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Chapter 6
Molecular Structure
F Cl
F
F
F
Xe
F F
F
F
Cl
F F
F
F
1. Draw the Lewis structure of each of the following ions, showing all nonzero formal charges. Indicate whether each
ion is linear or bent. If the ion is bent, what is the bond angle?
a) NO 2 1-^ b) N 3 1-^ c) ClO 2 1-
A lone pair on the N results in three electron groups and a bent structure with ~120 o bond angle.
~120 o
O N O (^) N N N
No lone pair on the central N result in two electron groups and a linear structure with a 180 o^ bond angle.
O Cl O
O
Cl ~109 O o
Two bonding and two lone pairs results in four electron groups and a ~109 o^ bond angle
O N O
3. Draw Lewis structures for the following molecules. Indicate nonzero formal charges and whether each is trigonal
planar or trigonal pyramidal.
a) PF 3 b) COCl 2 c) BF 3
F P
F
F (^) Cl C
O
Cl F B
F
F
P F (^) FF
O
C Cl Cl
F
B F F One lone pair on P results in four electron groups and trigonal pyramidal structure with ~109o bond angles.
No lone pairs on central atom, so it has only three electron groups. The molecules are trigonal planar with 120o^ bond angles.
~109o ~120o^120 o
Note that BF 3 does not have a double bond because a double bond to a halogen puts positive formal charge on the halogen. Instead, the boron is electron deficient having only six electrons.
5. Describe the shapes of the following:
a) ClF 3
SP = ½(32-28) = 2 shared pair, which is not sufficient for three Cl-F bonds. Cl uses an expanded valence shell, so use Eq. 6.1 to get lone pairs. Cl is in Group 7 and is in the + oxidation state, so LP = ½ (7-3) = 2. There are three bonds and two lone pairs, so the five electron regions are situated in a trigonal bipyramid arrangement about Cl. The lone pairs go into the equatorial positions, so ClF 3 is T-shaped.
b) XeF 4
SP= ½ (40-36) = 2 shared pair, which is insufficient for four Xe-F bonds. Xe is in Group 8 and has a +4 oxidation state, so Eq 6.1 yields LP = ½ (8 – 4) = 2 lone pairs. Six electron regions assume octahedral geometry with the lone pairs opposite one another. XeF 4 is square planar.
c) ClF 5
SP= ½ (48-42) = 3 shared pair, which is insufficient for five Cl-F bonds. Cl is in Group 7 and has a +5 oxidation state, so Eq 6.1 yields LP = ½ (7 – 5) = 1 lone pair. Six electron regions assume octahedral geometry with the lone pair in any position. ClF 5 is a square pyramid.
7. What is the hybridization on the central atom in each of the ions in Exercise 1?
Central atom electron regions hybridization
a) NO 2 1-^ 2 bonding + 1 lone pairs = 3 sp^2
b) N 3 1-^ 2 bonding + 0 lone pairs = 2 sp
c) ClO 2 1-^ 2 bonding + 2 lone pairs = 4 sp^3
9. What is the hybridization on the central atom in each of the molecules in Exercise 3?
Central atom electron regions hybridization
a) PF 3 3 bonding + 1 lone pairs = 4 sp^3
b) COCl 2 3 bonding + 0 lone pairs = 3 sp^2
c) BF 3 3 bonding + 0 lone pairs = 3 sp^2
11. Rank the following in order of increasing H-N-H bond angles: NH 2
1-
, NH 3 , and NH 4
1+
H N
H
H N
H
H H N
H
H
H
< (^) <
two lone pairs on N < one lone pair on N < no lone pairs on N
13. The nitrite ion (NO 2
1-
) can add an H
1+
ion (proton) to become nitrous acid. Based on the formal charges in the
Lewis structure of the anion (Exercise 1a), draw the Lewis structure of nitrous acid, HNO 2.
The positively charged proton attacks the site of negative formal charge. The H is written first in the formula because HNO 2 is an acid. We will discuss acids in detail in Chapter 11.
15. Explain why molecules A and C are planar while molecule B is not. Hint: which p orbitals are used to make each of
the π bonds?
C C
H
H
H
H
C C C
H
H
C C C C
H
H
H
H
H
H A B C
The p orbitals that combine to form a π bond must both be either px or they must both be py orbitals if the bonding axis is the z axis. A. If the two p orbitals forming the π bond are the py orbitals, then the two sets of sp^2 orbitals must use the px and pz orbitals and lie in the xz plane. Since the two sets of sp^2 orbitals are in the same plane, molecule A is planar. B. The central carbon atom must use different p orbitals to form two different π bonds. We arbitrarily assume that the carbon on the left uses its py orbital to form the π bond. In this case, the sp^2 orbitals on the carbon on the left lie in the xz plane. If the central carbon uses its py orbital to π bond with the carbon on the left, then it must use its px orbital to π bond with the carbon on the right. The carbon on the right must then use its px orbital to π bond to the central carbon. Since the px is used to form a π bond, the sp^2 orbitals on the carbon on the right must lie in the yz plane and are, therefore, perpendicular to those on the carbon to the left. C. If the π bond to the left most carbon results from the py orbitals then the central π bond must use the px orbitals and the one on the right must use the py orbitals. Since both terminal carbon atoms use the py orbitals to construct their π bonds, their sp
2 hybrid orbitals must both be constructed from the p (^) x and pz orbitals and lie in the xz plane. Thus, molecule C is planar.
