Sheets Page 1 Lecture 16

Lecture 16: Molecular geometry & bonding 2

Read: BLB 9.1–9.3 HW: BLB 9.16,17,21,23,25,27 Sup 9:1–7 Know: • VSEPR theory electron pair (electron domain) & molecular geometry • molecules with more than one central atom • molecules with multiple bonds BST #7 Molecular formulas & structure: Feb 19: TOMORROW! 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)

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Molecules with two electron domains (AB2)

• e– pair geometry will be , • molecular geometry will be ,

Molecules with three electron domains

(AB3) • e– pair geometry will be , • molecular geometry will be or • multiple bonds & lone pairs need more space!

Cl Be Cl

180 o

O C O

180 o

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Molecules with five electron domains (AB5)

• e– pair geometry will be • angles near

# of bonded atoms

(# lone pairs) placement of bonded atoms

(molecular geometry)

5 (0) PF5 4 (1) SF4 3 (2) BrF3 2 (3) XeF2

• 2 types of sites 1. axial 90° (to equatorial) 180° (to axial) 2. equatorial 120° (to equatorial) • equatorial positions are crowded, so lone pairs prefer to be here seesaw T-shaped

Br

F

F

F

89o

S

F

F

F

F

186o

116o

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Molecules with six electron domains (AB6)

• e– pair geometry will be • angles near or

# of bonded atoms

(# lone pairs) placement of bonded atoms

(molecular geometry)

6 (0) SF6 5 (1) BrF5 4 (2) XeF4

• for 5 bonded atoms: where does lone pair go? • for 4 bonded atoms: where does the second lone pair go?

square planar square pyramidal

82o

I

F

F

FF

FXeF

FF

F

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Molecular shape summary (fill on your own)

# of

electron domains

ED geometry

# of bonded atoms

molecular geometry example

2 2 CO2

3 3 BF3 3 2 O3

4 4 CH4 4 3 NH3 4 2 H2O

5 5 PCl5 5 4 SF4 5 3 BrF3 5 2 XeF2

6 6 SF6 6 5 BrF5 6 4 XeF4

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VSEPR rules applied to other structures • molecules without a single central atom (e.g., organic molecules) • molecules with multiple bonds

trigonal bipyramidal

S

F

FF

FO

HH

H

H

H

H

H

H

H

H

122o

117o

H3C

O

OCH3

tetrahedral

tetrahedral (bent)

tetrahedral

trigonal planar

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Example: Determine the approximate bond angles indicated.

angle 1 angle 2 A. 109° 109° B. 180° 120° C. 120° 90° D. 109° 120° E. 120° 109°

NCH2CH2CH2C

H

H

O

OH

1 2

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Example: Which molecule/ion shown below is not linear?

A. XeF2 B. CO2 C. OF2 D. IF2

– E. C2H2

What are the geometries & approximate bond angles of the species in the reaction: (verify for yourself; will not go over in class) 2HF + SbF5 → SbF6

– + H2F+ SbF5 trigonal bipyrimidal; 90°, 120° SbF6

– octahedral; 90° H2F+ bent; < 109.5°

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Review of what weʼve learned so far… • determine the Lewis structure • use VSEPR to determine e– domain geometry • determine molecular geometry

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Consequences of electronegativity • type of bond (ionic, covalent); review Lect 9 • polarity: the greater the difference in electronegativity (ΔEN) between two bonded atoms, the more polar the bond This is a rough guide… • ionic: large difference in electronegativity (ΔEN > 2); e– shared e.g. NaF: 4.0 (F) – 0.9 (Na) = 3.1 • polar covalent: some difference in electronegativity (ΔEN 0.5 – 2.0); e– shared e.g. H–Cl: 3.0 (Cl) – 2.1 (H) = 0.9 δ+ δ– • nonpolar covalent: small difference in electronegativity (ΔEN < 0.5); e– shared e.g. H2, F2, C–H (hydrocarbons)

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Molecular polarity • a molecule is polar if there is a net charge separation between two “ends” of the molecule: molecule has a distinct negative “end” and a distinct positive “end” • to have a net dipole:

1. bonds 2. geometry where bond dipoles DO NOT cancel out

• to determine the polarity of a molecule that has more than 2 atoms:

1. find molecular shape (VSEPR) 2. find “bond” dipoles 3. use vector “analysis” to find net molecular dipole

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Molecular polarity • polarity of entire molecule • two factors are involved:

1. ΔEN: polarity of individual bonds in a molecule 2. VSEPR: geometric arrangement of atoms & bonds in a molecule; that is, molecular geometry

example: SeF4 1. which structure is best? & why? 2. ED? MG? 3. dipole moments & polarity?

Se

F

F

F

FSe

F

F

F

F

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Dipole moments CO2 electronegativity: C = 2.5 O = 3.5 ΔEN = 3.5 – 2.5 = 1 O=C=O

CF4 electronegativity: C = 2.5 F = 4.0 ΔEN = 4.0 – 2.5 = 1.5

F

CF

F

F

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Dipole moments (cont.) H2O electronegativity: H = 2.1 O = 3.5 ΔEN = 3.5 – 2.1 = 1.4 NH3 electronegativity: H = 2.1 N = 3.0 ΔEN = 3.0 – 2.1 = 0.9 CH3Cl electronegativity: H = 2.1 C = 2.5 Cl = 3.0 ΔEN(C-H) = 2.5 – 2.1 = 0.4 ΔEN(C-Cl) = 3.0 – 2.5 = 0.5

O

H

H

NH

H

H

Cl

CH

H

H

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Before next class: Read: BLB 9.3 HW: BLB 9.33,35,38 Sup 9:8–11 • molecular geometry • molecular polarity Answers: p. 7: D p. 8: C