chemistry - cooper union
TRANSCRIPT
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Chapter 7 CHEMICAL BONDING AND MOLECULAR GEOMETRYKevin Kolack, Ph.D.The Cooper Union
HW problems: 5, 8, 13, 17, 21, 25, 30, 35, 45, 47, 52, 58, 65, 70, 75, 81, 89, 93, 99, 105, 116
CHEMISTRY
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CH. 7 OUTLINE
7.3 Lewis Symbols
7.1 Ionic Bonding
7.2 Covalent Bonding
7.3 Lewis Structures
7.4 Formal Charges and Resonance
7.5 Strengths of Ionic and Covalent Bonds
7.6 Molecular Structure and Polarity
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1A 1ns1
2A 2ns2
3A 3ns2np1
4A 4ns2np2
5A 5ns2np3
6A 6ns2np4
7A 7ns2np5
Group # of valence e-e- configuration
VALENCE ELECTRONS
• The outer shell electrons of an atom. The valence electrons are the electrons that participate in chemical bonding.
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ns1
ns2
ns2 n
p1
ns2 n
p2
ns2 n
p3
ns2 n
p4
ns2 n
p5 ns2 n
p6
d1 d5 d10
4f
5f
GROUND STATE ELECTRON CONFIG.
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Na [Ne]3s1 Na+ [Ne]Ca [Ar]4s2 Ca2+ [Ar]Al [Ne]3s23p1 Al3+ [Ne]
Metals lose electrons so that the cation has a noble gas e- configuration.
H 1s1 H- 1s2 or [He]
F 1s22s22p5 F- 1s22s22p6 or [Ne]
O 1s22s22p4 O2- 1s22s22p6 or [Ne]
N 1s22s22p3 N3- 1s22s22p6 or [Ne]
Nonmetals gain electrons so that the anion has a noble gas e- configuration.
E- CONFIGS OF CATIONS AND ANIONS
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CHARGES OF CATIONS AND ANIONS+1 +2 +3 -1-2-3
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Na+: [Ne] Al3+: [Ne] F-: 1s22s22p6 or [Ne]
O2-: 1s22s22p6 or [Ne] N3-: 1s22s22p6 or [Ne]
Na+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne
WHY?
ISOELECTRONIC: having the same number of electronsBeing isoelectronic with a noble gas imparts added stability to a cation or anion
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LEWIS DOT SYMBOLS
• Lewis symbol (Lewis structure) - a way to represent atoms using the element symbol and valence electrons as dots
• As only valence electrons participate in bonding, this makes it much easier to work with the octet rule
• The number of dots used corresponds directly to the number of valence electrons located in the outermost shell of the atoms of the element
dots
dots
dots
dots symbol
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LEWIS SYMBOLS (CONT’D)• The four sides around the atomic symbol can each have two dots for a
maximum of eight (octet).
• Drawing Lewis symbols
• Place one dot on each side until there are four dots around the symbol• Now add a second dot to each side in turn• The number of valence electrons limits the number of dots placed• Each unpaired dot (unpaired electron of the valence shell) is available to
form a chemical bond• Ex: carbon and chlorine
C
Cl
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CH. 7 OUTLINE
7.3 Lewis Symbols
7.1 Ionic Bonding
7.2 Covalent Bonding
7.3 Lewis Structures
7.4 Formal Charges and Resonance
7.5 Strengths of Ionic and Covalent Bonds
7.6 Molecular Structure and Polarity
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IONIC BONDING• The electrostatic force that holds ions together in an ionic compound.
• Electrons are TRANSFERRED from element that becomes cation to element that becomes anion, maintaining ELECTRONEUTRALITY.
Li + F Li+ F -
1s22s1 1s22s22p5 1s2 1s22s22p6
[He] [Ne]
Li Li+ + e-
e- + F F -
FLi + Li+ F -
LiF
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IONIC BONDING
• Remember: representative (main group) elements form ions that obey the octet rule
• Ions of opposite charge attract each other resulting in an ionic bond
• Electrons are lost by a metal and they are gained by a nonmetal
• Each atom achieves a noble gas configuration
• 2 ions are formed; a cation and anion, which are attracted to each other
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IONIC BONDING EXAMPLE
Consider the formation of NaCl
Na + Cl NaCl
Sodium has a low ionization energy; itreadily loses an electron
Na Na+ + e−
When sodium loses the electron, it attains a [Ne] configuration
Chlorine has a high electron affinity
When chlorine gains an electron, it attains an [Ar] configuration
:
..
