Drawing Lewis Structures
Chapter 10
Theories of Covalent Bonding
Molecular Geometry and Hybridization of Atomic Orbitals
1) Place least electronegative element as the central atom.
Recognize that C,S,P and N are often central atoms. H and
halogens are often bonded to central atoms.
2) Sum the total of valence electrons contributed by each
atom in the molecule. Look at the Group number to help you.
3) Place bonds to central atoms using 2-electrons per bond.
4) Place an octet of electrons (octet rule) around bonded
atoms remembering that H only has 2 electrons---no octet.
5) Place remaining electrons around central atom which
should have an octet if period 2 or less, but could be more
than octet if period 3 or higher.
6) Some “rules of thumb” to have at your fingertips.
H forms 1-bond, C forms 4-bonds, N forms 3-bonds, O forms
2-bonds.
Write the Lewis dot and skeletals structure of nitrogen
trifluoride (NF3).
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F --> N is central atom!
Step 2 - Count valence electrons = A; Nitrogen = 5, Fluorine = 3 X 7 = 21
Write the Lewis dot and skeletal structures of the
carbonate ion (CO32-).
Write the Lewis dot and skeletal structures structure of
the carbonate ion (BrO3-).
A = 5 + 21 = 26 valence electrons
Step 3 - Write structure with N central and three bonds and rest nonbonding octet electrons around the central atom.
Step 4 - Write structure with N central and three bonds and rest nonbonding octet electrons.
octet
octet
Write the Lewis dot and skeletal structures structure of
the carbonate ion HCN?
Write the Lewis structure of the carbonate ion (CO32-).
F
N
F
F
octet
octet
Write the Lewis structure of the carbonate ion (BrO3-).
Look at the formula sometimes it gives clues to the central atom
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count octet and valence electrons (N and A respectively)
Valence electrons = 4 + 6 + 6 + 6 + 2 = 24 valence electrons
Step 3 - Arrange the atoms draw bonds between C and O atoms
and complete octet on C and O atoms. 24 - 8 = 16 non-bonding
electrons.
[
O
C
O
O
]
[BrO ] –
Valence e- = 7 + 3(6) + 1 = 26
3
–
O
Br
O
O
2-
HCN
Valence e- = 1 + 4 + 5 = 10
Carbon is central atom, watch for hydrogen--1 bond
H
C
N
A concept called “resonance” is used when
more than one plausible Lewis structure can be
drawn.
Example: Ozone, O3
••
••
••
••
••
••
Carbonate Ion- [CO3] 2-
Measured
bond lengths
show they are
equal!
••
O
O O
••
••
O
O O
••
••
••
2 equally good
Lewis
structures
Resonance Structures
Benzene, C6H6
Which structure is correct?
Both are!
••
••
••
••
C - C single bond: 1.54 Å
C = C double bond: 1.34 Å
••
O
O O
••
••
••
••
O
O O
••
••
••
••
••
••
O
O O
••
••
C - Bond in C6H6: 1.40 Å
a resonance
hybrid
structure
Write resonance structures for the nitrate ion,
NO3-.
Write resonance structures for the nitrate ion,
NO3-.
PLAN:
A book-keeping method called “formal charge”
is used to determine the “best” Lewis structure
when multiple structures appear plausible.
Valence e- = 5 + (3X6) + 1 = 24 e-
To use the concept of formal charge, we draw the
plausible Lewis structures and then for each atom
determine it’s formal charge.
Atom Formal charge = # valence e- - Assigned e- to Atom
Assigned Atoms = all from lone pair e! + ! ( bonded e! )
These are different two plausible structures...how
do we decide?
O
C
O
O
C
O
O
C
O
O
Valence e# of Assinged eFormal Charge
C
O
6
4
6
6
4
6
0
0
0
This structure wins!
O
C
O
6
5
+1
4
4
0
6
7
!1
1. The best structure is one that minimizes total formal charge.
Less or no charge is better--and it must equal net charge of ion
or molecule.
2.! The best structure is one that places negative charge on the
more electronegative atom.
What if more than one structure works?
Formal Charge For Multiple Structures
0
]– [ S
C
1
N
]– [ S
C
2
N
0
There are three major exceptions to the octet rule.