17. Aspartame is the active ingredient in the sweetener Nutrasweet. What are the approximate bond angles α , β , γ , δ
and ε in the aspartame structure?
There must be one lone pair on the N at α and two on the O at ε
α = ~
o β = ~
o γ = ~ 120
o δ = ~
o ε = ~
o
O
N O (^) O
N O
H
H C
H
H
C C
H
H
H
H C
H
H
N O
O
C C
H
H
H
H
O
H
C N N C
H
H
H
H
H
H
H
33. What change in hybridization (if any) occurs on the carbon in each of the following reactions?
a) CO 2 + H 2 O → H 2 CO 3 b) CH 4 + 2O 2 → CO 2 + 2H 2 O c) C 2 H 2 + 2H 2 → C 2 H 6
sp → sp
2 sp
3 → sp sp → sp
3
35. Draw the result of mixing a px and a p y orbital.
Mixing two p orbitals rotates them by 45o^.
37. Add multiple bonds, lone pairs, and hydrogen atoms to the following:
The bond lengths indicate one single and one double bond. Carbon must have four bonds in order to have no formal charge, so C-H bonds are added to each carbon to bring the number of σ and π bonds to four.
1.5 Å indicates a C-N single bond while the 1.3A bond length is intermediate between the N-O single and double bonds, so the bonds have bond orders of 1.5 (indicated by dotted lines). The carbon requires three C-H bonds in order to have zero formal charge.
39. Add multiple bonds, lone pairs, and hydrogen atoms to the following:
The C-C and C-O bond lengths of are those of single bonds. To maintain zero formal charge, the oxygen has two bonds and each carbon has four. Lone pairs are added to the oxygen to give it an octet.
The bond lengths indicate that the C-N bonds are each single bonds, but the N-N bond is a double bond. Nitrogen has no formal charge when it is involved in three bonds, so there are no N-H bonds. Lone pairs are required on the nitrogens to obey the octet rule. C-H bonds are added to give each carbon atom four bonds (no formal charge).
41. An amide is a compound that has a nitrogen next to a C=O group.
a) What are the predicted values of α and β and what is the predicted hybridization on the nitrogen atom?
α ~ 109
o (Four regions around N, so it is sp^3 ) β ~ 120
o (Three electron regions around C, so it is sp^2 )
b) The experimentally C-N-C bond angle is 120 o^ and the amide skeleton is planar. What is the hybridization on
the nitrogen atom in this structure?
120o implies three electron groups and sp^2 hybridization.
c) Draw another Lewis structure that is important in the bonding description as determined
from the structural data. Indicate all nonzero formal charge in the structure.
N C
H
H 3 C
O
CH 3
43. Classify each of the following π MO’s as bonding, nonbonding, or antibonding.
a) one nonbonding interaction makes the MO a nonbonding MO
b) two antibonding plus one bonding interaction makes this MO antibonding
c) two nonbonding interactions makes this MO nonbonding
d) two bonding and one antibonding interaction makes this MO bonding.
45. Which of the above molecules are not predicted to form?
Bonds with zero bond order are predicted not to form. Thus, the Be-Be and Ne-Ne bonds do not form, so Be 2 and Ne 2 would not form.
47. Which has the strongest bond. O 2 1-^ , O 2 , or O 2 1+^?
See Figure 6.23 for the MO diagram of O 2. An additional electron in O 2 would enter a π∗(2p). Since the added electron is antibonding, the O-O bond in the O 2 1-^ ion would be weaker than the O-O bond in O 2. A removed electron would come from a π∗(2p) orbital. Since the removed electron is antibonding, O 2 1+^ would have a stronger bond than O 2. Thus, O 2 1+^ has the strongest bond.
49. Assume that the MO diagram for NO is the same as for O 2. Is NO more likely to gain or to lose an electron?
NO has 5 + 6 = 11 valence electrons. Placing 11 electrons into the O 2 diagram puts two electrons into the σ(2p), four electrons into the π(2p), and one in the π(2p) orbital. The NO bond order is then ½ (6-1) = 2.5. The additional electron in NO1-^ would also go into the π(2p) orbital, so it is less stable than NO with a bond order of ½ (6-2) = 2. The highest energy electron in NO is antibonding, so removing it makes NO1+^ more stable than NO (BO = ½ (6-0) = 3), so NO is more likely to lose an electron.
