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Understanding Covalent Bonds: Bond Length, Polarity, and Lewis Structures, Summaries of Chemistry

The concept of covalent bonds, discussing the factors determining bond length, the nature of different types of covalent bonds, and the use of lewis structures to represent them. Topics include polar covalent, ionic, and nonpolar bonds, as well as the electronegativity difference and bond dipoles.

Typology: Summaries

2021/2022

Uploaded on 09/12/2022

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Chapter 5
The Covalent Bond
5-1
1. What two opposing forces dictate the bond length? (Why do bonds form and what keeps the bonds from getting
any shorter?)
The bond length is the distance at which the repulsion of the two nuclei equals the attraction of the valence
electrons on one atom and the nucleus of the other.
3. List the following bonds in order of increasing bond length: H-Cl, H-Br, H-O.
H-O < H-Cl < H-Br. The order of increasing atom size.
5. Use electronegativities to describe the nature (purely covalent, mostly covalent, polar covalent, or ionic) of the
following bonds.
a) P-Cl Δχ = 3.0 – 2.1 = 0.9 polar covalent b) K-O Δχ = 3.5 – 0.8 = 2.7 ionic
c) N-H Δχ = 3.0 – 2.1 = 0.9 polar covalent d) Tl-Br Δχ = 2.8 – 1.7 = 1.1 polar covalent
7. Name the following compounds:
a) S2Cl2 disulfur dichloride b) CCl4 carbon tetrachloride
c) PCl5 phosphorus pentachloride d) HCl hydrogen chloride
9. Write formulas for each of the following compounds:
a) dinitrogen tetroxide N2O4 b) nitrogen monoxide NO c) dinitrogen pentoxide N2O5
11. Consider the U-V, W-X, and Y-Z bonds. The valence orbital diagrams for the orbitals involved in the bonds are
shown in the margin. Use an arrow to show the direction of the bond dipole in the polar bonds or indicate “not
polar” for the nonpolar bond(s). Rank the bonds in order of increasing polarity.
The bond dipole points toward the more electronegative atom in the bond, which is the atom with the lower
energy orbital. The magnitude of the dipole increases as the electronegativity difference (orbital energy
difference) increases. The W-X bond is not polar because the orbital energies of W and X are identical, and
the Y-Z bond in more polar than the U-V bond because the energy separation of the bonding orbitals is
greater. W-X < U-V < Y-Z
13. Use an arrow to indicate the bond dipole direction in each of the following bonds:
a) S-O b) C-H c) O-H d) C-O
→ ← Note the C-H bond is essentially nonpolar.
15. Use only the position of the atoms on the periodic table to list the following bonds in order of decreasing polarity:
a) S-O, Se-O, As-O As-O > Se-O > S-O Increasing electronegativity of non-oxygen atom
b) F-F, H-F, N-F H-F > N-F > F-F Increasing electronegativity of non-fluorine atom
c) P-Cl, Sb-Cl, Sn-Cl Sn-Cl > Sb-Cl > P-Cl Increasing electronegativity of non-chlorine atom
17. For each of the species listed below, indicate the number of electrons required to give each atom an octet or duet
(ER), the number of valence electrons (VE), and the number of shared pairs (SP) in the Lewis structure.
a) N2O4 b) CH4O c) HBrO2 d) S2O82-
ER 6(8) = 48 2(8) + 4(2) = 24 3(8) + 2 = 26 10(8) = 80
VE 2(5) + 4(6) = 34 4 + 4(1) + 6 = 14 1 + 7 + 2(6) = 20 10(6) + 2 = 62
SP 1/2 (48 - 34) = 7 1/2 (24 - 14) = 5 1/2 (26 - 20) = 3 1/2 (80 - 62) = 9
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Chapter 5

The Covalent Bond

1. What two opposing forces dictate the bond length? (Why do bonds form and what keeps the bonds from getting

any shorter?)

The bond length is the distance at which the repulsion of the two nuclei equals the attraction of the valence electrons on one atom and the nucleus of the other.

3. List the following bonds in order of increasing bond length: H-Cl, H-Br, H-O.

H-O < H-Cl < H-Br. The order of increasing atom size.

5. Use electronegativities to describe the nature (purely covalent, mostly covalent, polar covalent, or ionic) of the

following bonds.

a) P-Cl Δχ = 3.0 – 2.1 = 0.9 polar covalent b) K-O Δχ = 3.5 – 0.8 = 2.7 ionic

c) N-H Δχ = 3.0 – 2.1 = 0.9 polar covalent d) Tl-Br Δχ = 2.8 – 1.7 = 1.1 polar covalent

7. Name the following compounds:

a) S 2 Cl 2 disulfur dichloride b) CCl 4 carbon tetrachloride

c) PCl 5 phosphorus pentachloride d) HCl hydrogen chloride

9. Write formulas for each of the following compounds:

a) dinitrogen tetroxide N 2 O 4 b) nitrogen monoxide NO c) dinitrogen pentoxide N 2 O 5

11. Consider the U-V, W-X, and Y-Z bonds. The valence orbital diagrams for the orbitals involved in the bonds are

shown in the margin. Use an arrow to show the direction of the bond dipole in the polar bonds or indicate “not

polar” for the nonpolar bond(s). Rank the bonds in order of increasing polarity.

