Announcements
Exam III (Chapter 7-10)
Wednesday, October 3, 2012
Time: 6:00PM - 7:30PM
SEC A 214A and 215A
Chem 7 Final Exam
Wednesday, Oct 10 1:30-3:30AM
Chapter 1-12
70 or 75 multiple choice questions
Chapter 10
The Shapes of Molecules
Please note the following topics will be excluded from assessment. The page
numbers refer to the 2nd Edition of the textbook.
1. Numerical problems involving the Rydberg equation (Chapter 7)
2. Spectral analysis in the laboratory (Chapter 7 p. 226-227)
3. Trends among the transition elements (Chapter 8 p. 261)
4. Trends in electron affinity (Chapter 8 pp. 265-266)
5. Pseudo-noble gas configuration (Chapter 8 p. 269)
6. Lattice energy (Chapter 9 pp. 283-285)
7. IR spectroscopy (Chapter 9 p. 292)
8. Numerical problems involving electronegativity (Chapter 9 p. 296)
9. Electronegativity and oxidation number (Chapter 9 p. 297)
10. Section 11.3: MO theory and electron delocalization (Chapter 11 p.343)
11. All sections in chapter 12 except 12.3 (types of intermolecular forces).
Chemical bonds and the chemistry of an element is related
to the number of valence electrons are in the outer shell
(highest value of n quantum number) of the atom.
Group
e- configuration
1A
ns1
1
2A
ns2
2
3A
ns2np1
3
4A
ns2np2
4
5A
ns2np3
5
6A
ns2np4
6
7A
ns2np5
7
Lewis dot structures are used to depict valence
electrons and bonding between atoms.
+1
# of valence e-
+2
-3
+3
•
GOAL: Draw and connect Lewis structures with
geometry of a molecule (VSEPRT) and in Chapter 11
connect the geometry with how orbitals bond (VBT).
Molecular
formula
1
2
3
VSEPRT
Lewis
structure
2
Geometry
3
-1
LOOK FOR
Group #
A chemical symbol
represents the nucleus
and all core e-.
•
A single dot around
the symbol represents
one valence e-.
• For B group elements, the valence electrons are in
the highest value ns orbital and the (n-1)d orbitals.
1
-2
4
Valence Bond theory explains how bonding occurs
between atoms using “hybridized orbitals”.
Molecular
formula
VSEPRT
Lewis
structure
Geometry
Hybrid
orbitals
Hybrid
orbitals
4
Linear
VB
Theory
sp
Trigonal
Pyramidal
Tetrahedral
sp2
sp3
Valance Bond Theory
Trigonal
Bipyramidal
sp3d
Octahedral
sp3d2
Must Learn How To Draw Lewis Structures
1) Put least electronegative element as the central atom.
C,S,P and N are often central atoms. H and halogens are
often bonded to central atoms.
Electronegativity is an element’s inherent ability to
draw electrons to itself when chemically bonded to
another atom in a molecule (relative to Li).
F, O, N, Cl, Br, C
are highly
electronegative
with F being
the most
electronegative
2) Sum all valence e- from each atom in the molecule (careful
with ions add or subtract e-). Use the Group numbers!
3) Place bonds to central atoms using 2 e- per bond.
4) Place an octet of electrons (octet rule) around bonded
atoms remembering that H--bonds have no octet. Know the
incomplete and expanded octet exceptions.
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) Helpful Rules of Thumb: H forms 1-bond, C forms 4bonds, N forms 3-bonds, O forms 2-bonds.
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
N
5eF 7e- x 3 = 21e!
Total
26e!
Step 3 - Write Lewis structrue with N central and three
bonds and rest non-bonding octet electrons around the
central atom.
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.
We count and “code” the bonding/non-bonding
information into shorthand called AXnEm classification.
X = # of Bonded
Domains
octet
octet
F
N
F
F
octet
AX2E0 = AX2
A = Central Atom
shorthand
E = # Non-Bonded
Domains
octet
F
N
F
There are three major exceptions to the octet rule.
