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Chapters 1 & 2. General Chemistry Review Electronic Structure and Bonding. Molecular Representations. ~ 0.1 nm. Anders Jöns Ångström (1814-1874) 1 Å = 10 picometers = 0.1 nanometers = 10 -4 microns = 10 -8 centimeters. Nucleus = 1/10,000 of the atom. 1 nm = 10 Å

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Chapters 1 & 2

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Chapters 1 2

Chapters 1 & 2

General Chemistry Review

Electronic Structure

and

Bonding

Molecular

Representations


Chapters 1 2

~ 0.1 nm

Anders Jöns Ångström

(1814-1874)

1 Å = 10 picometers = 0.1 nanometers =

10-4 microns = 10-8 centimeters

Nucleus =

1/10,000

of the atom

  • 1 nm = 10 Å

  • An atom vs. a nucleus

  • ~10,000 x larger


Question 1 1

Question 1.1

  • What is the electronic configuration of carbon?

  • A)1s2 2s2 2px2

  • B)1s2 2s2 2px1 2py12pz0

  • C)1s2 2s2 2px12py12pz1

  • D)1s2 1px1 1py12s2


Electron configurations noble gases and the rule of eight

Electron Configurations Noble Gases and The Rule of Eight

  • When two nonmetalsreact to form a covalent bond: They share electrons to achieve a Noble gas electron configuration.

  • When anonmetaland a metal react to form an ionic compound: Valence electrons of the metalare lost and the nonmetal gains these electrons.


Chapters 1 2

Notes from Lewis’s notebook and his “Lewis” structure.

G.N. Lewis

Photo Bancroft Library, University of California/LBNL Image Library

Footnote:

G.N. Lewis, despite his insight and contributions to chemistry, was never awarded the Nobel prize.

http://chemconnections.org/organic/Movies%20Org%20Flash/LewisDotStructures.swf


Chapters 1 2

Ionic Compounds

  • Ionic compounds are formed when electron(s) are transferred.

  • Electrons go from less electronegative element to the more electronegative forming ionic bonds.

Worksheet 1: Bonds, Formulas, Structures & Shapes

http://chemconnections.org/organic/chem226/226assign-12.html#Worksheets


Chapters 1 2

Covalent Compounds

  • Share electrons.

  • 1 pair = 1 bond.

  • Octet rule (“duet” for hydrogen)

  • Lewis structures:

Notice the charges:

In one case they balance, can you name the compound?

In the other they do not, can you name the polyatomic ion?

More about “formal” charge to come.


Covalent bonding simple lewis structures

Covalent Bonding – Simple Lewis Structures

  • For simple Lewis structures:

    • Draw the individual atoms using dots to represent the valence electrons.

    • Put the atoms together so they share PAIRSof electrons to make complete octets.

  • Take NH3, for example:

  • Practice with SKILLBUILDER 1.3.


Question 1 2

Question 1.2

  • Select the correct Lewis structure for methyl fluoride (CH3F).

  • A)B)

  • C) D)


Chapters 1 2

Important Bond Numbers

(Neutral Atoms / Normal electron distribution / Free electron pairs not shown)

O

two bonds

three bonds

N

four bonds

C

one bond

H

F

Cl

Br

I


Question 1 3

Question 1.3

  • What is the best Lewis structure for formaldehyde (H2CO)?

  • A) B)

  • C)D)


Question 1 4

Question 1.4

  • Which of the following contains a triple bond?

  • A)SO2

  • B)HCN

  • C)C2H4

  • D)NH3


Formal charge

Formal Charge

Formal charge is the charge of an atom in a Lewis structure which has a different than normal distribution of electrons.

