Chemistry 121
Mines, Spring 2010
Answer Key, Problem Set 11 – Full
1. NT0 / MP; 2. MP; 3. MP; 4. NT1; 5. MP; 6. NT2; 7. NT3; 8. MP; 9. NT4; 10. NT5; 11. MP; 12. MP; 13. MP; 14. NT6;
15. 9.120 [9.126 in Mast]; 16. MP; 17. MP; 18. NT7; 19. NT8; 20. NT9; 21. NT10; 22. NT11; 23. NT12; 24. MP; 25. NT13; 26. NT14;
27. NT15. 28. 10.60 [10.60 in Mast]; 29. MP; 30. MP; 31. 10.59 [10.59 in Mast]; 32. NT16; 33. NT17; 34. NT18; 35. NT19
-------------------------------------------------------
Equivalent Resonance Structures
1. NT0 / MP Draw the Lewis structure for the nitrite ion, NO2-. Include resonance structures.
v = 5 + 2(6) + 1 = 18 e- (the “+ 1” is because of the -1 charge)
O
(2)
N
O
(2)
……a lone pair from one of the
O’s is made into a bonding pair (to
make a double bond)
O
N
(3)
O
(3)
O
(1)
N
O
after all 18 electrons are put into
the structure, the central N
doesn’t have an octet, so……
But the lone pair could have been
donated by the other O atom. That would
lead to a second, equally “valid” Lewis
structure. As such, the two structures are
called “resonance structures” of NO2-, and
since they are the same in every possible
way (except that it is a different O atom
making the db), these two structures are
NOTE: This species is isoelectronic with SO2, but unlike SO2, which can have two double bonds
(and 10 electrons around the S), NO2- cannot have that third resonance structure because N
does not have d orbitals and so it cannot have more than 8 electrons around it.
2. MP. No answer to this problem in the key at this time.
Assessing Relative Importance (Weight) of Nonequivalent Resonance Structures by Assigning Formal Charges
3. MP. No answer to this problem in the key at this time. But see my PowerPoint, as well as the videos,
text, and graphics in this MP problem as well as in the MP that is Item 5 on this problem set. I urge you
not to memorize the formula given in Tro, but think of comparing how many valence electrons the atom
in the given LDS “has” to the amount that kind of atom would have if it were neutral. It makes sense! 
4. NT1. 9.67.
How important is this resonance structure [the one with a triple bond to the left O and a single bond to the
right O] to the overall structure of carbon dioxide? Explain.
Answer: Not very important. As shown in class (and below), the formal charges for all the atoms in the
three resonance structures for CO2 are as shown below:
0
0
0
O
C
O
e ‘s in this structure*
6
6
4
6
6
6
6
5
FORMAL CHARGE:
0
0
0
+1
“BEST”
v e-‘s if neutral
-
1
O
0
1
1
0
1
C
O
O
C
O
4
4
6
7
6
7
4
4
6
5
0
-1
-1
0
+1
*Where both electrons in a lone pair are assigned to the given atom, but only one electron from each covalent bond is assigned to
the given atom.
Resonance structures are a way to patch up an inadequate bonding model. When more than one (at
least initially, “reasonable”) LDS can be drawn for a species, the actual bonding (i.e., the actual
PS11-1
Answer Key, Problem Set 11
“structure”) is assumed to be a “blending” of all the resonance structures. However, when the
resonance structures are not equivalent, each one is (informally) “weighted” before the “blending”.
Those that are considered “better” are given more weight and are considered to better represent the
actual structure. When assessing “goodness” using formal charges, the ones that have as many
atoms with a formal charge of zero as possible are considered to be the “best” (all other things being
equal, of course). By that criterion, the best structure for CO2 is the one with two double bonds (all
three atoms with FC = 0). The LDS asked about in this problem not only has two atoms with nonzero formal charges, but also has a formal charge of +1 on an O atom, which is particular unfavorable
since O is one of the most electronegative atoms (with a value of 3.5 out of 4.0). Thus, this LDS is
not considered to be very important in representing the actual structure of CO2.
