Lewis Structure

Transcription

Lewis Structure

Chapter 9. Chemical Bonding I: The Lewis Model
Modified by Dr. Cheng-Yu Lai
Chemicals Bonds
 Forces that hold groups of atoms together and make them
function as a unit.
 Ionic bonds – transfer of electrons
 Covalent bonds – sharing of electrons
• Explain how and why atoms attach together to form molecules
• Explain why some combinations of atoms are stable and others are not
– Why is water H2O, not HO or H3O?
• Can be used to predict the shapes of molecules
• Can be used to predict the chemical and physical properties of
compounds
© 2014 Pearson Education, Inc.
Electronegativity Difference and Bond Type
Polar-Covalent bonds
 Electrons are unequally shared
 Electronegativity difference between .5
and 1.9
Nonpolar-Covalent bonds
 Electrons are equally shared
 Electronegativity difference of
0 to 0.4
Inoic bonds
 Electrons are unequally shared
 Electronegativity difference above 2.0
Electronegativity
The ability of an atom in
a molecule to attract
shared electrons to
itself.
Linus Pauling
1901 - 1994
Table of Electronegativities
Polar Covalent Bonds
A polar covalent bond
• occurs between nonmetal atoms.
• is an unequal sharing of electrons.
• has a moderate electronegativity difference (0.5 to 1.7).
Examples:
Atoms
O-Cl
Cl-C
O-S
Electronegativity
Difference
3.5 - 3.0 = 0.5
3.0 - 2.5 = 0.5
3.5 - 2.5 = 1.0
Type of Bond
Polar covalent
Polar covalent
Polar covalent
6
Polar Covalent Bonds:
Electronegativity
Chapter 7/7
Polar Covalent Bonds and Dipole Moment:
The dipole moment is a measure of polarity in a chemical bond or
molecule. It is a drawing made on a Lewis Structure composed of an
arrow pointing from a "slightly-positive" molecule to a "slightlynegative" molecule.
H --- Cl
+------>
Partial positive
is represented
by this symbol
Partial negative
is represented
By this symbol
Ionic Bonds
An ionic bond
• occurs between metal and nonmetal ions.
• is a result of electron transfer.
• has a large electronegativity difference (1.8 or more).
Examples:
Atoms
Cl-K
N-Na
S-Cs
Electronegativity
Difference
3.0 – 0.8
= 2.2
3.0 – 0.9
= 2.1
2.5 – 0.7
= 1.8
Type of Bond
Ionic
Ionic
Ionic
9
Example 9.3 Classifying Bonds as Pure Covalent, Polar Covalent,
or Ionic
Determine whether the bond formed between each pair of atoms is covalent, polar covalent, or ionic.
a. Sr and F
b. N and Cl
c. N and O
Solution
a. In Figure 9.8, find the electronegativity of
Sr (1.0) and of F (4.0). The electronegativity
difference (ΔEN) is ΔEN = 4.0 – 1.0 = 3.0.
Using Table 9.1, classify this bond as ionic.
b. In Figure 9.8, find the electronegativity of
N (3.0) and of Cl (3.0). The electronegativity
difference (ΔEN) is ΔEN = 3.0 – 3.0 = 0.
Using Table 9.1, classify this bond as
covalent.
c. In Figure 9.8, find the electronegativity of
N (3.0) and of O (3.5). The electronegativity
difference (ΔEN) is ΔEN = 3.5 – 3.0 = 0.5.
Using Table 9.1, classify this bond as polar
covalent.
FIGURE 9.8 Electronegativities of the Elements
Electronegativity generally increases as we move across a row
in the periodic table and decreases as we move down a
column.
Predicting Bond Types
12
Bond Length and Energy
It is the energy required to break a bond.
It gives us information about the strength of a bonding
interaction.
Bond length
(pm)
Bond Energy
(kJ/mol)
Bond
Bond type
C - C
Single
154
347
C = C
Double
134
614
C  C
Triple
120
839
C - O
Single
143
358
C = O
Double
123
745
C - N
Single
143
305
C = N
Double
138
615
C  N
Triple
116
891
Bonds between elements become shorter and stronger
as multiplicity increases.
Bond Energy and Enthalpy
H   Dbondsbroken   Dbonds formed
Energy required
Energy released
D = Bond energy per mole of bonds
Breaking bonds always requires energy
Breaking = endothermic
Forming bonds always releases energy
Forming = exothermic
Example 9.11 Calculating ΔHrxn from Bond Energies
Hydrogen gas, a potential fuel, can be made by the reaction of methane gas and steam.
Use bond energies to calculate ΔHrxn for this reaction.
Solution
Begin by rewriting the reaction using the Lewis structures of the molecules involved.
Determine which bonds are broken in the reaction and sum the bond energies of these.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
Example 9.11 Calculating ΔHrxn from Bond Energies
Continued
Determine which bonds are formed in the reaction and sum the negatives of their bond energies.
Find ΔHrxn by summing the results of the previous two steps.
Nivaldo J. Tro
Ionic Bonding
 Electrons are transferred
 Electronegativity differences are
generally greater than 1.7
 The formation of ionic bonds is
always exothermic!
Determination of
Ionic Character
Electronegativity
difference is not the
final determination
of ionic character
Compounds are ionic
if they conduct
electricity in their
molten state
Coulomb’s Law
“The energy of interaction between a pair of
ions is proportional to the product of their
charges, divided by the distance between
their centers”
 Q1Q2 
E  (2.31 x 10 J  nm) 