1) Incomplete Octet - Occurs with Be, B and Al as
central atoms.
F
4 electron groups
3 bonding
1 non-bonding
AX3E1
1. Incomplete Octet - no “octet” around
central atom. Occurs with Be, B and Al as
central atoms.
Draw Lewis structures for the following
2) Expanded Octet (the largest class of octet
exceptions)-occurs mostly with Period 3 non-metals
like P, S and halogens.
BeH2
3) Odd-number electrons highly reactive species
called radicals that have an odd number of electrons
(uneven).
AlCl3
BF3
Incomplete Octet: Occurs With Group 2A (Be) and
3A (Boron and Aluminum)
2. Expanded Octet (the largest class of octet
exceptions)-occurs mostly with Period 3 nonmetals like P, S and halogens.
Draw Lewis structures for the following
Draw Lewis structures for the following
Be – 2e2H – 2x1e4e-
BeH2
H
Al –
3Cl – 3x7e24e-
Cl
Phosphorous
trichloride
Al
Cl
AX3
Cl
B – 3e3F – 3x7e24e-
BF3
AX2
H
SF6
3e-
AlCl3
Be
F
B
F
PCl5
AX3
[ICl4]-1
F
Expanded Octet (the largest class of octet
exceptions)-occurs mostly with Period 3 nonmetals like P, S and halogens.
F
••
F
P
••
••
[ICl4]-1
Cl •
••
••
••
••
••
••
Cl
••
••
••
•
Phosphorous trichloride
PCl3
Cl
P
••
••
Cl •
••
••Cl
••Cl
F
F
••
•• Cl
P – 5e5Cl – 35e40e-
••
S
F
PCl5
Cl
••
SF6
••
F
••
S – 6e6F – 42e48e-
Draw Lewis structures for PCl3 PCl5 and the
carbonate anion. Determine the AXE
designation for each.
•
Phosphorous
pentachloride
Draw Lewis structues and determine the AXE
designation for each.
••
MUST look to see if its an ion and add the necessary electron!
••
••
Cl
4 electron groups
3 bonding
1 non-bonding
••
P
••
Cl •
••
••Cl
••
••
••
••
Electron Domains
••
5 electron groups
5 bonding
0 non-bonding
AX5E0
Br
O
4 electron groups
3 bonding
1 non-bonding
O
Cl •
••
••
••
Cl
••
••
[
O
Cl
P
Valence e- = 7 + 3(6) + 1 = 26
3
–
••
•• Cl
••Cl
••
[BrO ] –
AX3E1
•
••
••
Write the Lewis structure of the carbonate ion (BrO3-)
and hydrogen cyanide, give AXE designation.
O
C
O
HCN
•
O
]
2-
3 electron groups
3 bonding
0 non-bonding
AX3E0
2 electron groups
2 bonding
0 non-bonding
Valence e- = 1 + 4 + 5 = 10
Carbon is central atom, watch for hydrogen--1 bond
H
C
N
AX3E1
AX2
Write resonance structures for the carbonate ion,
CO3-.
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons (C and O)
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.
Which structure is correct?
Electron Domains
3 electron groups
3 bonding
0 non-bonding
A concept called “resonance” is used when
more than one plausible Lewis structure can be
drawn.
Example: Ozone, O3
••
••
••
••
••
••
O
O O
••
••
••
O
O O
••
••
••
2 equally good
Lewis
structures
All equally good and plausible
“resonance structures”
AX3
Write resonance structures for the nitrate ion, NO3-.
Measured
bond lengths
show they are
equal!
Which structure is correct?
Both are!
••
••
••
••
••
••
••
••
••
O
O O
••
••
O
O O
••
••
••
••
O
O O
••
••
a resonance
hybrid
structure
Write resonance structures for the nitrate ion, NO3-.
