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Transcript of CH 3- CHEMICAL BONDING Jun 2014 Pt1.pdf
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Chemical Bonding
Chapter 3
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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Chemical Bonding
3.0 Introduction
3.1 Lewis Structure
3.2 Ionic Bond
3.3 Covalent Bond
3.4 Molecular Geometry and VSEPR theory
3.5 Valence bonds theory and hybridizations
3.6 Intermolecular Forces
3.7 Metallic Bonding
2
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3.0 Introduction to Chemical Bonds
Ionic Bonds
When a metal atom loses electrons it becomes
a cation
metals relatively easy to remove electrons
from them When a nonmetal atom gains electrons it
becomes an anion
Nonmetals are easier to add electrons to
these atoms
The oppositely charged ions are then attracted
to each other, resulting in an ionic bond
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5
Li + F Li+ F -
The Ionic Bond
1s22s1 1s22s22p5 1s2 1s22s22p6
[He] [Ne]
Li Li+ + e-
e- + F F -
F -Li+ + Li+ F -
LiF
Ion ic bond:the electrostatic force that holds ions together in
an ionic compound.
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3.0 Introduction to Chemical BondsCovalent Bonds
6
Covalent bond:chemical bond in which two
or more electrons are shared by two atoms.
Covalent compounds are compounds that
contain only covalent bonds.
In a covalent bond, each electron in a shared
pair is attracted to the nuclei of both atoms.
This attraction holds the two atoms together.
H + H H:H or H-H
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8
8e-
H HO+ + OH H O HHor
2e- 2e-
Lewis structure of water
Double bondtwo atoms share two pairs of electrons
single covalent bonds
O C O or O C O
8e- 8e-8e- double bonds
Triple bondtwo atoms share three pairs of electrons
N N
8e-8e-
N N
triple bond
or
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3.0 Introduction to Chemical Bonds
Three Models of Chemical Bonding.
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Topic
3.1 Lewis Structure
G.N. Lewis (1875-1946)
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3.1 Lewis Structure
Lewis Theory- using valence electrons to explain
bonding of atoms
Using Lewis Theory, we can draw models called
Lewis Structures or known as Electron-DotStructures.
Lewis structure represent the valence electrons of
main-group elements as dots surrounding the
symbol for the particular element.
13
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Valence Electrons
Because chemical bonding involves the transferor sharing of electrons between two or more
atoms, valence electrons are most important in
bonding
Lewis theory focuses on the behavior of thevalence electrons
3.1 Lewis Structure
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Determining the Number of Valence
Electrons in an Atom The column number on the Periodic Table will tell you
how many valence electrons a main group atom has Transition Elements all have two valence electrons. Why?
3.1 Lewis Structure
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Lewis Electron-Dot Symbols
Nitrogen, N, is in Group 5A, therefore has 5 valence electrons.
To draw the Lewis symbol for any main-groupelement:
Note the A-group number, which gives the
number of valence electrons.
Place one dot at a time on each of the four sidesof the element symbol.
Keep adding dots, pairing them, until all are
used up.
N
N
N
N
or or or
3.1 Lewis Structure
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Lewis Electron-Dot Symbols for Elements
in Periods 2 and 3.
3.1 Lewis Structure
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19
Lewis Dot Symbols for the Representative Elements &
Noble Gases
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20
1. Draw skeletal structure of compound showing
what atoms are bonded to each other. Put leastelectronegative element in the center.
2. Count total number of valence e-. Add 1 for each
negative charge. Subtract 1 for each positivecharge.
3. Complete an octet for all atoms excepthydrogen.
4. If structure contains too many electrons, formdouble and triple bonds on central atom as
needed.
Writing Lewis Structures
E l 1
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Example 1
Write the Lewis structure for nitrogen trifluoride (NF3) in which
all three F atoms are bonded to the N atom.
NF3is a colorless, odorless, unreactive
gas.
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E l 1
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Example 1
Step 3: We draw a single covalent bond between N and each
F, and complete the octets for the F atoms. We place
the remaining two electrons on N:
Because this structure satisfies the octet rule for all the atoms,
step 4 is not required.
CheckCount the valence electrons in NF3(in bonds and in
lone pairs). The result is 26, the same as the total number of
valence electrons on three F atoms (3 7 = 21) and one N
atom (5).
E l 2
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Example 2
Write the Lewis structure for nitric acid (HNO3) 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.
HNO3is a strong electrolyte.
E l 2
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Example 2
Solut ionWe follow the procedure already outlined for writing
Lewis structures.
Step 1: The skeletal structure of HNO3is
Step 2: The outer-shell electron configurations of N, O, and H
are 2s2
2p3
, 2s2
2p4
, and 1s1
, respectively. Thus, thereare 5 + (3 6) + 1, or 24, valence electrons to
account for in HNO3.
E l 2
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Example 2
Step 3: We draw a single covalent bond between N and each
of the three O atoms and between one O atom and
the H atom. Then we fill in electrons to comply withthe octet rule for the O atoms:
Step 4: We see that this structure satisfies the octet rule for
all the O atoms but not for the N atom. The N atom
has only six electrons. Therefore, we move a lone
pair from one of the end O atoms to form another
bond with N.
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Topic
3.2 Ionic Bond
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3.2 Ionic Bond
Lewis Structures of Ions
Cations have Lewis symbols withoutvalence
electrons
electrons lost in the cation formation
Anions have Lewis symbols witheight valence
electrons
electrons gained in the anion formation
Example:
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The Octet Rule for Ionic Bonding
Metals form cations by losing enough electrons
until they get the same electron configuration as
the previous noble gas
Nonmetals form anions by gaining enough
electrons until they get the same electron
configuration as the next noble gas
The noble gas electron configurations are very
stable because they have 8 electrons (an octet) in
their outermost shells (except He)
3.2 Ionic Bond
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Stable Electron Arrangements
and Ion Charge
3.2 Ionic Bond
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Lewis Theory and Ionic Bonding
Lewis symbols can be used to represent the
transfer of electrons from metalatom to nonmetal
atom, resulting in ions that are attracted to each
other and therefore bond
+
3.2 Ionic Bond
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Orbital diagram
Another Way to Depict Electron Transfer in
the Formation of Li+and F-.
