Chem 1050
Chemical Bonding I and II
Fall 2010
Chemical Bonding I Basic Concepts
Petrucci Herring Madura Bissonnette 10th Edition
Chapter 10
Exercises: 1, 3, 11, 13, 17, 21, 25, 27, 35, 37, 47, 49, 51, 53, 55, 59, 61, 63, 65, 67, 69,
73, 79, 81, 83, 85, 91, 93, 95, 97, 99, 101, 109
Petrucci Harwood Herring Madura 9 th Edition Chapter 10
Exercises: 1, 3, 11, 13, 15, 17, 21, 27, 35, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65,
69, 71, 73, 81, 83, 85, 95, 123, 127
1.
2.
Draw Lewis structures for the following ions Na+, Ca2+, O2G, P3G
List the following bonds (A-B) in order of increasing ionic character. Assume that %
ionic character is directly proportional to )EN (ENA-ENB ). Do not use the table in the
text but predict the answers based on the general periodic trends and other data given
in class.
(a)
(c)
C!F C!Br C!I C!Cl (b)
H!H C!N Na!F C!O C!F
Si!F Cl!F S!F P!F
3.
Define or explain coordinate covalent bond, resonance structures, resonance hybrid,
free radical, incomplete octet and expanded valence shell. An example is a valuable
addition to each explanation.
4.
For the following molecules and ions: OF2, CCl4, PF6!, BF2Cl, HCN, ICl5, SnBr2, I3!,
SeCl6, CO2, NF3, COS, SiCl3H, NOF, BrF 4+, ClNO2, BeH2, , ICl4G, AsCl5
1.
2.
3
4.
5.
6.
5.
Draw a Lewis structure. (Include formal charges)
Determine the electron group geometry and compound class.
Sketch and name the shape.
Indicate the approximate bond angles.
Indicate whether or not each species has a permanent dipole moment (i.e. is
polar or non!polar).
Determine the hybridization of the central atom. (Chemical Bonding II topic,
See Table 11.1 page 433 in 9th Edition or Table 12.1 page 445 in 8 th Edition)
Draw three plausible structures for XeF2. Account for the fact that the linear structure
is correct.
1
6.
7.
8.
Draw plausible resonance structures for N2O (NNO), HNO3 (HONO2), CO32!.
For each species, predict the bond order of each pair of bonding atoms.
For N2O, predict the length of the NN bond and its bond energy given the following
Bond
bond length, pm
bond
energy/kJ mol!1
N!N
145
163
N=N
123
418
N/N
109.8
946
Sketch the following molecules using VSEPR to determine the shape about each
multiply bonded atom. Indicate the ideal bond angles in each structure. Complete the
Lewis structures first: C 3H3ON (H3C!CO!CN), CH4N2 (H2N!CH!NH).
Using bond dissociation energies in Table 10.3 on page 435 in Petrucci 10th Edition
or Table 10.2 on page 407,Petrucci, 9 th Edition.
(a)
estimate )H of the following reaction: C2H4(g) + 3 O2(g) ÿ 2 CO2(g) + 2
H2O(g)
(b)
estimate the bond dissociation energy of the nitrogen-to-oxygen bond in NO(g)
if the enthalpy change for the reaction below is !756.3 kJ mol!1.
2 NO(g) + 5 H2(g) ÿ 2 NH3(g) + 2 H2O(g)
Calculate the % error given the measured value of 631 kJ mol!1.
9.
10.
(a)
Draw octet rule structures for SO2, XeOF 2, SOCl2, PO43G, SO32G and ClO3G.
Assign formal charges. Draw expanded valence shell structures for each
species in which the central atom has no formal charge. Draw all equivalent
resonance structures if applicable and indicate the sulfur to oxygen bond order,
chlorine to oxygen bond order, or phosphorus to oxygen bond order as
appropriate. Predict the shape of each structure and the hybridization of the
central atom. Explain why the octet rule structures are not considered as
reasonable representations of the bonding vs the expanded valence shell
structures.
(b)
What is wrong with the Lewis structure, O O O ?
Draw Lewis structures for free radicals, CH3, ClO2 and NO2.
2
Chemical Bonding II, Additional Aspects
Petrucci Herring Madura Bissonnette 10th Edition Chapter 11
Exercises: 3, 5, 7, 9, 11, 13, 15, 17, 21, 25, 33, 35, 37, 39, 43, 68
Petrucci Harwood Herring Madura 9 th Edition Chapter 11
Exercises: 3, 5, 9, 11, 13, 15, 17, 21, 23, 31, 33, 35, 37, 39, 62
11.
Use the simple valence bond method to describe bonding in H2Se and SnCl2.
Step#1:
Draw the Lewis structure
Step#2:
Draw the valence shell orbital diagrams of the separated atoms
Step#3:
Sketch only the orbitals of each atom that are involved in the bonding
showing how they overlap to make the bonds.
Step#4:
Indicate the predicted bond angles.
12.
Use valence bond theory and hybridization to describe bonding in BeH2, AlBr3, CH3Cl,
NF3, SF4, XeF2, XeF4, (H2CO, HCN in which there are multiple bonds).
