1
Bonding: General Concepts8※ Types of chemical bonds
Covalent bonding
Ex. H2E
0
458
(kJ/mol)
0.074 r (nm)
Repulsions of nucleus and es
Zero interaction at long distance
H H
r
H-H bond length
Two es shared by two Hs: covalent bonding
Ionic bondingEx. Na+ Cl: ionic forces operate
Coulomb attraction:
rQQ
E 2119 )nmJ1031.2(
charges
Polar covalent bondEx. H–F
Unequal sharing of e
Charge distribution occurs
H F
Stronger attraction for e
Fractional charge
A dipole
2
※ Electronegativity
The ability of an atom in a molecule to attract shared es
Linus Pauling (1901-1995)
A–B Expected BE =2
BEBE BBAA
expectedB-AactualB-A )BE()BE(
The greater EN difference, the greater value
Define EN(A) – EN(B) =Assign EN(F) = 4.0
102.0
Fluorine atom with the highest EN
EN for all elements can be determined
EN difference for A–B Bond type0 covalent
polar covalent
large ionic (A+, B)
A B
Higher EN
F
Increasing EN
DecreasingEN periodic table
3
※ Bond polarity and dipole moment
A dipole has a dipole moment ()
Will orient in an electric field
= Q·R
charge (c)
distance (m) Unit: debye (D)1D = 3.336 × 1030 cm
For polyatomic molecules
OH
H
net dipole
C OO
The center of all positive atoms and the center of all negative atoms do not merge
: net dipole = zero
※ Ions: Electron configurations and sizes
Ionic compoundsIn general A M+
non-metal main group metal
Non-metal: gains e
Metal: loses ereach noble gas configuration
Ex. Ca: [Ar]4s2
O: [He]2s22p4Ca2+: [Ar]O2: [He]2s22p6 = [Ne]
4
◎ Sizes of ionsdetermines the structure and stability of ionic compounds
Negative ions: larger than parent (gain e)Positive ions: smaller than parent (lose e)
Trend: Down a group larger
Ex. Li+ Be2+ O2 F
60 31 140 136 (pm)Na+
95
◎ The energy
M+(g) + X(g) M–X(s) energy released = lattice E
OverallEx.
Formation of an ionic compound energy depends on many factors difficult to predict qualitatively
Lattice E = )( 21
rQQ
k
Li(s) + 1/2 F2(g) LiF(s)
Li(g)
F(g)
Li+(g)ionization E
F(g)electron affinity
+
latticeenergy1/2 BE of F2
= 78.9160.7
520.5
328.0
1049.0
616.9 Unit:kJ/mol
At 298 K
LEtheo
= 966Born-Habercycle
5
※ Partial ionic character of covalent bonds
In the gas phase: no true ionic bond
Polar covalent bond
percent ioniccharacter
%100moment dipole calculatedmomentdipolemeasured
X+ Y
Ex. NaCl(g): 75%
In general: > 50% ionic solid
Ambiguity: NH4+Cl, Na2SO4
covalent bonded group
Definition of ionic compoundconducts electric current when melted
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※ The covalent bond: A model
Ex. CH4 C + 4H requires 1652 kJ
CH4 is more stable than the discreet atomsby 1652 kJ
Consider forces exist between C and 4H – bonding
Consider localized C–H bonds
To break each bond = = 413 kJ 4
1652
Ex. CH3Cl C + 3H + Cl requires 1578 kJ
3 C–H bonds, 1 C–Cl bond
1578 – 3(413) = 339 kJ/mol
Strength of C–Cl bond
Assumption: bond strength does not vary with structures
★ The covalent bond modelConsidering electrons are localized between two atoms
7
※ Covalent bond energiesand chemical reactions
Fact 1 HCH4(g) CH3(g) + H(g) 435 kJ/molCH3(g) CH2(g) + H(g) 453 CH2(g) CH(g) + H(g) 425 CH (g) C(g) + H(g) 339
total = 1652Average: 413
BEs are structurally dependent
Average BEs are used
Bond types
es sharedsingle 2double 4triple 6
Ex. BE (kJ/mol) bond length (Å)C–C 347 1.54C=C 614 1.34CC 839 1.20
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◎ Bond energy and enthalpy
Ex. H2(g) + F2(g) 2HF(g)
2H 2F 2H, 2F
identical
Bond E difference = H
H = D (bonds broken) – D (bonds formed)
Bond E (+)
H = DH-H + DF-F – 2DH-F
= 432 + 154 – 2(565)= –544 kJ
Calculated from Hf: H = 2(–271) = –542 kJHf
o(HF)
※ The localized electron bonding model
(Valence bonding model)
A covalent bond is formed between two atomssharing electrons using atomic orbitals
Pairs of es localized on an atom: lone pairsPairs of es localized between atoms: bonding pairs
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※ Lewis structures
Describe the arrangement of valence electrons
G. N. Lewis (18751946) observed:In most stable compounds the atoms achieve noble gas electron configuration
Molecules with covalent bonds
Duet ruleH, He
H:H He:
Octet ruleC, N, O, F: second row nonmetals
F F F Ne
Ex. H2O
H· H·
1 1 6
O
Total: 8 es
O HH
10
Ex. CO2
:C:
4 6 6
O
Total: 16
C OO
O
Does not obey octet rule
Try and error
C OO = C OO
C N
Ex. CN One extra charge
4 + 5 + 1 = 10 e
Not good
C N Not good
C N Not good
C N Good
C N
