~ 0.1 nm
Chapter 1
Structure
and
Bonding
Anders Jöns Ångström
(1814-1874)
1 Å = 10 picometers = 0.1 nanometers =
10 -4 microns = 10 -8 centimeters
Acids and Bases
Nucleus =
1/10,000
of the atom
• 1 nm = 10 Å
• An atom vs. a nucleus
~10,000 x larger
Question 1.1
• What is the electronic configuration of
carbon and how many valence electrons
does carbon have?
A) 1s2 2s2 2px2 (6 valence e-)
B) 1s2 2s2 2px1 2py12pz0 (4 valence e-)
C) 1s2 2s2 2px12py12pz1 (3 valence e-)
D) 1s2 1px1 1py12s2 (4 valence e-)
E) 1s2 2s2 2px2 (4 valence e-)
Electron Configurations
Noble Gases and The Rule of Eight
•
When two nonmetals react to form a
covalent bond: They share electrons to
achieve a Noble gas electron
configuration.
•
When a nonmetal and a metal react to
form an ionic compound: Valence
electrons of the metal are lost and the
nonmetal gains these electrons.
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.
Notes from Lewis’
Lewis ’s notebook and his “Lewis
Lewis”” structure.
Ionic Compounds
• Ionic compounds are formed when electron(s) are
transferred.
• Electrons go from less electronegative element to the
more electronegative forming ionic bonds.
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.
Question 1.2
• Select the correct Lewis structure for
methyl fluoride (CH3F).
•
A)
B)
•
•
C)
D)
Important Bond Numbers
(Neutral Atoms!)
one bond
two bonds
three bonds
H
F
Cl
O
N
four bonds
Question 1.3
• What is the correct Lewis structure of
formaldehyde (H2CO)?
•
A)
B)
•
C)
D)
Br
C
Question 1.4
• Which of the following contains a triple
bond?
•
A) SO2
•
B) HCN
•
C) C2H4
•
D) NH3
I
Formal Charge
•
•
•
HNO3 Nitric Acid
Equals the number of valence electrons 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.
See examples on the board.
Formal charge = number of valence electrons –
(number of lone pair electrons +1/2 number of bonding electrons)
Question 1.5
• What is the formal charge of the sulfur
atom in the Lewis structure?
•
A) -1
•
B) 0
•
C) +1
•
D) +2
Resonance Theory
Whenever a molecule or ion can be represented by more
than 1 valid Lewis structure where the difference is
only in the positions of the electrons not atoms, then:
1. No single resonance structure is a correct one for it.
2. Instead, the actual molecule or ion will be a hybrid or
weighted average of these structures which are not
all equal. There contribution varies.
3. See Table 1.6 pp. 27-28
Therefore, Resonance Structures exist only in theory and
are mental constructs, which are very important and
useful for predicting chemical behavior and chemical
reactivity nevertheless.
Resonance(
Resonance
)
≠
Equilibrium (
)
Question 1.6
•How many resonance structures can be
written for the NO3- ion in which the nitrogen
atom bears a formal charge of +1?
A) 1
B) 2
C) 3
D) 4
E) 5
VSEPR Model
Valence Shell Electron Pair Repulsion
Question 1.7
• Which resonance structure contributes
more to the hybrid?
VSEPR
•
A)
B)
VSEPR Model
Predicting a VSEPR Structure
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.
Orbital
Geometry
Molecular
Geometry
Bond Angle
# of lone pairs
Linear
Linear
0
Trigonal Planar
Trigonal Planar
0
Trigonal Planar
Bent
1
Tetrahedral
0
Tetrahedral
Trigonal Pyramidal
1
Tetrahedral
Bent
2
Trigonal Bipyramidal
Trigonal Bipyramidal
0
Trigonal Bipyramidal
Seesaw
1
Trigonal Bipyramidal
T-shape
2
Trigonal Bipyramidal
Linear
3
Octahedral
Octahedral
0
Octahedral
Square Pyramidal
1
Octahedral
Square Planar
2
Chem 226 Tetrahedral
• 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.
