Unit 11: Chemical Bonding - PowerPoint PPT Presentation

1 / 87
About This Presentation
Title:

Unit 11: Chemical Bonding

Description:

Unit 11: Chemical Bonding Chemical Bonding Overview Chemical Bond: attractive force is between the nucleus of one atom and the valence electrons of another atom. – PowerPoint PPT presentation

Number of Views:504
Avg rating:3.0/5.0
Slides: 88
Provided by: rach2152
Category:

less

Transcript and Presenter's Notes

Title: Unit 11: Chemical Bonding


1
Unit 11 Chemical Bonding
2
Chemical Bonding Overview
  • Chemical Bond
  • attractive force is between the nucleus of one
    atom and the valence electrons of another atom.

3
Review
  • How do you find valence electrons?
  • Hint there are two ways!
  • Examples
  • Mg ___
  • O ___
  • Ar ___
  • Si ___
  • Examples
  • Mg
  • O

2
6
8
4
1s2 2s2 2p6 3s2
1s2 2s2 2p4
4
Electron Dot Structures
  • Depicts element symbol w/ valence e- shown as
    dots.

Na
Mg
Al
Si
Cl
Ar
O
N
5
Purpose of Bonding
  • Chemical bonding is driven by the octet rule
    atoms will lose, gain, or share electrons in
    order to achieve the electron configuration of
    the closest noble gas.
  • Unfilled or partially filled valence orbitals are
    inherently unstable. (Unstable means possessing
    high potential energy.) Nature always strives for
    the lowest energy conditions

6
Types of Chemical Bonds
  • There are 3 types of chemical bonds
  • Ionic Covalent Metallic
  • Depends on the number of valence electrons of
    bonding atoms.
  • Metallic bonds form between metal atoms only.
  • Ionic bonds involve the transfer of electrons
    (lose/gain).
  • Covalent bonds involve the sharing of electrons.
  • Metallic bonds involve a mobile electron sea.

Flower Time
7
Movie
8
Ionic Bonds
  • Occurs when ions of opposite charge (,-) attract
    each other.
  • Metal ion Nonmetal ion
  • Simplest attraction
  • NaCl MgF2
  • Polyatomic ions
  • AlPO4 (NH4)2SO4

9
Formation of Ionic Bond
  • Cation- positive ion ()
  • Forms when a metal atom loses e- to become
    stable.
  • Anion- negative ion (-)
  • Forms when a nonmetal atom gains e- to become
    stable
  • An ionic bond is formed when e- are transferred
    between atoms and the resulting ions stick
    together.

10
Examples
  • Formation of NaCl
  • Na Cl ? Na Cl - NaCl
  • Formation of MgF2
  • Mg F F ? Mg2 F - F - MgF2
  • How would aluminum oxide form?

11
Electron Configuration of Ions
  • Cation example (metal)
  • Ca atom 1s2 2s2 2p6 3s2 3p6 4s2
  • Ca2 ion 1s2 2s2 2p6 3s2 3p6 Lost 2 electrons
    to obtain noble gas configuration (octet)

12
Electron Configuration of Ions
  • Anion example (nonmetal)
  • N atom 1s2 2s2 2p3
  • N3- ion 1s2 2s2 2p6
  • Gained 3 electrons to obtain noble gas
    configuration (octet)

13
Properties of Ionic Compounds
IONIC
e- transferred from metal to nonmetal
Bond Formation
Type of Structure
Crystal lattice
Physical State
Solid (hard and rigid)
Melting Point
high
Boiling Point
high
yes (solution or liquid)
Electrical Conductivity
Other Properties
brittle
14
Electrolyte
  • A substance that conducts electricity
  • Because of ionic bonds ionic (charged) nature,
    ionic compounds conduct electricity in the molten
    or aqueous forms.