 r 
 Q1Q2 
E

 r 
19
Example 9.2 Predicting Relative Lattice Energies
Arrange these ionic compounds in order of increasing magnitude of lattice energy: CaO, KBr, KCl, SrO.
Solution
KCl should have lattice energies of smaller magnitude than CaO because of their lower ionic charges (1+, 1–
compared to 2+, 2–.)
Order of increasing magnitude of lattice energy:
KCl < CaO
Actual lattice energy values:
For More Practice 9.2
Which compound has a lattice energy of higher magnitude, NaCl or MgCl2?
Nivaldo J. Tro
Electron-Dot Structures
Electron-Dot Structure (Lewis Structure): Represents an
atom’s valence electrons by dots and indicates by the
placement of the dots the way the valence electrons are
distributed in a molecule.
Chapter 7/21
Lewis Structures
Lewis structures show how valence electrons
are arranged among atoms in a molecule.
Lewis structures Reflect the central idea
that stability of a compound relates to
noble gas electron configuration.
Shared electrons pairs are covalent bonds
and can be represented by two dots (:) or
by a single line ( - )
From Atomic Valance Shell Electrons
To Electron Dot Structures
Symbols of atoms with dots to represent the valence-shell
electrons only
1
2
13
14
15
16
17
H
18
He:

Li Be


B 


C


Na Mg


Al

N



O




 Si 
 P
S





: F  :Ne :




:Cl  :Ar :


The Octet Rule
Combinations of elements tend to form so that
each atom, by gaining, losing, or sharing
electrons, has 8 electrons in its outermost
occupied energy level.
Monatomic chlorine
Diatomic chlorine
Octet Rule = atoms tend to gain, lose or share electrons so as to have 8
electrons
C would like to
N would like to
O would like to
Gain 4 electrons
Gain 3 electrons
Gain 2 electrons
The HONC Rule for Octet Rule
Hydrogen (and Halogens) form one covalent
bond
Oxygen (and sulfur) form two covalent bonds
One double bond, or two single bonds
Nitrogen (and phosphorus) form three covalent
bonds
One triple bond, or three single bonds, or one
double bond and a single bond
Carbon (and silicon) form four covalent bonds.
Two double bonds, or four single bonds, or a triple
and a single, or a double and two singles
The Octet Rule:
The Diatomic Fluorine Molecule
F
F
1s
2s
2p
1s
2s
2p
F F
Each has
seven valence
electrons
A Single Bond is when
2 electrons are shared
they are represented by
a single line in bond
diagrams
The Octet Rule:
The Diatomic Oxygen Molecule
O
O
1s
2s
2p
1s
2s
2p
O O
Each has six
valence
electrons
A Double bond is when 4
electrons are shared they
are represented by two
lines in bond diagrams
The Octet Rule:
The Diatomic Nitrogen Molecule
N
N
1s
2s
2p
1s
2s
2p
N N
Each has five
valence
electrons
A Triple bond is when
6 electrons are shared
they are represented by
three lines in bond
diagrams
LEWIS
STRUCTURES
 More complex Lewis structures can be drawn by
following a stepwise method:
1. Count the number of electrons in the structure.
2. Draw a skeleton structure.
- most metallic element generally central
- halogens and hydrogen are generally
terminal
- many molecules tend to be symmetrical
- in oxyacids, the acid hydrogens are attached
to an oxygen
31
LEWIS
STRUCTURES
 More complex Lewis structures can be drawn by
following a stepwise method:
3. Connect atoms by bonds (dashes or dots).
4. Distribute electrons to achieve Octet rule.
5. Form multiple bonds if necessary.
32
Example 1:
Write Lewis structure for H2O
Step 1:
Step 2:
Step 3:
Step 4:
H2O
= 8 electrons
2 (1) + 6 = 8

