General Chemistry
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ChMSt0611.pptxGeneral Chemistry I CHM 1010
Instructor:
Dr. William Antonio Boyle
Moore/Stanitski, Chapter 6 Covalent Bonding
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Covalent Bonding
G.N. Lewis (1916):
Some atoms share e- to form bonds.
Molecules with shared e- are covalent compounds with covalent bonds.
Number of bonds = Number shared e- pairs.
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Covalent Bonding
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Single Covalent Bonds
Single bond: one shared pair of e-.
H − H
H
H
Lewis structures: dot = 1 e-. Line = 1 pair of e-
Octet rule
To form bonds, elements gain, lose, or share e- to achieve 8 valence e-.
Doesn’t apply if Z ≤ 5 (H, He, Li, Be and B)
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Single Covalent Bonds
Bonding pairs = e- pairs shared between atoms.
Lone pairs = unshared e- pair.
lone pair
lone pair
H
O
H
bonding
bonding
H
O
H
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2F = 2(7) = 14 valence e-
Share 2 e- to form octets
F
F
H
O
H
O + 2H = 6 + 2(1) = 8 val. e-
Two O-H bonds
H
N
H
H
N + 3H = 5 + 3(1) = 8 val. e-
3 N-H bonds
Single Covalent Bonds
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Rules for Writing Lewis Structures of Molecules
These are rules for molecules of non-metals.
We begin with the formula for the molecule.
1a. Add all the valence electrons for all the atoms in the formula.
For anions or cations, add or subtract electrons.
1b. Place the atom with the smallest electronegativity in the center (exception: Hydrogen)
2. Use a pair of electrons to form a bond between each pair of bound atoms; which atoms are connected to which.
.
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Rules for Writing Lewis Structures of Molecules
3. Arrange the remaining electrons to form octets around each atom, or duets around each hydrogen (this is a trial & error procedure!)
4. Place any leftover electrons on the central atom, even if this results in more than an octet.
5. If there are not enough electrons to give the central atom an octet, try forming multiple bonds.
6. C, N, O, and F always follow the octet rule!
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Rules for Writing Lewis Structures of Molecules
7. B and Be can have less than an octet, because they are very small atoms.
8. Third and higher period elements can exceed an octet, because they can use their “d” orbitals.
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Writing Lewis Structures
Phosphorus trifluoride, PF3
Skeleton (X is central in XYn ).
1. PF3 = 5 + 3 (7) = 26 valence e-
P (group 5A)
3 x F (group 7A)
F
P
F
F
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F
P
F
F
Writing Lewis Structures
3. Build octets – start with terminal atoms.
6 + 20 = 26 e-
26 e- to begin with
6 e- used in 3 bonds, 20 e- remain (10 pairs)
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Writing Lewis Structures
Phosphate ion, PO43-
2. Skeleton; P is central
P (grp 5A)
4 x O (grp 6A)
charge (-3) 3 e-
1. PO43- = 5 + 4(6) + 3 = 32
O
O
P
O
O
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Writing Lewis Structures
3. Add e- pairs:
32 e- used.
8 e- used in 4 bonds, 24 e- remain (12 pairs)
O
P
O
O
O
Add brackets and overall charge to show this is an ion.
O
P
O
O
O
3-
32 e- to start with.
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Alkanes = hydrocarbons with C-C single bonds.
Methane
e- density: red = high density ; blue = low density
Saturated hydrocarbons: each C is bonded to its maximum number of H.
Single Covalent Bonds in Hydrocarbons
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Larger alkanes:
butane
2-methylpropane
Single Covalent Bonds in Hydrocarbons
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Multiple Covalent Bonds
Methanal (formaldehyde) H2CO
1. Valence e- = 2(1) + 4 + 6 = 12
2. Skeleton
O
C
H
H
6 e- in bonds. Add the other 3 pairs to O (outer atom).
Each H shares 2 e-
C only “has” 6 e-.
Too few “dots” to complete all the octets?
Convert lone pairs to shared pairs.
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Convert lone pairs to bond pairs.
O
C
H
H
Each H shares 2 e-
C shares 8
O “has” 8
Multiple Covalent Bonds
O
C
H
H
4 shared + 2 lone pairs
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Multiple Covalent Bonds
Carbon dioxide CO2
1. 4 + 2(6) = 16 e-
2. Skeleton
O
C
O
4 e- in bonds. Add 3 pairs to each O.
