Practice Questions Set 11.1 Bonding
1. Determine which ion will form
2. Sodium
8. Nitrogen
3. Magnesium
9. Fluoride
4. Lithium
10. Chloride
5. Aluminum
11. Sulfur
6. Boron
12. Phosphorous
7. Oxygen
13. Bromine
14. Explain why the ion formed by a transition metal cannot be predicted
15. Give the kind of bonding in the following compounds
16. NaCl
21. SO3
17. NaSO4
22. SiCl2
18. Fe2O3
23. SiO2
19. MgCl2
24. CO2
20. Cl2O
25. An element, X, with 6 valence electrons bonds with an element, Y, with 2 valence
electrons. Write the formula, showing the charges of the ionic compound that
forms.
26. An element Q, with 4 valence electrons covalently bonds with an element, R,
with 6 valence electrons. What will the formula of the covalent compound be?
27. Describe the main difference between a covalent and an ionic bond.
28. Describe the relationship between bond order, bond length, and bond strength
29. What is the difference between a pi bond and a sigma bond?
30. SKIP Give the number of sigma and pi bonds in
31. O2
32. N2
33. Based on electronegativity values, which bond is the most polar?
A. B-C
B. C-O
C. N-O
D. O-F
34. Draw the Lewis structures for the following and give the shape and bond angles
and determine if the molecules are polar.
35. F2
41. SiH4
47. SO2
36. O2
42. CO2
48. C2H2
N
SCl
37.
2
43.
2
49. C2H4
50.
38. H2O
CO3244. C2Cl2
51.
NO2-1
39. NH3
45. SiCl4
40. PH3
46. CO2
52. Which of the above molecules have resonance structures?
53. Why is it important to draw resonance structures?
54. Why are Lewis structures considered to be a localized electron model of
molecules?
55. What are the limitations of Lewis structures in showing electron arrangement?
56. Which elements can break the octet rule?
57. SKIP Why can these elements break the octet rule?
58. What is formal charge and how can it be used to evaluate Lewis structures?
59. Give the formal charge for the atoms in
60. SO3
61.
SO3262.
CO3263.
PO4364. Draw the 3-D structures for the following and explain the difference in bond
angle:
65. SiH4
66. PH3
67. Outline the principles of VSEPR theory
68. Why are multiple bonds treated as single bonds when determining geometry?
69. List two factors that determine molecular polarity
70. When the Lewis structure for HCOOCH3 is drawn, how many bonding and how
many lone pairs of electrons are present?
71. The carbon-carbon-carbon bond angle in is closest to CH3CHCH2
A. 180
B. 120
C. 109
D. 90
72.
What molecular shapes are associated with the following electron pairs
around the central atom?
73.
3 bonding pairs and 2 lone pairs
74.
4 bonding pairs and 1 lone pair
75.
3 bonding pairs and 1 lone pair
76.
2 bonding pairs and 2 lone pairs
77.
2 bonding pairs and 3 lone pairs
78.
5 bonding pairs and 1 lone pair
79.
Estimate, using the bond energies in your data booklet the enthalpy
change H° for the conversion of propene to isopropanol:
CH3CH=CH2 + H2O  CH3CH(OH)CH3
80.
When ethanol (CH3CH2OH)burns in air, heat is released. Estimate the
enthalpy of combustion, H° of ethanol vapor from the average bond energies
listed in your data booklet. Use heat of formation data in data booklet to calculate
the same thing. Compare both values to the enthalpy of combustion listed in data
booklet.
Problem Set 11.1 Answers
1. Determine which ion will form
a.
Na+1
b.
Mg+2
c.
Li+1
d.
Al+3
e.
B+3
f.
O-2
g.
N-3
h.
F-1
i.
Cl-1
j.
S-2
k.
P-3
l.