PLAN:
Count valence e- of atoms = 5 + (3X6) + 1 = 24 eMost electronegative atom in center
Surround and get an octet around N
When more than one Lewis structure is plausible,
we apply the concept of FORMAL CHARGE to
figure out the best Lewis structure!
Both are two plausible structures for CO2
O
C
O
VS
O
C
Which is the best one?
USE FORMAL CHARGE
Electron Domains
3 electron groups
3 bonding
0 non-bonding
AX3
O
To use the concept of formal charge, we determine
the formal charge for each atom.
Example: Write 3 plausible Lewis structures for the
thiocyanate ion [SCN]–
Atom Formal charge = # valence e- - Assigned e- to Atom
Assigned Atoms = all from lone pair e! + ! ( bonded e! )
O
e-
# Valence
# 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.
Net charge of ion or molecule must equal total formal charge.
2.! Also, the best Lewis structure places negative charge on the
most electronegative atom.
Example: Write 3 plausible Lewis structures for the
thiocyanate ion [SCN]–
Example: Write 3 plausible Lewis structures for
the thiocyanate ion [SCN]–
0
1. Count the Valence
S
C
N
# of Valence = 6 e- + 4 e- + 5 e- + 1 e- = 16 e-
[S
C
1
N
] [S
C
2
N
]– [ S
C
3
3-plausible Lewis structures which one is best?
Write resonance structures for the nitrate ion, NCOand determine the most plausible Lewis structure.
-1
N
-1
N
]
–
0
C
]– [ S
FCS = 6 - 4 -2 = 0
FCC = 4 - 0 - 4 = 0
FCN = 5 - 6 - 2 = -1
2. Draw the Lewis Structures With Least Electronegative Atom as
central atom.
–
0
C
[S
it’s a -1 ion
e-
0
N
+1
]– [ S
FCS = 6 - 6 -1 = -1
FCC = 4 - 0 - 4 = 0
FCN = 5 - 2 - 3 = 0
0
C
-2
N
]–
FCS = 6 - 2 -3 = 1
FCC = 4 - 0 - 4 = 0
FCN = 5 - 6 - 1 = -2
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.
0
[S
0
C
-1
N
]–
Structure on the left is “best” structure!
Write resonance structures for the nitrate ion, NCOand determine the most plausible Lewis structure.
EXAMPLE: NCO- has 3 possible resonance forms -
A
B
C
formal charges
-2
0
+1
-1
0
0
0
0
-1
Forms B and C have negative formal charges on N and O; this makes them
more preferred than form A.
Form C has a negative charge on O which is the more electronegative
element, therefore C contributes the most to the resonance hybrid.
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.
VSEPRT
Lewis
structure
Hybrid
orbitals
Geometry
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 electron geometry is the geometry of all
electron groups, 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.
Valence Bond
Theory
VSEPRT
VSEPRT
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
Electron Group
Geometry
Molecular
Geometry is
trigonal
pyramidal bond
angles <109.5˚
Molecular
Geometry
The 3-D geometry of a molecule is one of five basic
arrangements of electron groups (domains).
Linear
Trigonal
Pyramidal
Tetrahedral
sp2
sp3
Trigonal
Bipyramidal
Octahedral
VB
Theory
sp
sp3d
sp3d2
The total number of electron groups (domains)
defines one of the five basic geometries.
3 EG
4 EG
2 EG
Linear
Trigonal Planar
Tetrahedral
5 EG
Trigonal Bipyramidal
Octahedral
6 EG
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.
How Predict Geometry Using VSEPRT
1. Draw a plausible Lewis structure for the molecule.
2. Determine the total number of electron domains and
AX3E1 =
Tetrahedral
electron
geometery with
109.5˚ bond
angles.
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.
4. Establish the AXnEm designation to establish the
molecular geometry (or do both electron and molecular
geometry together simultaneously)
Molecular
Geometry is
trigonal
pyramidal bond
angles <109.5˚
2 Electron Groups = Linear Electron Geometry
and 1-Possible Molecular Geometry
Cl
Be
5. Remember bond angles in molecules are altered by lone
pairs of electrons (repulsion forces reduce angles).