1) Incomplete Octet - rare situation that occurs with
Be, B and Al as central atoms.
2) Expanded Octet (the largest class of octet
exceptions)-occurs mostly with Period 3 non-metals
like P, S and halogens.
3) Odd-number electrons highly reactive species
called radicals that have an odd number of electrons
(uneven).
Expanded Octet (the largest class of octet
exceptions)-occurs mostly with Period 3 nonmetals like P, S and halogens.
••
••
••
••
Cl
••
••
••
P
••
••
BeH2
Cl •
••
•
Phosphorous
pentachloride
]–
0
C
-2
N
]–
FCS = 6 - 2 -3 = 1
FCC = 4 - 0 - 4 = 0
FCN = 5 - 6 - 1 = -2
Structure on the left is “best” structure!
AlCl3
Al – 3e3Cl – 3x7e24e-
BF3
B – 3e3F – 3x7e24e-
H
Be
H
Cl
Al
Cl
Cl
F
B
F
F
Odd-Electron Molecules: Radicals are highly
reactive species that have an odd number of
electrons (uneven).
N – 5eO – 6e11e-
N
O
An Odd Number of Valence e- = No octet and radical
H
H—C—H
[ICl4]-1
FCS = 6 - 6 -1 = -1
FCC = 4 - 0 - 4 = 0
FCN = 5 - 2 - 3 = 0
Be – 2e2H – 2x1e4e-
••
Cl
••
••Cl
Phosphorous trichloride
PCl3
••
P–
5Cl – 35e40e-
•
••
PCl5
••
Cl •
+1
]– [ S
Draw Lewis structures for the following
NO
••
•• Cl
••Cl
F
F
••
-1
N
P
S
F
5e-
F
0
N
Incomplete Octet: Occurs With Group 2A (Be) and
3A (B and Al)
Cl
••
F
F
0
C
••
SF6
S – 6e6F – 42e48e-
–
[S
3
••
]
0
C
1. Formal charge must sum to charge of ion or molecule.
2. N is more electronegative than C or S, it should have a the most
negative charge in the “best structure”.
3. The most plausible structure has the least amount of formal
charge.
•
N
-1
]– [ S
Methyl radical
••
••
N=O
••
Nitrosyl radical
••
O—H
••
•
3-plausible Lewis structures which one is best?
C
-1
N
FCS = 6 - 4 -2 = 0
FCC = 4 - 0 - 4 = 0
FCN = 5 - 6 - 2 = -1
N
S
C
Valence = 6 e- + 4 e- + 5 e- + 1 e- = 16 e-
[S
0
C
[S
•
Example: Write 3 plausible Lewis structures for the
thiocyanate ion [SCN]–
Hydroxide radical
Chemists use Valence Shell Electron Pair Repulsion
Theory to predict the shapes of molecules using
these five electron group geometries.
1. Draw Lewis Structure
from chemical formula.
VSEPRT explains the geometry of molecules but
NOT how covalent bonds are formed with that
geometry.
Molecular
formula
2. Count all electron
domains to get AXE code.
Lewis
structure
VSEPRT
Geometry
Hybrid
orbitals
3. Group domains into
bonding and non-bonding
pairs of electrons.
4. Match the number of
bonding and non-bonding
domains to the proper
VSEPRT geometry.
Lewis Structure
The goal is to understand geometry (via VSEPRT) and to
relate it to a picture of covalent bonding in molecules.
Molecular
formula
VSEPRT
Lewis
structure
Geometry
Hybrid
orbitals
Valence BondTheory
sp
Linear
sp2
Trigonal
Pyramidal
sp3
sp3d
Tetrahedral
Trigonal
Bipyramidal
sp3d2
Octahedral
The total number of electron groups (domains)
defines one of the five basic geometries.
3 EG
4 EG
VSEPRT
VSEPRT
Valence Bond
Theory
The 3-D geometry of a molecule is one of five basic
arrangements of electron groups (domains).
Valence Shell
Electron Pair
Repulsion Theory: the
optimum arrangements
of a given number of
electron domains is the
one that minimizes
repulsion among them.