Electron configuration
Li 1s2
2s1
+ F 1s2
2p5
Li+
1s2
+ F-
1s2
2s2
2p6
Li
1s 2p
2s
1s 2p
2s
F
+
1s 2p2s
Li+
1s 2p
2s
F-
3.2 Ionic Bond
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Predicting Ionic Formulas
Using Lewis Symbols Electrons are transferred until the metal loses all
its valence electrons and the nonmetal has an
octet
Numbers of atoms are adjusted so the total
number of electrons lost balance up with number
of electrons gained
Lewis structure Formula
Li2O
3.2 Ionic Bond
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CHECKPOINT 1:
Depicting Ion Formation
SOLUTION:
Use partial orbital diagrams and Lewis symbols to
depict the formation of Na+and O2ions from the
atoms, and determine the formula of the compoundformed.
2Na+ +
2-O
Na
O
Na
3.2 Ionic Bond
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CHECKPOINT 2
Using Lewis theory to predict chemical formulas of ionic
compounds
Predict the formula of the compound that forms between
calcium and chlorine.
Draw the Lewis dot symbolsof the elements.
Ca
Cl
Transfer all the valence electrons
from the metalto the nonmetal,
adding more of each atom as you go,until all electrons are lost from the
metal atoms and all
nonmetal atoms have eight electrons.
CaCl
Cl
Ca2+
CaCl2
3.2 Ionic Bond
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Periodic Trends in Lattice Energy
Latt ice energyis the energy required to separate 1 mol of an
ionic solid into gaseous ions.
Lattice energy is a measure of the strength of the ionic bond.
3.2 Ionic Bond
Lattice energy is affected by ionic sizeand ionic charge.
As ionic size increases, lattice energy decreases. Lattice
energy therefore decreases down a group on the periodictable.
As ionic charge increases, lattice energy increases.
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Lattice Energy vs. Ion Size
3.2 Ionic Bond
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Lattice Energy vs Ion Charge
Lattice Energy =
910 kJ/mol
Lattice Energy =
3414 kJ/mol
3.2 Ionic Bond
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3.2 Ionic Bond
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Example 3: Order the following ionic compounds in
order of increasing magnitude of lattice energy:
CaO, KBr, KCl, SrO
First examine the ion charges
and order by sum of the charges
Ca2+& O2-,
K+& Br,
K+& Cl,
Sr2+& O2
(KBr, KCl) < (CaO, SrO)
Then examine the ion sizes of
each group and order by
radius; larger < smaller
(KBr, KCl) same cation,
Br> Cl(same Group)
KBr < KCl < (CaO, SrO)
(CaO, SrO) same anion,
Sr2+> Ca2+(same Group)
KBr < KCl < SrO < CaO
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1. High melting and boiling point Lewis theory predicts ionic compounds should have high
melting points and boiling points because breaking down
the crystal should require a lot of energy the stronger the attraction (larger the lattice energy), the higher the
melting point
41
Properties of Ionic Compounds3.2 Ionic Bond
NaCl (s) NaCl (liquid)
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CHECKPOINT 3
Which ionic compound below has the highest
melting point? KBr (734 C)
CaCl2(772 C)
MgF2(1261 C)
KBr
CaCl2
MgF2
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Properties of Ionic Compounds
2. Hard and brittle crystalline solid Lewis theory implies that the positions of the ions
in the crystal lattice are critical to the stability ofthe structure
Lewis theory predicts that moving ions out ofposition should therefore be difficult, and ionicsolids should be hard
hardness is measured by rubbing two materials
together and seeing which streaks or cuts the other the harder material is the one that cuts or doesnt
3.2 Ionic Bond
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Properties of Ionic Compounds
Lewis theory implies that if the ions are displacedfrom their position in the crystal lattice, that repulsiveforces should occur
This predicts the crystal will become unstable and
break apart. Lewis theory predicts ionic solids will bebrittle.
Ionic solids are brittle. When struck they shatter.
+ - + + + ++ + + +- -
--
--
--
+ - + + + ++ + + +
- -
-
-
-
-
-
-
+ - + + + ++ + + +- - - - - - - -
3.2 Ionic Bond
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Properties of Ionic Compounds
3. Good electrical conductor in liquid state
Lewis theory implies that, in the ionic solid, the ions
are locked in position and cannot move around
does not conduct electricity Lewis theory implies that, in the liquid state or when
dissolved in water,the ions will have the ability to
move around
conduct electricity
4. Many ionic solids are soluble in waterbut not inorganic solvents
3.2 Ionic Bond
Example 4
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Example 4
StrategyWe use electroneutrality as our guide in writing
formulas for ionic compounds, that is, the total positive charges
on the cations must be equal to the total negative chargeson the anions.
Solut ionThe Lewis dot symbols of Al and O are
Because aluminum tends to form the cation (Al3+) and oxygen
the anion (O2) in ionic compounds, the transfer of electrons is
from Al to O. There are three valence electrons in each Al
atom; each O atom needs two electrons to form the O2ion,
which is isoelectronic with neon.
Example 4
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Example 4
Thus, the simplest neutralizing ratio of Al3+to O2is 2:3; two
Al3+ions have a total charge of +6, and three O2ions have a
total charge of 6. So the empirical formula of aluminum oxideis Al2O3, and the reaction is
CheckMake sure that the number of valence electrons (24) is
the same on both sides of the equation. Are the subscripts in
Al2O3reduced to the smallest possible whole numbers?
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3.3 Covalent Bond
Lewis theory implies that another way atoms can
achieve an octet of valence electrons is to share
their valence electrons with other atoms
The shared electrons would then count toward
each atoms octet (Octet rule applied)
The sharing of valence electrons is called
covalent bonding
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....OSO.. .... .... .. ..