H O H

Skeleton has
Hydrogen
structure
doublet
4
electrons
used
Octet rule is satisfied should be
4 electrons remaining
symmetrical
33
Example 2:
Write Lewis structure for CO2
Step 1:
CO2
= 16 electrons 4 + 2(6) = 16
Step 2:
Step 3:
Step 4:
Step 5:






O C O 
Skeleton
structure
Octet
10
4
electrons
rule
is
NOT
used
Octet rule is satisfied should be
612electrons
electrons
satisfied
remaining
remaining
symmetrical
34
Writing Lewis Structures for
Polyatomic Ions
• the procedure is the same, the only
difference is in counting the valence
electrons
• for polyatomic cations, take away one
electron from the total for each positive
charge
• for polyatomic anions, add one electron to
the total for each negative charge
Example 3:
Write Lewis structure for CO32Step 1:
CO32-
= 24 electrons 4+3(6)+2 = 24
12
18
06 electrons
electrons
remaining



O

Step 5:

O C O

Step 4:


Step 3:

Step 2:
Octet
Octetrule
ruleisissatisfied
NOT
satisfied
36
Example 4:
Determine if each of the following Lewis structures
are correct or incorrect. If incorrect, rewrite the
correct structure.
Structure
has
14 electrons
Only
12
electrons
2(5) + 4(1) = 14
shown
2
4
2
2
2
Structure is
incorrect
Octets are complete
37
Exceptions to the Octet Rule
• H & Li, lose one electron to form cation
– Li now has electron configuration like He
– H can also share or gain one electron to have
configuration like He
• Be shares 2 electrons to form two single bonds
• B shares 3 electrons to form three single bonds
• expanded octets for elements in Period 3 or
below
– using empty valence d orbitals
• some molecules have odd numbers of electrons
– NO


