structure prediction, chemical bonding · structure prediction, chemical bonding 2 lewis structures...
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structure prediction, chemical
bonding
2
Lewis structures
• Atoms listed in order of increasing EN, no
connectivity implied
• CSPF
• (PNF2)4
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after the Lewis
Structure
• determine the steric number
number of bonds (single and multiple)
number of unshared pairs
or sometimes single unpaired electrons such as in
NO2
• then assign geometry using
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the following guidelines
SN geometry of electrongroups
2 linear (180°)3 trigonal planar (120°)4 tetrahedral (109.5°)5 trigonal pyramidal6 octahedral (90°)7 pentagonal bipyramid
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Cartoon representations
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8will need to refine
bond angles
considering
• the different sizes of
single bonds vs. multiple bonds
lone pairs vs. single bonds
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today in dj
• One chemical fad is/was the isolobal analogy.
Draw a chemical species or fragment that is
isolobal with the methylene group, CH2,
derived from methane by loss of two (2) H
atoms.
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• Repulsions between electron groups are in the order
LP-LP > LP-BP > BP-BP
• when lone pairs are present the bond angles are smaller than those
predicted for the idealized geometry in rule 1.
• lone pairs choose the site with the most volume available to them, or the
least sterically hindered site.
• if all sites are equivalent, the lone pairs will be trans to one another.
• multiple bonds occupy more space than single bonds.
• bonding pairs to electronegative substituents occupy less space than those
to less electronegative substituents
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• If the central atom is in third row or below in the periodic chart there are
two possibilities:
if the substituents are oxygen atoms or halogens, the above rules hold.
if the substituents are less electronegative than halogens, the lone pairs will
occupy a non-bonding s orbital and the bonding to the substituents will be
primarily through p orbintals with resulting bond angles of about 90°.
• Describe Structures by the arrangement of atoms. Unshared electrons
(lone pairs) influence the structure (arrangement of atoms about central
atom) but are not considered when describing the molecular structure.
We “see” atoms’ positions, but not the position of the unshared electrons.
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some examples
• SeBr5-
• ICl3
• CoF63-
• OF2
• COCl2
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Today’s stuff
• Next time, 63 thru 72
• Look at 76-82 as well…
• Exam 1, 20 October.. Up to Matrix
representations.
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Today’s q
• What would be the biggest bond angle in
BClBrI ?
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some reality
Molecule Angle(°) Molecule Angle(°) Molecule Angle(°)
H-X-H H-X-H H-X-H
H2O 104.5 NH3 107.3 CH4 109.5
H2S 92.2 PH3 93.3 SiH4 109.5
H2Se 91.0 AsH3 91.8 GeH4 109.5
H2Te 89.5 SbH3 91.3 SnH4 109.5
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more reality
Molecule X-O-Y H-C-H
H2O 104.5 CH4 109.5
F2O 103.2 CH3Cl 110.5
Cl2O 111 CH2Cl2 112.0
(CH3)2O 111 CH3Br 111.2
CH3OH 109 CH3I 111.4
CH3OH 109.3
H3C-CH3 109.3
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any refinement?
• SeBr5-
• ICl3
• CoF63-
• OF2
• COCl2
18 experimental
confirmation of the
postulated
hybridizations?
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19for s-p hybrids, bond
angle can give % s
character
• the relationship to right
can give relative s and p
character for equivalent
spx hybridscos =
s
s 1=
p 1
p
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consider O=CF2
cos108° =s
s 1= 0.309
s = 0.24• note that the orbitals
have more “p”
character due to either
steric effects or
electronegativity of F
substituents
• Remember that p
orbitals have greater
spacial extension!
More e- density to F
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hybridization and bond
angles
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hybridization
energetics
• orbital mixing in
hybridization usually
requires energy
• consider C hybridizing to
sp3
• this promotion costs
400 kJ/mole
• Payback? Stronger
bonds, more stable
product!