Bonding Pairs and Lone Pairs
Bonding pairs Lone pairs
Electrons that are shared by atoms are calledbonding pairs
Electrons that are not shared by atoms butbelong to a particular atom are called lone pairs
ornonbonding pairs
3.3 Covalent Bond
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Bonding Pairs and Lone Pairs
3.3 Covalent Bond
Generally, elements follow certain common
bonding patterns
The # of bonds = Group # or 8 Group #
C = 4 bonds & 0 lone pairs, N = 3 bonds & 1 lone pair,
O= 2 bonds & 2 lone pairs, H and halogen = 1 bond,
Be = 2 bonds & 0 lone pairs,
B = 3 bonds & 0 lone pairs
B C N O F
3 3 C l B d
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Single Covalent Bonds
F
F
F
F
HH O
HH O
F F
When two atoms share one pair of electrons it iscalled a single covalent bond One atom may use more than one single bond to
fulfill its octet
H only duet
3.3 Covalent Bond
3 3 C l t B d
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Double Covalent Bond
When two atoms share two pairs of electrons the
result is called a double covalent bond
four electrons
OO
O
O
3.3 Covalent Bond
3 3 C l t B d
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Triple Covalent Bond
When two atoms share three pairs of electrons theresult is called a triple covalent bond
six electrons
N
N
N N
3.3 Covalent Bond
3 3 C l t B d
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Lewis Structures
3.3 Covalent Bond
Lewis theory predicts that atoms will be most stablewhen they have their octet of valence electrons.
It does not require that atoms have the same number
of lone pair electrons they had before bonding.
First use the octet rule
3 3 C l t B d
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The Steps in Converting a Molecular
Formula into a Lewis Structure.Molecular
Formula
Atom
placement
Place atom with lowest
EN in center.
Step 1
Add A-group numbers.Step 2
Sum of
valence e-
Draw single bonds, and
subtract 2e-for each bond.
Step 3
Remaining
valence e-
Lewis
structure
Step 4 Give each atom8e-
(2e-for H).
3.3 Covalent Bond
3 3 C l t B d
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Writing Lewis Structures for NF3
Atom
placement
Sum of
valence e-
1 x N = 1 x 5 = 5e-
3 x F = 3 x 7 = 21 e-
Total = 28 e-
Molecular
FormulaN has a lower EN than F, so N is placed in the center.
Lewis
structure
Remaining
valence e-
3.3 Covalent Bond
3 3 C l t B d
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Writing Lewis Structures for Molecules
with One Central AtomExample 4 : Write a Lewis structure for CCl2F2
Step 1: Carbon has the lowest EN and is the
central atom. The other atoms areplaced around it.
Step 2: [1 x C(4e-)] + [2 x F(7e-)] + [2 x
Cl(7e-)] = 32 valence e-
Step 3-4: Add single bonds, then give
each atom a full octet.
3.3 Covalent Bond
3 3 C l t B d
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Writing Lewis Structures for Molecules with
More than One Central AtomExample 5 :Write the Lewis structure for
methanol (molecular formula CH4O)
Step 1: Place the atoms relative to each
other. H can only form one bond, so Cand O must be central and adjacent
to each other.
Step 2: [1 x C(4e-)] + [1 x O(6e-)] + [4 x
H(1e-)] = 14 valence e-
Step 3-4: Add single bonds, then give
each atom (other than H) a full
octet.
3.3 Covalent Bond
3 3 C l t B d
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Multiple Bonds
If there are not enough electrons for the centralatom
to attain an octet, a multiple bond is present.
Step 5: If the central atom does not have a full octet,change a lone pair on a surrounding atom into another
bonding pair to the central atom, thus forming a
multiple bond.
3.3 Covalent Bond
3 3 C l t B d
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Writing Lewis Structures for Molecules with
Multiple BondsExample 6: Write Lewis structures for the
following: Ethylene (C2H4)PLAN: After following steps 1 to 4 we see that the central
atom does not have a full octet. We must thereforeadd step 5, which involves changing a lone pair to a
bonding pair.
SOLUTION:
C2H4has 2(4) + 4(1) = 12 valence e-
. H can have only one bond peratom.
3.3 Covalent Bond
3 3 Co alent Bond
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Step 1: Place the atoms relative to each
other. H can only form one bond, so C
must be central and N adjacent to
carbon
Step 2: [1 x C(4e-)] + [1 x N(5e-)] + [1 x
H(1e-)] = 10 valence e-
Step 3-4: Add single bonds, then give
each atom (other than H) a full
octet.
Example 7: Write Lewis structure for HCN
NH C
H C N
3.3 Covalent Bond
3 3 Covalent Bond
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Writing Resonance Structures
SOLUTION:
Example 8 : Write resonance structures for the nitrate ion, NO3
and find the bond order.
Nitrate has [1 x N(5e-)] + [3 x O(6e-)] + 1e-] = 24 valence e-
After Steps 1-4:
3.3 Covalent Bond
A resonance structureis one of two or more Lewis structures
for a single molecule that cannot be represented accurately by
only one Lewis structure.
3 3 Covalent Bond
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Step 5. Since N does not have a full octet, we change alone pair from O to a bonding pair to form a double bond.
Writing Resonance Structures
3.3 Covalent Bond
3 3 Covalent Bond
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Example 9: Writing Lewis structures of
molecules, HNO3
1. Write skeletal structure H always terminal
in oxyacid, H outside attached to Os make least electronegative atom
central N is central not H
2. Count valence electrons sum the valence electrons for each
atom
add one electron for each charge
subtract one electron for each + charge
ONOH
O
N = 5H = 1
O3= 36 = 18
Total = 24 e
3.3 Covalent Bond
3 3 Covalent Bond
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3. Attach atom together with pairs of electrons, and
subtract from the total
dont forget, a line represents 2 electrons
Electrons
Start 24
Used 8Left 16ONOH
O..
..
.. ....
.
.
.
.
.
.
.
.
.
.
.
.
.
.