15
Bonding Types
  • Dog Analogy

16
Bond Polarity
  • Nonpolar
  • Polar
  • Ionic

17
NON-POLAR COVALENT BONDS
18
Polar Covalent Bonds Unevenly matched, but
willing to share.
19
Ionic Bonds One Big Greedy Thief Dog!
20
(No Transcript)
21
Bond Polarity
  • Nonpolar Covalent Bond
  • e- are shared equally
  • symmetrical e- density
  • usually identical atoms

22
Bond Polarity
  • Polar Covalent Bond
  • e- are shared unequally
  • asymmetrical e- density
  • results in partial charges (dipole)

23
Bond Polarity
  • Most bonds are a blend of ionic and covalent
    characteristics.
  • Difference in electronegativity determines bond
    type.

If DEN is Bond type is
lt 0.4 Nonpolar covalent
0.4 lt DEN lt 1.7 Polar covalent
gt 1.7 Ionic
24
  • A large electronegativity difference leads to an
    ionic bond.
  • A small electronegativity difference leads to a
    polar covalent bond.
  • No electronegativity difference between two atoms
    leads to a nonpolar covalent bond.

25
  • Cesium and sulfur
  • Carbon and oxygen
  • Fluorine and fluorine

If ?EN is Bond type is
lt 0.4 Nonpolar covalent
0.4 lt ? EN lt 1.7 Polar covalent
gt 1.7 Ionic
Ionic Polar
Nonpolar
26
Covalent Bonds
  • Occurs when 2 nonmetals share pairs of electrons
    to become stable. Molecular compounds are
    formed.
  • Examples
  • H2O CO2 C6H12O6 PCl5

27
Covalent Bonds
  • Covalent bonds can be
  • single (1 shared pair)
  • double (2 shared pairs)
  • or triple (3 shared pairs)
  • Bond strength triple gt double gt single
  • Bond length single gt double gt triple

28
Lewis Structures
  • Creating Lewis Structures
  • Lewis structures are depictions of molecules that
    show valence electrons as __dots________.
  • Shared pairs of electrons (i.e. _Valence
    electrons_) are drawn between the atoms sharing
    them.
  • Unshared or ___lone_______ pairs of electrons are
    represented by dots located on one atom only

29
8
2
NEED
4
8
8
8
8
6
8
30
8
1
Available
2
4
5
6
7
3
Figure out through e config
31
Creating Lewis Structures
  • Follow this system
  • Example H2O
  • 1) Draw a skeleton of the molecule. It
    generally works to place the different atom in
    the center.
  • H O H

32
Creating Lewis Structures
  • Find the needed electrons (N) for each atom and
    add them up. N will be 8 for most elements, with
    these exceptions
  • N 12
  • H 2
  • O 8
  • H 2
  • H gets 2 valence e-
  • Be gets 4 valence e-
  • B gets 6 valence e-

33
  • 3) Find the available (valence) electrons (A)
    for each atom and then add them up.
  • A
  • special note when completing a Lewis structure
    for a polyatomic ion, you will need to correct A
    by adding the absolute value of the charge if
    negative, and subtracting the charge if positive.
    For example, for the ion PO43-, you would add 3
    to A. For the ion NH4, you would subtract 1
    from A. (You do the opposite of the charge.)

H 1 O 6 H 1 Total A 8
N 12 A 8
34
  • 4) Find the shared (S) electrons for the entire
    molecule by this formula S N A
  • S

S 12 8 4
N 12 A 8 S 4
35
  • 5) The shared electrons are the bonding
    electrons. Place all of the shared electrons
    between the atoms.
  • H O H
  • 6) You must place all of the available (A)
    electrons in the picture. The shared electrons
    are part of the available. See how many of the
    available electrons still need to be placed, and
    put them in the picture as lone pairs (unshared
    pairs) so that every atom gets an octet (remember
    H only needs 2).
  • H O H

N 12 A 8 S 4
N 12 A 8 S 4
4
36
(No Transcript)
37
(No Transcript)
38
(No Transcript)
39
  • CH4

H H C H H
N A S
8(2x4) 16
4(1x4) 8
0
8
40
  • CF4

F F C F F
N A S
8(4x8) 40
4(4x7) 32
24
8
41
Cl Cl P Cl
  • PCl3

N A S
8(3x8) 32
5(3x7) 26
20
6
42
  • CO2

O C O
N A S
8(2x8) 24
4(2x6) 16
8
8
43
Polyatomic Ions
  • To find total of valence e- (A)
  • Add 1e- for each negative charge.
  • Subtract 1e- for each positive charge.
  • Place brackets around the ion and label the
    charge.