Step#1:
Draw the Lewis structure
Step#2:
Determine the number of electron groups of the central atom and the
VSEPR shape. ( This determines the type of hybrid orbitals of the
central atom and their spacial geometry)
Step#3:
Write the valence shell orbital diagrams of the end atoms and the
appropriate hybrid orbital diagram of the central atom.
Step#4:
Sketch all the hybrid orbitals of the central atom but only the valence
orbitals of the end atoms that are involved in the bonding showing how
they overlap to make the bonds. Indicate the location of any lone pairs
of the central atom. Indicate the type of bonds formed (A or F) and the
sign of the wave functions (+ or !).
Step#5:
Indicate the predicted bond angles.
13.
Distinguish between the bonding and antibonding molecular orbitals formed by
combining a pair of atomic orbitals.
Distinguish between A (pi) and F (sigma) bonding molecular orbitals.
Distinguish between normal B bonding and delocalized B bonding.
Draw valence molecular orbital diagrams (i.e. omitting inner shell orbitals) for the
following species, H2!, He2+, O2, N2+, C22!, Ne2+, Li2, C2, O22+, F2, Be2 and the following
heteronuclear diatomic species, BN, NF, and NO using Figure 11!27 page 473 as
a guide. Determine the bond order of each species. Indicate any species which will be
unstable in the gas phase. Determine whether each species is diamagnetic or
paramagnetic.
Draw two resonance structures for benzene C6H6. Use VSEPR to determine the shape
of the molecule about each carbon. Determine the hybridization of each carbon atom
and draw an valence orbital diagram to represent the carbon atoms. Sketch how the
orbitals of carbon and hydrogen overlap to form sigma bonds. Sketch separately the
delocalized A bonding and draw an energy level diagram to show how the A electrons
occupy the available A molecular orbitals.
14.
15.
16.
17.
3
18.
19.
20.
Pick which of the following species possess delocalized molecular orbitals:
CO2, NO3–, C 2H4, and O3.
Explain what are non-bonding delocalized A molecular orbitals.
Use a molecular orbital diagram similar to Figure 11!27 page 473 in Petrucci 10th
Edition for OF+ (assume O+ and F). Hint: which has the higher Zeff for its valence
electrons, O or F?
4
Answers: See Dr. Flinn’s website for additional answers for questions 1,4, 5, 6, 7, 9 10, 11
and 12.
2.
Least ionic: C!I < C!Br < C!Cl < C!F :most ionic
Least ionic: Cl!F < S!F < P!F < Si!F :most ionic
Least ionic: H!H < C!N < C!O < C!F < Na!F :most ionic
4.
formula
electron group
geometry
Compound
class
name of
shape
bond
angles
polar
yes or no
hybrid.
OF2
tetrahedral
AX2E2
V-shaped
~109.5
yes
sp3
CCl4
tetrahedral
AX4
tetrahedral
~109.5
No
sp3
PF6!
octahedral
AX6
octahedral
90, 180
no
sp3d2
BF2Cl
trigonal planar
AX3
trigonal planar
120
yes
sp2
HCN
linear
AX2
linear
180
yes
sp
ICl5
octahedral
AX5E
square
pyramidal
~90, ~180
yes
sp3d2
SnBr2
trigonal planar
AX2E
V-shaped
~120
yes
sp2
I3!
trigonal
bipyramidal
AX2E3
linear
180
no
sp3d
SeCl6
octahedral
AX6
octahedral
90, 180
no
sp3d2
CO2
linear
AX2
linear
180
no
sp
NF3
tetrahedral
AX3E
trigonal
pyramidal
~109.5
yes
sp3
COS
linear
AX2
linear
180
yes
sp
SiCl2H2
tetrahedral
AX4
tetrahedral
~109.5
yes
sp3
NOF
trigonal planar
AX2E
V-shaped
~120
yes
sp2
BrF 4+
trigonal
bipyramidal
AX4E
sawhorse
~90,
~180,
~120
yes
sp3d
ClNO2
trigonal planar
AX3
trigonal planar
~120
yes
sp2
BeH2
linear
AX2
linear
180
no
sp
ICl4!
octahedral
AX4E2
square planar
90, 180
no
sp3d2
AsCl5
trigonal
bipyramidal
AX5
trigonal
bipyramidal
90, 120,
180
no
sp3d
5
5.
The V!shaped and right angled structures have 90° lone pair!lone pair repulsions. They
are of much higher energy and are much less stable than the linear structure which has
no 90°lone pair!lone pair repulsions.
6.
N2O, B.Order NN bond = 2.5, B. order NO bond = 1.5, NN bond length = 116 pm, NN
bond energy = 682 kJ@mol!1 HNO3 B. Order OH bond = 1, B. order for NO bond (O
bonded to H) = 1, B. Order other two NO bonds = 1.5 CO32! B. order for each CO bond
= 4/3
7.
H3C!CO-CN, C1(sp3), C2(sp2), C3(sp), and N(sp)
H2N!CH!NH, N1(sp3), C(sp2) and N2(sp2)
8.
(a) -1039 kJ@mol!1 (b) 627 kJ@mol!1, % error = 0.6 %
9.