:NO:
Some molecules, such as SF6 and PCl5 have more
than 8 electrons around a central atom in their
Lewis structure.
Draw an electron-dot structure for SF6.
Step 1: 6 + 4(7) = 34 valence electrons
F
F
F
F
S
Step 2:
F
F
S
Step 3:
F
F
F
F
F
F
SF6 and PCl5 can violate the octet rule through the use of empty d orbitals:
both S and P can utilize empty d orbitals to hold pairs of electrons that help
bond halogen atoms.
Electron-Dot Structures of Compounds Containing
Elements Below the Second Row
Draw an electron-dot structure for ICl3.
Step 1: 7 + 3(7) = 28 valence electrons
Cl
Cl
I
Step 2:
Step 4:
Cl
Step 3:
Cl
Cl
Cl
Cl
Cl
I
Cl
I
Worked Example 7.3 Drawing an Electron-Dot Structure
Draw an electron-dot structure for the deadly gas hydrogen cyanide, HCN.
Strategy
First, connect the carbon and nitrogen atoms. The only way the carbon can form four bonds and the nitrogen
can form three bonds is if there is a carbon–nitrogen triple bond.
Solution
Draw an electron-dot structure for carbon dioxide, CH3CN and CH3OH
Instructor Resource DVD for Chemistry, 6th Edition
John McMurry & Robert C. Fay
Resonance
• we can often draw more than one valid
Lewis structure for a molecule or ion
• in other words, no one Lewis structure
can adequately describe the actual
structure of the molecule
• the actual molecule will have some
characteristics of all the valid Lewis
structures we can draw
Resonance- Same length
• Lewis structures often do not accurately represent
the electron distribution in a molecule
– Lewis structures imply that O3 has a single (147 pm) and
double (121 pm) bond, but actual bond length is
between, (128 pm)
• Real molecule is a hybrid of all possible Lewis
structures
• Resonance stabilizes the molecule
– maximum stabilization comes when resonance forms
contribute equally to the hybrid
O
O
+
O
O
O
+
O
Electron-Dot Structures of Compounds Containing
Elements Below the Second Row
Draw an electron-dot structure for O3.
Step 1: 3(6) = 18 valence electrons
Step 2:
O
O
O
Step 4:
O
O
O
Check octet rule
Step 3:
O
O
O
Step 5:
O
O
O
Resonance in Polyatomic Ions
Resonance in a carbonate ion:
The bond lengths in the structures are identical, and
between those of single and double bonds.
Resonance in an acetate ion:
Electron-Dot Structures and Resonance
Move a lone pair from this oxygen?
Step 4:
O
O
O
Or move a lone pair from this oxygen?
O
O
O
O
O
O
Resonance
Oxygen bond lengths are identical, and intermediate to
single and double bonds
Formal Charges
# of
# of
# of
1
Formal
–
bonding – nonbonding
= valence e –
Charge
2
in free atom
e–
e–
Calculate the formal charge on each atom in O3.
O
1
6 – (4) – 4 = 0
2
O
O
1
6 – (6) – 2 = +1
2
1
6 – (2) – 6 = -1
2
Chapter 7/47
Example 9.8 Assigning Formal Charges
Assign formal charges to each atom in the resonance forms of the cyanate ion (OCN–). Which resonance form is
likely to contribute most to the correct structure of OCN–?
Solution
Calculate the formal charge on each atom by finding the number of valence electrons and subtracting the number
of nonbonding electrons and one-half the number of bonding electrons.
The sum of all formal charges for each structure is –1, as it should be for a 1– ion. Structures A and B have the
least amount of formal charge and are therefore to be preferred over structure C. Structure A is preferable to B
because it has the negative formal charge on the more electronegative atom. You therefore expect structure A to
make the biggest contribution to the resonance forms of the cyanate ion.
Nivaldo J. Tro
Resonance – Same length
Resonance is invoked when more than one valid Lewis
structure can be written for a particular molecule.
H
Benzene, C6H6
H
H
H
H
H
H
H
H
H
H
H
The actual structure is an average of the resonance
structures.
The bond lengths in the ring are identical, and
between those of single and double bonds.
Resonance Bond Length and Bond Energy
Resonance bonds are shorter and stronger
than single bonds.
H
H
H
H
H
H
H
H
H
H
H
H
Resonance bonds are longer and weaker than double
bonds.
Isomers: Same composition, two different
constitutional Lewis structures
HCN = atomic compositional structure
HCN possesses 10 VE = Lewis compositional structures
Two possible Lewis constitutional structures:
H-C-N or H-N-C
Both need to have 10 VE in their Lewis structure
Problem: Try to achieve an acceptable Lewis structure
(duet and octet rule followed) for both.
51
HCN = atomic compositional structure
HCN: 10 VE = Lewis compositional structures
Two possible Lewis constitutional structures
H-C-N or H-N-C
Any acceptable Lewis structure for HCN needs to show 10 VE
Try to achieve an acceptable Lewis structure (duet and octet rules
obeyed) for all isomeric structures.
Two acceptable Lewis structures. Which is better?
H
C
N
H
N
C
52
Use formal charges to decide on the stability of
isomeric Lewis structures
H
C
N
H
N
C
VE (atom)
1
4
5
1
5
4
1/2 BE (molecule)
-1
-4
-3
-1
-4
-3
UE (molecule)
0
0
-2
0
0
-2
________________________________________________________
FC on atom
0
0
0
0
+1
-1
H
C
N
H
N
C
Important: the net charge of composition HCN = 0, so the
sum of the formal charges in any acceptable Lewis structure
must be = 0 also.
53
Which is more stable?
0
0
H
C
0
N
0
+1
-1
H
N
C
Rule:
For two isomeric acceptable Lewis structures, the one
with the least separation of formal charges is more
stable.
Therefore, H
C N
stable isomer of the pair.
is the more
54
Molecular Structure
Molecular geometry is the general
shape of a molecule or the
arrangement of atoms in three
dimensional space.
Physical and chemical properties
depend on the geometry of a molecule.
55
Chapter 10.Chemical Bonding II: Molecular
Shapes, Valence Bond Theory, and Molecular
Orbital Theory
It is a method for predicting the shape of a molecule from
the knowledge of the groups of electrons around a central
atom.
VSEPR Model
The Valence Shell Electron Pair Repulsion model
predicts the shapes of molecules and ions by
assuming that the valence shell electron pairs are
arranged as far from one another as possible to
minimize the repulsion between them.
57
Predict Molecular Shape-VSEPR

