Sheets Page 1 Lecture 17 Lecture 17: VSEPR & polarity 2 Read: BLB 9.3 HW: BLB 9.33,35,38 Sup 9:8–11 • molecular geometry • molecular polarity Exam #2: Monday, March 2 @ 6:30; review previous material, so you UNDERSTAND what we’ve done and what we are doing now, and start preparing now! Last day to sign up for conflict exam is Wed, Feb. 25. Final Exam: Monday, May 4 @ 12:20; MUST register on elion for a final exam conflict or overload by March 8. See http://www.registrar.psu.edu/exams/exam_overload.cfm http://www.psu.edu/dus/handbook/exam.html#conflict Need help?? Get help!! TAs in CRC (211 Whitmore) and Supplemental Instruction (SI)—hours on Chem 110 website; Sheets office hours: Mon 12:30-2; Tue 10:30-12 in 324 (or 326 Chem Bldg)Sheets Page 2 Lecture 17 Calculating a dipole moment • dipole moment (µ): an observable measure of the overall polarity of a molecule H–F ΔEN = 4.0 – 2.1 = 1.9 • µ is a : it has magnitude & direction; debye (D) = 3.33 × 10−30 C m • for a diatomic molecule: µ = Qr • r is separation distance • Q is charge; –Q and +Q are equal in magnitude for diatomic molecule; units of Q ⇒ 1.6 × 10−19 C = charge of an e– • for two charges (+1 and –1) separated by 1 Å, µ = 4.8 D • in polyatomic molecules, µ depends on: 1. bond polarities 2. molecular geometriesSheets Page 3 Lecture 17 Dipole trends in HX HX bond length [r] (Å) ΔEN µ (D) (expt.) µ (D) (ionic)* % ionic HF 0.92 1.9 1.82 4.42 41% HCl 1.27 0.9 1.08 6.10 18% HBr 1.41 0.7 0.82 6.77 12% HI 1.61 0.4 0.44 7.74 6% * if full charge separation occurs; µ = 4.80 r D (where r is bond length); see prev page X = halogen; µ(expt.) = experimentally observed µ; % ionic = µ(expt)/µ(ionic) × 100 • consider HCl: full charge separation (ionic): µ= 6.10 D but HCl molecule has an actual µ = 1.08 D thus, the charge separation in HCl is only (1.08 D / 6.10 D) = 0.18 of the full charge therefore, itʼs polar covalent H⎯X δ+ δ–Sheets Page 4 Lecture 17 Oxidation numbers, formal charges & partial charges • oxidation number: a charge an atom would have if its bonds were completely ionic: i.e., assign e– to the more electronegative atom (Chap 4 ~Lecture 34; FYI right now…); electronegativity is overemphasized Oxidation number—FYI only for now… 1. atoms in elemental form: 0 2. monoatomic ions: charge on the ion 3. nonmetals are usually negative • O is –2, except in peroxides where it is –1 • H is +1 when bonded to nonmetals, and –1 when bonded to metals • F is –1, other halogens –1 in binary compounds, but positive in oxyanions 4. sum must equal the charge on the species • formal charge: a charge an atom would have if all bonding e– were shared equally; FC = VE – LSE (review Chap 8); electronegativity is deemphasized • partial charge: “real” charges on atoms in molecules; most accurate way of representing charges; electronegativity is explicitSheets Page 5 Lecture 17 Oxidation numbers, formal charges & partial charges (cont.) • consider HCl: oxidation numbers formal charges partial charges (real) H–Cl H–Cl H–Cl polar covalent µ = 0.18 • consider HCl: full charge separation (ionic): µ= 6.10 D but HCl molecule has an actual µ = 1.08 D thus, the charge separation in HCl is only (1.08/6.10) = 0.18 of the full charge • dipole moment is one measure of the actual charges on atoms in a moleculeSheets Page 6 Lecture 17 Molecular geometry & dipole moment Do not memorize this table; think things out! NOTE: this table is presented in terms of molecular formula not in terms of ED geometry as weʼve seen before. Draw the generic structures & think about them to verify! NOTE: X is the same atom on all positions; A is the central atom formula molecular geometry dipole moment AX linear ? HCl AX2 linear 0 CO2 bent ? H2O AX3 trigonal planar 0 BF3 trigonal pyramidal ? NH3 T-shaped ? AX4 tetrahedral 0 CH4 square planar 0 seesaw ? AX5 trigonal bipyramidal 0 square pyramidal ? AX6 octahedral 0Sheets Page 7 Lecture 17 Dipole moments of cis & trans isomers • which molecule is polar and why??? cis trans • this is an example of a geometric isomer; atoms are connected in the same order & with same bonds BUT spatial orientation is different; this is NOT the same as structural isomers (see Lecture 13) electronegativity: H = 2.1 C = 2.5 Cl = 3.0 bonds: C—H C—Cl ΔEN = 2.5 – 2.1 = 0.4 ΔEN = 3.0 – 2.5 = 0.5 what about symmetry of the molecule? NET dipole? C CClHClHC CHClClHSheets Page 8 Lecture 17 Example: CO2 is not polar Is SO2 polar? (Lect 16, p 13) CF4 is not polar Is SF4 polar? (Lect 16, p 13)Sheets Page 9 Lecture 17 Yet another example: Rank by increasing polarity ClClClA BC DE FCl FF FClFFFFFFSheets Page 10 Lecture 17 • Lewis structures: accounts for e– pairs involved in bonding & lone pairs (non-bonding) • VSEPR: accounts for ED & molecular geometry • so, how are bonds made anyway? & where are the electrons???Sheets Page 11 Lecture 17 Molecular orbitals • e– distribution in atoms: atomic orbitals (s, p, d …); orbitals give us the probability of finding an electron (review Chap 6) • two models describe e– distribution in molecules: • valence bond theory: valence orbitals on one atom with valence orbitals on another atom, resulting in a covalent bond (what weʼre talking about now) • molecular orbital theory: a better model that uses wave theory, BUT not covered in Chem 110 [will see in organic chemistry]Sheets Page 12 Lecture 17 Covalent bonding • covalent bonding: overlap or combining of two singly occupied atomic orbitals to form a new doubly occupied molecular orbital, allowing for sharing of electrons by the two nuclei; e– of opposite spin share the common space between nuclei • two forces operating: ⇒ increased overlap of atomic orbitals (better sharing) ⇒ closer distance between nuclei increases e– density between nuclei thereby lowering energy (see next page) balance of forces ⇒ bond length (0.74 Å for H2) recall: E !Q1Q2dSheets Page 13 Lecture 17Sheets Page 14 Lecture 17 Two bond types 1. σ (sigma) bond: results from overlap of orbitals; e– density is symmetric about the internuclear axis examples: s-s s-p s-spxp-p spx-spxinternuclearradius where spx and spy are hybrid
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