6. Molecules with more than one central atom can be
handled individually.
3 Electron Groups = Trigonal Planar Electron
Geometry and 2-Possible Molecular Geometries
Cl
Examples:
SO3, BF3,
NO3-, CO32-
Bond Angle
S
O
C
C
N
O
AX2E0 = AX2
A = Central Atom
X = # of Bonded
Domains
4 Electron Groups = Tetrahedral Electron
Geometry and 3-Possible Molecular Geometries
Bond
Angle
Examples:
SO2, O3,
PbCl2, SnBr2
E = # Non-Bonded
Domains
AX4
3-Electron Domain
Other Examples:
CS2, HCN, BeF2
Examples:
CH4, SiCl4,
SO42-, ClO4-
AX3
A
AX2E1
A
5 Electron Groups = Trigonal Bipyramial Electron
Geometry and 4-Possible Molecular Geometries
PF5
AsF5
AX5
AX4E1
XeO2F2
SOF4
IF4+
IO2F2-
Equatorial
Position
AX3E1
NH3
PF3
ClO3
H3O+
AX2E2
H 2O
OF2
SCl2
SF4
Axial
Position
ClF3
AX2E3
BrF3
AX3E2
XeF2
I3 IF2-
6 Electron Groups = Octahedral Electron
Geometry and 3-Possible Molecular Geometries
AX6
BrF5
TeF5XeOF4
SF6
Predicting Molecular Shapes
Draw the molecular shape and predict the bond
angles (relative to the ideal bond angles) of (a)
PF3 and (b) COCl2.
IOF5
AX5E1
XeF4
AX4E2
Predicting Molecular Shapes
Draw the molecular shape and predict the bond
angles (relative to the ideal bond angles) of (a)
PF3 and (b) COCl2.
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
Determine the molecular shape and predict the bond
angles (relative to the ideal bond angles) of (a) SbF5
and (b) BrF5.
ICl4-
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.
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
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?
ethane
CH3CH3
Take one atom at a time and apply the
rules of electron domains.
Predicting the Molecular Shape With Multiple Central Atoms
Determine the shape around each of the central
atoms in acetone, (CH3)2C=O.
ethanol
CH3CH2OH
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).
Find the shape of one atom at a time after writing the Lewis
structure.
tetrahedral
Rank
F
O
N
Cl
Br
tetrahedral
trigonal planar
>1200
<1200
Differences in elements electronegativity between
bonding atoms result in the formation of polarcovalent bonds and net dipole moments in
molecules.
Polar Bond
d
on
Polar Bond
B
lar
Po
Draw a Lewis structure, show the AXE
designation, determine electron and molecular
geometry and whether polar or non-polar of:
CCl3H
Po
lar
B
on
d
CCl4
No Net Dipole Moment
Net Dipole Moment
Think of the dipole moment as a molecule with separated
charges + and -.
CH4
Draw a Lewis structure, show the AXE
designation, determine electron and molecular
geometry and whether polar or non-polar of:
AX4
Tetrahedral EG
Tetrahedra MG
CH3Cl
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
CCl4
CCl4
Polar bonds
Not Polar Molecule
No Dipole Moment
CH3Cl
Polar bond
Polar Molecule
Has Dipole Moment
Dipole
Moment
Dipole
Moment
No Net
Dipole
Moment
No Net
Dipole
Moment
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
From electronegativity (EN) values (button) and
their periodic trends, predict whether each of the
following molecules is polar and show the
direction of bond dipoles and the overall
molecular dipole when applicable:
(a) Ammonia, NH3
Predicting the Polarity of Molecules
PROBLEM:
From electronegativity (EN) values (button) and their periodic
trends, predict whether each of the following molecules is polar
and show the direction of bond dipoles and the overall
molecular dipole when applicable:
(a) Ammonia, NH3
(b) Boron trifluoride, BF3
(c) Carbonyl sulfide, COS (atom sequence SCO)
PLAN: Draw the shape, find the EN values and combine the concepts to
determine the polarity.