O
C
O
4. Convert lone pairs to bond pairs.
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Multiple Covalent Bonds in Hydrocarbons
Alkenes: hydrocarbons with C=C bonds
ethene C2H4
(ethylene)
H
C
H
C
H
H
propene C3H6
(propylene)
H
C
C
H
H
C
H
H
H
–ene ending
CnH2n (unless cyclic)
Unsaturated (can add more H)
H
C
C
H
H
C
H
H
C
H
H
H
butene C4H8
(1 isomer shown)
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Multiple Covalent Bonds in Hydrocarbons
Alkynes: hydrocarbons with C≡C bonds
propyne
H
H-C≡C-C-H
H
C3H4
–yne ending
CnH2n-2
Unsaturated
ethyne (acetylene)
H-C≡C-H
C2H2
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Bond Length
More bonds = greater e- density between atoms
= atoms pulled together more strongly
Multiple bonds (pm): N-N 140 N=N 120 N≡N 110
C-C 154 C=C 134 C≡C 121
C-N 147 C=N 127 C≡N 115
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Bond Enthalpy
Table 6.2 Average bond enthalpies (kJ/mol)
| I | Br | Cl | S | P | Si | F | O | N | C | H | |
| H | 299 | 366 | 431 | 347 | 322 | 323 | 566 | 467 | 391 | 416 | 436 |
| C | 213 | 285 | 327 | 272 | 264 | 301 | 486 | 336 | 285 | 356 | |
| N | 193 | ~200 | 335 | 272 | 201 | 160 | |||||
| O | 201 | 205 | ~340 | 173 | 190 | 146 | |||||
| F | 255 | 326 | 490 | 582 | 158 | ||||||
| Si | 234 | 310 | 391 | 226 | 226 | ||||||
| P | 184 | 264 | 319 | 209 | |||||||
| S | 213 | 255 | 226 | ||||||||
| Cl | 209 | 217 | 242 | ||||||||
| Br | 180 | 193 | |||||||||
| I | 151 |
Shorter bond = stronger bond.
Multiple bonds: N-N 160 N=N 418 N≡N 946
C-C 356 C=C 598 C≡C 813
C-N 285 C=N 616 C≡N 866
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N2H4(g) + O2(g) → N2(g) + 2H2O(g)
Bond Enthalpies
Can be used to estimate ΔH for a reaction:
ΔH =(bonds broken) – (bonds formed)
= [4(N-H) + (O=O) + (N-N)] – [4(O-H) + (N≡N)]
≈ 4(391) + 498 + 160 – 4(467) – 946
≈ -592 kJ/mol
Using ΔfH: ΔH = 2ΔfH (H2O) – ΔfH(N2H4) = 2(-241.8)-(95.4) = -579.0 kJ/mol
H H
N-N
H H
O=O
N≡N
H
O
H
..
..
..
..
..
..
..
..
..
..
exact value
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Bond Properties: Polarity
Bonding pairs are not always equally shared.
d+
d-
F has stronger e- attraction. Polar bond.
F
F
Nonpolar = equally shared (identical atoms).
Polar = unequal sharing (different attraction for e-)
F
H
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Bond Properties: Electronegativity
Electronegativity = ability of an atom in a covalent bond to attract shared e- to itself.
Linus Pauling developed the first scale.
Based on bond energies.
Pauling’s scale: F = 4.0 (arbitrary).
Unitless.
Polar bond: one atom is more electronegative.
Linus Pauling
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Differences, ΔEN, determine bond polarity:
| EN | 4.0 4.0 | 2.1 2.6 | 2.1 4.0 | 0.9 4.0 | ||
| F–F | H–Br | H–F | Na+ F- | |||
| ΔEN | 0 | 0.5 | 1.9 | 3.1 |
Bond Properties: Electronegativity
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Bond Properties: Electronegativity
For the following bond pairs, indicate the δ+ and δ- atoms and choose the more polar bond:
Cl–F and Br–F; Si–Br and C–Br
ClF vs BrF
ENF > ENCl > ENBr.
In each case F is δ-, the other end is δ+ .
ΔEN (F and Br) > ΔEN (F and Cl)
BrF is more polar than ClF.
(Electronegativities: Br = 2.8, Cl = 3.0, F = 4.0
increasing
electronegativity
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Bond Properties: Electronegativity
Indicate δ+ and δ- and choose the more polar bond: Si–Br and C–Br.
SiBr vs CBr
ENSi < ENC < ENBr. Si is δ+ in SiBr, C is δ+ in CBr.
Largest ΔEN between Si and Br
SiBr is more polar than SiC.
(Electronegativities: Si = 1.8, C = 2.5, Br = 2.8
increasing
electronegativity
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Molecular Polarity Affects Solubility in Water
© 2014 Pearson Education, Inc.