Note each shape has
a specific “bond
angle”
The electron geometry is the geometry of all
electron domains, whereas the “molecular
geometry” describes the geometry of only the
atoms bonded to the central atom.
AX3E1 =
Tetrahedral
electron
geometery with
109.5˚ bond
angles.
2 EG
5 EG
6 EG
Molecular
Geometry is
trigonal
pyramidal bond
angles <109.5˚
An electron group (domain) is either a pair of bonding
electrons or a pair of non-bonding electrons
surrounding a central atom. Multiple bonds only
count as 1-group or domain.
••
••
P
••
Cl •
••
4 electron groups ••Cl
3 bonding
1 non-bonding
••
••
F
N
F
4 electron groups
3 bonding
1 non-bonding
F
••
••
••
••
Cl
••
Cl •
••
••Cl
••
••
••
P
••
•• Cl
••
Cl
••
5 electron groups
5 bonding
0 non-bonding
[
•
O
C
O
O
]
2-
3 electron groups
3 bonding
0 non-bonding
How Predict Geometry Using VSEPRT
1. Draw a plausible Lewis structure for the molecule.
2. Determine the total number of electron domains and
identify them as bonding or lone pairs.
3. Use the total number of electron domains to establish the
electron geometry from one of the five possible
geometric shapes.
•
F
N
F
Cl
Be
Cl
Be
Cl
S
C
N
2 electron groups
bonding
C
O
AX2E0 = AX2
A = Central Atom
X = # of Bonded
Domains
handled individually.
A
Other Examples:
CS2, HCN, BeF2
E = # Non-Bonded
Domains
4 Electron Groups = Tetrahedral Electron
Geometry and 3-Possible Molecular Geometries
AX3
Examples:
AX2
Bond Angle
6. Molecules with more than one central atom can be
SO3, BF3,
NO3-, CO32-
Cl
AX3E1
2 Electron Groups = Linear Electron Geometry
and 1-Possible Molecular Geometry
5. Remember bond angles in molecules are altered by lone
3 Electron Groups = Trigonal Planar Electron
Geometry and 2-Possible Molecular Geometries
4 electron groups
3 bonding
1 non-bonding
F
molecular geometry (or do both electron and molecular
geometry together simultaneously)
pairs of electrons (repulsion forces reduce angles).
E = # Non-Bonded
Domains
It’s implied that E = 0
O
4. Establish the AXnEm designation to establish the
AX2E0 = AX2
X = # of Bonded
Domains
••
Cl
A = Central Atom
••
Be
Cl
Cl
2 electron groups
bonding
We count and “code” the bonding/non-bonding
information into shorthand called AXE classification.
AX4
Bond
Angle
Bond
Angle
Examples:
CH4, SiCl4,
SO42-, ClO4-
3-Electron Domain
Examples:
SO2, O3,
PbCl2, SnBr2
A
AX2E1
AX3E1
NH3
PF3
ClO3
H3O+
AX2E2
H 2O
OF2
SCl2
5 Electron Groups = Trigonal Bipyramial Electron
Geometry and 4-Possible Molecular Geometries
PF5
AsF5
AX5
AX4E1
6 Electron Groups = Octahedral Electron
Geometry and 3-Possible Molecular Geometries
SF4
SF6
AX6
XeO2F2
SOF4
IOF5
IF4+
IO2F2-
Equatorial
Position
Axial
Position
ClF3
AX2E3
BrF3
XeF2
BrF5
I3 -
TeF5-
IF2
AX3E2
-
Non-bonding electrons repulse bonding
electrons and alter the bond angles in molecules.
AX5E1
XeF4
AX4E2
XeOF4
ICl4-
Double-bonds and/or triple bonds in molecules
also decrease bond angles in molecules (think
repulsion by electron rich regions).
Electron lone pairs render the normal 109˚ tetrahedral
angle less than 109!
H
120°
H
C
H
H
116°
C
H
bonding-pair vs. bonding
pair repulsion
<
lone-pair vs. bonding
pair repulsion
<
lone-pair vs. lone pair
repulsion
The electron geometry is the geometry of all
electron domains, whereas the “molecular
geometry” describes the geometry of only the
atoms bonded to the central atom.
