1. Chemical Bonding I: The Covalent Bond
Chapter 9

2. Classify the following bonds as ionic, polar covalent,
or covalent: The bond in CsCl; the bond in H2S; and
the NN bond in H2NNH2.
Cs – 0.7
Cl – 3.0
3.0 – 0.7 = 2.3
Ionic
H – 2.1
S – 2.5
2.5 – 2.1 = 0.4
Nonpolar covalent
N – 3.0
N – 3.0
3.0 – 3.0 = 0
Covalent
9.5
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3. Lewis structures
A Lewis structure is a representation of covalent bonding in which shared electrons pairs are shown either as lines or as pairs of dots between two atoms, and lone pairs are shown as pairs of dots on individual atoms. O H
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4. Writing Lewis Structures
Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center.
Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge.
Use valence e- determined in (2) to complete octets of atoms bonded to central atom (except hydrogen)
If central atom has less than an octet, form double and triple bonds on central atom as needed.
9.6
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5. 5 + (3 x 7) = 26 valence electrons
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on N and F atoms.
Step 4 - Does central atom have an octet? yes
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons F N
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6. 4 + (3 x 6) + 2 = 24 valence electrons
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e-
4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on O atoms.
Step 4 - Does central atom have an octet? no
Step 5 - Form double bond to give central atom an octet O C
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7. 4 + (3 x 6) + 2 = 24 valence electrons
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e-
4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on O atoms.
Step 4 - Does central atom have an octet? no
Step 5 - Form double bond(s) to give central atom an octet O C O C
9.6
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8. ( ) - - Two possible skeletal structures of formaldehyde (CH2O) H C O
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.
# bonds
formal charge on an atom in a Lewis structure =
total number of valence electrons in the free atom -
total number of nonbonding electrons - 1 2
total number of bonding electrons ( )
The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.
9.7
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9. ( ) - -1 +1 H C O = 1 2 = 4 - 2 - ½ x 6 = -1 = 6 - 2 - ½ x 6 = +1
C – 4 e-
O – 6 e-
2H – 2x1 e-
12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12 H C O
# bonds
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
formal charge on C
= 4 -
2 -
½ x 6 = -1
formal charge on O
= 6 -
2 -
½ x 6 = +1
9.7
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10. ( ) - H C O = 1 2 = 4 - 0 - ½ x 8 = 0 = 6 - 4 - ½ x 4 = 0 C – 4 e-H C O
C – 4 e-
O – 6 e-
2H – 2x1 e-
12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
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
formal charge on C
= 4 -
0 -
½ x 8 = 0
formal charge on O
= 6 -
4 -
½ x 4 = 0
9.7
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11. 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 large formal charges are less plausible than those with small 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.
Which is the most likely Lewis structure for CH2O? H C O -1 +1 H C O
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12. Back to CO2- O C
Figure out which atoms have the charge.
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13. Write Lewis structures for the following:
Draw skeletal structure of compound showing what atoms are bonded to each other. In general, least electronegative element goes in the center.
Hydrogen and fluorine usually occupy the periphery (terminal positions in the Lewis structure). Sometimes the higher symmetry structure is the correct one.
HCN
BF3
CHFO
SnO2
These correct Lewis structures should help you answer 9.23.
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14. 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. O + - O + -
Expect one O-O bond to be longer than the other if ozone really does contain
one double bond and one single bond.
O-O : 148 pm
O=O : 121 pm
Experimental evidence shows that bond oxygen-to-oxygen bonds are equal
in length (128 pm).
9.8
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15. What are the resonance structures of the carbonate (CO32-) ion?
9.8
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16. Exceptions to the Octet Rule
The Incomplete Octet
Be – 2e-
2H – 2x1e-
4e-
BeH2 H Be
B – 3e-
3F – 3x7e-
24e-
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24 F B
BF3
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17. Exceptions to the Octet Rule
Odd-Electron Molecules
Odd-electron molecules are
sometimes called radicals.
Many radicals are highly reactive.
N – 5e-
O – 6e-
11e- NO N O
The Expanded Octet (central atom with principal quantum number n > 2) S F
S – 6e-
6F – 42e-
48e-
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
SF6
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18. Write Lewis structures for the following:
HCN
BF3
CHFO
SnO2
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19. Exceptions to the Octet Rule
The Incomplete Octet
BeH2
BX3
AlX3
Odd-Electron Molecules NO
NO2
The Expanded Octet
(central atom with principal quantum number n > 2)
SF6
PF6
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20. Stability of molecules based on their covalent bond energies…
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21. Single bond < Double bond < Triple bond
The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.
Bond Energy
H2 (g)
H (g) +
DH0 = kJ/mol +
Cl2 (g)
Cl (g)
DH0 = kJ/mol
HCl (g)
H (g) +
Cl (g)
DH0 = kJ/mol
O2 (g)
O (g) +
DH0 = kJ/mol O
N2 (g)
N (g) +
DH0 = kJ/mol N
Bond Energies
Single bond < Double bond < Triple bond
9.10
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22. Average bond energy in polyatomic molecules
H2O (g)
H (g) +
OH (g)
DH0 = 502 kJ/mol
OH (g)
H (g) +
O (g)
DH0 = 427 kJ/mol
Average OH bond energy = 2
= 464 kJ/mol
9.10
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23. Bond energies in thermochemistry
Energy is always required to break a bond.
Energy is always released when a bond is formed.
Whether a reaction is exothermic depends on the relative strengths of the bonds in the product(s) compared to those in the starting material(s).
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24. Bond Energies (BE) and Enthalpy changes in reactions
Imagine a reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.
DH0 = total energy input – total energy released
= SBE(reactants) – SBE(products)
endothermic
exothermic
9.10
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25. H2 (g) + Cl2 (g) HCl (g)
2H2 (g) + O2 (g) H2O (g)
9.10
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26. Use bond energies to calculate the enthalpy change for:
H2 (g) + F2 (g) HF (g)
DH0 = SBE(reactants) – SBE(products)
Type of bonds broken
Number of bonds broken
Bond energy (kJ/mol)
Energy change (kJ/mol) H 1
436.4 F 1
156.9
Type of bonds formed
Number of bonds formed
Bond energy (kJ/mol)
Energy change (kJ/mol) H F 2
568.2
1136.4
DH0 = – = kJ
9.10
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27. Ex. 9.11
Estimate the enthalpy change for the combustion of hydrogen gas.
-469 kJ/mol
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28. 9.50 (a) For the reaction 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)
Predict the enthalpy of reaction from the average bond energies in Table (b) Calculate the enthalpy of reaction from the standard enthalpies of formation (see Appendix 2) of the reactant and product molecules, and compare the result with your answer for (a).
(a) kJ/mol (b) kJ/mol
slightly different b/c (a) used average bond energies
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