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Formal Charge
The charge a bonded atom would have if its bonding e- were shared equally.
Used to study charge distribution in a molecule.
Method
Find the number of e- assigned to each atom:
e- “on” an atom = (lone pair e-) + ½ (bonding e-)
Formal charge of each atom
= (# of valence e-) – (e- “on” the atom).
Note: sum of formal charges = molecular charge
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Check: (formal charges) = ion charge = -1
N
C
O
[ ]
–
O C N
Valence e- 6 4 5
Lone pair e- 6 0 2
½ shared e- 1 4 3
Formal Charge -1 0 0
Formal Charge
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Formal Charge
If there is choice between Lewis structures:
Smaller formal charges are favored.
Negative formal charges should be on the most EN atoms
Like charges should not be on adjacent atoms
Formal charges:
Which N2O structure is preferred?
N
N
O
-1
+1
0
0
+1
-1
N
N
O
Preferred. ENO > ENN
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Which ClO2- structure is preferred?
Preferred
(Smaller charges)
O
Cl
O
O
Cl
O
Formal charges:
-1
0
0
-1
+1
-1
Formal Charge
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Lewis Structures & Resonance
Ozone has 2 equivalent structures:
Both:
obey the octet rule
have the same number and types of bonds
have the same formal charges
O
O
O
O
O
O
Experiments show that the OO bonds are identical.
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O
O
O
O
O
O
A resonance hybrid best represents O3. Each bond is ≈1½ bonds. Ozone does NOT “flip” back and forth.
Resonance structures are used to show O3 is a mixture of both:
Lewis Structures & Resonance
Valid resonance structures:
Bonding and e- pair positions are changed.
Atom positions must not change.
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Lewis Structures & Resonance
Resonance in CO32-
Experiment: All three CO bonds = 129 pm
Typical bond lengths
C-O =143 pm; C=O = 122 pm.
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Lewis Structures and Resonance
Benzene, C6H6, is a ring compound best described using resonance ideas:
Experiments: All C-C bonds have identical lengths.
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Resonance & Structure of Benzene
or
Solid ring shows resonance
Benzene is usually drawn without its hydrogen atoms shown, or with a ring representing six delocalized electrons spread evenly over the carbon atoms:
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Exceptions to the Octet Rule
Be and B form e- deficient compounds:
H
Be
H
2 + 2(1) = 4 valence e-
3 + 3(7) = 24 valence e-
B
F
F
F
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Fewer than Eight Valence Electrons
Often very reactive
+
N
H
H
H
B
F
F
F
B
F
F
F
N
H
H
H
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Odd Number of Valence Electrons
Some stable molecules have an odd number of e-.
Examples
NO 5 + 6 = 11 valence e-
O
N
O
N
O
NO2 5 + 2(6) = 17 valence e-
Free radical = atom or molecule with unpaired e-.
Very reactive. Most stable molecules have paired e-.
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More Than Eight Valence Electrons
“Expanded octets” are relatively common.
Low-lying d orbitals can accept extra e-
(only 3rd period and beyond).
Examples
5 bonds (5 e- pairs) around P in PF5
NF5 does not exist
4 bonds and 1 lone-pair (5 e- pairs) around S in SF4
OF4 does not exist.
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More Than Eight Valence Electrons
More Than Eight Valence Electrons
ClF3 7 + 3(7) = 28 val. e-
Make octets on F
24 e- used, 4 remain
[bonds (3 x 2); Lone pairs (3 x 6)]
Add 2 lone pairs to Cl – the 3rd period element
Cl
F
F
F
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Aromatic Compounds
Aromatic compounds contain benzene (or benzene-like) rings:
They typically have strong (often pleasant) odors.
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Isomers of Aromatic Compounds
Naming benzene rings:
p-xylene
m.p.= 13.2 °C
b.p. = 138.4 °C
para
CH3
CH3
CH3
CH3
meta
m-xylene
m.p.= -47.8 °C
b.p. = 139.1 °C
CH3
CH3
ortho
o-xylene
m.p.= -25.2 °C
b.p. = 144.5 °C
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Aromatic Compounds
Systematic names:
1,2 –dimethylbenzene
CH3
CH3
1
2
3
4
5
6
CH3
CH3
1,3 dimethylbenzene
CH3
CH3
1,4 dimethylbenzene
1,2,4–trichlorobenzene
1,3,5–trichlorobenzene
Cl
Cl
Cl
Cl
Cl
Cl
1,2,3–trichlorobenzene
Cl
Cl
Cl
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C
C
C
C
H
H
H
H
H
H
H
H
H
H
C
C
C
H
H
H
C
H
H
H
H
H
H
H
