen may achieve is four spa hybrids (nitrogen may also hybridize sp and spz). For nonmetals in the third period and higher d orbitals are present in the valence level and may also be hybridized (five sp3d and six spad2 hybrids are possible). b) Oxygen also is a period 2 element, and has only one 25 orbital and three 2p orbitals in its valence level and therefore may use only these four orbitals for bonding which precludes formation of six bonds required of OFE. Sulfur, on the other hand, is a third period element whose valence electrons are in the n = 3 level. Included in this level are one 35 orbital, three 3p orbitals, and five 3d orbitals. This allows the formation of six spadz hybrid orbitals used in the bonding within SFE. All three species are sp3 hybridized yielding approximate 'l' tetrahedral orbital geometry. All the o electron pairs KN around the nitrogen in the ammonium ion, NHJ,+ , are H' l \ equivalent bonding pairs with equal repulsions and 105V H therefore achieve pure tetrahedral angles of 109.5“. Amide ion, NHZ' , has two 0 bonding pairs (BP's) and two -I- o lone pairs (LP's) occupying the sp3 orbital set, but N because electron repulsion is greatest between the two \H LP's, followed by the BP- LP repulsions, the angle 107'\_/1H between the two BP's is compressed to less than 109.5”, specifically 105'. in ammOnia, NH3 , only one LP is present along with three BP‘s. all located within an sp3 ill hybrid orbitals. While LP—BP repulsions are greater than ‘N *' BP-BP repulsions and result in compression of the angle Ht’ \ H between the two ElP's (to 107“), the compression is less 1095;} H than that encountered with the two LP's found in NHZ‘. Five sp3d hybrids with 120° angles between adjacent equatorial 7. a) cc:4 :6“ =é—c-@= CI 4 0 BP'S g) BeHz H- Be-H 2 D" BP'B sp i) (3313‘ Br- (ES- Br Br 3 o BP'S 1 0 LP sp3 0: I ,0 CI'{ \Cl C Tetra hedral 'I' .N 01"! \CI 0] Net trigonai pyramid Net trigonal planar H-Be-H Linear Net trigonal pyramid 2 U BP'S 2 o LP'S j) SCI: §_@: :9]: 2 0 BP'S 2 a LP‘s Sp .1. ,0 F"! \s F Net angular .I. +1 H"70\H H Net trigonal pyramid F I -1 F'Ff\F Tetrahedral Tetrahedral .1. Gig? ‘9 Net Angular mscl4 m—é—m d”b1 1 0 LP 4 a BP's sp3d 3 o LP'S 2 o BP's spad q} F20" Net see-saw 1 Iflf\l Net trigonaf pyramid Net linear Net linear 2 o BP's 2 0 LP'S sp3 Cl Cl." 11,,CI s m/iWu m Octahedral Trigonal bipyramid FKT,F ' Kr- F’L‘F Net square planar +1 .1. . t P? \I. Net angular Y) N92+ O=N=O 2 a BP'S '.-. 5p cr9'4 ‘ n—Cl 01/81 C Trigonal bipyramid Net linear F 4 ‘xBe-F ,1 Net trigonal planar O=N=O Linear 1) BrF5 5 U BP'S ‘i 0 LF’ 5p3d2 BH2+ H—euH 2 O BP'S sp cs2 ='s'=c=é= 2 a BP'S sp HONO H—Q—fi=6= On N: 1 0 LP 2 0 BP's Sp F._ 'i' (,F ‘81: / i N F F F Net square pyramid H—e-H Linear S=C=S Linear 'N=6: HO, Net anguiar around N 8. 10. Sigma versus Pi Bonding b) b) a) A 0 bond is formed from the head-to-head overlap of an atomic orbital of one atom with an atomic orbital of another atom (Usually these atomic orbitals are hybridized.) The electron density along the axis connecting the two nuclei of the bond is symmetrical. 9X? 00 CO A n bond results from the side-to-side overlap of two unhybridized p atomic orbitals. Orbital overlap occurs in two different regions in space and therefore the electron density is unsymmetrical about the bond axis. or may: 0 Bond 11 Bond A 0 bond is generally strongerthan a n bond for a number of reasons. The electron density of a 0 bond is concentrated in a region between the two nuclei. This not only allows both nuclei to experience the stabilizing effect cf electron sharing close up, but also provides for more efficient screening of the positive charges of each nucleus from one another and therefore decreases internuclear repulsion. The n bond, on the other hand, concentrates its electron density in two more distant regions above and below the internuclear axis. The rt bonding electron pairtherefore is less efficient than a o bonding electron pair at screening internuclear repulsion. As an example, the 0 bond in a typical C=C double bond has a bond energy of 