..Cl: e ....Cl:
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FEATURES OF IONIC BONDING• Metals tend to form cations because they have low
IE and low EA• Nonmetals tend to form anions because they have
high IE and high EA• Ions are formed by the transfer of electrons• The oppositely charged ions formed are held
together by an electrostatic force• Reactions between metals and nonmetals tend to
form ionic compounds
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ION ARRANGEMENT IN A CRYSTAL
• As a sodium atom loses one electron, it becomes a smaller sodium ion
• When a chlorine atom gains that electron, it becomes a larger chloride ion
• Attraction of the Na cation with the Cl anion forms NaCl ion pairs that aggregate into a crystal, an infinite array of alternating Na and Cl ions in 3 dimensions
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EXAMPLE
Use Lewis dot symbols to show the formation of aluminum oxide (Al2O3), aka the mineral corundum.
Corundum (Al2O3)
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CH. 7 OUTLINE
7.3 Lewis Symbols
7.1 Ionic Bonding
7.2 Covalent Bonding
7.3 Lewis Structures
7.4 Formal Charges and Resonance
7.5 Strengths of Ionic and Covalent Bonds
7.6 Molecular Structure and Polarity
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COVALENT BONDING – AN INTRODUCTION
• Let’s look at the formation of H2:
• H + H H2
• Each hydrogen has one electron in its valance shell
• If it were an ionic bond it would look like this:
• However, both hydrogen atoms have an equal tendency to gain or lose electrons
• Electron transfer from one H to another will not occur under normal conditions...
:H H H H
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COVALENT BONDING (CONT’D)
H:H H H Instead, each atom attains a noble gas configuration by sharingelectrons
The shared electron
pair is called a covalent bond
Each hydrogen atom now has two electrons around it and has attained a [He] configuration
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BOND LENGTH
Sharing (covalent bonding) is due to overlap of orbitalsChange in potential energy of two hydrogen atoms as a function of their distance of separation
Closer together; attraction increases.
At 74 pm, attractive forces are at a
maximum, energy is at a minimum.
Closer than 74 pm, repulsion
increases.
Hydrogen nuclei far apart; just a little
attraction.
~Figure 7.4
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Why should two atoms share electrons?
F F+
7e- 7e-
F F
8e- 8e-
F F
F F
Lewis structure of F2
lone pairslone pairs
lone pairslone pairs
single covalent bond
single covalent bond
COVALENT BONDING DEFINED
• A chemical bond in which two or more electrons are shared by two atoms.
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FEATURES OF COVALENT BONDS
• Covalent bonds form between atoms with similar tendencies to gain or lose electrons
• ie, near each other on the periodic table• Compounds containing covalent bonds are called “covalent
compounds” or “molecules”
• The diatomic elements have completely covalent bonds (totally equal sharing)
• H2, N2, O2, F2, Cl2, Br2, I2
:....F:
..
..F: :....F
..
..F: Each fluorine is surrounded by 8 electrons…a [Ne] configuration
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Bond Lengthstriple bond < double bond < single bond
BOND LENGTH
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IONIC VS. COVALENT PROPERTIES
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POLAR COVALENT BONDING
• A polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms
e- riche- poor
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ELECTRONEGATIVITY
• The ability of an atom to attract the electrons in a chemical bond. Attraction for shared electrons.
• Similar to electron affinity, but NOT the same
Electronegativity - relative (F is highest)
Electron affinity - measurable (Cl is highest)X(g) + e- X-(g)
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PAULING ELECTRONEGATIVITY VALUES
• Increases up to the right
• (F is the most electronegative element on the periodic table)
Linus Pauling (1901–1994) made many important contributions to the field of chemistry. He was also a prominent activist, publicizing issues related to health and nuclear weapons.
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PERIODIC VARIATION IN EN VALUES
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Difference* Bond Type
< “0.4” Covalent
“1.8” Ionic“0.4” < and < “1.8” Polar Covalent
ELECTRONEGATIVITY DIFFERENCES
*(These values vary from book to book.)
share e-
partial transfer of e-
transfer e-
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CH. 7 OUTLINE
7.3 Lewis Symbols
7.1 Ionic Bonding
7.2 Covalent Bonding
7.3 Lewis Structures
7.4 Formal Charges and Resonance
7.5 Strengths of Ionic and Covalent Bonds
7.6 Molecular Structure and Polarity
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WRITING LEWIS STRUCTURES
• (I do not like your book method- surprise!)
