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Chem 6A Michael J. Sailor, UC San Diego

Chapter 10:The Shapes of Molecules

Chem 6A, Section D Nov 10, 2011

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Announcements:• Quiz next Thursday (Nov 17) is on Chapter 9.

That quiz will also review concepts from chapters 7-8

• Practice Final is posted on the web:http://sailorgroup.ucsd.edu/Chem6A_sailor/Final_Exam_MASTER.pdf

• Tues Nov 22 (Thanksgiving week) is review day• No office hours Weds Nov 23


N

H

H

H

PROBLEM: Drawing Lewis structuresDraw the Lewis structure of NH31. Count total valence electrons in the molecule2. Make bonds to satisfy octet rule

(a) Electrons left over? Make lone pairs(b) Too few electrons? Make double (or triple) bonds

Lone pair


PROBLEM: Lewis Structures(see problem 10.5, 10.7, 10.11)

Which of the following is the most plausible Lewis structure for the nitrate ion:

(a) (b) (c) (d) (e)


Resonance in Lewis Structures(see problem 10.15)

Resonance occurs when more than valid Lewis dot structure exists. Draw all the resonance structures for the nitrate ion:




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Announcements:• Thurs Nov 17 quiz (#8) will be on Chapter 9,

with some review from chapters 6-8


Quiz 7 score histogram


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Draw the Lewis Structure for XeF2

Draw the Lewis Structure for BF3

F

Xe

F

B

F

FF


Violations of the Octet Rule

EXCEPTIONS TO THE OCTET RULE

Too few electrons/atom: BF3number of valence electrons: 3(7e-) + 3e- = 24

Too many electrons/atom: XeF2number of valence electrons: 2(7e-) + 1(8e-) = 22


Violations of the Octet Rule

How can you exceed the octet? Use d-orbitals

Some Guidelines for Octet Rule Violations:• Never violate: C, N, O, F

• Often have too few electrons: B, Be

• Never exceed octet: all 2nd row elements (Li, Be, B, C, N, O, F), because there are no 2d orbitals, and the 3d orbitals are too high in energy).


Pauling Electronegativity ScaleFig. 9.19


Using Electronegativity to Classify Bonds Fig. 9.21

Things that lead to increased covalent character:For anions: Polarizable large, highly negativeFor cations: Large polarizing power, small, highly positiveExamples: Ba-Cl (Δ electronegativity) = 3.0 – 0.9 = 2.1 ionicBi-I (Δ electronegativity) = 2.5 – 1.9 = 0.6 polar covalentSi-H (Δ electronegativity) = 2.1 – 1.8 = 0.3 covalentO-H (Δ electronegativity) = 3.5 – 2.1 = 1.4 polar covalent


Formal Charges(see problem 10.15)

•Count number of electrons around each atom and compare it to the number of valence electrons for the atom. If the numbers are different, you must assign formal charges

•Molecules generally try to minimize the number and magnitude of formal charges. The most electronegative atoms will hold negative formal charges.



Example: Which is the best Lewis structure for BF3?

F

BF F

F

BF F

F is too electronegative to hold a positive formal charge.



Example: Which is the best Lewis structure for sulfate?

S

O

O

O

O

2-

S

O

O

O

O

2-

S

O

O

O

O

2-

Molecules try to minimize the number and size of formal charges. The most electronegative atoms will

hold negative formal charges.


PROBLEM: Formal Charges(see problem 10.15)

Which Lewis structure of chlorate shown below has the most stable distribution of formal charges?

(a) (b) (c) (d) (e)

For all Lewis structures, you must consider: (1) Number of electrons(2) Octet rule(3) Distribution of formal charges


Valence Shell Electron Pair Repulsion (VSEPR)

Valence Shell Electron Pair Repulsion Model

Used to predict structures of main group compounds.

The main concept is that electrons in bonds and lone electron pairs repel each other and will arrange themselves around the molecule to maximize their distance apart.



Example: What is the shape of BF3?

1.Draw Lewis dot structure, figure out how many lone pairs + bonding pairs you have (the coordination number).

2.Use idealized structures for number of atoms + lone pairs

3.Give lone pairs more room


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PROBLEM: Drawing VSEPR structures from Lewis structures

Recall the Lewis structure of NF3:

N

F

F

FLone pairs

Total electrons: 26Total pairs of electrons: 13

Total bonding pairs: 3Total lone pairs: 10




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Announcements:• Thurs Dec 1 quiz (#9) will be on Chapter 10• Final exam will be Fri Dec 9, 7-10 pm

- Final will have 50 multiple choice questions- You will have to calculate numeric answers


Final ExamFinal Exam: Fri Dec 9, 7 pmBring:• Calculator (no iPhones, PDAs, anything that can

transmit or receive RF)• #2 Pencils• Binder or other hard surface to write on• 5x7 note card (your handwritten notes can be on both

sides)• Your UCSD ID card


N

H

H

H

PROBLEM: Drawing VSEPR structures from Lewis structures

the Lewis structure of NH3:

Lone pair

What is the structure?