Klein: 2.4, 2.9


Chapters 1 2

O

two bonds

three bonds

N

four bonds

C

Important Bond Numbers

(Neutral Atoms / Normal electron distribution)

one bond

H

F

Cl

Br

I


Chapters 1 2

Important Bond Numbers

(Neutral Atoms / Normal electron distribution)


Formal charge1

Formal Charge

Formal charge = number of valence electrons –

(number of lone pair electrons +1/2 number of bonding electrons)

  • Equals the number of valence electrons (Group Number of the free atom) minus [the number of unshared valence electrons in the molecule + 1/2 the number of shared valence electrons in the molecule].

  • Moving/Adding/Subtracting atoms and electrons.


Chapters 1 2

HNO3 Nitric Acid


Chapters 1 2

Complete the following table. It summarizes the formal charge on a (“central”) atom for the most important species in organic chemistry.  


Formal charge of valence e s 1 2 of bonding e s of non bonding e s

Formal Charge = # of valence e-s -[1/2(# of bonding e-s ) + # of non-bonding e-s ]

=?

=?

Worksheet #1 & Worksheet #4

http://chemconnections.org/organic/chem226/226assign-12.html#Worksheets


Question 1 5

Question 1.5

  • What is the formal charge of the carbon atom in the Lewis structure?

  • A)-1

  • B)0

  • C)+1

  • D)+2

C


Question 1 6

Question 1.6

What is the formal charge of the oxygen atom in the Lewis structure?

A)-1

B)0

C)+1

D)+2


Resonance

Resonance

Klein: 2.7 – 2.12


Chapters 1 2

Resonance

Resonance is a very important intellectual concept that was introduced by Linus Pauling in 1928 to explain experimental observations.

Eg. SO2

Bond order  1.5

Bond length > double bond; Bond length< single bond

TUTORIAL

http://www.nku.edu/~russellk/tutorial/reson/resonance.html


Resonance1

Resonance

  • Two or more Lewis structures may be legitimately written for certain compounds (or ions) that have double bonds and/or free pairs of non-bonded electrons

  • It is a mental exercise in “pushing” or moving electrons.


Rules of resonance

Rules of Resonance

  • Step 1: The atoms must stay in the same position. Atom connectivity is the same in all resonance structures. Only electrons move.

  • NON-Example: The Lewis formulas below are not resonance forms. A hydrogen atom has changed position.


Chapters 1 2

Rules of Resonance

  • Step 2: Each contributing structure must have the same total number of electrons and the samenetcharge.

  • Example:All structures have 18 electrons and a net charge of 0.


Chapters 1 2

Rules of Resonance

  • Step 3: Calculate formal charges for each atom in each structure.

  • Example:None of the atoms possess a formal charge in this Lewis structure.


Chapters 1 2

Rules of Resonance

  • Step 4: Calculate formal charges for the second and third structures.

  • Example:These structures have formal charges.

NOTE: They are less favorable Lewis structures.


Pushing electrons

H

H

..

..

..

+

:

:

H

C

O

N

O

H

C

O

N

O

..

..

..

..

..

H

H

“Pushing” Electrons

  • same atomic positions

  • differ in electron positions

more stable Lewis structure

less stable Lewis structure


Pushing electrons1

H

H

..

..

..

+

:

:

H

C

O

N

O

H

C

O

N

O

..

..

..

..

..

H

H

“Pushing” Electrons

  • same atomic positions

  • differ in electron positions only

more stable Lewis structure

less stable Lewis structure


Why use resonance structures

Why use Resonance Structures?

  • Delocalization of electrons and charges between two or more atoms helps explain energetic stability and chemical reactivity.

  • Electrons in a single Lewis structure are insufficient to show electron delocalization.

  • A composite of all resonance forms more accurately depicts electron distribution. (HYBRID) NOTE: Resonance forms are not always evenly weighted. Some forms are better than others.


