Transcript Chemical Bonding Chapters 9 & 10 1

Chemical Bonding
Chapters 9 & 10
1
Review

Draw Lewis dot structures for


SO2
BrF41+
ClO2
Use formal charge to determine whether N or O is
the more likely central atom in NOF
2
How equally do atoms share e– ?
Hydrogen and fluorine share one pair
of e– in a single covalent bond
 In the Lewis dot structure, they appear
to share the electrons equally, but do
they?

H F
3
How equally do atoms share e– ?
When you put HF in an
electric field, the molecules
line up, as if the F end were
negative and the H end
positive.
 The electrons are shared
unequally.

4
How equally do atoms share e– ?

When bond e– are shared unequally, the bond is said to be
a polar covalent bond.

A polar covalent bond has a dipole moment: one end is
more negative than the other end
5
Electronegativity

Electronegativity is the ability to attract bond e–

The higher the EN, the “greedier” the atom
6
Electronegativity

Electronegativity increases across a period and
decreases down a group
7
Electronegativity and bond polarity

Bond polarity depends on the difference in EN
8
Evaluating bond type

F–F
12
Evaluating bond type

F–F
F = 4.0, F = 4.0
 ∆EN = 0

13
Evaluating bond type

F–F
F = 4.0, F = 4.0
 ∆EN = 0
 Bond is nonpolar covalent (zero dipole moment)

F–F
14
Evaluating bond type

C=O
15
Evaluating bond type

C=O
C = 2.5, O = 3.5
 ∆EN = 1.0

16
Evaluating bond type

C=O
C = 2.5, O = 3.5
 ∆EN = 1.0
 Bond is polar covalent (has a dipole moment) with O
end more negative



C=O
–
Doesn’t matter whether bond is single or double
17
Evaluating bond type

KCl
18
Evaluating bond type

KCl
K = 0.8, Cl = 3.0
 ∆EN = 2.2

19
Evaluating bond type

KCl
K = 0.8, Cl = 3.0
 ∆EN = 2.2
 Bond is ionic; e– transferred from K to Cl

1+
K
1–
Cl
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Order, length, energy

Bond order = type of bond


Bond length = distance between bonded atoms


1 = single, 2 = double, 3 = triple
Higher bond order => shorter bond length
Bond energy = energy needed to break bond

Higher bond order => higher bond energy
21
Estimating ∆H from bond energy
Breaking bonds is endothermic (energy in)
 Making bonds is exothermic (energy out)
 ∆H is overall energy change

² H rxn  ² H bonds broken  ² H bonds


formed
∆H negative (exothermic) when weak bonds break & strong
bonds form
∆H positive (endothermic) when strong bonds break & weak
bonds form
22
CH4 + 2 O2 → CO2 + 2 H2O
4 C – H  414kJ  1656kJ
2 C  O  –799kJ  –1598kJ
2 O  O  498kJ  996kJ
Total  2652kJ
4 H – O  –464kJ  –1856kJ
Total  –3454kJ
² H  2652kJ   3454kJ   802kJ
Observed ∆H for this reaction is –890 kJ
(values from bond energy are approximate)
23
Shapes of Molecules
Lewis dot structures do not reveal the threedimensional shape of a molecule
 Molecular shape can be predicted from the dot
structure using the Valence Shell Electron Pair
Repulsion (VSEPR) model

24
VSEPR

The VSEPR model focuses on electron groups in
the valence shell
A bond (single, double, or triple) is one electron group
 A lone pair is one electron group

VSEPR proposes that electron groups will take
positions around the central atom that are as far
away from each other as possible, to minimize
repulsions
 This gives a set of 5 possible geometries

25
Terminology
A = central atom
 X = atom or group of atoms bonded to central atom
 E = lone pair of e– on central atom

H
O
C
AX2
O
H
C
H
H
H
O
H
AX2E2
AX4
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Terminology