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Structure of NH3 (trigonal pyramidal):

NHH

H

NHH

HIDEAL ACTUAL

107.3°109.5°



Why give lone pairs more room? Lone pairs require more space than bonding pairs. Example: H2O

Draw the VSEPR structure of the H2O molecule. Is the H-O-H angle greater than or less than 109.5°, and why?

H

H

O



ANSWER:

experimentally determined angle is 104.5°

H-O-H angle is < 109.5 because the lone pair electrons require more space than bonding pairs

H

O H


PROBLEM: Molecular Shape and Polarity (VSEPR)

Is ClF3 polar? Draw the VSEPR diagram and show the direction of the molecular dipole.

Atom Valence electrons

Cl 73 x F 3x7=21

Total: 28

T-shaped

F

Cl

F

F




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Announcements:• Thurs Dec 1 quiz (#9) will be on Chapter 10• Final exam will be Fri Dec 9, 7-10 pm

- Final will have 50 multiple choice questions- No penalty for guessing- You will have to calculate numeric answers


Final ExamFinal Exam: Fri Dec 9 from 7-10 pmBring:• Calculator (no iPhones, PDAs, anything that can

transmit or receive RF)• #2 Pencils• Binder or other hard surface to write on• 5x7 note card (your handwritten notes can be on both

sides)• Your UCSD ID card Location: Mandeville Auditorium


Final Exam location

You are here

MandevilleAuditorium


Final Exam location

You are here

MandevilleAuditorium


Quiz 8 score histogram


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Final ExamRecommendations for studying for

the final:• Do practice final (http://sailorgroup.ucsd.edu/Chem6A_sailor/Final_Exam_MASTER.pdf)• Do all (3) versions of all quizzes posted online• Make 5x7 card


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PROBLEM: Drawing VSEPR structures

Draw the VSEPR structures for XeF3+, SbBr3, and GaCl3.

Indicate the dipole moment of each molecule (if it has one) with a dipolar arrow.






Xe 83 x F 3x7=21

charge Less 1 electron

Total: 28

T-shaped

F

Xe

F

F

+






Sb 53 x Br 3x7=21

Total: 26

Trigonal pyramidal

SbBrBr

Br






Ga 33 x Cl 3x7=21

Total: 24

Trigonal (planar)

Not polarGa

Cl

Cl

Cl


Extra Credit(Was Due Nov 17)

The vial contains NO2(g) 1. Draw the Lewis structure of the clear gas

that forms when the tube is cool.2. Draw the Lewis structure of the brown gas

that forms when the tube is warm.3. Why do the colors change?


Extra Credit

(1) Draw the Lewis structure for NO2(g)Valence electrons: N = 5, 2O = 2x6 = 12, so total

number of electrons = 17e-

A molecule with an odd number of electrons is called a free radical

ONO

ONOor


Extra Credit

(2) Draw the Lewis structure of the clear gas that forms when the tube is cooled.

N N

O O

OO

N2O4


The Energy Spectrum


The Solar Spectrum


Environmental Chemistry

1995 NOBEL PRIZE in CHEMISTRY:

http://nobelprize.org/nobel_prizes/chemistry/laureates/1995/

UCSD"for their work in atmospheric chemistry, particularly concerning the formation and decomposition of ozone".


Upper atmosphere ozone depletion

Crutzen, Molina, and Rowland won the Nobel prize for identifying the important man-made sources of chemicals that can deplete ozone, and for realizing that those chemicals are removing ozone faster than Nature can supply it.

O O

Oozone

Cl

CF

F

F

Cl

ONO

N O

Chem 6A Michael J. Sailor, UC San Diego44

The Ozone Layer

Image: www.nasa.gov

Ozone (O3) found between 10 and 30 km above earth's surface.(a 747 cruises at an altitude of ~10 km)

O3 + UV light (λ< 320 nm) → O2 + OOzone is depleted by reaction with light:

O2 + UV light (λ< 240 nm) → 2 OO2 + O + M → O3 + M(M is some other air molecule (N2 or O2)

Ozone is also regenerated by reaction with light:


Ozone absorbs UV-C light

Molina’s UV-C absorption spectrum of ozone(from “Chemistry of the Environment,” R.A. Bailey)

O3 + UV light (λ< 320 nm) → O2 + O


Upper atmosphere ozone depletion by chlorofluorocarbons

1974, Mario Molina and Sherwood Rowland identified the threat to the ozone layer from chlorofluorocarbon (CFC) gases - "freons" - from spray bottles, refrigerators, air conditioners, and plastic foam manufacture. Based on 2 prior observations: •James Lovelock (England) showed that man-made, chemically inert, CFC gases had spread globally throughout the atmosphere.