Only five basic shapes.
Electron Pair Geometry –
is determined by the number and
n m
arrangement of all electron pairs
(bonding and lone) around the central
AXE shorthand notation:
atom.
• A - central atom
Molecular geometry –
• X - terminal atoms
is determined by the arrangement of
• n – bonding pairs
atoms (or bonding electron pairs only)
around the central atom.
• E - lone pair
• m – lone pair electrons
AX E
AX3E0
AXnEm Type
SO32－
1. Central atom S=A
2. Terminal atomes O=X, n=3
3. Calculate Total valence electrons
6+3*6+2 electrons = 26 e
4. bonding pairs n = (total valence electrons)/8 = 26 /8= 3 bonding pairs
+ 2 remainder electrons
5. Lone pair m = (remainder electrons)/2 =2/2=1 lone pairs
6. SO32－ = AX3E1
AXnEm Type
NH4+
1. Central atom N=A
2. Terminal atomes H=X, n=4
3. Calculate Total valence electrons – In VSEPR calculation, for H fullfills
octet rule, we only need to add one more electron into 1S orbital. But
for our calculation here, we make H=7 pseudo valance electrons
3+4*7-1 electrons = 32 e
4. bonding pairs n = (total valence electrons)/8 = 32/8= 4 bonding pairs
+ 2 remainder electrons
5. Lone pair m = (remainder electrons)/2 =0/2=0 lone pairs
6. NH4+ = AX4E0
More AXnEm Types – 13 students
Example 1: BeCl2
Example 2: BF3
AX2E0
Example 3: SO2
AX3E0
Example 4: CH4
AX4E0
Example 5: NH3
AX4E0
AX3E1
Example 6: H2O
AX4E0
AX2E2
AX3E0
AX2E1
More AXnEm Types
Example 7: PF5
AX5E0
Example 8: SF4
Example 9: BrF3
AX5E0
AX5E0
AX4E1
AX3E2
Example 10: XeF2
AX5E0
AX2E3
Example 11: SF6
Example 12: IF5
AX6E0
Example 13: XeF4
AX6E0
AX6E0
AX5E1
AX5E1
Molecular Shapes
Predicting Molecular Geometries
AX2E0
AX3E0
AX2E1
Chapter 9
65
Molecular Shapes
Predicting Molecular Geometries
NH4+
AX4E0
AX3E1
AX2E2
66
Molecular Shapes
Molecules with Expanded Valence Shells
AX5E0
PF4-
AX4E1
AX3E2
AX2E3
67
Molecular Shapes
Molecules with Expanded Valence Shells
AX6E0
AX5E1
AX4E2
Chapter 9
68
VSEPR – Valence Shell Electron Pair
Repulsion
X+E
Overall Structure
Forms
2
3
4
5
6
Linear
AX2
Trigonal Planar
AX3, AX2E
Tetrahedral
AX4, AX3E, AX2E2
Trigonal bipyramidal
AX5, AX4E, AX3E2, AX2E3
Octahedral
AX6, AX5E, AX4E2
A = central atom
X = atoms bonded to A
E = nonbonding electron pairs on A
More AXnEm Types
Example 1: BeCl2
Example 2: BF3
AX2E0 linear
Example 3: SO2
AX2E1 angular (bent)
Example 4: CH4
AX4E0 tetrahedral
Example 5: NH3
AX3E1 triangular pyramidal
Example 6: H2O
AX2E2 angular (bent)
AX3E0 triangular planar
Example 10.4 Predicting the Shape of Larger Molecules
Predict the geometry about each interior atom in methanol (CH 3OH) and make a sketch of the molecule.
Solution
Begin by drawing the Lewis structure of CH3OH. CH3OH contains two interior atoms: one carbon atom and one
oxygen atom. To determine the shape of methanol, determine the geometry about each interior atom as follows:
Using the geometries of each of these, draw a three-dimensional sketch of the molecule as shown here:
Nivaldo J. Tro
Geometry and Polarity of Molecules
• For a molecule to be polar it must
1) have polar bonds, symmetrical shape, and
different terminal atoms
2) have polar bonds
• electronegativity difference - theory
• bond dipole moments – measured
3) have an unsymmetrical shape
• using vector addition
• polarity effects the intermolecular forces
of attraction
Dipole moment is the measured polarity of a polar covalent bond.
It is defined as the magnitude of charge (electrons) on the atoms and the
distance between the two bonded atoms.