3.3 Covalent Bond
Example 9: Writing Lewis structures of
molecules, HNO3
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4. Complete octets, outside-in
H is already complete with 2
1 bond
and re-count electrons
N = 5H = 1
O3= 36 = 18
Total = 24 e
ElectronsStart 24
Used 8
Left 16
ElectronsStart 16
Used 16
Left 0
ONOH
O
....
..
....
.
.
.
.
.
.
Example 9: Writing Lewis structures of
molecules, HNO3
Example 9: Writing Lewis structures of
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5. If all octets complete, giveextra electrons to the centralatom
elements with dorbitals canhave more than eight electrons
Period 3 and below
6. If central atom does not haveoctet, bring in electrons from
outside atoms to share follow common bonding patterns if
possible
ONOH
O
ONOH
O
..
..
..
..
.
.
.
.
.
.
..
..
..
..
.
.
.
.
.
.
..
ONOH
O
..
...
.
.
.
.
.
..
..
Example 9: Writing Lewis structures of
molecules, HNO3
CHECKPOINT 4
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CHECKPOINT 4
Draw Lewis Structures of the Following
CO2
SeOF2
NO2
H3PO4
SO32
P2H4
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CO2
SeOF2
NO2
H3PO
4
SO32
P2H4
16 e
26 e
18 e
26 e
32 e
14 e
Two possible skeletal structures of formaldehyde (CH2O)
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70
H C O HH
C O
H
An atoms formal chargeis 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.
formal charge
on an atom in
a Lewis
structure
=1
2
totalnumber
ofbonding
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.
C 4 e- 2 single bonds (2x2) = 4-1 +1
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71
H C O H
C4 e
O6 e-
2H2x1 e-
12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
formal charge
on C= 4 - 2- x 6 = -1
formal charge
on O= 6 - 2- x 6 = +1
formal charge
on an atom in
a Lewis
structure
=1
2
total number
of bonding
electrons( )total number
of valence
electrons in
the free atom
-total number
of nonbonding
electrons-
1 1
C 4 e- 2 single bonds (2x2) = 4H 0 0
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72
C4 e
O6 e-
2H2x1 e-
12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
HC O
H
formal charge
on C= 4 - 0- x 8 = 0
formal charge
on O= 6 - 4- x 4 = 0
formal charge
on an atom in
a Lewis
structure
=1
2
total number
of bonding
electrons( )total number
of valence
electrons in
the free atom
-total number
of nonbonding
electrons-
0 0
Example 10
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StrategyThe Lewis structure for the carbonate ion was
developed as below:
The formal charges on the atoms can be calculated using the
given procedure.
Solut ionWe subtract the number of nonbonding electrons and
half of the bonding electrons from the valence electrons of each
atom.
Example 10
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The C atom: The C atom has four valence electrons and there
are no nonbonding electrons on the atom in the
Lewis structure. The breaking of the double bondand two single bonds results in the transfer of four
electrons to the C atom. Therefore, the formal
charge is 4 4 = 0.
The O atom in C=O: The O atom has six valence electrons and
there are four nonbonding electrons on
the atom. The breaking of the double
bond results in the transfer of two
electrons to the O atom. Here the formal
charge is 6 4 2 = 0.
Example 10
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The O atom in CO: This atom has six nonbonding electrons
and the breaking of the single bond
transfers another electron to it.Therefore, the formal charge is
6 6 1 = 1.
Thus, the Lewis structure for with formal charges is
CheckNote that the sum of the formal charges is 2, the same
as the charge on the carbonate ion.
Formal Charge and Lewis Structures
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76
g
1. For neutral molecules, a Lewis structure in which there
are no formal charges is preferable to one in which
formal charges are present.
2. Lewis structures with large formal charges are less
plausible than those with small formal charges.
3. Among Lewis structures having similar distributions offormal charges, the most plausible structure is the one in
which negative formal charges are placed on the more
electronegative atoms.
Example 11
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Formaldehyde (CH2O), a liquid with a disagreeable odor,
traditionally has been used to preserve laboratory specimens.
Draw the most likely Lewis structure for the compound.
Example 11
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StrategyA plausible Lewis structure should satisfy the octet
rule for all the elements, except H, and have the formal charges
(if any) distributed according to electronegativity guidelines.
Solut ionThe two possible skeletal structures are
Example 11
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First we draw the Lewis structures for each of these possibilities
To show the formal charges, we follow the procedure before. In
(a) the C atom has a total of five electrons (one lone pair plus
three electrons from the breaking of a single and a double
bond). Because C has four valence electrons, the formal
charge on the atom is 4 5 = 1. The O atom has a total offive electrons (one lone pair and three electrons from the
breaking of a single and a double bond). Since O has six
valence electrons, the formal charge on the atom is 6 5 = +1.
Example 11
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In (b) the C atom has a total of four electrons from the breaking
of two single bonds and a double bond, so its formal charge is
4 4 = 0. The O atom has a total of six electrons (two lonepairs and two electrons from the breaking of the double bond).
Therefore, the formal charge on the atom is 6 6 = 0. Although
both structures satisfy the octet rule, (b) is the more likely
structure because it carries no formal charges.
CheckIn each case make sure that the total number of
valence electrons is 12. Can you suggest two other reasons
why (a) is less plausible?
CHECKPOINT 5
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CO2
SeOF2
NO2
H3PO4
SO32
P2H4
CHECKPOINT 5
Assign formal charges
CHECKPOINT 5 (answers)
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CO2
SeOF2
NO2
all 0
Se = +1
H3
PO4
SO32
P2H4
CHECKPOINT 5 (answers)
P = +1
rest 0
S = +1
all 0
3.3 Covalent Bond
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Exceptions to Octet Rule
Some elements violate the octet rule1. Incomplete octet
Be forms twobonds with no lone pairs in its compounds
B. Al forms threebonds with no lone pairs in itscompounds
2. Expanded octet
many elements may end up with more than eight valence
electrons in their structure if they can use their empty dorbitals for bonding
3. Odd number electron speciese.g., NO
- Will have one unpaired electron, free radical & very
reactive
3.3 Covalent Bond
3.3 Covalent Bond
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Molecules with Electron-Deficient Atoms
B and Be are
commonly electron-
deficient.
