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Chapter 8:
Ionic and Covalent Bonding
RVCC Fall 2009CHEM 103 – General Chemistry I
Chemistry: The Molecular Science, 3rd Ed. by Moore, Stanitski, and Jurs
2
Bonding – What holds atoms together?
Octet rule: Octet rule: To form bonds, atoms gain, lose, or share e- to achieve a valence shell of 8 (or isoelectronic with a noble gas).
Ionic bond – an electrostatic attraction between a cation and an anion that forms when electrons transfer from one atom to another.
Covalent (Molecular) bond – the net attractive force that results from the sharing of electrons between atoms.
3
Ionic Bonds
An ionic bond is formed by the transfer of electrons from one atom (metal with low EA) to another (nonmetal with high EA). The resultant ions are held together by electrostatic attraction.
Na.:
Cl: . :
Na+ Cl-
[Ne]3s1 [Ne]3s23p5
[Ne] [Ne]3s23p6 = [Ar]
Each atom has satisfied the octet rule.
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Ionic Compounds - Properties
•crystalline•high melting point•high boiling point•soluble in water•electrolytes
•hard•brittle
Crystal Lattice
6
Covalent Bonding - G.N. Lewis (1916)
Some atoms shareshare e- to form bonds. When two nonmetals bond, they often share electrons since they have similar attractions (EA) for them. This sharing of valence electrons is called the covalent bond.
Attraction Stable bond Repulsion
Number of bonds = Number shared e- pairs.
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single bond- one shared pair of e-
H − HHH
Lewis structures:
show ALL valence electrons
dot = 1 e- line = 1 pair of e-
Single Covalent Bonds
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# of e- sharedGroup # of to form an octet Example valence e- (8 - A group#)
4A 4 4 C in CH4
5A 5 3 N in NF3
6A 6 2 O in H2O
7A 7 1 F in HF H – F ....
Single Covalent Bonds..
H
H – C – H
H
H – O – H ....
H – F ....
F – N – F
F ..
..
......
..
...... ..
# of e- sharedto form an octet(8-A group#)
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Lewis Structures
Lewis electron-dot formulas or Lewis structures.
An electron pair is either a bonding pair (shared between two atoms) or a lone pair (an electron pair that is not shared).
bonding pair
lone pairs::H Cl
::
:H Cl
::bonding pair
lone pair:
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Multiple Bonds
In the molecules described so far, each of the bonds has been a single bond, that is, a covalent bond in which a single pair of electrons is shared.
It is possible to share more than one pair. A double bond involves the sharing of two pairs between atoms.
CC
H
H
H
H
orC:CH
H
H
H: : ::
:
C has octet.H OK with 2.
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Triple bonds are covalent bonds in which three pairs of electrons are shared between atoms.
Multiple Bonds
CC orHH
::
CC HH
:::
Elements that form multiple bonds: C, O , N, S
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The Procedure Using the molecular formula, count the total
number of valence electrons available (bonding + lone pairs). Valence electrons for each atom corresponds to group # Adjust for charge (add electron for each minus, delete
electron for each plus)
H2O
8 valence electrons
H3O+
8 valence electrons
2(1) + 6 3(1) + 6 - 1
16
The Procedure Make a skeleton by connecting the atoms with single bonds
only. When connecting atoms, remember… Put the least electronegative atom in the center. (Usually the first listed
in the chemical formula.) Hydrogen is ALWAYS a terminal atom. More electronegative atoms
are terminal (F, O…) Make the structure symmetric.
H2O
8 valence electrons
H3O+
8 valence electrons
H O H H O H
H
4 electrons left over two electrons left over
4 pairs 4 pairs
2 pairs left 1 pair left
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The Procedure Put the left over electrons as lone pairs, preferably on
the more electronegative atoms Is the octet rule satisfied? If YES, then you’re done…
H2O
8 valence electrons
H3O+
8 valence electrons
H O H H O H
H
4 electrons left over two electrons left over
H O H H O H
H
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The Procedure
If you are electron-deficient (not enough electrons to complete an octet), then some atoms must share more than two electrons. “If you have a lone pair, make those two atoms share.”