347 lemol, whereas the n bond has a bond energy of 264 lemol. A double bond contains one o and one n bond. A triple bond contains one o and two 11 bonds. If an atom is sp2 hybridized, it has already used two of its three p atomic orbitals to form the three sp2 hybrid orbital set, leaving one unhybn‘dized p atomic orbital available to form E n bond with a p orbital on an adjacent atom. if an atom is sp hybridized, one atomic p orbital has been used to form the hybrid set, leaving two unhybridized p orbitals availabte to form two 11 bonds. If an atom is sp3 hybridized. all three p atomic orbitals have been hybridized and are unavailable to form n bonds. (Note that elements in the third period have d orbitals in their valence level and which have the potential of forming their own kind of n bond.) All O-to-H bonds are n bonds. False: H has only one s orbital (and no p orbitals) in its valence level and therefore may only form one 0 bond. All C-to-H bonds are 0 bonds. True All C—to-C bonds consist of a o and a n bond. False: A 0-0 single bond consists of only one a bond and a C C triple bond is made of one o and m n bonds. Only the C=C double bond contains one a and one 11 bond. All C-to-C are n bonds. False: All C-to-C bonds contain one a bond. 11. a) HOl'\l=C)HCH3 b) acrolein Atomic hybridizations: Atomic hybridizations: mam3 (2m:2 (3}le2 (4)893 (nspz (asp? Angles: Angles: (a) 1095- (b) 120° {3) 120= (b) 120"- (c) 120- (d) 120° c) tabun Atomic hybddizafidns: (1) $93 (2) 593 (3) 5p (4) sp“ Angles: (a) 1095- (b}1BD’ (c) 109_5= d) aspirin e) urea f) histamine Atomic hybridizations: Atomic hybridizations: Atomic hybridizatlons: (1)5132 (2) Sp2 (3)5133 (1)893 (2)592 (Uspa (2) sp3 (4)5133 (3)592 (4) Sp2 Angles: Angles: Angles: (3)120” (b)120' (3)1095" (b) 120- @1095- (b)‘120' (c) 1095- @1035- (c) 109.9- 12. a) HzNNHZ b) HONO H K: N H r~ll FJ-l H 6 N—é- fa l'll H}? \N’ " ° "‘0’ - ‘r l ./ \H Each N: 2 o LP‘s 2 a BP's ‘IoLP 2UBP'5 10LP 3 o BP's - sp3 - sp2 sp3 2% 3 a BP'S. sp2 h H __ :O=N=O: N: 2 o BP's, sp 0: 2 ULP'S, 1 0 BP, sp2 9) NO+ N: 1 0 LP. 1 0 BP, 5p 0: 1oLP. ‘l UBP, sp N: loLP, ‘l aBP, 5p C: 2 o BP‘s, sp g} :o=cr-1—r'_:'_|: o; 2 o LP‘s. 1 0 BP, sp2 h) cron=cncr c: 3 o BP's, sp2 C: 3 o BP's. sp2 sag—0’s a H.‘O“O..H HJfli ,cr . _C. t _C, ‘ _C,. -. \CI cV’M‘m Clm\H cis trans Note that the 11 bond prevents rotation from one planar form to the other. These molecules are geometric isomers. 13. a) The n bond in NOZ‘ is delocalized. This is clearly indicated by the fact that two equivalent resonance contributors may be written: b) Remember that resonance contributors represent the actual electronic hybrid of the molecule or ion. which is also being represented by the delocalization of the 11 system by the extended overlap of p orbitals. We are describing the same system. +1 +1 +1 ' a r_i s . o m D ‘ \ 1 " ‘\ ‘I 4’ \p: :9.” \pj- _ e of \0 fr 0/ \o isomers 14. a) Which of the following pairs of molecules are structural isomers? (1) HOCN and cho (2) CHBCHz-o-CH3 and CH3-O-CHZCH3 Structural isomers Identical (3) CH3CHZCHZCH3 and CH3CH(CH3)2 (4) CH3CH2-O-CH3 and CH30H20H2—OH Structural isomers Structural isomers b) Which of the following pairs of molecules are geometric isomers? {1) (2) CH3 CH3 OH H CH CH3 )c=c< 3;: c< :>D=CH2 H20: cf: H H H CH3 CH3 CH3 Geometric isomers Identical (3) Pt(NH3)2Cl2. is sguare Dianar [4) HZN.‘ “NH; H2N“ ,CI klsz ,NHz CL‘ ,NHZ .P, .P.- t . .Pt- ci’ 1‘CI of Kmil-i2 Cl/PKCI of ‘NH2 cis trans identical c) Which of the following pairs of molecules are optical isomers? (1) '(2) as to ’i ‘i wf‘mrt HO” "‘H Cl" “C! CVCQCI c l _ H OH Optical isomers Identical (Note four difierent groups on tetrahedral carbon (Note two identical groups on carbon yield nonsuperimposible mirror images via asymmetric carbon) yieId symmetric carbon} {3) (4) *i i‘ a it‘s CH? COZH HOZC"'Q{‘ CH3 H ’7‘C‘CHZCI CICHz/Cx‘g +1 H2 NH2 HZN iii-12 Optical isomers Optical isomers ...
View Full Document

Page1 / 9

Hw9K-G1-MolStruct-PDF - HW SET 9 - KEY MOLECULAR STRUCTURE...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online