44
Polyatomic Ions
  • ClO4-

N A S
8(4x8) 40
O O Cl O O
7(4x6) 31
1 32
24
8
45
Resonance
  • Resonance structures are structures that occur
    when it is possible to write 2 or more valid
    Lewis structures for the same molecule or ion.

46
  • Experimental data indicate that the 2 bonds in
    ozone are the same __length_____, BUT double
    bonds are shorter than single bonds! So the
    explanation for this is that the actual bonds are
    __combination_______ of those in the 2 resonance
    structures. The extra electron pair in ozone is
    delocalized over the two bonding regions. So
    each bond spends about half the time being single
    and half being double.

47
Expanded Octets
  • Some molecules do not follow the octet rule. The
    central atom has more than 8 electrons, called an
    expanded octet.
  • In these cases, the standard method for
    determining the Lewis structure will fail.

48
How to figure out a Lewis structure for an
expanded octet
  • 1. Calculate the available (A) number of
    electrons.
  • 2. Give the surrounding atoms an octet, and
    assume only single bonds to the central atom
  • 3. Place any remaining electrons as lone pairs
    on the central atom so that A electrons are
    included.
  • not always the case, but you should still get
    the right answer

49
Examples
F F P F F F
  • PF5

N A S
5(5x7) 40
50
VSEPR Theory
  • Valence Shell Electron Pair Repulsion Theory
  • Electron pairs orient themselves in order to
    minimize repulsive forces.

51
A. VSEPR Theory
  • Types of e- Pairs
  • Bonding pairs - form bonds
  • Lone pairs - nonbonding e-

52
A. VSEPR Theory
  • Lone pairs reduce the bond angle between atoms.

53
B. Determining Molecular Shape
  • Draw the Lewis Diagram.
  • Tally up e- pairs on central atom.
  • double/triple bonds ONE pair
  • Shape is determined by the of bonding pairs and
    lone pairs.

54
Common Molecular Shapes
2 bond 0 lone
LINEAR 180
55
Common Molecular Shapes
2 bond 1 lone
BENT lt120
56
Common Molecular Shapes
2 bond 2 lone
BENT 109.5
57
Common Molecular Shapes
2 bond 3 lone
Linear
58
Common Molecular Shapes
3 bond 0 lone
TRIGONAL PLANAR 120
59
Common Molecular Shapes
3 bond 1 lone
TRIGONAL PYRAMIDAL 107
60
Common Molecular Shapes
3 bond 2 lone
T-shaped
61
Common Molecular Shapes
4 bond 0 lone
TETRAHEDRAL 109.5
62
Common Molecular Shapes
4 bond 1 lone
See-Saw
63
Common Molecular Shapes
4 bond 2 lone
Square planar
64
Common Molecular Shapes
5 bond 0 lone
Trigonal bipyramidal
65
Common Molecular Shapes
5 bond 1 lone
Square pyramidal
66
Common Molecular Shapes
6 bond 0 lone
Octahedral
67
Examples
  • PF3

3 bond 1 lone
PYRAMIDAL 107
68
Examples
  • CO2

2 total 2 bond 0 lone
LINEAR 180
69
Examples
  • Use VSEPR Theory to predict the shape of
  • CO2 b) ClO3- c) NO3-
  • d) SCl2 e) PCl3