AX3E1 =
Tetrahedral
electron
geometery with
109.5˚ bond
angles.
Molecular
Geometry is
trigonal
pyramidal bond
angles <109.5˚
Double-bond vs. Single-bond
repulsion
C
H
Predicted Bond Angles
122°
H
C
H
Actual Bond Angles
>
Single-bond vs, Single-bond
repulsion
Predicting Molecular Shapes
Draw the molecular shape and predict the bond
angles (relative to the ideal bond angles) of (a)
PF3 and (b) COCl2.
Predicting Molecular Shapes
Draw the molecular shape and predict the bond
angles (relative to the ideal bond angles) of (a)
PF3 and (b) COCl2.
Predicting Molecular Shapes with Two, Three, or Four Electron Groups
(b) For COCl2, C has the lowest EN and will
be the center atom.
1. Count the valence electrons and draw Lewis structure
for PF3: VE = 5 + 3(7) = 26 e2. Count the electron domains and find electron
geometry and molecular from core 5 electron domain
shapes (using AXE designation and sub-shapes)
3. There are 4 electron domains so the electron
geometry is tetrahedral
4. The designation is AX3E1 so the molecular geometry is
trigonal pyramidal.
5. The F-P-F bond angles should be <109.50 due to
the repulsion of the nonbonding electron pair.
<109.50
Predicting Molecular Shapes with Two, Three, or Four Electron Groups
(b) For COCl2, C has the lowest EN and will
be the center atom.
There are 24 valence e-, 3 atoms attached to
the center atom.
Determine the molecular shape and predict the bond
angles (relative to the ideal bond angles) of (a) SbF5
and (b) BrF5.
1. Draw the Lewis structure
2. Count the electron domains and establish
electron geometry from 5 shapes
3. There are 3 electron domains so the electron geometry
is trigonal planar
4. The molecular geometry designation is AX3E0 so the
molecular geometry is also trigonal planar (no lone
Type AX3
pairs).
5. The Cl-C-Cl bond angle will
be less than 1200 due to the
electron density of the C=O.
124.50
1110
Determine the molecular shape and predict the bond
angles (relative to the ideal bond angles) of (a) SbF5
and (b) BrF5.
(a) SbF5 - 40 valence e-; all electrons
around central atom will be in bonding
pairs; shape is AX5 - trigonal bipyramidal.
(b) BrF5 - 42 valence e-; 5 bonding pairs
and 1 nonbonding pair on central atom.
Shape is AX5E, square pyramidal.
More Than One Central Atom
• In acetic acid, CH3COOH, there are three central atoms.
• We assign the geometry about each central atom
separately.
What is the geometry
around these atoms?
Take one atom at a time and apply the
rules of electron domains.
Predicting the Molecular Shape With Multiple Central Atoms
More Than One Central Atom
Determine the shape around each of the central
atoms in acetone, (CH3)2C=O.
Find the shape of one atom at a time after writing the Lewis
structure.
tetrahedral
ethane
CH3CH3
tetrahedral
electron
domain and
molecular
geometry
tetrahedral
trigonal planar
>1200
ethanol
CH3CH2OH
<1200
Electronegativity is an element’s inherent property to
draw electrons to itself when chemically bonded to
another atom in a molecule. The units are
dimensionless (all relative measurements to Li).
Rank
F
O
N
Cl
Br
Differences in elements electronegativity
between bonding atoms result in the formation
of polar-covalent bonds and net dipole
moments in molecules.
Polar Bond
d
Polar Bond
P
No Net Dipole Moment
on
rB
ola
Po
lar
B
on
d
Net Dipole Moment
Think of the dipole moment as a molecule with separated
charges + and -.
For a poly-atomic molecule we must consider the
vector sum of polar bonds in the molecule to see if
there is a net dipole moment.
Dipole
Moment
Dipole
Moment
No Net
Dipole
Moment
Valence Bond Theory explains covalent bonding
by the spatial overlap of atomic orbitals on
bonding atoms and the sharing of electron pairs.
Bonding in H2
1s1 + 1s1
Electrons that must have opposite spins.