Exceptions to Octet Rule
3.3 Covalent Bond
Exceptions to the Octet Rule
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85
Exceptions to the Octet Rule
The Incomplete Octet
H HBeBe2e-
2H2x1e-
4e-
BeH2
BF3
B3e-
3F3x7e-
24e-
F B F
F
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18Total = 24
3.3 Covalent Bond
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Expanded Valence Shells
An expanded valence shell is only possible for nonmetals
from Per iod 3 or higherbecause these elements have
available d o rbi tals.
Exceptions to Octet Rule
Odd-Electron SpeciesA molecule with an
odd number of
electrons is called a
free radical.
Exceptions to the Octet Rule
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87
Odd-Electron Molecules
N5e-O6e-
11e-
NO N O
The Expanded Octet
(central atom with principal quantum number n > 2)
SF6S6e-
6F42e-
48e-S
F
F
F
FF
F
6 single bonds (6x2) = 1218 lone pairs (18x2) = 36
Total = 48
Example 13
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Draw the Lewis structure for aluminum triiodide (AlI3).
AlI3has a tendency to
dimerize or form two units as
Al2I6.
Example 13
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StrategyWe follow the procedures used in Examples 11 and
12 to draw the Lewis structure and calculate formal charges.
Solut ionThe outer-shell electron configurations of Al and I are
3s23p1and 5s25p5, respectively. The total number of valence
electrons is 3 + 3 7 or 24. Because Al is less electronegative
than I, it occupies a central position and forms three bonds with
the I atoms:
Note that there are no formal charges on the Al and I atoms.
Example 13
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CheckAlthough the octet rule is satisfied for the I atoms, there
are only six valence electrons around the Al atom.
Thus, AlI3is an example of the incomplete octet.
Example 14
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Draw the Lewis structure for phosphorus pentafluoride (PF5), in
which all five F atoms are bonded to the central P atom.
PF5is a reactive
gaseous compound.
Example 14
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StrategyNote that P is a third-period element. We follow the
procedures given in Examples 11 to draw the Lewis structure
and calculate formal charges.
Solut ionThe outer-shell electron configurations for P and F
are 3s2
3p3
and 2s2
2p5
, respectively, and so the total number ofvalence electrons is 5 + (5 7), or 40.
Phosphorus, like sulfur, is a third-period element, and therefore
it can have an expanded octet.
Example 14
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The Lewis structure of PF5is
Note that there are no formal charges on the P and F atoms.
CheckAlthough the octet rule is satisfied for the F atoms,there are 10 valence electrons around the P atom, giving it an
expanded octet.
Example 15
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Draw a Lewis structure for the sulfate ion in which all
four O atoms are bonded to the central S atom.
Example 15
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StrategyNote that S is a third-period element. We follow the
procedures given in Example 11 to draw the Lewis structure
and calculate formal charges.
Solut ionThe outer-shell electron configurations of S and O
are 3s23p4and 2s22p4, respectively.
Step 1: The skeletal structure of is
Example 15
St 2 B h O d S G 6A l d h
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Step 2: Both O and S are Group 6A elements and so have
six valence electrons each. Including the two
negative charges, we must therefore account for atotal of 6 + (4 6) + 2, or 32, valence electrons in
Step 3: We draw a single covalent bond between all thebonding atoms:
Example 15
N t h f l h th S d O t
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Next we show formal charges on the S and O atoms:
Note that we can eliminate some of the formal charges for
by expanding the S atoms octet as follows:
Example 15
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At this stage of learning, you should realize that bothrepresentations 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 atoms octet, only
add enough double bonds to make the formal charge
on the central atom zero.
Example 15
Th th f ll i t t ld i f l h
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Thus, the following structure would give formal charges on
S(2) and O(0) that are inconsistent with the electronegativities
of these elements and should therefore not be included torepresent the ion.
Example 16
D L i t t f th bl d
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Draw a Lewis structure of the noble gas compound xenon
tetrafluoride (XeF4) in which all F atoms are bonded to the
central Xe atom.
Example 16
St t N t th t X i fifth i d l t W f ll th
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StrategyNote that Xe is a fifth-period element. We follow the
procedures in Example 11 for drawing the Lewis structure and
calculating formal charges.
Solut ion
Step 1: The skeletal structure of XeF4is
Step 2: The outer-shell electron configurations of Xe and F
are 5s25p6and 2s22p5, respectively, and so the total
number of valence electrons is 8 + (4 7) or 36.
Example 16
St 3 W d i l l t b d b t ll th
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Step 3: We draw a single covalent bond between all the
bonding atoms. The octet rule is satisfied for the F
atoms, each of which has three lone pairs. The sumof the lone pair electrons on the four F atoms (4 6)
and the four bonding pairs (4 2) is 32. Therefore,
the remaining four electrons are shown as two lone
pairs on the Xe atom:
We see that the Xe atom has an expanded octet.
There are no formal charges on the Xe and F atoms.
3.3 Covalent Bond
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Determine BestLewis Structures
BestLewis structure have:
All atoms obey the octet rule
Zero or fewer formal charges
smaller formal charges
negative formal charge on the moreelectronegative atom
Like charges on adjacent atoms are notdesirable
Sum of formal charges equal zero, or equalthe ionic charge for polyatomic ions
CHECKPOINT 6
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CHECKPOINT 6
Show the resonance forms of NCO-and
predict which resonance structure is the
most stable structure.
104
Answer:
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NCOhas 3 possible resonance forms:
N C O
A
N C O
B
N C O
C
N C O N C O N C O
formal charges
2 0 +1 1 0 0 0 0 1
Forms B and C have negative formal charges on N and O; thismakes 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.