1. Go for symmetry
• (usually, least electronegative element at the center)2. Treat every element like a 4-sided box and add dots
3. Connect the dots
• Go for octetsa) (there MUST be an octet for C, N, O, F)b) (3p and larger elements can have more than 8)c) (B and other small elements may have fewer than 8)d) (H and He get a “duet”)
4. Clean up
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EXAMPLE
• Write the Lewis structure for nitrogen trifluoride (NF3, a colorless, odorless, unreactive gas).
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8e-
H HO+ + OH H O HHor
2e- 2e-
Lewis structure of water
Double bond – two atoms share two pairs of electrons
single covalent bonds
O C O or O C O
8e- 8e-8e- double bonds
Triple bond – two atoms share three pairs of electrons
N N8e-8e-
N N
triple bond
or
SINGLE, DOUBLE, AND TRIPLE BONDS
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EXAMPLE
• Write the Lewis structure for nitric acid (HNO3, a strong acid/strong electrolyte) in which the three O atoms are bonded to the central N atom and the ionizable H atom is bonded to one of the O atoms.
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EXCEPTIONS TO THE OCTET RULE
• The incomplete octet
• While H and He get a “duet,” other elements have neither 2 nor 8 in their structures (generally unstable)
H HBe
BeH2
BF3
F B F
F
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EXCEPTIONS (CONT’D)
• Odd electron structures (also highly reactive)
• Expanded octet
• (central atom with principal quantum number n > 2 ONLY)
NO N O
SF6 S
F
F
F
FF
F
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O3 VS SO2
The internet is wrong…
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EXAMPLE
• Draw the Lewis structure for aluminum triiodide (AlI3).
AlI3 has a tendency to dimerize (form two units)
forming Al2I6 (similar to many boron compounds).
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EXAMPLE
• Draw the Lewis structure for phosphorus pentafluoride (PF5).
PF5 is a reactive gaseous compound.
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EXAMPLE
• Draw a Lewis structure of the noble gas compound xenon tetrafluoride (XeF4).
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CH. 7 OUTLINE
7.3 Lewis Symbols
7.1 Ionic Bonding
7.2 Covalent Bonding
7.3 Lewis Structures
7.4 Formal Charges and Resonance
7.5 Strengths of Ionic and Covalent Bonds
7.6 Molecular Structure and Polarity
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formal charge on an atom in a Lewis structure
=12
total number of bonding electrons( )
total number of valence electrons in the free atom
-total number of nonbonding electrons
-
The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.
FORMAL CHARGE
• An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons “assigned” to that atom in a Lewis structure.
• Overly complicated method:
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H C O H
formal charge on C = 4 -2 -½ x 6 = -1
formal charge on O = 6 -2 -½ x 6 = +1
-1 +1
Two possible skeletal structures of formaldehyde (CH2O)
FORMAL CHARGE IN FORMALDEHYDE
HC O
H
0 0
formal charge on C = 4 -0 -½ x 8 = 0
formal charge on O = 6 -4 -½ x 4 = 0
The best Lewis structure minimizes formal charges.
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FORMAL CHARGE AND LEWIS STRUCTURES
• For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.
• Lewis structures with small formal charges are preferable to those with large formal charges.
• Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.
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EXAMPLE
Write the Lewis structure for the carbonate ion (CO32-).
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O O O+ -
OOO+-
RESONANCE STRUCTURES
• A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.
• Ozone (O3):
• Benzene (C6H6):
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EXAMPLE
Write formal charges for one of the resonance forms of the carbonate ion.
Check Note that the sum of the formal charges is −2, the same as the charge on the carbonate ion.
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EXAMPLE
• Draw three resonance structures for the molecule nitrous oxide, N2O (the atomic arrangement is NNO).
• Indicate formal charges.
• Rank the structures in their relative importance to the overall properties of the molecule.
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SOLUTION
Solution The three resonance structures are
Structure (b) is the most important one because the negative charge is on the more electronegative oxygen atom.
Structure (c) is the least important one because it has a larger separation of formal charges. Also, the positive charge is on the more electronegative oxygen atom.
How about ?NOO N
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EXAMPLE
• Draw a Lewis structure for the sulfate ion (SO42-).
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SOLUTIONThe question of which of these two structures is more important, that is, the one in which the S atom obeys the octet rule but bears more formal charges or the one in which the S atom expands its octet, has been the subject of some debate among chemists. In many cases, only elaborate quantum mechanical calculations can provide a clearer answer.
At this stage of learning, you should realize that both representations are valid Lewis structures and you should be able to draw both types of structures. One helpful rule is that in trying to minimize formal charges by expanding the central atom’s octet, only add enough double bonds to make the formal charge on the central atom zero.