Ex. C2H4
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The Procedure
If you have excess electrons, at least one atom must have an expanded valenceMust be element from third period or lowerUsually the central atom
e.g. SF4
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The total number of valence electrons on an atom (from bonds & lone pairs) cannot exceed that atom’s maximum valence. First period: 2 electrons (s) Second period: 8 electrons (s,p) Third period & below: prefer to have 8, but can expand
when necessary (s,p,d)
H O H H C H
H
H
P
Cl
Cl Cl
ClCl
General Rule
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Cl
C
Cl
O
Writing Lewis Dot Formulas
COCl2
24 e- total 12 pairs
:
:: :
::
0 left
:
: :
Note that the carbon has only 6 electrons.
9 left
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Cl
C
Cl
O
Writing Lewis Dot Formulas
COCl2 12 pairs9 e- left
:
:: :
::
0 e- left
:: :
To fulfill the octet rule…“If you have a lone pair, make those two atoms share!”
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Writing Lewis Dot Formulas
COCl224 e- total
Cl
C
Cl
18 e- left
:
:: :
::
0 e- leftO: :
Note that the octet rule is now obeyed.
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Question
First evaluate the total valence electrons:24 e-
a. 26 e-, wrongb. 24 e-, looks OKc. 24 e-, one F has too manyd. 24 e-, N not enoughe. 24 e-, but least electronegative has to be in the centerf. 24 e-, no bond between two N.
no no
no no no
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Exceptions to the Octet Rule
Although many molecules obey the octet rule, there are exceptions where the central atom has less or more than eight electrons.
Incomplete octet – B, H
BF3Boron has 3 valence electrons
:F – B – F: .. .... ..
:F:..
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Exceptions to the Octet Rule
: F :
::: F :
F ::
:
: F
:: PF :
::
If a nonmetal is in the third period or greater it can accommodate as many as twelve electrons as the central atom.
PF5
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Exceptions to the Octet Rule
In sulfur tetrafluoride, SF4, the sulfur atom must accommodate two extra lone pairs for a total of 5 electron pair (10 electrons)
F ::
:
: F :
:
SF :
::
: F
:: :
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Formal Charge and Lewis StructuresIn certain instances, more than one feasible Lewis structure can be
illustrated for a molecule. For example,
H C N CNHor: :
The concept of “formal charge” can help us decide which structure is correct.
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Formal Charge and Lewis Structures
formal charge = valence e- before bonding– valence e- after bonding
= valence e- - [1/2 bonding e- + lone pair e-]
H C N CNHor: :
H: 1-½(2) = 0
C: 4 - ½(8) = 0
N: 5 – (½(8) + 2) = 0
H: 1-½(2) = 0
C: 4 – (½(6)+2) = -1
N: 5 – (½(8)) = +1
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Formal Charge and Lewis Structures
Smaller formal charges are more favorable More electronegative (or higher EA) atom should have
negative formal charges Like charges should not be on adjacent atoms Net formal charge should be the overall charge on the
molecule/ion.
orH C N:0 0 0
CNH :formal charges
0 +1 -1
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Practice
Determine the most stable structure for dinitrogen Determine the most stable structure for dinitrogen oxide. (All structures have 16 valence electrons.)oxide. (All structures have 16 valence electrons.)
N=N=ON=N=O N-N≡ON-N≡O N ≡ N - ON ≡ N - O-1 +1 0 -2 +1 +1 0 +1 -1
formal charge= valence e- - [1/2 bonding e- + lone pair e-]
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Delocalized Bonding: Resonance
The structure of ozone, O3, can be represented by two different Lewis electron-dot formulas.
O O
O:: :
::
:
OO
O
:::
::
:
The bond lengths for the above structures are:
O – O 132 pm O = O 112 pm
However, experiments show that both bonds are identical.
36
Delocalized Bonding: Resonance
According to theory, one pair of bonding electrons is spread (delocalized) over the region of all three atoms.
In fact, the actual bond length is 127.8 pm (in between 132 and 112pm).
The actual molecule is a hybrid or composite structure and not different structures that change back and forth… although, we often represent it that way.
OO
O
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Delocalized Bonding: Resonance
Lewis resonance structures, have the same atoms in the same positions. Only an electron pair position is different.