Dipole
Moment
No Net
Dipole
Moment
Bonding in HF
1s1 + 2p5
Bonding in F2
2p5 + 2p5
Major Points and Themes of Valence Bond Theory
1. Pauli Exclusion Principle Holds: 2-electrons per overlapped
bond with opposing spins.
Connect the dots and it becomes easy to see and
understand.
Molecular
formula
VSEPRT
Lewis
structure
2. Greater orbital overlap gives stronger bonds. Depends
on orbital shapes and how they overlap.
Valence Bond Theory
explains how bonds
are made
3. Bonding is accounted by mixing or blending or
“hybridization of pure valance atomic orbitals”.
4. The number of hybrid orbitals formed equals the
number of atomic orbitals combined.
5. The types of hybrid orbitals combined varies with the
types of orbitals mixed or blended together. USE VSEPRT
to help!
We use “pure atomic orbitals” (think ground
state electronic structure and those orbitals) to
describe bonding in some molecules.
Hybrid
orbitals
Geometry
sp
Linear
sp2
Trigonal
planar
sp3
Tetrahedral
sp3d
Trigonal
Bipyramidal
1s1 + 1s22s22p5
Bonding in F2
only two bond are possible in the ground
state but we don’t observe CH2
By combing or mixing different numbers of pure atomic
orbitals we make “hybrids” that match one of the VSEPRT
geometries. For example 1 pure s orbital + 1 p-orbital
combine to give and two “sp hybrids” that when
superimposed form a linear geometry for bonding.
Pure
atomic
orbitals
(valence
orbitals)
hybridization
1s22s22p5 + 1s22s22p5
2p5 + 2p5
Octahedral
Bonding in carbon presents a problem as combining
atomics orbitals fails. Valance Bond Theory solves
this by allowing the blending or mixing of pure atomic
orbitals in a process called hybridization.
Bonding in HF
1s1 + 2p5
sp3d2
sp3
hybridized
orbitals
By hybridizing 4
bonds are possible.
The process of combining pure atomic orbitals to
form “hybrid orbitals” on central bonding atoms
in a molecule is called hybridization.
sp3 hybrid orbitals
s-orbital + p-orbital --> 2 sp hybrid orbitals --> 2-superimposed sp
hybrid orbitals
s-orbital + Three p-orbitals -> Four sp3 hybrids = Tetrahedral
s-orbital + Two p-orbital --> 3 sp2 hybrids = Trig Planar
Some generalized rules and comments on VBT
and the formation of hybridized orbitals.
1. The number of hybrid orbitals obtained equals the
number of atomic orbitals mixed.
The logic is connected all the way to Lewis and VSEPRT
Molecular
formula
VSEPRT
Lewis
structure
Hybrid
orbitals
Geometry
2. The name of and shape of a “hybrid orbital” varies
with the types of atomic orbitals mixed. (s + p vs s +
two p)
3. Each hybrid orbital has a specific geometry that
matches one of five VSEPRT shapes (show below).
sp
Linear
sp2
Trigonal
Planar
sp3
Tetrahedral
sp3d
sp3d2
Trigonal
Bipyramidal
sp
Linear
sp2
sp3
Tetrahedral
Trigonal
Pyramidal
sp3d
sp3d2
Trigonal
Bipyramidal
Octahedral
Octahedral
Electron
Geometry
Molecular
Geometry
AXnEm
Hybridization
Electron
Geometry
Molecular
Geometry
AXnEm
Hybridization
Linear
Linear
AX2
sp
Linear
Linear
AX2
sp
Trigonal
planar
Trigonal planar
V-shaped bent
AX3
AX2E1
sp2
Trigonal
planar
Trigonal planar
V-shaped bent
AX3
AX2E1
sp2
Tetrahedral
Tetrahedral
Trigonal pyramidal
V-shaped bent
AX4
AX3E1
AX2E2
sp3
Tetrahedral
AX4
AX3E1
AX2E2
sp3
Trigonal
bipyramidal
Trigonal bipyramidal
Seesaw
T-shaped
Linear
AX5
AX4E1
AX3E2
AX2E3
sp3d
Trigonal
bipyramidal
Trigonal bipyramidal
Seesaw
T-shaped
Linear
AX5
AX4E1
AX3E2
AX2E3
sp3d
Octahedral
Octahedral
Square pyramidal
Square planar
AX6
AX5E1
AX4E2
sp3d2
Octahedral
Octahedral
Square pyramidal
Square planar
AX6
AX5E1
AX4E2
sp3d2
Determine the VSEPRT geometry, the bond
angles and the hybridization of each indicated
atom in the following molecule? How many
sigma and pi bonds are in the molecule?