16 electrons is…unusual.
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CH. 7 OUTLINE
7.3 Lewis Symbols
7.1 Ionic Bonding
7.2 Covalent Bonding
7.3 Lewis Structures
7.4 Formal Charges and Resonance
7.5 Strengths of Ionic and Covalent Bonds
7.6 Molecular Structure and Polarity
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H2 (g) H (g) + H (g) H0 = 436.4 kJ
Cl2 (g) Cl (g)+ Cl (g) H0 = 242.7 kJ
HCl (g) H (g) + Cl (g) H0 = 431.9 kJ
O2 (g) O (g) + O (g) H0 = 498.7 kJ O O
N2 (g) N (g) + N (g) H0 = 941.4 kJ N N
Bond Enthalpy
Bond Enthalpies
Single bond < Double bond < Triple bond
BOND ENERGY
The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond enthalpy. (I know we skipped Chapter 6...)
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H2O (g) H (g) + OH (g) H0 = 502 kJ
OH (g) H (g) + O (g) H0 = 427 kJ
Average OH bond enthalpy = 502 + 427
2= 464 kJ
BOND ENERGIES
• Tabulated bond energy values are average bond enthalpies in polyatomic molecules
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H0 = total energy input – total energy released= BE(reactants) – BE(products)
endothermic exothermic
HEAT OF REACTION• Imagine a reaction proceeding by breaking all bonds
(takes E) in the reactants and then using the gaseous atoms to form all the bonds (releases E) in the products.
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EXAMPLE
Calculate the enthalpy of reaction for the process
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SOLUTIONStart by counting the number of bonds broken and the number of bonds formed and the corresponding energy changes:
Next, obtain the total energy input and total energy released:
And finally, find the heat of reaction:
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EXAMPLE
Estimate the enthalpy change for the combustion of hydrogen gas:
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SOLUTION
Obtain the total energy input and total energy released:
Thus,
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ELECTROSTATIC (LATTICE) ENERGY
• Lattice energy (U) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.
Lattice energy increases as Q increases and/or
as r decreases.
Compound Lattice Energy (kJ/mol)MgF2
MgO
LiFLiCl
29573938
1036853
Q: +2,-1Q: +2,-2
r F- < r Cl-
E = kQ+Q-r
Q+ is the charge on the cationQ- is the charge on the anionr is the distance between the ions
E is the potential energy
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The Born-Haber cycle shows the relative energies of each step involved in the formation of an ionic solid from the necessary elements in their reference states.
FIGURE 7.13
Hoverall = H1 + H2 + H3 + H4 + H5o ooooo
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Al2O3 14,910 2072
LATTICE ENERGIES
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CH. 7 OUTLINE
7.3 Lewis Symbols
7.1 Ionic Bonding
7.2 Covalent Bonding
7.3 Lewis Structures
7.4 Formal Charges and Resonance
7.5 Strengths of Ionic and Covalent Bonds
7.6 Molecular Structure and Polarity
63
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Bond distances (lengths) and angles are shown for the formaldehyde molecule, H2CO.
MOVING FROM 2D LEWIS TO 3D SHAPE
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VSEPR THEORY
Valence Shell Electron Pair Repulsion theory allows us to predict the geometry of a molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs
AB2 2 0
Class
# of atomsbonded to
central atom
# lonepairs on
central atomGeometry of
electron groupsShape of the
molecule
linear linear
B B
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Cl ClBe
2 atoms bonded to central atom
0 lone pairs on central atom
EXAMPLEBeryllium chloride
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3 BONDED ATOMS
AB3 3 0 trigonal planar
trigonal planar
AB2 2 0
Class
# of atomsbonded to
central atom
# lonepairs on
central atom
linear linear
Geometry ofelectron groups
Shape of themolecule
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Boron trifluoride
EXAMPLE
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4 BONDED ATOMS
AB4 4 0 tetrahedral tetrahedral
AB3 3 0trigonal planar
trigonal planar
AB2 2 0
Class
# of atomsbonded to
central atom
# lonepairs on
central atom
linear linear
Geometry ofelectron groups
Shape of themolecule
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Methane
EXAMPLE
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AB5 5 0 trigonalbipyramidal
trigonalbipyramidal
5 BONDED ATOMS
AB4 4 0 tetrahedral tetrahedral
AB3 3 0trigonal planar
trigonal planar
AB2 2 0
Class
# of atomsbonded to
central atom
# lonepairs on
central atom
linear linear
Geometry ofelectron groups
Shape of themolecule
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Phosphorus pentachlorideEXAMPLE
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6 BONDED ATOMS
AB6 6 0 octahedraloctahedral