OH O N
O
OH O N
O
OH O N
O
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Resonance Structures
O=S OO=S O O S=OO S=O
S=S OS=S O S O=SS O=S
Which pair does NOT represent resonance structures?
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Resonance StructuresDraw resonance structure(s) for the following:
O O C
O
-2
O O C
O
-2
O O C
O
-2
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“All covalent bonds are created equal but some are more equal than others.”
(We assumed equal sharing when we calculated formal charge.)
41
Electroegativity vs Atomic Number
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 10 20 30 40 50 60
Atomic Number
Ele
ctro
neg
ativ
ity
(Pau
ling
)
He NeAr
KrXe
F
Cl Br
I
Electronegativity……is a measure of the ability of an atom in a molecule
to draw bonding electrons to itself when bonded.
decreasesdown a group
increasesacross a period
Periodic Trend - Electronegativity
43
Types of Bonds
Ionic: ΔEN >1.8 electron transfer
Covalent: ΔEN <1.8 electron sharing
Metallic: electron-sea model or band theory
0.9 3.0
Na+ Cl-
2.5 2.1
C - H
1.6 1.6
Zn Zn
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Covalent Bonds (EN<1.8)
Non-polar covalent - ΔEN = 0 – 0.5
Examples: H-H, Cl-Cl, C-H bonds
Polar covalent - ΔEN = 0.5 – 1.8 Examples: H-O, C-Cl, C-O bonds
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Bond Polarity
In HCl we have a partial negative charge on the chlorine (denoted -) and a partial positive charge on the hydrogen (denoted +)
The bond is polar covalent.
H :Cl::
:
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Bond PolarityArrange the following bonds from the most to the least polar:
HH, HCl, HF, HI, HBr
Compare the electronegativity of Cl, F, I and Br:Least Most Electronegative ElectronegativeI Br Cl F
Determine polarity:HH HI HBr HCl HFnon polar most polar
49
PracticeWhich of the following bonds in each pair are more polar?
C-S or C-OCl-Cl or O=O N-H or C-H
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Bond LengthBond length (or bond distance) is the distance between the
nuclei of two bonded atoms (the sum of atomic radii).
51
The size of atoms that form the bond determine the length.
C - N 147 pmC - C 154 pmC - P 187 pm (P, period 3)
Bond Length
52
Bond Length – Multiple Bonds
As the electron density between atoms increases the bond lengths decrease; the atoms are pulled together more strongly.
decrease
53
Bond EnthalpyBond Enthalpy – the enthalpy change that occurs when the bond
between two bonded atoms in the gas phase is broken and the atoms are separated completely at constant pressure.
As the electron density between two atoms increases, the bond
gets shorter and stronger.
Bond length Bond enthalpy
C
C
C
C
CC
154 pm
134 pm
120 pm 835 kJ/mol
602 kJ/mol
346 kJ/mol
NaCllattice energy-786 kJ/mol
MgOlattice energy-3791 kJ/mol
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Bond Enthalpy
Hº = ∑[(moles of bonds) × D(bonds broken)] -
∑[(moles of bonds) × D(bonds formed)]
Hº - standard enthalpy of reaction
Bond Enthalpies can be used to calculate the standard enthalpies of reaction (gas phase, STP)
55
Bond Enthalpy
Estimate the Hº for the following reaction:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
+ 2 :O = O: : O = C = O: + 2H – O - H
:: :: ::
4 C – H bonds 1 O = O bond 2 C = O bond 2 H – O bondper molecule per molecule per molecule per molecule
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Hº = ∑[(moles of bonds) × D(bonds broken)] -
∑[(moles of bonds) × D(bonds formed)]
+ 2 :O = O: : O = C = O: + 2H – O - H
:: :: ::
4 C – H bonds 1 O = O bond 2 C = O bonds 2 H – O bondsper molecule per molecule per molecule per molecule
= [4 × D(C-H) + 2 × D (O=O)] – [2 × D (C=O) + 4 × D(H-O)] =
[4 × 416 + 2 × 498] – [2 × 803 + 4 × 467] = -814 kJ
4 C – H bonds 2 O = O bonds 2 C = O bonds 4 H – O bonds