:OCO:
O
H

H
polar bonds,
and unsymmetrical
shape causes molecule
to be polar
polar bonds,
but nonpolar molecule
because pulls cancel
Cl
Cl
Cl
Cl
C
C
Cl
CCl4
m = 0.0 D
H
H
Cl
CH2Cl2
m = 2.0 D
Adding Dipole Moments
76
Molecular Geometry
Dipole Moment and Polarity
CO, PCl3, BCl3, GeH4, CF4
• Which compound is the most polar?
• Which compounds on the list are non-polar?
77
Orbitals Consistent with
Molecular Shape
Lewis dot + VSEPR gives the correct
shape for a molecule. BUT…
How do atomic orbitals (s, p, d …) produce
these shapes?
Valence bond theory describes a bond as
an overlap of atomic (hybrid) orbitals.
78
Molecular Shapes

Atomic orbitals (AOs) can be hybridized (mixed).
• Sets of identical hybrid orbitals form identical
bonds.
• # AOs that hybridize = # hybrids orbitals .
s+p
sp + sp
s+p+p
sp2 + sp2 + sp2
etc….
79
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/hybrv18.swf
sp Hybrid Orbitals
Energy, E
AX2E0, Ex: BeCl2,
2p 2p 2p
2p 2p 2p
Promotion
Two unhybridized
p orbitals
Orbital
hybridization
2s
2s
Two sp hybrid
orbitals on Be in BeF2
Isolated Be atom
sp hybridization occurs around the central atom whenever there
are two regions of high e- density.
Two equivalent covalent bonds form (180° apart) LINEAR.
80
sp Hybrid Orbitals
sp2 Hybrid Orbitals
AX3E0, Ex: BF3
The result is THREE equivalent
hybrid orbitals, in a VSEPR basic
shape of trigonal planar.
82
p. 396
sp3 Hybrid Orbitals
AX4E0, Ex: CH4
TETRAHEDRAL
83
sp3 Hybrid Orbitals
AX3E1 ( NH3)
and
AX2E3 ( H2O)
84
Molecular Shapes
Describe bonding in PCl5 using hybrid orbitals.
:
:
P
: :
:
: Cl :
Cl:
:
: Cl
Cl :
trigonal bipyramidal
We need 5 orbitals.
:
:
:Cl :
AX5E0
85
sp3d Hybrid Orbitals
3d
valence
shell
hybridization
3p
five equal sp3d
hybrid orbitals
X
3s
P atom (ground state)
86
sp3d Hybrid Orbitals
3d
sp3d
P atom (hybridized state)
87
Molecular Shapes
Describe the bonding in SF6 using hybrid orbitals.
:
:
:
:F
:F:
F:
:F
: :
: :
S
F:
AX6E0
Octahedral
We need 6 orbitals.
:
:
: F:
:
88
sp3d 2 Hybrid Orbitals
3d
X
hybridization
3p
six equal sp3d2
hybrid orbitals
X
3s
S atom (ground state)
89
3
2
sp d
Hybrid Orbitals
3d
sp3d2
S atom (hybridized state)
90
Hybridization
Mixed
s+p
s+p+p
s+p+p+p
Hybrids (#) Remaining Geometry
sp (2)
p+p
Linear
sp2 (3)
p
Triangular planar
sp3 (4)
Tetrahedral
Mixed
Hybrids (#) Remaining Geometry
s+p+p+p+d
sp3d (5)
d+d+d+d
Triangular bipyramid
s+p+p+p+d+d
sp3d2 (6)
d+d+d
Octahedral
91
Summary - Hybrid Orbitals
Hybrid
Orbital
Geometric
Arrangements
Number of
Orbitals
Example
sp
Linear
2
Be in BeF2
sp2
Trigonal planar
3
B in BF3
sp3
Tetrahedral
4
C in CH4
sp3d
5
P in PCl5
sp3d2
Octahedral
6
S in SF6
92
Hybridization and Molecular Geometry
Forms
Overall Structure
Hybridization
of “A”
AX2
Linear
sp
AX3, AX2E
Trigonal Planar
sp2
AX4, AX3E, AX2E2
Tetrahedral
sp3
AX5, AX4E, AX3E2, AX2E3
sp3d
AX6, AX5E, AX4E2
Octahedral
sp3d2
A = central atom
X = atoms bonded to A
E = nonbonding electron pairs on A
Practice
What are the
hybridization and
approximate bond
angles for each C, N, O
in the given molecules?
94
What about…
multiple bonding!
According to valence bond theory hybrid orbitals
include:
– single bonds