AB5 5 0trigonal
bipyramidaltrigonal
bipyramidal
AB4 4 0 tetrahedral tetrahedral
AB3 3 0trigonal planar
trigonal planar
AB2 2 0
Class
# of atomsbonded to
central atom
# lonepairs on
central atom
linear linear
Geometry ofelectron groups
Shape of themolecule
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Sulfur hexafluoride
EXAMPLE
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SUMMARY – VSEPR - NO LONE PAIRS
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AB2E 2 1 trigonal planar
bent
AB3 3 0trigonal planar
trigonal planar
Class
# of atomsbonded to
central atom
# lonepairs on
central atom
2 BONDED PAIRS; 1 LONE PAIR
Geometry ofelectron groups
Shape of themolecule
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AB3E 3 1
AB4 4 0 tetrahedral tetrahedral
tetrahedral trigonal pyramidal
AB2E2 2 2 tetrahedral bent
4 ELECTRON GROUPS
Class
# of atomsbonded to
central atom
# lonepairs on
central atomGeometry of
electron groupsShape of the
molecule
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lone-pair vs. lone-pair repulsion
bonded-pair vs. bonded-pairrepulsion
lone-pair vs. bonded-pair repulsion< <
MODIFIED BOND ANGLES
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AB5 5 0trigonal
bipyramidaltrigonal
bipyramidal
AB4E 4 1 trigonalbipyramid
al
(sawhorse)seesaw
5 ELECTRON GROUPS
Class
# of atomsbonded to
central atom
# lonepairs on
central atomGeometry of
electron groupsShape of the
molecule
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AB3E2 3 2 trigonalbipyramidal
T-shaped
5 ELECTRON GROUPS (CONT’D)
Class
# of atomsbonded to
central atom
# lonepairs on
central atom
AB2E3 2 3 trigonalbipyramidal
linear
Geometry ofelectron groups
Shape of themolecule
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AB6 6 0 octahedraloctahedral
AB5E 5 1 octahedral square pyramidal
6 ELECTRON GROUPS
Class
# of atomsbonded to
central atom
# lonepairs on
central atom
AB4E2 4 2 octahedral square planar
Geometry ofelectron groups
Shape of themolecule
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VSEPR – SUMMARY WITH LP (7.19)
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PREDICTING MOLECULAR SHAPE
1. Draw the Lewis structure for molecule.
2. Count the number of lone pairs on the central atom and the number of atoms bonded to the central atom.
3. Use VSEPR to predict the geometry around the central atom (total number of “things”) and the shape of the molecule (total number of atoms).
• For multi-center molecules, you’ll be describing the geometry and shape around a specific atom.
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STRUCTURE DETERMINES FUNCTIONOxycodone Codeine
Need to take ~35mg to have the same effect as 5mg of oxycodone
H3C
CH3
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STRUCTURE DETERMINES FUNCTION
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POLAR BONDS REVISITED
H F
electron richregionelectron poor
region
= Q * rQ is the charger is the distance between charges1 D = 3.36 x 10-30 C m
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(a) There is a small difference in electronegativity between C and H, represented as a short vector.
(b) The electronegativity difference between B and F is much larger, so the vector representing the bond moment is much longer.
FIGURE 7.26
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A molecule can have polar bonds (if the bonded atoms have different electronegativities), but it may not possess a dipole moment if it has a highly symmetrical geometry.
POLAR MOLECULES
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field off field on
BEHAVIOR OF POLAR MOLECULES
(a) Molecules are always randomly distributed in the liquid state in the absence of an electric field.
(b) When an electric field is applied, polar molecules like HF will align to the dipoles with the field direction.
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The overall dipole moment of a molecule depends on the individual bond dipole moments and how they are arranged.
(a) Each CO bond has a bond dipole moment, but they point in opposite directions so that the net CO2 molecule is nonpolar.
(b) In contrast, water is polar because the OH bond moments do not cancel out.
FIGURE 7.27
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MORE EXAMPLES
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NH3 AND NF3
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MICROWAVE OVENS
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QUESTION
Predict whether each of the following molecules has a dipole moment:
(a)BrCl
(b)BF3 (trigonal planar)
(c)CH2Cl2 (tetrahedral)
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HW problems:
5, 8, 13, 17, 21, 25, 30, 35, 45, 47, 52, 58, 65, 70, 75, 81, 89, 93, 99, 105, 116
This file is copyright 2017, Rice University, and adapted by Kevin Kolack, Ph.D.
All Rights Reserved.