Resonance example

+

••

••

O

O

O

••

••

••

••

Resonance Example

  • Ozone (O3)

    • Lewis structure of ozone shows one double bond and one single bond

Expect: one short bond and one long bond

Reality: bonds are of equal length (128 pm)


Resonance example1

+

••

••

O

O

O

••

••

••

••

+

+

••

••

••

O

O

O

O

O

O

••

••

••

••

••

••

••

••

••

Resonance Example

  • Ozone (O3)

    • Lewis structure of ozone shows one double bond and one single bond

Resonance:


Resonance example2

+

+

••

••

••

O

O

O

O

O

O

••

••

••

••

••

••

••

••

••

Resonance Example

  • Ozone (O3)

    • Electrostatic potentialmap shows both endcarbons are equivalentwith respect to negativecharge. Middle carbonis positive.


Detailed resonance examples

Detailed Resonance Examples


Question 1 7

Question 1.7

  • Which resonance structure contributes more to the hybrid?

  • A) B)


Vsepr

VSEPR

Klein: 1.10


Vsepr model valence shell electron pair repulsion

VSEPR ModelValence Shell Electron Pair Repulsion


Vsepr model

VSEPR Model

The molecular structure of a given atom is determined principally by minimizing electron pair (bonded &free) repulsions through maximizing separations.

Some examples of minimizing interactions.


Predicting a vsepr structure

Predicting a VSEPR Structure

  • 1.Draw Lewis structure.

  • 2.Put pairs as far apart as possible.

  • 3.Determine positions of atoms from the way electron pairs are shared.

  • 4.Determine the name of molecular structure from positions of the atoms.


Chapters 1 2

Orbital

Geometry

Molecular

Geometry

Bond Angle

# of lone pairs

Chem 226

Linear Linear

Trigonal PlanarTrigonal Planar

Trigonal Planar Bent

Tetrahedral Tetrahedral

TetrahedralTrigonal Pyramidal

TetrahedralBent

Trigonal BipyramidalTrigonal Bipyramidal

Trigonal BipyramidalSeesaw

Trigonal BipyramidalT-shape

Trigonal BipyramidalLinear

OctahedralOctahedral

OctahedralSquare Pyramidal

OctahedralSquare Planar

0

0

1

0

1

2

0

1

2

3

0

1

2


Molecular geometry summary

Molecular Geometry –Summary

Practice: SKILLBUILDER 1.8.


Lewis structures vsepr molecular models

Lewis Structures / VSEPR / Molecular Models

  • Computer Generated Models

    Ball and stick models of ammonia, water and methane.

Worksheet 1: Bonds, Formulas, Structures & Shapes

http://chemconnections.org/organic/chem226/226assign-12.html#Worksheets

http://chemconnections.org/COT/organic1/VSEPR/


Covalent bond polarity molecular polarity dipole moment

CovalentBond PolarityMolecular PolarityDipole Moment

Klein: 1.11


Chapters 1 2

Covalent Compounds

  • Equal sharing of electrons: nonpolar covalent bond, same electronegativity (e.g., H2)

  • Unequal sharing of electrons between atoms of different electronegativities: polar covalent bond (e.g., HF)


Polar covalent bonds

Polar Covalent Bonds

  • Electrons tend to shift away from lower electronegativity atoms to higher electronegativity atoms.

  • The greater the difference in electronegativity, the more polar the bond.

Practice: SKILLBUILDER 1.5


Question 1 8

Question 1.8

  • Which of the following bonds is the most polar?

  • A) B)

  • C)D)


Chapters 1 2

(e) : magnitude of the charge on the atom

(d) : distance between the two charges

Bond Dipole & Dipole Moment

  • Dipole moments are experimentally measured.

  • Polar bonds have dipole moments.

dipole moment: D = m = exd

  • 1 debye = 10-18 esu ∙ cm An electrostatic unit of charge (esu) is a unit of charge. One electron has a charge of 4.80 x 10-10esu.


Question 1 9

Question 1.9

  • Which of the following bonds have the greatest dipole moment ()?