3.3 Covalent Bond
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Properties of Covalent Compounds
1) Low melting points and boiling points
involves breaking the attractions between the molecules,
but not the bonds between the atoms
the covalent bonds are strong, but the attractions
between the moleculesare generally weak
intermolecular forces
2) Do not conduct electricity in the solid or liquid state
there are no charged particles (no ions) around to allow
the material to conduct
3.3 Covalent BondP ti f C l t C d
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3) Electronegativity The ability of an atom to attract bonding electrons to itself is
called electronegativity
Increases across period (left to right) and
Decreases down group (top to bottom) fluorine is the most electronegative element
francium is the least electronegative element
opposite of atomic size trend
The larger the difference in electronegativity, the more polarthe bond
negative end toward more electronegative atom
Properties of Covalent Compounds
The Electronegativities of Common Elements
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108
3.3 Covalent Bond
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Polar Covalent Bonding
Covalent bonding between unlike atoms results in unequalsharingof the electrons
The more electronegative atom pulls the shared electronscloser to its side
one end of the bond has larger electron density than theother
The result is a polar covalent bond
bond polarity
the end with the larger electron density gets a partialnegative charge ()
The other end that is electron deficient gets a partialpositive charge (+)
3.3 Covalent Bond
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Polar Covalent Bonding
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B d P l it
3.3 Covalent Bond
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HF
H Fd+ d-
H F
EN = 2.1 EN = 4.0
Bond Polarity
Classification of bonds by difference in electronegativity
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113
Covalent
share e-
Polar Covalent
partial transfer of e-
Ionic
transfer e-
Increasing difference in electronegativity
Difference Bond Type
0 Covalent
2 Ionic
0 < and
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Classify the following bonds as ionic, polar covalent, or
covalent:
(a) the bond in HCl
(b) The bond in KF
(c) the CC bond in H3CCH3
Excercise
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StrategyWe follow the 2.0 rule of electronegativity difference.
Solut ion(a) The electronegativity difference between H and Cl is 0.9,
which is appreciable but not large enough (by the 2.0 rule)
to qualify HCl as an ionic compound. Therefore, the bond
between H and Cl is polar covalent.(b) The electronegativity difference between K and F is 3.2,
which is well above the 2.0 mark; therefore, the bond
between K and F is ionic.
(c) The two C atoms are identical in every respectthey arebonded to each other and each is bonded to three other H
atoms. Therefore, the bond between them is purely
covalent.
3.3 Covalent Bond
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In covalent bonding, the more electronsin two atoms share,
the stronger the bond
Bond strength is measured by how much energy needed
to break the bond
triple bonds are stronger than double bonds, and double
bonds are stronger than single bonds
The more electrons two atoms share, the shorter the bond
Bond length is determined by measuring the distance
between the nuclei of bonded atoms
In general, triple bonds are shorter than double bonds,
and double bonds are shorter than single bonds
4) Covalent Bond Strength and Bond Length
3.3 Covalent Bond
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Trends in Bond Strength
In general, the more electrons two atoms share,the stronger the covalent bond must be comparing bonds between like atoms
CC (837 kJ) > C=C (611 kJ) > CC (347 kJ) CN (891 kJ) > C=N (615 kJ) > CN (305 kJ)
In general, the shorter the covalent bond, thestronger the bond must be comparing similar types of bonds
BrF (237 kJ) > BrCl (218 kJ) > BrBr (193 kJ)
bonds get weaker down the column
bonds get stronger across the period
3.3 Covalent Bond
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Bond Lengths
The distance between the nuclei ofbonded atoms is called the bond
length Because the actual bond length
depends on the other atoms aroundthe bond we often use the average
bond length averaged for similar bonds from many
compounds
3.3 Covalent Bond
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Bond Lengths
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121
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122
Copyright The McGraw-Hill Companies, Inc. Permission required for
Chemical Bonding II:Molecular Geometry and
Hybridization of Atomic
Orbitals
Molecular Geometry
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Molecular Geometry
3-dimensional arrangement of atoms in amolecules
Molecular geometry affects its physical and
chemical properties We often describe the shape of a molecule
with terms that relate to geometric figures
The geometric figures also havecharacteristic angles that we call bondangles
Using Lewis Theory toPredict
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PredictMolecular Shapes
In general, bond lengths and bondangles are determined by
experiment By knowing the number of electrons
surrounding a central atoms (Lewisstructure), we can simply predict the
overall geometry of the molecules.
Lewis theory says that these regions of
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125
electron groups should repel each other
This idea can then be extended topredict the shapes of molecules
the position of atoms surrounding a
central atom will be determined by
where the bonding electron groupsare
the positions of the electron groups
will be determined by trying tominimize repulsions between them
VSEPR Theory
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The geometric arrangement of electron pairs around acentral atom by minimize the electrostatic repulsionbetween electron pairs.
Electron groups around the central atom of a molecule willbe most stable when they are separated as far apart aspossiblewe call this valence shell electron pair
repulsiontheory (VSEPR) The resulting geometric arrangement will allow us to predict
the shapes and bond angles in the molecule
VSEPR formula: ABaEb
A = central atom; B = surrounding atoms
E = lone pairs on central atom; a, b = integers (1,2,3)
Electron Groups
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The Lewis structure predicts the number ofvalence electron pairs around the central atom(s)
Each lone pair of electrons constitutes oneelectron group on a central atom
Each bonding pair constitutes one electron group
on a central atom regardless of whether it is single, double, or
triple
O N O
there are three electron
groups on Nthree lone pair
one single bond
one double bond
Valence shel l electro n pair repu lsion(VSEPR) model:
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128
Predict the geometry of the molecule from the
electrostatic repulsions between the electron(bonding and nonbonding) pairs.