5. MP. No answer to this problem in the key at this time. But see the comments for problem 3.
6. NT2. 9.66.
H
Use formal charges to determine which Lewis structure is better:
H
The one on the right is better
because the structure has more
atoms with FC’s that are zero
+2
2
S
C
H
whereas the one on the left has
more atoms with large-magnitude
formal charges (+2 and -2).
H
S0
C0
H
H
H
H
S
C
S
C
-
# of v e ‘s a neutral isolated atom has:
6
4
6
4
# of v e-‘s the atom "has" in this
structure*
4
6
6
4
+2
-2
0
0
FORMAL CHARGE:
*Where both electrons in a lone pair are assigned to the given atom, but only one electron from each covalent bond is assigned to
the given atom.
In the left structure, the S has no lone pairs and makes four bonds, thus it is assigned four electrons
“formally”. The C has two lone pairs (4 e-‘s) and makes two bonds (2 more e-‘s) for a total of six
formally. In the right structure, the S is the one that has two lone pairs and makes two bonds, so it
gets six electrons formally. The C makes four bonds (with no lone pairs) and thus has four electrons
formally.
Expanding an Octet by Making Multiple Bonds to Lower Formal Charges in an LDS
7. NT3. 9.72(bd).
Write a Lewis structure for each ion. Include resonance structures if necessary and assign formal
charges to all atoms. If necessary, expand the octet on the central atom to lower the number of atoms with non-zero
formal charges (the authors say “lower formal charge”, but they mean what I wrote).
(b) HSO4-
v = 1(6) + 4(6) + 1(1) + 1 = 32 e-
1
O

1
O
2
S
0
O
0
S
0
O
0
O
H
e ‘s in this structure*
6
4
1
1
FORMAL CHARGE:
-1
0
+2
0

0
O
S
6
6
-
1
O
O
6
7
v e-‘s if neutral
H
O
1
O
(terminal) (center)
0
H
0
PS11-2
Answer Key, Problem Set 11
O
O
(term, sb) (term, db)
O
(center)
S
H
e-‘s in this structure*
6
7
6
6
6
6
6
6
1
1
FORMAL CHARGE:
-1
0
0
0
0
v e-‘s if neutral
The initial LDS has all octets where possible, but the formal charges are non-zero for four atoms, and
one of those is a +2. Making a double bons “inward” from an outer O atom increases the FC by one
on the O and lowers it by one on the (inner) S. Thus, making two double bonds is “best”, yielding all
zero FC’s except for one O (which is the best you can do since the overall charge on the ion is -1).
NOTE: There are two equivalent resonance structures to the 2nd one above, so the four structures
would be:

O
O
S
O

O
O
H
S
O
O

O
H
O
S
O
O

O
O
H
S
O
O
O
Technically, there are three additional resonance structures having just one double bond, but I am
not going to show those here. If you included those, then more power to you! 
(d) BrO2v = 1(7) + 2(6) + 1 = 20 e+1
1
O
1
0
0
O
Br
O
Br

O
e ‘s in this structure*
Br
7
6
FORMAL CHARGE:
-1
+1
v e ‘s if neutral

1
O
6
7
-
-
0
0
-1
O
Br
O

v e ‘s if neutral
-
e ‘s in this structure*
-
O (db)
6
6
O (sb)
6
7
Br
7
7
0
-1
0
FORMAL CHARGE:
The first LDS has all octets, but no atom has a zero formal charge. Making one double bond “inward”
from either (outer) O raises the FC on O by one and lowers the FC on Br by one. That creates the two
equivalent LDS’s below, each of which has only one atom with a non-zero formal charge.
Predicting Relative Strength and Length of Bonds Using Bond Order Concept
8. MP. No answer to this problem in the key at this time. Read the text info in the introduction to the
problem in Mastering!
9. NT4. 9.76 Which of these compounds has the stronger N-N bond?
The shorter N-N bond?
H2NNH2 or HNNH
Answer: HNNH has the stronger and shorter N-N bond because it has a double bond rather than a
single bond (see below). The greater the bond order, the stronger and shorter the bond.