The bond dipole points toward the more electronegative atom in the bond, which is the atom with the lower energy orbital. The magnitude of the dipole increases as the electronegativity difference (orbital energy difference) increases. The W-X bond is not polar because the orbital energies of W and X are identical, and the Y-Z bond in more polar than the U-V bond because the energy separation of the bonding orbitals is

greater. W-X < U- V < Y- Z

13. Use an arrow to indicate the bond dipole direction in each of the following bonds:

a) S-O b) C-H c) O-H d) C-O

← → Note the C-H bond is essentially nonpolar.

15. Use only the position of the atoms on the periodic table to list the following bonds in order of decreasing polarity:

a) S-O, Se-O, As-O As-O > Se-O > S-O Increasing electronegativity of non-oxygen atom

b) F-F, H-F, N-F H-F > N-F > F-F Increasing electronegativity of non-fluorine atom

c) P-Cl, Sb-Cl, Sn-Cl Sn-Cl > Sb-Cl > P-Cl Increasing electronegativity of non-chlorine atom

17. For each of the species listed below, indicate the number of electrons required to give each atom an octet or duet

(ER), the number of valence electrons (VE), and the number of shared pairs (SP) in the Lewis structure.

a) N 2 O 4 b) CH 4 O c) HBrO 2 d) S 2 O 8

2-

ER 6(8) = 48 2(8) + 4(2) = 24 3(8) + 2 = 26 10(8) = 80 VE 2(5) + 4(6) = 34 4 + 4(1) + 6 = 14 1 + 7 + 2(6) = 20 10(6) + 2 = 62 SP 1 / 2 (48 - 34) = 7 1 / 2 (24 - 14) = 5 1 / 2 (26 - 20) = 3 1 / 2 (80 - 62) = 9

H O C O H H C

O

O H

A B

O N

O

N

O

O

19. Draw Lewis structures for each of the following organic compounds and indicate all nonzero formal charges.

a) C 2 H 2 b) C 3 H 4 c) C 3 H 6 d) COF 2

H C C H

H C

H

C C

H

H

H C C C

H

H

H

H C

H

H

C

H

C

H

H F C

O

or F

ER=20 VE=10 SP=5 ER=32 VE=16 SP=8 ER=36 VE=18 SP=9 ER=32 VE=24 SP=

21. Draw Lewis structures for each of the following ions and indicate all nonzero formal charges.

a) NO 3

1-

b) NO

1+

c) N 3

1-

d) NO 2

1-

O N

O

O N^ O

N N N

N N N

or O^ N^ O

ER=32 VE=24 SP=4 ER=16 VE=10 SP=3 ER=24 VE=16 SP=4 ER=24 VE=18 SP=

23. For which ions in Exercises 20 and 21 are more than one resonance form important?

CO 3

2- , CHO (^2) 1- , NO (^3) 1- , and NO (^2) 1- all have more than one important resonance form. Although two resonance structures can be drawn for N 3

1- , the one that places a -2 formal charge on a nitrogen is not expected to be important.

25. Which of the Lewis structures of formic acid is preferred? Justify your answer.

Structure A involves charge separation, but structure B does not. Structure B is preferred.

27. In the Lewis structure of N 2 O 4 , each nitrogen is bound to one nitrogen and two oxygens. What are the formal

charges and oxidation states of the nitrogen atoms?

The formal charges are shown in the Lewis structure. Oxygen is more electronegative than nitrogen so all bonding N-O electrons are assigned to oxygen when determining oxidation states. The electron pair in the N-N bond is split between the two nitrogen atoms. Nitrogen is in group 5A, but only the one electron in the N-N bond is assigned to in determining the oxidation state: OXN = 5 - 1= +4. The nitrogen atoms are each in the +4 oxidation state. Each oxygen is assigned eight electrons for an oxidation state of -2. Note that the same results could have been obtained by applying the rules given in Section 4.3.

29. List the following in order of increasing carbon-oxygen bond lengths and bond energies:

O C

O

O H C

O

H C O H C

H

H

O

H A B C D

CO bond order

4 / 3 2 3 1 Bond length increases as the bond order decreases: C < B < A < D Bond energy increases as the bond order increases: D < A < B < C