2s
2p
sp3 hybrids
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factors influencing
hybridization
• energetics of mixing
how much energy needed for promotion?
• bond energies
depends on the system
will BE’s return energy required for hybridization
• electron pair repulsions
less important in larger atoms
24consider
rehybridization and
repulsions
• for PH3 vs NH3 similar rehybrization energies
are required
• electron pair in ns orbital must be promoted to
sp3 orbital energy (for P, E 600 kJ/mole)
• sizes of atoms are different
75 vs 110 pm
• Therefore orbital mixing is not advantageous
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Sterics vs. Electronics
What is more important?
Steric effects or electronic effects?
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Lone Pairs can be
insignificantSn(C5Ph5)2
Has two electrons in valence shell..
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27 or stereochemically
active (Janiak, et al.,
Chem. Ber.(1988) 121
p1745)
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consider reality
Molecule Angle(°) Molecule Angle(°) Molecule Angle(°)
H-X-H H-X-H H-X-H
H2O 104.5 NH3 107.3 CH4 109.5
H2S 92.2 PH3 93.3 SiH4 109.5
H2Se 91.0 AsH3 91.8 GeH4 109.5
H2Te 89.5 SbH3 91.3 SnH4 109.5
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results
• smaller bond angles in PH3 and lower
congeners indicate low degrees of hybridization
• lone pair is in “s”orbital
• Much lower basicity!!
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more reality
Molecule X-O-Y H-C-H
H2O 104.5 CH4 109.5
F2O 103.2 CH3Cl 110.5
Cl2O 111 CH2Cl2 112.0
(CH3)2O 111 CH3Br 111.2
CH3OH 109 CH3I 111.4
CH3OH 109.3
H3C-CH3 109.3
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bond energies,
electronegativities
• this factor can be difficult to deconvolute since
electronegativity difference leads to stronger
bonding because of ionic resonance forms
A-B A+-B-
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Bent’s Rule
• More electronegative substituents prefer hybrid orbitals
of less “s” character and the less elctronegative
substituents prefer orbitals having more “s” character.
• the “p” character tends to concentrate in orbitals of
weaker covalency and “s” character concentrates in
orbitals of stronger covalency
covalency depends on electronegativity difference and orbital
overlap
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for example
• PMeCl4
• PMe2Cl3
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VSEPR is a lie that
works
• we think of H2O as
having this structure:
H
O
H
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the lone pairs in H2O
look like this:
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Valence Bond Theory
• developed by Pauling
• invokes overlap
• molecular orbitals are formed from products of
1 electron atomic orbitals
= (1) (2)
• hybridization is required to obtain the correct
geometry
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today
• Introduction to the chemical bond- yet another
time…
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Valence Bond Model
and refinements
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(a) = 1sA• 1sB
(b) delocalization = 1sA(1) 1sB(2) +1sA(2) 1sB(1)
(c) shielding by electrons
(d) ionic resonance forms H-H H+- H- H-- H+ = 1sA(1) 1sB(2) +
1sA(2) 1sB(1)+ ( 1sA(1)1sB(1) + 1sA(2) 1sB(2))
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aspects of VB theory
• hybridization
• delocalization is not inherent
• ionic forms are not inherent
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Linear Combinations
of Atomic Orbitals
• a molecular orbital theory
• can use symmetry to make building MO’s
easier
• start with diatomics, O2 and NO
• finish with several triatomics, BeH2 H2O, CO2
and NO2-
some these latter will be “solved” using group theory
42
LCAO assumptions
• same size/energy orbitals equal distribution of electrons, therefore each atomic
orbital makes an equal contribution to the molecular orbital. This condition does
not hold in heterodiatomic molecules.
• use the orbital approximation
the wave function of the N electrons in an atom can be written as the product of the 1
electron wavefunctions, each a fcn of n, l , ml and ms.
for electrons in low-lying orbitals (near the nucleus) in a molecule the wavefunction
resembles that of the lowlying corresponding orbital in the atom.
a reasonable first approximation to the molecular wave functioncan be obtained from
linear combinations of the atomic orbitals, and specifically, the valence orbitals of the
atoms involved.