H2NNH2
HNNH
v = 4(1) + 2(5) = 14 eH
N
N
H
H
H
v = 2(1) + 2(5) = 12 e-
H
PS11-3
N
N
H
H
Answer Key, Problem Set 11
(b) NO+ or NO [Weaker bond? Longer bond?] (Hint: Draw the LDS’s);
Answer: NO has the weaker and longer (N-O) bond because it has a double bond whereas NO+ has a
triple bond (see below). The smaller the bond order, the weaker and longer the bond.
v = 5 + 6 - 1 = 10 eN
O

v = 5 + 6 + 1 = 12 eN
O

10. NT5.
Based on the three Lewis Structures of OCN- shown in Example 9.8 (p.382) and the data in Table 9.4 (p. 390),
would you expect the C-N bond length to be:
(a) greater than 147 pm,
(b) between 128 pm and 147 pm,
(c) between 116 pm and 128 pm, or
(d) less than 116 pm.
Give your reasoning.
Answer: (c), between 116 pm and 128 pm, because the bond order is expected to be between two and
three, and so the bond length is expected to be between the length of a C-N double bond and a C-N
triple bond (128 pm and 116 pm, respectively [Table 9.4]).
The best Lewis structure is the one with the triple bond, but the second best (not all that far behind) is
the one with the double bond (see Example 9.8). That second structure should contribute at least a
little bit, making one predict that the bond order should be (at least a little bit) less than three. The
third Lewis structure is so bad (because of the -2 formal charge) that one can consider its contribution
essentially negligible (it is the only one with a C-N single bond [bond length 147 pm]).
Using Bond Energies to Calculate an Estimate of H of a Chemical Reaction Equation
11. MP. No answer to this problem in the key at this time.
12. MP. No answer to this problem in the key at this time.
13. MP. No answer to this problem in the key at this time.
PS11-4
Answer Key, Problem Set 11
14. NT6. 9.77.
Hydrogenation reactions are used to add hydrogen across double bonds in hydrocarbons and other
organic compounds. Use average bond energies to calculate [an estimate for] Hrxn for the hydrogenation
reaction. (i.e., the H for the thermochemical equation below).
H2CCH2 + H2 → H3CCH3
Answer: -128 kJ/mol
H = H1 + H2 = 1047 +(-1175) = -128 kJ
Reasoning:
H
C
C
H
H
H
H
H
H
(1)
(1) Break one C=C (611 kJ/mol)
and one H-H (436 kJ/mol)
(2)
H
= 611 + 436 = 1047 kJ
C
C
H
H
H
H
H
H
H
C
C
H
H
(2) Make one C-C (347 kJ/mol)
and two C-H (414 kJ/mol)
= -347 + 2(-414) = -1175 kJ
NOTE: If you do not want to try to figure out the “net” bonds broken and made, you can always just
break every bond in every reactant molecule and then make every bond in every product molecule.
For this problem, that approach would look like:
Break:
Make:
4 C-H (414)
6 C-H (414)
1 C=C (611)
1 C-C (347)
1 H-H (436)
4(414) + 1(611) + 1(436) = +2703 kJ

6(-414) + 1(-347) = -2831 kJ
H = 2703 + (-2831) = -128 kJ (as before, of course)
Relating Bond Strengths to Endo- or Exothermicity of a Chemical Reaction
15. 9.120 [9.126 in Mast].
Which statement is (generally) true of an endothermic reaction?
(a) Strong bonds break and weak bonds form.
(b) Weak bonds break and strong bonds form.
(c) The bonds that break and those that form are of approximately the same strength.
Answer: (a) Breaking bonds is “energy absorbing” and making them is “energy releasing”, so if the
bonds that break are the strong ones, then more energy will be used to break those (strong) bonds
than the amount of energy that will be released upon forming the new (weak) bonds.
Note: The savvy student will recognize that this need not always be the case (which is why I added the word “generally” up
above) because the amount of bonds made and broken need not be the same (i.e. the argument above implicitly
assumes that the number of bonds made equals the number of bonds broken). If more weak bonds are made than
the amount of (fewer) strong bonds broken, it is possible that the reaction could be exothermic. In fact, the prior
problem in this set (#6, 9.77) is a great example of this!! Both of the bonds that were broken were stronger than the
bonds made, but only two of them were broken and three of the weaker bonds were made. The net result was an
exothermic process!