(b) Boron trifluoride, BF3
SOLUTION:
The dipoles reinforce each
other, so the overall molecule
is definitely polar.
(a) NH3
ENN = 3.0
(c) Carbonyl sulfide, COS (atom sequence SCO)
ENH = 2.1
bond dipoles
molecular
dipole
10-
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 10.9
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Predicting the Polarity of Molecules
(b) BF3 has 24 valence e! and all electrons around the B will be involved in
bonds. The shape is AX3, trigonal planar.
1200
Chapter 11
F (EN 4.0) is more electronegative
than B (EN 2.0) and all of the dipoles
will be directed from B to F. Because
all are at the same angle and of the
same magnitude, the molecule is
nonpolar.
Theories of Covalent Bonding
(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is
quite polar ("EN), so the molecule is polar overall.
10-
11-
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Theories of Covalent Bonding
Valence Bond Theory explains covalent bonding
by the spatial overlap of atomic orbitals on
bonding atoms and the sharing of electron pairs.
This only!
Bonding in H2
1s1 + 1s1
11.1 Valence Bond (VB) Theory and Orbital Hybridization
Electrons that must have opposite spins.
11.2 The Mode of Orbital Overlap and the Types of Covalent Bonds
Bonding in HF
1s1 + 2p5
Bonding in F2
2p5 + 2p5
11.3 Molecular Orbital (MO)Theory and Electron Delocalization
11-
It’s natural to think of using “pure atomic
orbitals” to describe bonding in some
molecules. It works well for some....but fails
with carbon.
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 + 1s22s22p5
1s1 + 2p5
Bonding in F2
only two bond are possible with use of atomic
orbitals only....we don’t observe CH2
hybridization fixes the problem!
1s22s22p5 + 1s22s22p5
2p5 + 2p5
Hybridization combines or mixes different numbers of pure
atomic orbitals that match one of the VSEPRT geometries.
For example 1 pure s orbital + 1 p-orbital combine to give
and 2 “sp hybrids” that when superimposed form a linear
geometry for bonding.
Pure
atomic
orbitals
(valence
orbitals)
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.
Connect the dots and it becomes easy to see and
understand.
Molecular
formula
VSEPRT
Lewis
structure
2. The name of and shape of a “hybrid orbital” varies
with the types of atomic orbitals mixed. (s + p vs s +
two p)
Valence Bond Theory
explains how bonds
are made
3. Each hybrid orbital has a specific geometry that
matches one of five VSEPRT shapes (show below).
sp
sp2
Linear
sp3
sp3d
Tetrahedral
Trigonal
Planar
sp3d2
Trigonal
Bipyramidal
Octahedral
Electron
Geometry
Molecular
Geometry
AXnEm
Hybridization
Linear
Linear
AX2
sp
Trigonal
planar
Trigonal planar
V-shaped bent
AX3
AX2E1
sp2
AX4
AX3E1
AX2E2
sp3
Tetrahedral
Tetrahedral
Trigonal pyramidal
V-shaped bent
Trigonal
bipyramidal
Trigonal bipyramidal
Seesaw
T-shaped
Linear
AX5
AX4E1
AX3E2
AX2E3
sp3d
Octahedral
Octahedral
Square pyramidal
Square planar
AX6
AX5E1
AX4E2
sp3d2
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?
trig planar 120˚, sp2
2
sp2 sp
bent, <109.5, sp3
sp3
tetrahedral, 180,
sp3
Hybrid
orbitals
Geometry
sp
Linear
sp2
sp3
Trigonal
planar
sp3d
Tetrahedral
Trigonal
Bipyramidal
sp3d2
Octahedral
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?