Lewis Structures / VSEPR /
Molecular Models
• Computer Generated Models
Ball and stick models of ammonia, water and
methane. For many others see:
http://chemconnections.org/organic/pdb-lib/
http://chemconnections.org/organic/chem226/Smell-Stereochem.html
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)
Question 1.8
• Which of the following bonds is the most
polar?
•
A)
B)
•
C)
D)
Bond Dipole & Dipole Moment
• Dipole moments are experimentally measured.
• Polar bonds have dipole moments.
dipole moment (D) = µ = e x d
(e) : magnitude of the charge on the atom
(d) : distance between the two charges
Question 1.9
• Which of the following bonds have the
greatest dipole moment?
•
A)
B)
•
C)
D)
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 δ and
the polarity with a vector arrow.
H F
δ+
Question 1.10
• In which of the compounds below is the δ+
for H the greatest?
•
A) CH4
•
B) NH3
•
C) SiH4
•
D) H2O
δ−
Question 1.11
Electrostatic Potential Maps
Models that visually portray polarity and dipoles
• 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
Hydrogen Halides
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 δ.
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
An electrically
charged rod
attracts a stream
of chloroform but
has no effect on a
stream of carbon
tetrachloride.
Ammonia and in the Ammonium Ion
Water
The “Lotus Effect”
Polarity & Physical Properties
Biomimicry
Ozone and Water
http://www.bfi.org/Trimtab/spring01/biomimicry.htm
0.1278 nm
•
•
•
Resultant Molecular Dipoles > 0
Solubility: Polar molecules that
dissolve or are dissolved in like
molecules
•
•
•
The Lotus flower
Water & dirt repellancy
Wax
Lotus petals have micrometer-scale roughness, resulting
in water contact angles up to 170°
See the Left image in the illustration on the right.
The “Lotus Effect”
Question 1.12
Biomimicry
http://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°.
• Which molecule would have a dipole
moment equal to zero?
A)
B)
C)
D)
CCl4
CH3OH
CH3OCH3
CH3Cl
Science, 299, (2003), pp. 1377-1380, H. Yldrm Erbil, A.
Levent Demirel, Yonca Avc, Olcay Mert
Molecular Representations
VIDEO
Empirical Formula, Molecular Formula, Structure:
(Lewis, Kekule, Condensed, Line), Visual Model:
wireframe, stick, ball & stick, space filling, electrostatic,
energy surface
Question 1.13
Bond Line
OH
• The bond-line representation for
(CH3)2CHCH2CH2CHBrCH3 is
Molecular Formula
C7H16O
•
A)
B)
•
C)
D)
Question 1.14
Question 1.15
• The total number of bonded pairs of
electrons and of unshared pairs of
electrons in morpholine is:
•
A) 7, 0
•
B) 7, 1
•
C) 15, 0
•
D) 15, 1
•
E) 15, 3
• The molecular formula of morpholine is:
•
A) C2HNO
•
B)
C4HNO
•
C) C4H4NO
•
D) C4H5NO
•
E) C4H9NO
Question 1.16
C2H6O:
H
H
H
C
C
H
H
H
H
H
O
C
H
Ethanol (b.p. 78.5 C)
• How many constitutional isomers can have
the formula C3H7Cl?
•
A) one
•
B) two
•
C) three
•
D) four
H
O
C
H
H
Dimethyl ether (b.p. –23 C)
Constitutional or Structural Isomers:
These isomers have the same molecular formula, but different arrangement of atoms
in space (Bonding differences)
Question 1.17
O
O
H
O
OH
O
O
CH3
CHO
• How many constitutional aldehyde isomers
can have the molecular formula C4H8O?