Assessing Relative Bond Polarity (or Nonpolarity) Using Electronegativities of Atoms, and Representing
Bond Polarity in Structures
16. MP. No answer to this problem in the key at this time.
17. MP. No answer to this problem in the key at this time.
PS11-5
H
Answer Key, Problem Set 11
18. NT7. 9.56.
Determine whether a bond between each pair of atoms would be pure (nonpolar) covalent, polar covalent,
or ionic. Add Part II: For each polar covalent bond, indicate which atom would be partially negative.
(a) C and N
Polar covalent. EN = 3.0 – 2.5 = 0.5
(b) N and S
Polar covalent EN = 3.0 – 2.5 = 0.5
(c) K and F
Ionic
(d) N and N
Nonpolar covalent EN = 3.0 – 3.0 = 0
(e) C and H
Nonpolar covalent
EN values from Figure 9.8:
K(0.8), H(2.1); C(2.5); S(2.5); N(3.0); F(4.0)
EN = 4.0 –0.8 = 3.2
NOTE: You are required to know the trends
in EN values so that if a table is not
provided, you can still predict relative
polarities of bonds!!
EN = 2.5 – 2.1 = 0.4
19. NT8. 9.96
Draw a Lewis structure for urea, H2NCONH2, one of the compounds responsible for the smell of urine. (The
central carbon atom is bonded to both nitrogen atoms and to the oxygen atom.) Does urea contain polar bonds? Which
bond in urea is most polar? Add: Add arrows (with a “+” sign on the back end) to your LDS to indicate the polarity of each
polar bond.
Answers: See right and below. Yes, it has polar bonds.
In fact every bond is polar! (Below, the more
electronegative atom is listed first in each case.)
O-C: EN = 3.5 – 2.5 = 1.0 ← most polar
O
H
C
N
N
H
N-H: EN = 3.0 – 2.1 = 0.9
H
N-C: EN = 3.0 – 2.5 = 0.5
H
Using VSEPR to Predict ECG’s, AG’s (MG’s), and Bond Angles in Species with One or More Centers
20. NT9. 10.35(abd).
(i) Determine the electron geometry (electron-cloud geometry), molecular geometry (atom
geometry), and idealized bond angles for each molecule. In which cases do you expect deviations from the idealized
bond angle? (ii) Add: Sketch each one with wedges and dashes (and atoms and lone pairs, where applicable) as done in
the lab handout.
(a) PF3
v = 5 + 3(7) = 26 e-
F
P
F
P
F
tetrahedral, 109.5°
1, trigonal pyramidal
ECG & idealized bond angles
# lone pairs & AG
(b) SBr2
v = 6 + 2(7) = 20 e-
F
In these sketches, lone pairs on
terminal atoms are not shown—
just the central (interior) atoms.
4 (1 lone pair, 3 single bonds)
# electron groups (clouds)
Deviations from idealized
angle?
F
F
Yes. A bit less. Lone pair “compresses”
Br
S
Br
P
Br
# electron groups (clouds)
ECG & idealized bond angles
# lone pairs & AG
Deviations from idealized
angle?
Br
4 (2 lone pairs, 2 single bonds)
tetrahedral, 109.5°
2, bent
Yes. A bit less. Lone pairs “compress”
v = 4 + 2(6) = 16 e-
PS11-6
Answer Key, Problem Set 11
(d) CS2
S
C
C
S
1 (0 lone pairs, 2 double bonds)
# electron groups (clouds)
linear, 180°
0, linear
ECG & idealized bond angles
# lone pairs & AG
Deviations from idealized
angle?
(e) BrF5
S
S
No. No lone pairs (& all bonds “same”)
v = 7 + 5(7) = 42 e-
F
F
F
F
F
Br
Br
F
F
F
F
F
6 (1 lone pairs, 5 single bonds)
# electron groups (clouds)
octahedral, 90 (& 180°)
1, square pyramidal
ECG & idealized bond angles
# lone pairs & AG
Deviations from idealized
angle?
Yes. A bit less. Lone pair “compresses”
NOTE: You are responsible for all of the VSEPR atom geometries even though I could not “fit” all of them
into this problem set. Consider your lab handout to be “fair game” for the exam (and use it and your text
for further practice).