Linking VSEPRT To Valence Bond Theory Hybrids
Atomic
Orbitals
Mixed
#
Hybrid
Orbitals
Formed
Linear Trig
Tetrahedral Trig Bypyr Octahedral
AX2
Planar AX3 AX4
AX5
AX6
s+p
s+2p
s+3p
Two sp
Three sp2
Four sp3
Two p
one p
none
s+3p+d
Five sp3d
s + 3 p + 2d
Six sp3d2
Hybrid
Shape
linear 180˚, sp
Orbitals
Leftover
for Pi
bonds
Four d
Three 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
Example sp hybrid: Show the bonding scheme and
hybridized orbitals used in BeCl2
2 unhybridized unoccupied p-orbitals
2 “left-over” p-orbitals
Hybridization
s-orbital
p-orbital
Two sp hybrid sp hybrid orbitals
orbitals
superimposed
s + p Hybridization = 2 sp
--The number of hybrid orbitals formed is equal to the number of
“pure orbitals” combined!
After hybridization we
have on the central atom,
2 pure p-orbitals and two
sp hybrids.
--When superimposed the “sp-hybrid” give us bonding orbitals for
a linear molecules.
Show the bonding scheme and hybridized orbitals in BeCl2
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
hybridization
Hybridized Be Atom
Isolated Be Atom
two sp hybrids
on Be
Superimposed
Hybrid orbitals
form a triginal
planar geometry
two lone p-orbitals
sp2 = Triginal planar geometry, 120˚ bond angle
Example 2: sp2 hybridizaton scheme BF3.
Tetrahedral geometry = sp3 hybrid orbitals
combine to generate
four sp3 orbitals
Boron Hybrid Box Diagram
Boron Orbital
Box Diagram
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
Bonding of pure p-orbital in F
with sp2 hybridized orbitals in BF3
Example: sp3 orbital hybridization: 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 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.
This is the
ground state
configuration of
valence atomic
orbitals
sp3 is tetrahedral
shape. In water
we have AX2E2
the four sp3
hybrid
orbitals form
a tetrahedral
shape
CH4
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?
What is the electronic geometry?
What is the molecular geometry?
What orbitals contribute to bonding?
Example 2: sp3 hybridization in NH3.
Tetrahedral Electron Geometry AX3E1
Trigonal Pyramidal Molecular Geometry
sp3d hybridization in PCl5.
Trigonal Bipyramidal Electron Geometry AX5E0
Trigonal BiPyramidal Molecular Geometry
Isolated
P atom
The sp3d2 hybrid orbitals in SF6
Octahedral Electron Geometry AX6E0
Octahedral Molecular Geometry
Note the lone
pairs occupy 2-of
the sp3 orbitals
Electron
Geometry
Molecular
Geometry
AXnEm
Hybridizaton
Linear
Linear
AX2
sp
Trigonal
planar
Trigonal planar
V-shaped bent
AX3
AX2E1
sp2
AX4
AX3E1
AX2E2
sp3
Tetrahedral
Trigonal pyramidal
V-shaped bent
Tetrahedral
Trigonal
bipyramidal
Trigonal bipyramidal
Seesaw
T-shaped
Linear
AX5
AX4E1
AX3E2
AX2E3
sp3d
Octahedral
Octahedral
Square pyramidal
Square planar
AX6
AX5E1
AX4E2
sp3d2
Describe the types of bonds and orbitals in
acetone, (CH3)2CO and in CO2 and in HCN?
Step 1
Molecular
formula
Step 2
Lewis
structure
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.
Lone p orbitals that are not hybridized
sp3 hybridized
sp3 hybridized
sp2 hybridized
$ bond
sp hybrid:Ethylyne: HC!CH:Linear
sp2 hybrid orbitals on each
carbon atom use end to
end overlap to form a
sigma bond.
Hybrid
orbitals
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)
SOLUTION:
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.
VSEPRT
Geometry
sp hybrid:Ethylyne: HC!CH:Linear
Describe the types of bonds and orbitals in
acetone, (CH3)2CO.
# bonds
Step 3
# bonds overlap
side to side
sp hybrid orbitals