Electron group geometry = shape made by the e–
groups (bonds + lone pairs)
2 electron groups = linear
 3 electron groups = trigonal planar
 4 electron groups = tetrahedral
 5 electron groups = trigonal bipyramidal
 6 electron groups = octahedral

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Sketching the 5 basic geometries
Linear
Trigonal planar
Tetrahedral
Trigonal bipyramidal
Octahedral
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Terminology
Molecular geometry = shape made by joining
bonded nuclei with straight lines
 Each e– group geometry forms one or more
molecular geometries


Example: four electron groups (tetrahedral) could be
AX4
 AX3E
 AX2E2


tetrahedral
trigonal pyramidal
angular or bent
Bond angle = angle between adjacent bonds
29
Applying VSEPR Theory




Draw a plausible Lewis structure.
Determine the number of bonds and lone pairs, and assign
a VSEPR notation (AXE) to the molecule.
Establish the e– group geometry.
Determine the molecular geometry.


If there is more than one central atom, analyze each atom
individually.
Sketch the electron group geometry, indicating atoms with
circles and lone pairs with dots.
30
Molecular
Geometry
as a
Function
of Electron
Group
Geometry
31
Molecular
Geometry
as a
Function
of Electron
Group
Geometry
32
Beyond Lewis theory

Lewis theory
Describes what happens (dot structures)
 Offers a simple idea of why (octet rule)


but

Is not correlated to modern atomic theory (orbitals)
34
Valence bond theory

Chemical bond
forms when
half-filled
orbitals overlap
at optimum
balance of
attraction &
repulsion
35
Valence bond theory

For most molecules, molecular geometry does not
match orientation of atomic orbitals
36
Hybridization of atomic orbitals

When bonds form, central atom valence orbitals
combine → new set of wave functions called the
hybrid set
Number of hybrid orbitals = number of atomic orbitals
combined
 Hybrid orbitals have different shape and orientation
than original orbitals
 Shapes of hybrid sets correspond to VSEPR geometries

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sp3 hybrid set
One 2s + three 2p orbitals → four sp3 orbitals
 The orbitals in the sp3 set all have the same energy
 Four valence e– enter the four sp3orbitals according
to Hund’s Rule.

38
39
sp3 hybrid set

C has 4 valence e–
Each valence e– is in an sp3 orbital
 Four half-filled sp3 overlap with
four half-filled 1s H orbitals


N has 5 valence e–
Each valence e– is in an sp3 orbital
 Three half-filled sp3 overlap with
three half-filled 1s H orbitals
 Fourth sp3 contains lone pair

40
sp2 hybrid set
2s + 2px + 2py orbitals → three sp2 orbitals in xy
plane
 2pz orbital is unhybridized, remains along z axis

41
42
Single & double bonds in VB

The hybrid orbitals overlap head-to-head with
orbitals on other atoms to form single bonds

Head-to-head overlap is called a sigma () overlap

Sigma overlaps provide the skeleton structure and
VSEPR shape
43
Single & double bonds in VB

Unhybridized orbital overlaps side-to-side with
orbital on another atom to form another bond

Side-to-side overlap is called a pi () overlap

Pi overlap is the second bond in a double bond
44
Bonding in H2CO
45
46
sp hybrid set
2s + 2px orbitals → two sp orbitals on x axis
 2py + 2pz orbitals are unhybridized, remain along
y and z axes

47
48
sp hybrid set

The sp set can form two sigma bonds (s) and two
pi bonds (p)
Single bond = triple bond
 Two double bonds

49
Bonding in C2H2
50
sp3d hybrid set
One 3s + three 3p + one 3d orbital
→ five sp3d orbitals
 Valence e– must be on n ≥ 3 to form this set
 No double or triple bonding

51
sp3d2 hybrid set
One 3s + three 3p + two 3d orbitals
→ six sp3d2 orbitals
 Valence e– must be on n ≥ 3 to form this set
 No double or triple bonding

52
One  + two  bonds
Two  + one  bond
All  bonds
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