21. NT10. 10.38.
Which species has the smaller bond angle, ClO4- or ClO3-? Explain.
Answer: ClO3-. Both Lewis structures have 4 electron clouds around the central Cl, but ClO3- has a
lone pair which “pushes harder” on the bonding pairs, making the bond angles a bit less than
the ideal angle of 109.5° predicted for ClO4-, which has no lone pairs. See below.
ClO4-
-
O
v = 7 + 4(6) + 1 = 32 e-
ClO3v = 7 + 3(6) + 1 = 26 e-
O
Cl
O
Cl
O
O
O
O
22. NT11. 10.42(bc) Determine the molecular (atom) geometry about each interior atom (center) and make a sketch of the
molecule. NOTE: Use the LDS’s for these that you drew in problem #4! Add: State the HNN bond angle in each.
H2NNH2
H
N
N
H
H
HNNH
H
Each N: 4 clouds  tetrahedral ECG, 109.5°
idealized bond angles.
+ 1 lone pair  AG = trigonal pyramidal &
HNN angle is a bit less than 109°
PS11-7
H
N
N
H
Each N: 3 clouds  trigonal planar ECG, 120°
idealized bond angles.
+ 1 lone pair  AG = bent & HNN angle is
a bit less than 120°
-
Answer Key, Problem Set 11
23. NT12. 10.43 & 10.44.
Each ball-and-stick model shows the electron and molecular geometry of a generic molecule.
Explain what is wrong with each molecular geometry and provide the correct molecular geometry, given the number of
lone pairs and bonding groups on the central atom.
<Sketches from this problem not shown in this key at this time>
10.43: (a) With four electron clouds (one of them a lone pair), the ECG would be tetrahedral, not
“trigonal planar bonding pairs with the lone pair at 90° to all of them” as is shown. Basically,
the lone pair would push those three bonding pairs “below” the central atom to make the
tetrahedral ECG, opening up those 90° angles to about 109°. The correct AG would be
trigonal pyramidal.
(b) With five electron clouds (one of them a lone pair), the ECG would be trigonal bipyramidal,
making the four bonded atoms fall into two categories: axial and equatorial. This model
shows them all “equivalent” (without any three atoms being linear). The correct AG would
be see-saw.
(c) With six electron clouds, the ECG would be octahedral as shown. However, two lone pairs
in this ECG (where all positions are “equivalent”) will get as far apart from each other as
possible (180°) making the AG square planar. They will not be at 90° from each other as
shown (in a see-saw-like AG).
10.44
(a) Four electron clouds (even if two of them are lone pairs) will form a tetrahedral ECG. This
model shows the two bonding clouds at 180° and each lone pair at 90° to the bonding
clouds (forming a linear AG). The correct AG would be bent (with bond angles a bit less
than 109°).
(b) Five electron clouds (three of which are lone pairs) would form a trigonal bipyramidal ECG.
The model shows all the clouds in the same plane (I think), with the AG being bent (for
sure). In a trigonal bipyramidal ECG, three lone pairs would occupy the equatorial positions
(where there are only two 90° interactions instead of three), making the (correct) AG linear.
(c) This model shows five atoms around a central one in a “perfect” AG of trigonal bipyramidal,
except with a lone pair (sixth electron cloud) at 180° to one of the equatorial bonding clouds.
Six clouds would actually form an octahedral ECG, and with one of them a lone pair, the
correct AG would be square pyramidal.
Valence Bond Theory (VBT): Concept
24. MP. No answer to this problem in the key at this time.
25. NT13. 10.54.
The valence electron configurations of several atoms are shown below. How many bonds can each
atom make without hybridization?
Answers: (a) one bond; (b) three bonds; (c) two bonds [in each case, one bond for each unpaired e-]
Reasoning: In VBT, an atom forms a covalent bond with another by “donating” an unpaired electron (in
an orbital, obviously). Without hybridization, you must simply look at the ground-state electron
configurations (really, the orbital diagrams) and see how many unpaired electrons are present to
predict how many bonds it could form.