– lone pairs
– one of the bonds in a multiple bond.
The electrons in the unhybridized atomic orbitals
are used to form the additional multiple
bonds (from Molecular Orbital Theory.)
95
Multiple Bonding
• A s (sigma) bond is an overlap of orbitals
(hybrids) along the bond axis.
• A p (pi) bond is a overlap of parallel “p” orbitals,
creating an electron distribution above and below
the bond axis. ()
96
Multiple Bonding
(unhybridized)
2p
2p
Energy
sp2
2s
1s
C atom (ground state)
1s
(3 sp2 hybrid
+ 1 unhybridized p)
97
Multiple Bonding
σ
π
98
Sigma and Pi Bonds:
Double bonds
1 p bond
H
H
C
H
H
C
H
C
H
H
C
H
Ethene
1 s bond
Multiple Bonding
100
Sigma and Pi Bonds
Sigma (s) bonds exist in the region directly between
two bonded atoms.
Pi (p) bonds exist in the region above and below a line
drawn between two bonded atoms.
Single bond
1 sigma bond
Double Bond
1 sigma, 1 pi bond
Triple Bond
1 sigma, 2 pi bonds
The De-Localized Electron Model
Pi bonds (p) contribute to the delocalized model of
electrons in bonding, and help explain resonance
H
H
H
H
H
H
H
H
H
H
H
H
Electron density from p bonds can be distributed
symmetrically all around the ring, above and below
the plane.
Practice
Please draw the lewis structure,
identify the bond types and point
the types hybridization orbitals in
C and N.
CH3CN → CH3CH2NH2
• Identify the pi and sigma
bonds in the given
molecules.
σ
σ
σ
π
σ
σ
σ
π, π
103
Electron Dot Formula for CS2
1. Central atom C=A
2. Terminal atomes S=X, n=2
4+2*6 electrons = 16 e
4. bonding pairs n = (total valence
electrons)/8 = 16 /8= 2 bonding
pairs + 0 remainder electrons
5. Lone pair m = (remainder
electrons)/2 =0/2=0 lone pairs
6. CS2 = AX2E0 → Linear → S-C-S
7. Hybridization orbital → sp
8. Bond angle -180 degree
9. Apply octet rule to all atoms
10. Draw resonance structure
11. Calculate formal charge
12. Which lewis structure is more
table
Electron Dot Formula for NO21. Central atom N=A
2. Terminal atomes O=X, n=2
5+2*6+1 charge electrons = 18 e
4. bonding pairs n = (total valence electrons)/8 = 18 /8= 2 bonding
pairs + 2 remainder electrons
5. Lone pair m = (remainder electrons)/2 =2/2=1 lone pair
6. NO2- = AX2E1 → BENT
7. Hybridization orbital → sp2
8. Bond angle <120 degree
9. Apply octet rule to all atoms
10. Draw resonance structure
11. Calculate formal charge
12. Which lewis structure is more table
Electron Dot Formula for NO2+ , NO3- and O3
Example 10.1 VSEPR Theory and the Basic Shapes
Determine the molecular geometry of NO3−.
Solution
The molecular geometry of NO3− is determined by the number of electron groups around the central atom (N).
Begin by drawing a Lewis structure of NO3−.
NO3− has 5 + 3(6) + 1 = 24 valence electrons. The Lewis structure has three resonance structures:
The hybrid structure is intermediate between these three and has three equivalent bonds.
Use any one of the resonance structures to determine the number of electron groups around the central atom.
The nitrogen atom has three electron groups.
Based on the number of electron groups, determine the geometry that minimizes the repulsions between the groups.
Trigonal Planar
Nivaldo J. Tro
sp2