(a)
(b)
11.
SO2 (V!shaped, sp 2 sulfur), XeOF 2 (T!shaped, sp3d xenon), SOCl2 (trigonal
pyramidal, sp3 sulfur), PO43G (tetrahedral, sp3 phosphorus, PO bond order = 1.25),
SO32G (trigonal pyramidal, sp3 sulfur, SO bond order = 1a) ClO 3G (trigonal
pyramidal, sp3 chlorine, ClO bond order = 1b).
Central oxygen cannot have 10 valence electrons.
H2Se: H(1s1), Se([Ar]3d104s24p4 (2 unpaired 4p electrons for bonding).
SnCl2: Cl ([Ne]3s23p5, Sn([Kr] 4d105s25p2), (2 unpaired 5p electrons for bonding) ideal
bond angle for both molecules is 90 degrees.
12.
molecule
#electron
groups
electron
group
geometry
# hybrid
A.O.’s formed
hybrid
orbital
type
hybrid
orbital
geometry
bond
angles
BeH2
2
linear
2
sp
linear
180
AlBr3
3
trigonal
planar
3
sp2
trigonal
planar
120
CH3Cl
4
tetrahedral
4
sp3
tetrahedral
109.5
NF3
4
tetrahedral
4
sp3
tetrahedral
109.5
SF4
5
trigonal
bipyramidal
5
sp3d
trigonal
bipyramidal
90,
120,
180
XeF2
5
trigonal
bipyramidal
5
sp3d
trigonal
bipyramidal
180
XeF4
6
octahedral
6
sp3d2
octahedral
90,180
H2CO
3
trigonal
planar
3
sp2
trigonal
planar
120
HCN
2
linear
2
sp
linear
180
6
13.
A bonding molecular orbital places a high electron charge density between the two
nuclei. This reduces the repulsions between the positively charged nuclei, lowering the
energy and increasing the stability of the molecule.
An anti-bonding molecular orbital places a low electron charge density between the two
nuclei. The repulsions between the nuclei increase because they are poorly shielded from
each other, increasing the energy and decreasing the stability of the molecule.
14.
B M.O.
F M.O.
15.
electron density concentrated above and below internuclear axis ( 2 banana
shaped clouds) formed from constructive interference (overlap) of parallel
np orbitals. Overlap occurs in 2 locations.
electron density concentrated along the internuclear axis. Overlap occurs in
one location.
In normal B bonding, only two 2pz orbitals overlap to form B M.O.’s while in delocalized B
bonding, there is continuous overlap of 2pZ orbitals from an unbroken chain of 3 or more
atoms forming an equal number of delocalized B M.O.’s.
7
16.
H2
σ*
σ
1s
1s
–
σ
1s
N2
paramagnetic
bond order =
1
2
paramagnetic
σ *
2s
π
σ
2p
2p
π * σ*
2p
2p
+
bond order = 2
σ
2s
C2
1
2
σ*
1s
He 2+
σ
2s
bond order =
σ*
2s
π
σ
2p
2p
1
2
paramagnetic
π * σ*
2p
2p
2–
bond order = 3 diamagnetic
σ
2s
σ*
2s
π
σ
2p
2p
π * σ*
2p
2p
bond order = 2 diamagnetic
C2
σ
2s
σ*
2s
π
σ
2p
2p
π * σ*
2p
2p
Li2
bond order = 1 diamagnetic
σ
2s
σ*
2s
π
σ
2p
2p
π * σ*
2p
2p
Be 2
bond order = 0 diamagnetic
Be 2 is unstable
σ
2s
σ* σ
2s
2p
π
2p
π * σ*
2p
2p
O2
bond order = 2 paramagnetic
σ
2s
O2
σ* σ
2s
2p
π
2p
π * σ*
2p
2p
2+
bond order = 3 diamagnetic
σ
2s
Ne2
σ* σ
2s
2p
π
2p
π * σ*
2p
2p
+
bond order =
σ
2s
σ* σ
2s
2p
π
2p
1
2
paramagnetic
π * σ*
2p
2p
F2
bond order = 1 diamagnetic
BN:
NO:
NF:
note:
1F2 2F2 1B4
bond order = 3 diamagnetic
2
2
4
2
1
1F 2F 1B 3F 2B
bond order = 2.5 paramagnetic
1F2 2F2 1B4 3F2 2B2
bond order = 2 paramagnetic
the 2F, 2B and 4F are antibonding orbitals
8
17.
VSEPR shape: trigonal planar about each carbon giving benzene its flat hexagonal shape,
all C’s sp2 hybridized. Each carbon has three unpaired sp2 electrons for F bonding and
one 2p electron for B bonding. Three delocalized B bonding MO’s and three delocalized
B antibonding MO’s formed from six 2p AO’s.
18.
NO3! and O3
19.
Non-bonding delocalized A molecular orbitals are orbitals whose energy is the same as
the 2pz orbitals from which they are formed and hence do not contribute to bonding. One
non!bonding A molecular orbital is formed when odd numbers of 2pz orbitals combine.
20.
Developed by: Dr. Chris Flinn
9