(a) B:
(b) N:
2s
2p
(c) O:
2s
PS11-8
2p
2s
2p
Answer Key, Problem Set 11
26. NT14. 10.56.
Write orbital diagrams to represent the electron configurations—without hybridization—for all the atoms
in SF2. Circle the electrons involved in bonding. Draw a three-dimensional sketch of the molecule and show
orbital overlap. What bond angle do you expect from the unhybridized orbitals? How well does valence bond
theory agree with the experimentally measured bond angle of 98.2°?
v = 6 + 2(7) = 20 e-
F
S
Right F:
F
2s
Left F:
S:
2s
2p
3s
2p
F
: F(p)—S(p)
3p
: F(p)—S(p)
F
S
Answer / Reasoning: I find it easiest to analyze these kinds of problems starting from the Lewis
structure (even though it is not necessary). The LDS indicates two single bonds, which means two 
bonds. The orbital diagrams indicate that each F atom will use a p orbital to make its  bond with S,
and that the S will use its two singly occupied p orbitals (which are necessarily perpendicular to one
another since all the p orbitals in a given sublevel are perpendicular to one another in an atom) to
make those  bonds to the F’s. The result is that the bond angle is expected to be 90° (if one
does not hybridize) because of the two (perpendicular) p orbitals on S. This prediction is 8°
lower than the experimentally observed angle of ~98°, which is not great but is not
terrible either. Interestingly, the VSEPR / hybridization model would predict something that is
about equally “off” but on the high end (a little bit less than 109° which is roughly 10° higher than the
actual angle). Thus this molecule is not particularly consistent with either bonding model.
VBT: Hybridization of Atomic Orbitals
27. NT15. 10.58.
(a) Write orbital diagrams to represent the electron configuration of carbon before and after sp
hybridization.
Answer:
C:
2s
2p
Before hybridization
sp
2p
After hybridizing an s and a p to make two sp hybrids
(the 2p orbitals remain unhybridized)
(b) Rank the following orbitals (all in a C atom) from lowest to highest energy:
2s (before sp hybridization), 2p (before sp hybridization), sp (after sp hybridization)*, 2p (after sp hybridization)
If any orbitals have the same energy, use an equal sign (“=”) rather than a “less than” sign (“<”).
Answer:
lowest energy: 2s < sp < 2p (before)  2p (after)
Reasoning: Hybrid orbitals are a result of mathematical “blending”. As a result, the energy of any
hybrid orbital is necessarily intermediary between the energies of the orbitals from which it was
“created”. s orbitals are lower in energy than p orbitals (because of shielding and penetration; see
PS9). So any hybrid orbitals derived from s and p orbitals (i.e., sp, sp2, or sp3) will all have energies
that are higher than the (pure) s and lower than the (pure) p. The 2p orbitals after hybridization are
exactly the same orbitals as before the hybridization (because they were not involved in the
hybridization!), so of course their energies must be identical. This can be seen on the graphic on p.
427 in Tro, reproduced here:
PS11-9
Answer Key, Problem Set 11
28. MP. No answer to this problem in the key at this time.
VBT: Sigma () and Pi () Bonds and Hybridization
29. MP. No answer to this problem in the key at this time.
30. MP. No answer to this problem in the key at this time.
31. 10.59.
Which hybridization scheme allows the formation of at least one  bond? sp3, sp2, sp3d2
Answer: sp2, because it is the only one that leaves a p orbital unhybridized. -bonds are always (in this
class) formed by sideways overlapping p-orbitals (which are not hybridized!). Hybrid orbitals are
never used for making -bonds! They are only used for making -bonds or for housing lone pairs!!!
So if all of the p orbitals on an atom are used to make hybrid orbitals, there will not be any -bonds
made to that atom. (I think this is easier to see when you draw an orbital diagram. That is why I
wrote out the answer to 10.58 above.)
VBT: Application of Ideas / Visualization With Sketches
32. NT16. 10.63(ad). Write a hybridization and bonding scheme for each molecule or ion. Sketch the
structure, including overlapping orbitals, and label all bonds using the notation shown in Examples 10.6
and 10.7.
: O(p)—C(p)
: C(sp2)—Cl(p)
(a) COCl2 (C is the center)
v = 4 + 6 + 2(7) = 24 e-
Cl
O
O
Cl
C
Cl
C
Cl
O
C
Cl
: C(sp2)—Cl(p)
Cl
: O(p)—C(sp2)
Strategy / Approach:
1) Start from the LDS to figure out how many orbitals are “needed” for bonding. I like to first
determine how many  or  bonds are made to a center. (Remember, every double bond
equals exactly one  bond and one  bond. Every triple bond equals one  bond and two 
bonds.)