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For H2 one can form:
(1) = (1) + (2)
(2) = (1) (2)
44
Consider H2: by
overlap of 1s orbitals
• See the spreadsheets:
just overlapping wavefcns
VB formalism
And LCAO method
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an important
foundation
• orbitals of different symmetry cannot mix to
form molecular orbitals
• for diatomic molecules, the z axis contains the
line joining the nuclei.
• the px and py orbitals have the same symmetry
as do the pz and s orbitals.
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combinations
• “s” orbitals
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bonding and
antibonding orbitals
• antibonding
• bonding
+
+
+
+
48
And still more
• Other combos
• Which are bonding?
• Antibonding?
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Start with O2
• What are valence orbitals on O atoms?
2s and 2p orbitals
• What orbitals can overlap and form
bonding/antibonding combinations?
2s and 2pz can form bonds
2px and 2py form bonds
• What of relative energies?
50
Cache ab initio MO’s for
O2
• s orbitals give these sigma
bonding and antibonding
orbitals in O2
• (s- )= (2s) + (2s)
• (s- )= (2s) - (2s)
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pz orbitals give sigma
bonding and antibonding
orbitals
52px and py orbitals give
degenerate and *
orbitals
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MO diagram for O2
• s orbitals give bonding
and antibonding orbitals
• p orbitals give and
bonding and antibonding
orbitals
*not completely true!
np
p
p
ns
np
ns
54
differences between homo-
and hetero- binuclear
molecules
• expect that electron density
should favor proximity to
more electronegative atom
(bonding MO’s are located on
more en atom)
• antibonding MO’s should
have more character of less
en atom
np
ns
np
ns
N O
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compare orbitals for
O2 and NO
56compare p orbital
and * orbitals on NO
and O2
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compare p orbital
and * orbitals
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differences between homo-
and hetero- binuclear
molecules
• note that in most cases, the
bonding orbitals that are
occupied have more electron
density on the more
electronegative atom
• antibonding MO’s have more
character on the less en atom
np
ns
np
ns
N O
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VSEPR is a lie that
works
• we think of H2O as
having this structure:
H
O
H
60
the lone pairs in H2O
look like this:
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This is the orbital
mixing!
bonding
Non-
bonding
62
What is % s
• Hard to tell
• Not exactly25% like sp3, eh.
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Isolobal analogy
• Orbital similarities between molecular
fragments have predictive power of the types
of compounds formed.
64
What is meant by
isolobal?
To determine whether two molecules are isolobal one must
consider the frontier orbitals (the valence orbitals including the
HOMO and the LUMO).
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two molecular fragments are
considered isolobal if:
1) the number
2) the symmetry
3) the electron occupancy and
4) the approximate energy
of the frontier orbitals are the same.
66
What’s the use?
• Isolobal analogy:
compounds which have frontier orbitals which have the same
symmetry
and electron occupation
• will tend to form analogous complexes.
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Making fragments
H
H HH
-H•
H HH
-e-
H HH
•CH3 CH3+
68
What is isolobal with
•CH3 and CH3+?
H
H HH
-H•
H HH
-e-
H HH
•CH3 CH3+
H•
Cl•
H+
HB
HH
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Simple predictions 1.
• Since •CH3 forms a compound with the •:NH2
radical, so will the isolobal H• and Cl•
fragments.
70
Pictorially,
H
HH
•CH3 NH
H
+ H3C NH2
therefore:
H•
Cl•N
H
H
+
Cl NH2
H NH2
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however
• these fragments are isolobal to each other.
72
Unlike fragments, CH3+
H
HH H
NH
H+
exists, therefore
H
NH
H
H
NH
H
H
and probably existH
B
HH
+
+