Br-1
2. Explain why the ion formation by a transition metal cannot be predicted
Transition metals have more than one possible oxidation state
3. Give the kind of bonding in the following comounds
a. Ionic
b. Ionic (between Na and SO42-) and covalent (between S and O)
c. Ionic because between metal and non-metal. Some may classify as
polar covalent (metals with high oxidation states form polar bonds when
bonding with non-metals)
d. Ionic
e. Non-polar covalent (two non-metals with similar electronegativities)
f. polar covalent (bond is polar b/c two non-metals with different
electronegativities but molecule is non-polar)
g. polar covalent
h. polar covalent (however, molecule is non-polar)
i. polar covalent (however, molecule is non-polar)
4.
valence of 6 gives charge of -2 and valence of 2 gets a charge of +2 so formula
is Y2+X25. 4 valence electrons (needs 4 bonds or two double bonds) and 6 valence
electrons (needs two bonds or one double bond) so must be QR2.
a. Can’t be QR because R would have to donate four electrons to Q which
would give it too many electrons
6. Covalent bonds involve share electrons, ionic bonds involve a transfer of
electrons
7. bond order is the number of bonds (single = order of 1, double = order of 2). The
higher the order, the stronger and shorter the bond.
8. A sigma bond occurs when a single lobe of an orbital overlaps with a single lobe
of another orbital. A pi bond results when more than one lobe of an orbital
overlaps with more than one lobe of another orbital. All multiple bonds have one
sigma bond and all single bonds are sigma bonds.
9. O2 has a double bond so it has one sigma bond and one pi bond
a. N2 has a triple bond so it has one sigma bond and two pi bonds
10. C-O If you have a data booklet you can look up electronegativity values and
find the difference. An easier way, though is to see how far apart the elements
are on the periodic table
a. B-C EN = 0.5 (next to each other)
b. C-O EN = 1.0 (two away from each other)
c. N-O EN = 0.5 (next to each other)
d. O-F EN = 0.5 (next to each other)
11.
Molecule
Lewis
Shape
Bond
Angle
Polar?
Resonance
1s
3 lp on each
1d
2 lp on each
1t
1 lp on each
2s
2 lp on O
3s
1 lp on N
3s
1 lp on P
Linear
180
No
No
Linear
180
No
No
Linear
180
No
No
Bent
104.5
Yes
No
Trigonal
pyramidal
Trigonal
pyramidal
107
Yes
No
Yes
No
4s
0 lp
2d
2 lp on each O
2s
3 lp on each Cl
2 lp on S
1 d C=C
4 s C-Cl
3 lp on each Cl
4s
3 lp on each Cl
2d
1 lp on S
2 lp on each O
1t
2 s C-H
1 d C=C
4 s C-H
Tetrahedral
Predict 107
(but
actually 94)
109.5
No
No
Linear
180
No
No
Bent
104.5
No
No
Trigonal
planar
120
No
No
Tetrahedral
109
No
No
Bent
<120
Yes
No
Linear
180
No
No
Trigonal
planar
120
No
No
1 d w/ 2 lp on
O
2 s w/ 3 lp on
O’s
1 d w/ 2 lp on
O
1 s w/ 3 lp on O
1 lp on N
Trigonal
planar
120
Yes
Bent
<120
No, but it is an
ion so it has
an overall
charge
No, but it is an
ion so it has
an overall
charge
s = single bond
d = double
t = triple
lp = lone pairs
F2
O2
N2
H 2O
NH3
PH3
SiH4
CO2
SCl2
C2Cl2
SiCl4
SO2
C2H2
C2H4
CO32-
NO2-1
Yes
12. answered in table
13. It is important to draw resonance structures because only showing one structure is
misleading – it can lead to thinking that there are two kinds of bonds (single and
double) in the molecule when really there is only one. Lewis structures are based
14.
15.
16.
17.
18.
19.
on localized e- model but some molecules have electrons that are delocalized.
Resonance structures are meant to show the delocalization
Localized e- model means electrons belong to a specific atom or pair of atoms
whereas a delocalized model means electrons are shared by more than two
atoms
Lewis structures cannot show delocalization very well
Boron can break the octet rule – it can have only 6 electrons around it
a. Elements in the third row of the periodic table can also break the octet
rule
because they have d orbitals to hold the extra eFormal charge is used to compare the number of valence electrons an atom
donates with the number of electrons it “owns” when bonded. If it has a deficit it
will have a positive charge b/c it has more protons than electrons. If it has more
than it started with, it will have a negative charge. Formal charge is useful for
atoms that can break the octet rule. How do you know when to break the octet
rule and when not to? Consider the formal charge and find a way to minimize
the formal charge on each atom. The structure that minimizes formal charge is
the best structure (although, this must always be verified by experiment).