AB2 2 0
Class
# of atoms
bonded to
central
atom
# lone
pairs on
central
atom
Arrangement
of electron
pairs
Molecular
Geometry
linear linear
B B
# of atoms
bonded to
# lone
pairs on Arrangement
VSEPR
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129
AB2 2 0 linear linear
Class
bonded to
central
atom
pairs on
central
atom
g
of electron
pairs
Molecular
Geometry
AB3 3 0trigonal
planar
trigonal
planar
Boron Trifluoride
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130
# of atoms
bonded to
# lone
pairs on Arrangement Molecular
VSEPR
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131
AB2 2 0linear linear
Class
bonded to
central
atom
pairs on
central
atom
Arrangement
of electron
pairs
Molecular
Geometry
AB3 3 0trigonal
planar
trigonal
planar
AB4 4 0tetrahedral tetrahedral
Methane
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132
# of atoms
bonded to# lone
pairs on Arrangement Molecular
VSEPR
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133
AB2 2 0linear linear
Class
central
atom
pairs on
central
atom
g
of electron
pairs
Molecular
Geometry
AB3 3 0trigonal
planar
trigonal
planar
AB4 4 0tetrahedral tetrahedral
AB5 5 0trigonal
bipyramidal
trigonal
bipyramidal
Phosphorus Pentachloride
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134
# of atoms
bonded to
# lone
pairs on Arrangement
VSEPR
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135
AB2 2 0 linear linear
Class
bonded to
central
atom
pairs on
central
atom
of electron
pairs
Molecular
Geometry
AB3 3 0trigonal
planar
trigonal
planar
AB4 4 0tetrahedral tetrahedral
AB5 5 0trigonal
bipyramidal
trigonal
bipyramidal
AB6 6 0 octahedraloctahedral
Sulfur Hexafluoride
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136
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137
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138
bonding-pair vs. bonding-
pair repulsionlone-pair vs. lone-pair
repulsion
lone-pair vs. bonding-
pair repulsion> >
# of atoms
bonded to
# lone
pairs on Arrangement Molecular
VSEPR
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139
Class central
atom
p
central
atom
g
of electron
pairs
Geometry
AB3 3 0trigonal
planar
trigonal
planar
AB2E 2 1trigonal
planarbent
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# of atoms
bonded to
t l
# lone
pairs onArrangement
of electron Molecular
VSEPR
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141
Class
central
atomcentral
atom
of electron
pairsMolecular
Geometry
AB4 4 0tetrahedral tetrahedral
AB3E 3 1tetrahedral
trigonal
pyramidal
AB2E2 2 2tetrahedral bent
# of atoms
bonded to# lone
pairs onArrangement
of electron Molecular
VSEPR
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142
Class
central
atomcentral
atom
of electron
pairs Geometry
AB5 5 0 trigonalbipyramidal
trigonal
bipyramidal
AB4E 4 1 trigonal
bipyramidal
distorted
tetrahedron
Cl
# of atoms
bonded to
central
t
# lone
pairs on
central
t
Arrangement
of electron
pairsMolecular
G t
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143
Class atom atom pairs Geometry
AB5 5 0trigonal
bipyramidal
trigonal
bipyramidal
AB4E 4 1trigonal
bipyramidal
distorted
tetrahedron
AB3E2 3 2trigonal
bipyramidalT-shaped
# of atoms
bonded to
central
# lone
pairs on
central
Arrangement
of electron Molecular
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144
Class atom atom pairs Geometry
AB5 5 0 trigonalbipyramidal
trigonal
bipyramidal
AB4E 4 1 trigonalbipyramidal
distorted
tetrahedron
AB3E2 3 2 trigonalbipyramidal
T-shaped
AB2E3 2 3 trigonalbipyramidal
linear
# of atoms
bonded to
# lone
pairs on
t l
Arrangement
f l t M l l
VSEPR
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145
Classcentral
atom
central
atom
of electron
pairsMolecular
Geometry
AB6 6 0octahedraloctahedral
AB5E 5 1octahedral
square
pyramidal
# of atoms
bonded to
central
# lone
pairs on
central
Arrangement
of electronMolecular
Geometry
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146
Class atom atom pairsGeometry
AB6 6 0octahedraloctahedral
AB5E 5 1octahedral
square
pyramidal
AB4E2 4 2octahedral square
planar
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147
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Predicting Molecular
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149
gGeometry
1. Draw Lewis structure for molecule.
2. Count number of lone pairs on the
central atom and number of atoms
bonded to the central atom.3. Use VSEPR to predict the geometry of
the molecule.
Example 1Use the VSEPR model to predict the geometryof the following molecules and ions:
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of the following molecules and ions:
(a) AsH3
(b) OF2
(c)
(d)
(e) C2H4
Example 1StrategyThe sequence of steps indetermining molecular geometry is as follows:
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determining molecular geometry is as follows:
Solut ion
(a) The Lewis structure of AsH3is
There are four electron pairs around thecentral atom; therefore, the electron pairarrangement is tetrahedral
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Example 1Recall that the geometry of a molecule is determined onlyby the arrangement of atoms (in this case the O and F
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by the arrangement of atoms (in this case the O and Fatoms). Thus, removing the two lone pairs leaves us with
two bonding pairs and a bent geometry, like H2O. Wecannot predict the FOF angle accurately, but we know that itmust be less than 109.5 because the repulsion of thebonding electron pairs in the OF bonds by the lone pairson O is greater than the repulsion between the bonding
pairs.
(c) The Lewis structure of is
Example 1There are four electron pairs around the central atom;therefore, the electron pair arrangement is tetrahedral.
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Because there are no lone pairs present, thearrangement of the bonding pairs is the same as theelectron pair arrangement. Therefore, has atetrahedral geometry and the ClAlCl angles are all109.5.
(d) The Lewis structure of is
There are five electron pairs around the central I atom;therefore, the electron pair arrangement is trigonalbipyramidal. Of the five electron pairs, three are lonepairs and two are bonding pairs.
Example 1Recall that the lone pairs preferentially occupy theequatorial positions in a trigonal bipyramid. Thus,
i h l i l i h li
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removing the lone pairs leaves us with a linear geometryfor, that is, all three I atoms lie in a straight line.
(e) The Lewis structure of C2H4is
The C=C bond is treated as though it were a single bondin the VSEPR model. Because there are three electron
pairs around each C atom and there are no lone pairspresent, the arrangement around each C atom has atrigonal planar shape like BF3, discussed earlier.