70
UNDERSTANDING MOLECULAR GEOMETRIES
PREDICTED BY VSEPR
H
H
C
H
H
FROM THE PERSPECTIVE OF
HYBRIDIZATION OF ATOMIC ORBITALS
71
Electron Both wave and particle
72
Electron Both wave and particle
73
s, p, d Orbitals
p Orbitals
Px
Px
Pz
Pz
Py
Py
74
Hybrid Orbitals
  • __Hybrid_______ orbitals are formed when several
    atomic orbitals blend to form the same total
    number of equivalent orbitals. This process is
    known as hybridization.
  • Hybridization explains the tetrahedral structure
    of methane. The s and px, py, and pz orbitals of
    the carbon atom blend to form four identical
    hybrid orbitals

75
(No Transcript)
76
Number of regions of high electron density Orbitals that blend to make the hybrids Hybridization
2 s p sp
3 s p p sp2
4 s p p p sp3
5 s p p p d sp3d
6 s p p p d d sp3d2
77
Now see if you can use what you have learned to
predict the hybridization of some other
compounds. Lets start with water.
Lewis Diagram
of electron clouds
four
sp3 hybridization
Hybridization
78
Now try carbon dioxide
Lewis Diagram
of electron clouds
Two
sp hybridization
Hybridization
Remember double or triple bonds count as 1
electron cloud in VSEPR theory.
79
Now try the sulfate ion (SO4-2)
Lewis Diagram
of electron clouds
Four
sp3 hybridization
Hybridization
80
Now try the carbonate ion (CO3-2)
Lewis Diagram
of electron clouds
Three
sp2 hybridization
Hybridization
Remember double or triple bonds count as 1
electron cloud in VSEPR theory.
81
Now try the sulfur hexaflouride (SF6)
Lewis Diagram
of electron clouds
Six
sp3d2 hybridization
Hybridization
82
Sigma and Pi Bonds
  • When two atomic orbitals combine along the axis
    connecting the 2 atomic nuclei (end-to-end
    overlap), a __sigma________ bond is formed. One
    sigma bond is formed in all bonds (single,
    double, triple).
  • When two atomic orbitals combine with a
    side-to-side overlap, a ___pi________ bond is
    formed. A pi bond occurs in double and triple
    bonds only.

83
Ethylene (Ethene) , C2H4
84
Summary
  • single bond
  • composed of one _sigma___ bond
  • double bond
  • composed of one __sigma___ and one ____pi___ bond
  • triple bond
  • composed of one _sigma_____ and two ___pi____
    bonds

85
Polarity of Bonds
  • The bonding pairs of electrons in covalent bonds
    are located between the ____elements_______ of
    the atoms sharing the electrons.
  • When the pull of each nucleus for the electrons
    is equally strong, the electrons are shared
    equally and are located on average halfway
    between the two nuclei. Recall that this type of
    bond is a ___non polar__________ covalent bond.
  • When the pull of one nucleus for the electrons is
    stronger than the other, the electrons spend more
    time closer to the more electronegative nucleus.
    Recall that this type of bond is a
    __polar________ covalent bond.

86
  • The more electronegative atom acquires a partial
    ___negative___________ charge. The less
    electronegative atom therefore acquires a partial
    __positive___________ charge.
  • These partial charges are indicated by the
    following symbols
  • d - for the more electronegative element.
  • d for the less electronegative element.

87
Ex H2O
d
d
d -
The water molecule contains two dipole moments,
sometimes just called dipoles. The dipole
moments can be drawn in with arrow notation
88
Polarity of Molecules
  • Note that just because a molecule contains polar
    bonds, it may or may not be classifed as a polar
    molecule.
  • The polarity of a molecule will depend on
  • (1) existence of polar bonds
  • (if none of the bonds are polar, the molecule is
    nonpolar)
  • (2) shape of the molecule
  • (3) orientation of the polar bonds

89
  • Note to help determine if the molecule will be
    polar, look for lines of symmetry.
  • If the molecule is symmetrical
  • then the dipoles will cancel and the molecule is
    nonpolar.
  • If the dipoles do not cancel
  • the molecule is polar.

90
Examples
  • HCl CO2
  • CF4 NH3
Write a Comment
User Comments (0)
About PowerShow.com