Tetrahedral
Trigonal pyramidal
V-shaped bent
Determine the electron domain, molecular
geometry, the bond angles and the hybridization
of each indicated atom in the following molecule?
How many sigma and pi bonds are in the
molecule?
sp2
trig planar 120˚,
2
sp2 sp
linear 180˚, sp
bent, <109.5, sp3
sp3
tetrahedral, 180,
sp3
Linking VSEPRT To Valence Bond Theory Hybrids
Atomic
Orbitals
Mixed
Linear Trig
Tetrahedral Trig Bypyr Octahedral
AX2
Planar AX3 AX4
AX5
AX6
s+p
s+2p
s+3p
s+3p+d
An sp hybrid is formed from the combination of a one
pure 1s orbital and a one 2p orbital from a central
bonding atom producing two new orbitals called sp
orbitals. 2s
s + 3 p + 2d
Hybridization
#
Hybrid
Orbitals
Formed
Two sp
Three sp2
Four sp3
Five sp3d
Six sp3d2
p-orbital
Two sp hybrid sp hybrid orbitals
orbitals
superimposed
Hybrid
Shape
Orbitals
Leftover
for Pi
bonds
s-orbital
--The number of hybrid orbitals formed is equal to the number of
“pure orbitals” combined!
Two p
one p
Four d
none
Three d
Show the bonding scheme and hybridized orbitals
used in BeCl2
2 unhybridized unoccupied p-orbitals
--When superimposed the “sp-hybrid” give us bonding orbitals for
a linear molecules.
Show the bonding scheme and hybridized orbitals in BeCl2
2 “left-over” p-orbitals
hybridization
Hybridized Be Atom
Isolated Be Atom
two lone p-orbitals
two sp hybrids
on Be
s + p Hybridization = 2 sp
After hybridization we
have on the central atom,
2 pure p-orbitals and two
sp hybrids.
An sp2 hybrid is formed from the combination of a
one pure 1s orbital and a two 2p orbitals from a
central bonding atom producing two new orbitals
called sp2 orbitals.
3-atomic orbitals, s
and two p’s
combine to form 3sp2 hybrid orbitals
sp hybrid:Ethylyne: HC!CH:Linear
Sigma bonds (! bonds) and Pi bonds (" bonds)are two
different types of covalent chemical bonds that form as a
result of end to end spatial overlap of atomic orbitals or
hybridized orbitals (! bonds) or side to side overlap on
bonding atoms (" bonds)
Lone p orbitals that are not hybridized
Superimposed
Hybrid orbitals
form a triginal
planar geometry
sp2 = Triginal planar geometry, 120˚ bond angle
sp hybrid orbitals
sp hybrid:Ethylyne: HC!CH:Linear
Lone p orbitals that were not hybridized on
each carbon atom are able to form Pi bonds
in a “side to side” overlap. A pair of
electrons is shared in this region of space.
Example 2: sp2 hybridizaton scheme BF3.
" bonds overlap
side to side
Boron Hybrid Box Diagram
Boron Orbital
Box Diagram
sp2 hybrid orbitals on each
carbon atom use end to
end overlap to form a
sigma bond.
Bonding of pure p-orbital in F
with sp2 hybridized orbitals in BF3
Tetrahedral geometry = sp3 hybrid orbitals
combine to generate
four sp3 orbitals
Example: sp3 orbital hybridization: CH4.
sp3 hybridization mixes
one 2s orbital with three
2p orbitals to produce four
sp3 orbitals on each carbon
atom. End to end overlap
with a 1s orbital from H
gives four sigma bond in
CH4.