2) Deal with the  bonds first by counting clouds. The number of clouds equals the number of
hybrid orbitals needed for  bonding and lone pairs (if present).
PS11-10
Answer Key, Problem Set 11
3) Then deal with the  bonds. Are there any leftover (not hybridized) p orbitals? There had better
be if you have  bonds to that center! If there are no  bonds to that center, then there should
not be any leftover p orbitals (i.e., they should all have been used in the hybridization).
4) Terminal atoms in this textbook are generally never hybridized—there is no need for it. So just
use an unhybridized orbital (usually a p orbital for any atom other than H, which uses its s
orbital). You can check the orbital diagram just to make sure there is an unpaired electron.
5) To make the sketch, I’d actually draw the wedge and dash structure first so that I get the
“geometry” correct. (Try to get all  bonds so that they are in the plane of the paper!) Then
draw the sketch with the orbitals next to it (I used to do these “on top of one another” but it gets
pretty messy.) For convenience, you may use simple “lines” to connect sideways overlapping p
orbitals (rather than showing the overlap the way the text does).
6) Label each bond and lone pair (on the center). Ignore lone pairs on terminal atoms.
7) NOTE: You do not need to write out the orbital diagrams from any atoms in this approach! It
should always “work out” for simple molecules.
Execution of Strategy:
There are 3  bonds and one  bond made by the center C. Each bond “needs” a distinct orbital on
C.
3 clouds around C means 3 hybrid orbitals are needed to make the  bonds (there are no lone pairs
 use the first three orbitals (s, p, p) for hybridization (i.e., C is sp2 hybridized).
here)
That leaves one p orbital (not hybridized) to make the  bond to O.
O uses one p orbital for the  bond and a different p orbital for the  bond (as in Example 10.7).
The Cl atoms each use their singly occupied p orbital for the  bond.
The C has a trigonal planar ECG and AG. Put the C=O in the plane of the paper with the Cl’s
coming out and going back (wedge and dash) so the sideways-overlapping p orbitals can be in the
plane (to make the  bond)
(d) I3-
v = 3(7) +1 = 22 e-
I
I
I
: Central I(sp3d)—Terminal I(p)
I
I
I
I
I
Lone pairs (3) in sp3d orbitals
I
: Central I(sp3d)—Terminal I(p)
Explanation: (see strategy above)
From LDS, only two  bonds are made and no  bonds.
# of electron groups around central I is 5  use the first five orbitals available (s, p, p, p, d) to
make a set of five sp3d hybrids (i.e., I is sp3d hybridized). Two of these will be used to make the 
bonds to the terminal I’s; three will be used for the three lone pairs.
Each terminal I atom will just use its singly occupied p orbital to make the  bond to the center.
5 clouds means an ECG of trigonal bipyramidal. Use VSEPR to determine that the lone pairs will
go into the equatorial positions and make a wedge and dash sketch. Use that (as well as the
orbital info noted above) to make the orbital sketch.
PS11-11
Answer Key, Problem Set 11
Combining VSEPR and VBT (Prediction of ECG, AG, Angles, and Hybridization, # of  and  Bonds from a
Lewis Structure)
33. NT17. 10.85(c) Amino acids are biological compounds that link together to form proteins, the workhorse molecules
in living organisms. The skeletal structures of several simple amino acids are [cysteine is] shown here. Complete the
Lewis structure and determine the geometry and hybridization about each interior atom. [Don’t sketch it!] Also state the
value of each of the following bond angles in the molecule:
H
H
H
O
N
C2
C1
H
C3
H
S
H
O
H
(i) HOC: a bit less than 109°
Atom
#
clouds
Hybridization
ECG
# lone
pairs
AG
O, S
4
sp3
tetrahedral
2
bent
C1
3
sp2
trigonal
planar
0
trigonal
planar
C2, C3
4
sp3
tetrahedral
0
tetrahedral
N
4
sp3
tetrahedral
1
trigonal
pyramidal
(ii) OCC: ~120° (iii) CCC: 109.5° (iv) CCS: 109.5°
(v) HNC: a bit less than 109°
NOTE: To “complete the Lewis structure”, I merely added lone pairs in order to get octets around the N,
S, and O atoms. This “works” because the multiple bonds were already shown and this is a
simple organic molecule. Technically, I used the HONC patterns noted in class (C forms 4
bonds and has no lone pairs; N forms 3 bonds with one lone pair; O (and S, here) forms 2 bonds
with two lone pairs). If multiple bonds are not initially shown in a structure, you must be
particularly careful not to “just” add lone pairs to get octets!