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Ex. N2
10 e isoelectronic with CN
N N
Ex. NH3
8 e
CH4
8 e
NO+
5 + 6 1 = 10 e
N HH
H
N HH
H
C HH
H
H
C HH
H
H
N O+
CF4
32 eC FF
F
F
C FF
F
F
※ Exceptions to the octet rule
◎ Less than 8
B
F
FF
Only 6 e
How about
B
F
FF
Fact: B is very reactive toward electron rich molecules
H NH
H
+ B
F
FFN BH F
H
H
F
F
12
◎ More than 8
S
F
F
FF
FF
12 e
Classical explanationuse of low lying empty d orbitals
(3d for sulfur)
PCl55 + (5 × 7) = 40 e
P
Cl
Cl
ClCl
Cl
10 e
Allow the central atom to expand its valency
Ex. XeO3
8 + (3 × 6) = 26 e Xe
O
O O
obeys
BeCl22 + (2 × 7) = 16 e
Be ClCl
Only 4 e
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※ Resonance
Ex. NO3
N
O
O O
Predict: a short double bondtwo long single bonds
Fact: Equal length (between a single and a double)
A new theory must be formulated
Actually there are three possible structures:
N
O
O ON
O
O ON
O
O O
We can consider the real structure as the avg. of the three– a resonance hybrid
Any single structure can not represent the real structure Each Lewis structure is called resonance structure
(exists only on paper) Represented as
N
O
O ON
O
O ON
O
O O
★ double headed arrow (not )
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Only allow electrons to move(nuclei stay at the same position)
Must be proper Lewis structuresEx. Carbon can not have five bonds
◎ Odd-electron molecule
Nitric oxide NO5 + 6 = 11
N O N O N O
Not good A radical with unpaired e
*Keep the number of unpaired e the lowest
C CH
H H
HC C
H
H H
HNot a viable resonance structure
NO2
5 + 2(6) = 17
O N O ONO O N O ONO
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◎ Formal charge
Q: Structure of SO42 (32 e)
or
S
O
O
OO
2
S
O
O
OO
2
S
O
O
OO
2What about
10 e 12 e
How to decide?
Method: Estimate the charge
Oxidation state: O 2S +6
Over estimated
Only useful for bookkeeping
The concept of formal charge
Compare: number of e in a neutral atom·····Anumber of e in a bonded atom····B
When A = B no chargeA > B (+)A < B ()
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Define Formal charge = A – B
Assign lone pair es to the atom completelyDivide shared es equally
Ex.
S
O
O
OO
2 For OA = 6B = 6 + 2/2 = 7
lone pair es
shared es
Formal charge on O = 6 – 7 = –1
For S: A = 6 B = 0 + 8/2 = 4
lone pair esshared es
Formal charge on S = 6 – 4 = +2
S
O
O
OO
21
1
1
12 Overall charge
= (+2) + 4(1)
S O
S
O
O
OO
2 Single bonded O: 1Double bonded O: 6 – [4 + 4/2] = 0S: 6 – (10/2) = +1
S
O
O
OO
20
11
1
1
ok
17
S
O
O
OO
20
1 1
0
6 – (12/2) = 0
A better structure due to minimum formal charge
There are resonance structures
S
O
O
OO1 1
S
O
O
OO
1
1
S
O
O
OO
1
1etc.
Another school of thought:Keeping the octet rule ismore important S
O
O
OO1
1
1
1
2
Note: they are all resonance structures
Guidelines For 2nd row elements: never exceed octet rule Formal charges as low as possible Negative charge on more electronegative atoms
B
F
FF
1
1
Not good
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※ Molecular structures: The VSEPR model
To predict approximate 3-D structureA simple model:
Valence shell electron pair repulsion (VSEPR) model(useful for nonmetals)
The main postulation:minimizing electron pair repulsions
Bonding and nonbonding pairs
Ex. BeCl2Cl–Be–Cl : linear
(two pairs)
As far away as possible
BF3
B
F
FF
120o
C
H
HHH
109.5o
NHH
H
O
H
H
C
H
H
HH
90o
Trigonal planar(three pairs)
CH4
Tetrahedral(four pairs)
Cf.
Square planar(not good)
NH3
Based on the position of atoms: trigonal pyramid
H2OA bent structure(four pairs)
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Bond angles
C
H
HHH
109.5oN
HHH
107o
O
H
H
104.5o
lonepairs
0 1 2
Explanationlone pair is closer to the nucleus exerts large repulsion towards the other electron pairs
NHH
H
squeezed
Five pairs: PCl5
P
Cl
Cl
ClCl
Cl Trigonal bipyramidal
Six pairs: PCl6
P
Cl
Cl
Cl
Cl Cl
ClOctahedral
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Ex. XeF4
Xe
F
F
FF
Six pairs: octahedral arrangement
XeF
F F
F as far away as possible
Square planar shape
Ex. I3
I I I
Five pairs: trigonal bipyramidal arrangement
III
90o, 90o, 180o
I
I
I
120, 120, 120
I
I
I
90, 90, 120
Best(no 90o∠ repulsions)
linear
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Treat double or triple bonds as one effective pair
N
O
O O
Ex.
3 pairs Bent structure
SO2S
OOS
O O
In fact ∠O-S-O ~120o with little distortion long pair is satisfied with 120o∠