Note the number
of hybrids
formed is the
number of
atomic orbitals
combined!
which are represented
collectively as: sp3
sp3 = Tetrahedral geometry = 109.5˚ bond angle
This is the
ground state
configuration of
valence atomic
orbitals
the four sp3
hybrid
orbitals form
a tetrahedral
shape
CH4
Example 3: sp3 hybrid orbitals in H2O.
sp3 hybridization mixes
one 2s orbital with three
2p orbitals to produce four
sp3 orbitals. The e- are
distributed throughout the
hybrids ready for bonding.
End to end overlap with a
1s orbital from H gives four
sigma bond in CH4.
sp3 is tetrahedral
shape. In water
we have AX2E2
What is the electronic geometry?
What is the molecular geometry?
What orbitals contribute to bonding?
Note the lone
pairs occupy 2-of
the sp3 orbitals
What is the electron geometry, the molecular
geometry at each carbon atom? Use that
information to determine the hybridization around
each carbon atom in nicotinic acid? How many
sigma and pi bonds are in nicotinic acid?
Example 2: sp3 hybridization in NH3.
sp3d hybridization in PCl5.
Trigonal Bipyramidal Electron Geometry AX5E0
Trigonal BiPyramidal Molecular Geometry
Isolated
P atom
Tetrahedral Electron Geometry AX3E1
Trigonal Pyramidal Molecular Geometry
The sp3d2 hybrid orbitals in SF6
Describe the types of bonds and orbitals in
acetone, (CH3)2CO and in CO2 and in HCN?
Octahedral Electron Geometry AX6E0
Octahedral Molecular Geometry
Step 1
Molecular
formula
Lewis
structure
Step 3
VSEPRT
Geometry
Hybrid
orbitals
Postulating Hybrid Orbitals in a Molecule
Describe the types of bonds and orbitals in
acetone, (CH3)2CO.
PLAN:
Draw the Lewis structures to ascertain the arrangement of
groups and shape at each central atom. Postulate the
hybrid orbitals taking note of geometries predicted from
VSEPRT. Draw the orbitals and show overlap.
PROBLEM:
PLAN:
hybridized
sp3 hybridized
SOLUTION:
sp2 hybridized
Use partial orbital diagrams to describe mixing of the atomic
orbitals of the central atom leads to hybrid orbitals in each of
the following:
(a) Methanol, CH3OH
SOLUTION:
sp3
Step 2
(b) Sulfur tetrafluoride, SF4
Use the Lewis structures to ascertain the arrangement of
groups and shape of each molecule. Postulate the hybrid
orbitals. Use partial orbital box diagrams to indicate the hybrid
for the central atoms.
(a) CH3OH
The groups around C are
arranged as a tetrahedron.
O also has a tetrahedral
arrangement with 2 nonbonding
e- pairs.
" bonds
# bond
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Postulating Hybrid Orbitals in a Molecule
Postulating Hybrid Orbitals in a Molecule
(a) Methanol, CH3OH
SOLUTION:
(a) CH3OH
The groups around C are
arranged as a tetrahedron.
(b) SF4 has a seesaw shape with 4 bonding and 1
nonbonding e- pairs.
O also has a tetrahedral
arrangement with 2 nonbonding
e- pairs.
S atom
single C atom
hybridized
C atom
hybridized
S atom
hybridized
O atom
single O atom
11-
Bond order is the number of bonds between two
bonded atoms.
Higher bond orders give shorter bond lengths and
require more energy to break a bond.
– Single bond between 2 atoms = order = 1
– Double bond between 2 atoms = order = 2
– Triple bond between 2 atoms = order = 3
S
C
N
C-N:
Bond order = 2
S-C:
Bond order = 2
F
F
F
N
Note how bond
energies (energy
required to break
a bond) goes up
as bond order
increases.
N
S
F
F
F
Bond order = 3
S-F
Bond order = 1
Bond Lengths
Triple bond < Double Bond < Single Bond
sp hybridization showing " and # bonds in acetylene (C2H2).
Ethane (CH3CH3) sp3 hybrid
both C are sp3 hybrids
sp3-sp3 C overlap forms
a " bond
overlap in one position - "
H and C
are s-sp3
overlaps to
" bonds
p overlap - #