Formal charge
a. SO3 The structure of SO3 that follows the octet rule has single bonds
between two of the oxygen atoms and sulfur and a double bond
between another oxygen atom and sulfur. In this case the formal charges
are as follows:
i. S  +2
ii. O (single bond)  -1
iii. O (double bond)  0
iv. All the formal charges add up to zero (there are two single
bonded O’s) so that’s good but another structure might be better
b.
SO32i. The best structure consists of one double bond, and two single
bonds plus a lone pair on the sulfur (octet rule is broken).
ii. The structure that does not break the octet rule, has all single
bonds. Formal charges will not be minimized
Octet Rule Broken
Octet Rule Followed
S0
S  +1
O (single bond)  -1
O  -1
O (double bond)  0
All the formal charges add up to -2
All formal charges add up to -2
c. CO32- The structure consists of one double bond, and two single bonds
i. C  0
ii. O (single bond)  -1
iii. O (double bond)  0
iv. All the formal charges add up to -2
d. PO43- The structure consists of one double bond, and three single bonds
(octet rule is broken)
Octet Rule Broken
Octet Rule Followed
P0
P  +1
O (single bond)  -1
O  -1
O (double bond)  0
All the formal charges add up to -3
All formal charges add up to -3
20. SiH4 = tetrahedral and PH3 = trigonal pyramidal. The bond angles are different
even though they are both based on the tetrahedral electron arrangement
because the PH3 molecule has a lone pair that repels bonding pairs more.
21. Principles of VSEPR
a. Valence electron pairs repel each other – lone pairs repel more than
bonding pairs
b. To determine the geometry, the number of bonding and lone pairs
around the central atom are considered
c. The location of the nuclei of each atom determine the shape
22. Because it’s the location of the nuclei that determine the shape and double and
triple bonds do not increase the number of nuclei
23. polar bonds (difference in electronegativity) and symmetry of the molecule
24. 4 lone pairs (2 on each oxygen) 8 shared pairs
25. 120 degrees
26. shapes:
a. T-shaped
b. See-saw
c. Trigonal pyramidal
d. Bent
e. Linear
f. Square pyramidal
27. [2 (O-H) + 6(C-H) + (C=C) + (C-C)] – [(O-H) + 2(C-C) + 7 (C-H) + (C-O)] =
(O-H) – (C-H) + (C=C) – (C-C) -(C-O) = 463 – 412 + 612 – 348 -360 = -45 kJ
28. Combustion: CH3CH2OH + 3 O2 2 CO2 + 3 H2O
Bond enthalpy [5(C-H) + (C-C) + (C-O) + (O-H) + 3(O=O)] - [4 (C=O) + 6 (O-H)]
5(412) + (348) + (360) + (463) + 3(496)] - [4 (799*) + 6 (463)] = -1255 kJ
* this is different that data booklet b/c CO2 has higher enthalpy than other C=O
bonds
Heats of Formation: (had to look in book for CO2 and water)
(water is a gas) (2 CO2 + 3 H2O) – (ethanol) = 2(-393.5) + 3(-242) – (-278) = -1235 kJ
(water is a liquid) (2 CO2 + 3 H2O) – (ethanol) = 2(-393.5) + 3(-286) – (-278) = -1367 kJ
Heat of combustion = -1371 kJ
If all are rounded to 2 sig figs, they are about the same. Heat of formation and heat of
combustion are closest. Remember that bond enthalpies are average and will therefore
lead to less accuracy.
28. The diagrams shows that there is an ideal bond length such that both atoms are
at minimum energy. When atoms get closer together there is an increase in
potential energy b/c nuclei repel. As atoms get farther apart there is also an
increase in energy because neither atom is stable on its own.