•
A) one
•
B)
two
•
C) three
•
D) four
Line Drawing and Ball & Stick
More Molecular Representations
Empirical Formula, Molecular Formula, Structure:
(Lewis, Kekule, Condensed, Line), Visual Model:
wireframe, stick, ball & stick, space filling,
electrostatic, energy surface
8.16 Å (0.816 nm)
http://chemconnections.org/organic/chem226/Smell-Stereochem.html
Worksheet: Organic Molecules I
http://chemconnections.org/organic/chem226/
Very Large Molecules
Very Large Molecules:DNA
http://www.umass.edu/microbio/chime/beta/pe_alpha/atlas/atlas.htm
Views & Algorithms
http://info.bio.cmu.edu/courses/03231/ProtStruc/ProtStruc.htm
B-DNA: Size, Shape & Self Assembly
46 Å
10.85 Å
10.85 Å
Rosalind Franklin’s
Photo
12 base sequence
Several formats are commonly used
but all rely on plotting atoms in 3
dimensional space; .pdb is one of the
most popular.
(1953-2003)
http://molvis.sdsc.edu/pdb/dna_b_form.pdb
Atomic Orbitals
Atomic Orbitals
s orbitals
p orbitals
Molecular Orbitals
• Atomic orbitals mix to form molecular orbitals
• σ bond: formed by overlapping of two s orbitals
Molecular Orbitals (MO)
MO: a linear combinaiton of AOs
H2:
Chemical bonds formation---AO’s
overlapping
Ψ2,1 = Ψ1sA(1)Ψ1sB (2 ) + Ψ1sB (1)Ψ1sA(2 )
Bond formation:
Ψ1, 2 = Ψ1sA(1)Ψ1sB (2 ) − Ψ1sB (1)Ψ1sA(2 )
1.Closed energy level
2.Similar size and shape of AOs
3.Significant overlapping
In-phase overlap of s atomic orbitals form a bonding MO;
out-of-phase overlap forms an antibonding MO
A single bond is a sigma (σ) bond.
A sigma bond (σ) is formed by end-on overlap of two p orbitals
Double bonds have 1 π and 1 σ bond.
A π bond is weaker than a σ bond.
A double bond is shorter and stronger than a single bond.
A pi bond (π) is formed by sideways overlap of two parallel
p orbitals. A π bond is weaker than a σ bond.
A double bond is shorter and stronger than a single bond.
Bonding in Methane and Ethane:
Single Bonds
Hybridization of orbitals:
The orbitals used in bond formation determine the
bond angles
• Tetrahedral bond angle:
109.5°
Hybrid Orbitals of Ethane
The sigma bond (σ) between the carbon atoms
is part of the overall hybridization. The molecular
orbital allows rotation about the C-C single bond.
• Electron pairs spread themselves into space as far from
each other as possible
Bonding in Ethene
Carbon Double Bonds
Double bonds have 1 π and 1 σ bond.
A π bond is weaker than a σ bond.
A double bond is shorter and stronger than a single bond.
The pi bond (π) of the sp2 hybrid does not allow rotation
about the double bond. This produces a fixed geometry
about the double bond and results in cis-trans, (Z-E),
isomerism
• The bond angles of the sp2 carbon are about 120°
• The sp2 carbon is trigonal planar
• The atoms bonded to the sp2 carbon are all in the
same plane
Bonding in Ethyne: A Triple Bond
Question 1.18
• In which of the following compounds could
be found the shortest carbon- carbon
bond(s)?
• A triple bond consists of one σ bond and two π bonds
•Triple bonds are shorter and stronger than double bonds
• There is a bond angle of the sp carbon: 180 °
A) C3H8
B) C4H10
C) C3H4
D) C3H6
Question 1.19
• What is the molecular shape about the
carbon atoms of acetylene (HC≡CH)?
•
•
•
•
A)
B)
C)
D)
tetrahedral
bent
trigonal planar
linear
Reactive Intermediates
Carbocation
Reactive Intermediates
Radical
Reactive Intermediates
Carbanion