34. NT18 10.87.
The structure of caffeine, present in coffee and many soft drinks, is shown here. How many pi bonds
are present in [one molecule of!] caffeine? How many sigma bonds? Insert the lone pairs in the molecule. What
kinds of orbitals do the lone pairs on the N’s occupy?
NOTE: Don’t ignore the condensed notations “H3C-“ and “-CH3”! These groups have bonds that are not explicitly shown!
Perhaps expand these to make sure you don’t miss any bonds.
H
H
H
O
C
H
C
C
N
C
C
C
N
H
C
H
H
a) 25 sigma bonds (count the “first” bond of
a double or triple bond as a sigma)
H
N
C
O
Answers:
H
N
H
b) 4 pi bonds (count the second [or third]
bond of a double [or triple] bond as a pi
bond.
c) Each of the three N’s having only single
bonds has its lone pair in an sp3 orbital
(because there are four clouds)
Note: Lone pairs were assigned as in the prior problem (see #18).
PS11-12
Answer Key, Problem Set 11
VBT: Visualization with Sketches and Relation to Rotation Around Bonds (Free or Not?)
35. NT19. 10.94.
The compound C3H4 has two double bonds. (a) Describe its bonding and geometry using a valence
bond approach. Add: (b) Sketch the molecule using wedges and dashes as in lab. (Hint: Think about the bonding carefully
(particularly the  bonds) before making your sketch. Can all of the atoms in this molecule be in the same plane?) (c) Is this
molecule rigid or can it freely rotate around any of its bonds?
: C(sp)— C(sp2)
C3H4 v: 3(4) + 4(1) = 26 eH
C1
H
C2
C3
H
C
C
H
Answers
(a)
C
: C(p)—C(p)
H
C1 and C3 are trigonal planar, and C2 is linear.
C1 and C3 use sp2 hybrids to make sigma
bonds to the two H’s and to C2. C2 uses sp
hybrids to make sigma bonds to C1 and C3.
C1 and C3 each use their (single) leftover p
orbital to make a pi bond to C2. C2 uses one
leftover p orbital to make a pi bond with C1
and another p orbital to make a pi bond to C3.
C
C
C
H
H
H
(b)
See bottom sketch off to the right. Two of the H’s and all three C’s are in one plane, but the
other two H’s are in a plane perpendicular to the first. All 6 atoms cannot be in the same
plane because the two pi bonds are in planes perpendicular to one another. I’ve shown the
“orbital” sketch to try to show this (although I did not ask you to draw that for this problem)
(c)
This molecule is rigid. It cannot rotate freely around either of the C-C bonds because rotation
about either one of them would require the breaking of a pi bond, which takes energy (see
discussion in Tro, p. 425 and box on “vision” on p. 426)
Reasoning:
(a) (approach as in #17):
Carbons 1 and 3 each make three  and one  bond
C2 makes two  and two  bonds
Carbons 1 and 3 have three clouds  sp2 hybrids (p left over for  bond)
C2 has two clouds  sp hybrids (two p’s left over for two  bonds)
Carbons 1 and 3 each have a trigonal planar ECG and AG (3 clouds, no lone pairs)
C2 is linear (ECG and AG; 2 clouds, no lone pairs)
**Since C2 must use two different p orbitals for its two p bonds, those p orbitals (and
subsequent p bonds) must be at 90 from one another. That means the H atoms cannot all be
in the same plane! They are, in fact, in perpendicular planes to one another. That is why I
drew the left two in the plane, and the right two in front of and behind the plane of the
paper.**
PS11-13