Example 1Thus, the predicted bond angles in C2H4are all 120.
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Comment
(1) The ion is one of the few structures for which the
bond angle (180) can be predicted accurately eventhough the central atom contains lone pairs.
(2) In C2H4, all six atoms lie in the same plane. The overallplanar geometry is not predicted by the VSEPR model, but
we will see why the molecule prefers to be planar later. Inreality, the angles are close, but not equal, to 120because the bonds are not all equivalent.
Predict the geometries of the following species using the
CHECKPOINT 7
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Predict the geometries of the following species using theVSEPR method:
a) PCl3
b) H2O
c) CHCl3d) ClF3
e) TeCl4
trigonal pyramidal
bent shape
Tetrahedral
t-shape
seesaw
CHECKPOINT 8
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Which of the following species are tetrahedral?SiCl4, SeF4, XeF4, Cl4, CdCl4
2-
Answer: SiCl4 Cl4 and CdCl42-
Dipole Moments and Polar
Molecules
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H F
electron rich
regionelectron poorregion
d+ d-
ris the distance between charges
1 D = 3.36 x 10-30C m
Dipole moment, m, is ameasure of bond polarity
a dipole is a material with a + and end
it is directly proportional to the sizeof the partial charges and to thedistance between the charges
Generally, the more electronstwo atoms share and the largerthe atoms are, the larger thedipole moment
A dipole arrow is usedto the polarity of a bond
m= Qx rQis the charge
* Note to students: You will not be
tested on DP calculation
Behavior of Polar Molecules
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160
field off field on
Molecular Geometry and Polarity
3.3 Covalent Bond
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y y
For a molecule to be polar it must1. have polar bonds
electronegativity differencetheory
2. have an unsymmetrical shape and an overall DP
vector addition
Polarity affects the intermolecular forces ofattraction
therefore boiling points and solubilities Nonbonding pairs affect molecular geometry and
also molecular polarity
Molecule Polarity3.3 Covalent Bond
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The HCl bond is polar. The bonding electrons
are pulled toward the Cl end of the molecule. The
net result is a polar molecule.
Molecule Polarity
3.3 Covalent Bond
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The OC bond is polar. The bonding electrons arepulled equally toward both O ends of the molecule. The 2
dipole cancell out each other. The net result is a nonpolar
molecule.
Molecule Polarity
3.3 Covalent Bond
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Molecule Polarity
The HO bond is polar. Both sets of bonding electrons are pulled
toward the O end of the molecule. A net dipole moment (dotted dipole
arrow) appears. The net result is a polar molecule.
Predicting Polarity of Molecules
3.3 Covalent Bond
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1. Draw the Lewis structure and determine themolecular geometry
2. Determine whether the bonds in the moleculeare polar or non-polar
3. Determine whether the polar bonds add
together to give a net dipole moment
Predict whether NH3is a polar molecule
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1. Draw the Lewis structureand determine themolecular geometry
a) eight valence electrons
b) three bonding + one lonepair = trigonal pyramidalmolecular geometry
Predict whether NH3is a polar molecule
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2. Determine if the bonds arepolar
a) Find electronegativitydifference EN
b) if the bonds are not polar,we can stop here anddeclare the molecule will benonpolar
ENN = 3.0
ENH = 2.1
3.0 2.1 = 0.9
therefore the bonds
are polar covalent
Predict whether NH is a polar molecule
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Predict whether NH3is a polar molecule
3) Determine whetherthe polar bonds add
together to give anet dipole momenta) vector addition
b) generally, asymmetricshapes result inuncompensatedpolarities and a netdipole moment
The HN bond is polar. All
the sets of bondingelectrons are pulled toward
the N end of the molecule.
The net result is a polar
molecule.
Bond moments and resultant dipole moments
in NH3and NF3.
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169
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170
Example 2
Predict whether each of the following
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gmolecules has a dipole moment:
(a) BrCl
(b) BF3(trigonal planar)
(c) CH2Cl2(tetrahedral)
Example 2Strategy
Keep in mind that the dipole moment of a
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p pmolecule depends on both the difference in
electronegativities of the elements present andits geometry.
A molecule can have polar bonds (if the
bonded atoms have differentelectronegativities), but it may not possess adipole moment if it has a highly symmetricalgeometry.
Example 2Solut ion
(a) Because bromine chloride is diatomic, it has ali t Chl i i
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linear geometry. Chlorine is more
electronegative than bromine. so BrCl is polarwith chlorine at the negative end
Thus, the molecule does have a dipole moment.In fact, all diatomic molecules containingdifferent elements possess a dipole moment.
Example 2(b) Because fluorine is more electronegative than
boron, each BF bond in BF3(boron trifluoride) ispolar and the three bond moments are equal
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polar and the three bond moments are equal.However, the symmetry of a trigonal planar shapemeans that the three bond moments exactly cancelone another:
An analogy is an object that is pulled in the
directions shown by the three bond moments. If theforces are equal, the object will not move.Consequently, BF3has no dipole moment; it is anonpolar molecule.
B-F
Example 2(c) The Lewis structure of CH2Cl2(methylenechloride) is
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This molecule is similar to CH4in that it hasan overall tetrahedral shape. However,because not all the bonds are identical,there are three different bond angles: HCH,HCCl, and ClCCl. These bond angles areclose to, but not equal to, 109.5.
Example 2Because chlorine is more electronegative thancarbon, which is more electronegative thanhydrogen the bond moments do not cancel and
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hydrogen, the bond moments do not cancel and
the molecule possesses a dipole moment:
Thus, CH2Cl2is a polar molecule.
Example 3Explain why CO2is nonpolar, but OCS is polar.
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Ans: In CO2the two bond moments point inopposite directions and are of equalmagnitude. Therefore, they cancel. In OCS,even though the two bond moments point in
opposite directions, they are not of the samemagnitude and do not cancel.
CHECKPOINT 9
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List the following molecules in order ofdecreasing dipole moment: H2O, HF, CO2
Answer: HF, H2O,CO2