Example 9.9 Drawing Resonance Structures and Assigning Formal
Charge for Organic Compounds
Draw the Lewis structure (including resonance structures) for nitromethane (CH 3NO2). For each resonance structure,
assign formal charges to all atoms that have formal charge.
Solution
Begin by writing the skeletal structure. For organic compounds, the condensed structural formula (in this case
CH3NO2) indicates how the atoms are connected.
Calculate the total number of electrons for the Lewis structure by summing the number of valence electrons for
each atom.
Nivaldo J. Tro
Continued
Place a dash between each pair of atoms to indicate a bond. Each dash counts for two electrons.
(12 of 24 electrons used)
Distribute the remaining electrons, first to terminal atoms then to interior atoms.
(24 of 24 electrons used)
If there are not enough electrons to complete the octets on the interior atoms, form double bonds by moving lone
pair electrons from terminal atoms into the bonding region with interior atoms.
Nivaldo J. Tro
Continued
Draw any necessary resonance structures by moving only electron dots. (In this case, you can form a double bond
between the nitrogen atom and the other oxygen atom.)
Assign formal charges (FC) to each atom.
FC = # valence e– – (# nonbonding e– + ½ # bonding e–)
Carbon, hydrogen, and the doubly-bonded oxygen atoms have no formal charge. Nitrogen has a +1 formal charge
[5 – ½ (8)] and the singly bonded oxygen atom in each resonance structure has a –1 formal charge [6 – (6 + ½ (2))].
Nivaldo J. Tro
Molecular Orbital Theory For
Physics / Chem Major Students
Atomic Orbital: A wave function whose square gives
the probability of finding an electron within a given
region of space in an atom.
Molecular Orbital: A wave function whose square
gives the probability of finding an electron within a
given region of space in a molecule.
Chapter 7/111
The Octet Rule- Valence Bond Theory
The Diatomic Oxygen Molecule
O
1s
2s
2p
1s
2s
2p
O
1 s bond
1 p bond
O O
Each has six
valence
electrons
A Double bond is when 4
electrons are shared they
are represented by two lines
in bond diagrams
BUT NOT 2 s bonds
Molecular Orbital Theory: The
Hydrogen Molecule
s bonding orbital
s* antibonding orbital
Bond Order =
(# bonding e– – # antibonding e–)
2
Chapter 7/113
Hydrogen Molecule
2–0
Bond Order =
=1
2
Chapter 7/114
Hydrogen Molecule
Bond Order:
2–1
1
=
2
2
2–2
=0
2
Chapter 7/115
Molecular Orbital Theory:
Other Diatomic Molecules
Chapter 7/116
Example 10.10 Molecular Orbital Theory
Draw an MO energy diagram and determine the bond order for the N2− ion. Do you expect the bond to be stronger or
weaker than in the N2 molecule? Is N2− diamagnetic or paramagnetic?
The N2− ion has 11 valence electrons (5 for each nitrogen atom plus 1 for the negative charge). Assign the electrons
to the molecular orbitals beginning with the lowest energy orbitals and following Hund’s rule.
Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in
bonding orbitals and dividing the result by two.
Nivaldo J. Tro
Key learning for Ch9/Ch10 Exam
Nivaldo J. Tro

Lewis Structure

Transcription

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