  • A)B)

  • C)D)


Bond polarity

Bond Polarity

A molecule, such as HF, that has a center of positive charge and a center of negative charge is polar, and has a dipole moment. The partial charge is represented by  and the polarity with a vector arrow.


Question 1 10

Question 1.10

  • In which of the following is oxygen the positive end of the bond dipole?

  • A)O-F

  • B)O-N

  • C)O-S

  • D)O-H


Question 1 11

Question 1.11

  • In which of the compounds below is the d+ for H the greatest?

  • A)CH4

  • B)NH3

  • C)SiH4

  • D)H2O


Chapters 1 2

Molecular Polarity

Electrostatic Potential Maps

Models that visually portray polarity and dipoles


Chapters 1 2

Hydrogen Halides


Chapters 1 2

Molecular Polarity & Dipole Moment

When identical polar bonds

point in opposite directions,

the effects of their polarities

cancel, giving no net dipole

moment. When they do not

point in opposite directions,

there is a net effect and a net

molecular dipole moment,

designated d.


Chapters 1 2

Molecular Dipole Moment

The vector sum of the magnitude and the direction of the individual

bond dipole determines the overall dipole moment of a molecule


Chapters 1 2

An electrically

charged rod

attracts a stream

of chloroform but

has no effect on a

stream of carbon

tetrachloride.


Molecular polarity

Molecular Polarity

  • Consider the resultant dipole for CH3Cl

  • Dipole moment (μ) = charge (e) x distance (d)

    e= 1.056 x 10-10 esu;d = 1.772 x 10-8 cm

    • The amount of charge separation is less than if it were complete charge separation (4.80 x 10-10 esu).

  • μ = 1.87 x 10-18 esu ∙ cm

    • Conversion to debye

  • μ = 1.87 D


Molecular polarity electrostatic potential maps

Molecular PolarityElectrostatic Potential Maps

Visual depictions of polarity:


Chapters 1 2

Ammonia and in the Ammonium Ion


Chapters 1 2

Water


Chapters 1 2

  • The Lotus flower

  • Water & dirt repellancy

Polarity & Physical PropertiesOzone and Water

0.1278 nm

  • Resultant Molecular Dipoles > 0

  • Solubility: Polar molecules that dissolve or are dissolved in like molecules

Worksheet 4: Functions, Polarity, Formal Charge

http://chemconnections.org/organic/chem226/226assign-12.html#Worksheets


Solubility

Solubility

  • Generally likes dissolves like.

  • Polar compounds with other polar compounds:

    • If compounds that are mixed are capable of H-bonding and/or strong dipole–dipole interactions, then they should dissolve.

  • Nonpolar compounds with other nonpolar compounds:

    • If a compound has no strong intermolecular attraction with itself, then there are no strong forces which would have to be overcome to have it dissolve with a similar compound.


The lotus effect biomimicry http bfi org biomimicry

The “Lotus Effect” Biomimicryhttp://bfi.org/biomimicry

  • Lotus petals have non-polar wax on its micrometer-scale rough surface, resulting in water contact angles up to 170°

  • See the middle image above.

  • When it rains, water dissolves any polar materials on the surface and gravity takes care of any insoluble, non-polar “dirt” on the lotus surface, much like a snow ball.

Wax


The lotus effect biomimicry http www sciencemag org cgi content full 299 5611 1377 dc1

The “Lotus Effect” Biomimicryhttp://www.sciencemag.org/cgi/content/full/299/5611/1377/DC1

  • Isotactic polypropylene (i-PP) melted between two glass slides and subsequent crystallization provided a smooth surface. Atomic force microscopy tests indicated that the surface had root mean square (rms) roughness of 10 nm.

  • A) The water drop on the resulting surface had a contact angle of 104° ± 2

  • B) the water drop on a superhydrophobic i-PP coating surface has a contact angle of 160°.

Science, 299, (2003), pp. 1377-1380, H. Yldrm Erbil, A. Levent Demirel, Yonca Avc, Olcay Mert


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