•Richard Stolarski and Ralph Cicerone (USA): free chlorine atoms in air decompose ozone catalytically.


How do chlorofluorocarbons deplete ozone?

Initiation:Cl

CF

F

F

+ UV light CF

F

F

+ Cl

Propagation: Cl. + O3 → O2 + OCl.

OCl. + O → O2 + Cl.

Net: O3 + O → 2O2Catalytic reaction: Cl. is not used upAfter initiation, no more UV light is needed

Freon-13


Nitrogen oxides also deplete ozone by a catalytic reaction

Propagation:NO + O3 → NO2 + O2NO2 + O → NO + O2

Net: O3 + O → 2O2

Catalytic reaction: NO is not used upNo initiation step

Paul Crutzen (1970) showed that the nitrogen oxides NO and NO2 react catalytically with ozone:


Upper atmosphere ozone depletion

• 1957 Team of British scientists led by Joseph Farman began measuring Ozone at Halley Bay, Antarctica.

• They noticed a drop in Ozone level beginning in 1977, correlated with the appearance of CFCl3 and CF2Cl2

• NASA, using NIMBUS satellites with more sensitive instruments, did not report a similar drop.


Farman’s Halley Bay, Antarctica data

• Ozone at Halley Bay, Antarctica over 26 years.

• Drop in ozone level correlates with increase in appearance of CFCl3 and CF2Cl2

Increasing CFC

l3 and CF

2 Cl2


Upper atmosphere ozone depletionNature 315, 207 - 210 (16 May 1985)Large losses of total ozone in Antarctica reveal seasonal ClOx/NOx interactionJ. C. FARMAN, B. G. GARDINER & J. D. SHANKLIN

“…We suggest that the very low temperatures which prevail from midwinter until several weeks after the spring equinox make the Antarctic stratosphere uniquely sensitive to growth of inorganic chlorine, ClOX, primarily by the effect of this growth on the NO2/NO ratio. This, with the height distribution of UV irradiation peculiar to the polar stratosphere, could account for the O3 losses observed.”


The hole in the ozone layer

http

://to

ms.

gsfc

.nas

a.go

v


Chlorofluorocarbons (CFCs) release chlorine atoms that deplete ozone

CFCs have been used as refrigerants75% of atmospheric ClO comes from CFCs

Montreal Protocol: Phaseout of CFC production and use: December 31, 1995 marked the end of the production of CFCs in the industrial world.


Montreal Protocol of 1987Ch

lorin

e in

stra

tosp

here

Predicted change in chlorine content in the stratosphere for three different scenarios:

a) Without restrictions on release

b) Limitations according to the original Montreal Protocol of 1987

c) Release limitations now internationally agreed.

Chlorine content is a measure of the magnitude of ozone depletion.


Results of the ban on CFCs

“Currently, we are experiencing depletion of approximately 3 percent at Northern Hemisphere mid-latitudes and 6 percent at Southern Hemisphere mid-latitudes…if no action had been taken to limit CFCs, ozone depletion at mid-latitudes would eventually have reached 20 percent or more…If international agreements are adhered to, the ozone layer is expected to recover around 2050.” –US EPA, http://www.epa.gov/ozone/geninfo/benefits.html

“CFCs are no longer accumulating in the atmosphere at an accelerating rate”


What to use in place of chlorofluorocarbons (CFCs)?

hydrochlorofluorocarbons (HCFCs)hydrofluorocarbons (HFCs)

HCFCs and HFCs are more chemically reactive than CFCs and decompose in the troposphere before reaching the stratosphere.

C C

FH

F F

H F

HFC-134a

C C

ClH

H F

H F

HCFC-142b


Why are hydrofluorocarbons more reactive than chlorofluorocarbons?

C C

ClH

H F

H F

HCFC-142b

Bond EnergykJ/mol

C-C 347C-H 413C-F 453C-Cl 339O-Cl 203O-H 467O-F 190 C C

ClCl

F F

Cl F

Freon-113

HCFC-142b:3(C-H)-3(OH)3 x 413 – 3 x 467 = -162 kJ/molFreon-113:2(C-Cl) + 1(C-F) – 2(O-Cl) – 1(O-F)2 x 339 + 453 – 2(203) – 190 = +535


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