Download Chemical Bonding

Chem 1050
Chemical Bonding I and II
Text: Petrucci, Harwood, Herring 8th Edition
Chapter 11
Chemical Bonding I Basic Topics
Suggested Text Problems
Review problems: 4!8, 10 !18, 20, 21
Exercises:
27, 29, 33, 37, 39, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 71, 73, 79, 81,
83, 94, 99
1.
Draw Lewis structures for the following ionic compounds: NaF, Mg3P2, CaO
2.
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!F C!Br C!I C!Cl
(b)
(c)
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 octet. An example is a valuable addition
to each explanation.
4.
For the following molecules and ions: OF2, CCl4, PF6!, BF2Cl, HCN, ICl5, SnBr2, I3!,
XeOF2, SeCl6, CO2, NF3, COS, SiCl3H, SOCl2, PO43!, NOF, BrF4+, ClNO2, BeH2,
SO32!, ICl4!
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. (Chapter 12 topic, See
Table 12.1 page 445)
Draw three plausible structures for XeF2. Account for the fact that the linear structure
is correct.
6.
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
7.
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. Atom frameworks are H3C!CO!CN, H2N!CH!NH.
8.
Using bond dissociation energies in table 11.3 on page 422 in Petrucci and
Harwood, 8th 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.
Chapter 12
Chemical Bonding II Advanced Topics
Review problems: 4, 5, 6, 9!14, 16, 17, 18
Exercises: 23, 27, 33, 35, 43, 45, 47, 49, 70
9.
Draw Lewis structures for free radicals, CH3, ClO2 and NO2.
10.
Use the simple valence bond method to describe bonding in H2Se and SnCl2.
Step#1:
Step#2:
Step#3:
Step#4:
11.
Draw the Lewis structure
Draw the valence shell orbital diagrams of the separated atoms
Sketch only the orbitals of each atom that are involved in the bonding
showing how they overlap to make the bonds.
Indicate the predicted bond angles.
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
Step#3:
Step#4:
Step#5:
central atom and their spacial geometry)
Draw the valence shell orbital diagrams of the end atoms and the
appropriate hybrid orbital diagram of the central atom.
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 (Π or σ).
Indicate the predicted bond angles.
12.
Distinguish between the bonding and antibonding molecular orbitals formed by
combining a pair of atomic orbitals.
13.
Distinguish between Π (pi) and σ (sigma) bonding molecular orbitals.
14.
Distinguish between normal π bonding and delocalized π bonding.
15.
Draw valence molecular orbital diagrams (i.e. omitting inner shell orbitals) for the
following homonuclear diatomic species, H2!, He2+, O2, N2!, C22!, Ne2+, (Na2, Mg2,
P2, you can assume that third period diatomics form valence molecular orbitals
similar to second period diatomics but with n=3) and the following heteronuclear
diatomic species, CO, NO and BN (assume the same energy level diagram as Be2
to N2). Determine the bond order of each species. Indicate any species which will
be unstable in the gas phase. Use Figure 12-20 on page 405 in Petrucci and
Harwood, 7th Edition as a guide.
16.
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 Π bonding and draw an energy level diagram to
show how the Π electrons occupy the available Π molecular orbitals.
17.
Pick which of the following species possess delocalized molecular orbitals:
CO2, NO3!, C2H4, O3, and C4H6 (H2C=CH-CH=CH2).
18.
Explain what are non-bonding delocalized Π molecular orbitals.
19.
Explain metallic bonding using the electron sea model and band theory.
20.
Distinguish between conductors, semi-conductors and non-conductors using band
theory.
Answers: for most questions, partial answers given only.
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
polarity
hybrid.
OF2
tetrahedral
AX2E2
V-shaped
~109.5
Polar
sp3
CCl4
tetrahedral
AX4
tetrahedral
~109.5
Non
sp3
PF6!
octahedral
AX6
octahedral
90, 180
non
sp3d2
BF2Cl
triangular planar
AX3
triangular
planar
120
polar
sp2
HCN
linear
AX2
linear
180
polar
sp
ICl5
octahedral
AX5E
square
pyramidal
~90, ~180
polar
sp3d2
SnBr2
triangular planar
AX2E
V-shaped
~120
polar
sp2
I 3!
triangular
bipyramidal
AX2E3
linear
180
non
sp3d
XeOF2
triangular
bipyramidal
AX3E2
T-shape
~90, ~180
polar
sp3d
SeCl6
octahedral
AX6
octahedral
90, 180
non
sp3d2
CO2
linear
AX2
linear
180
non
sp
NF3
tetrahedral
AX3E
triangular
pyramidal
~109.5
Polar
sp3
COS
linear
AX2
linear
180
polar
sp
SiCl2H2
tetrahedral
AX4
tetrahedral
~109.5
polar
sp3
SOCl2
tetrahedral
AX3E
triangular
pyramidal
~109.5
Polar
sp3
PO43!
tetrahedral
AX4
tetrahedral
~109.5
Non
sp3
NOF
triangular planar
AX2E
V-shaped
~120
polar
sp2
BrF4+
triangular
bipyramidal
AX4E
sawhorse
~90,
~180,
~120
polar
sp3d
ClNO2
triangular planar
AX3
triangular
planar
~120
polar
sp2
BeH2
linear
AX2
linear
180
Non
sp
SO32!
tetrahedral
AX3E
triangular
pyramidal
~109.5
Polar
sp3
ICl4!
octahedral
AX4E2
square planar
90, 180
Non
sp3d2
5.
Highest energy: V-shaped (two 90 degree lone pair-lone pair repulsions) > right
angled (one 90 degree lone pair-lone pair repulsion) > linear (no 90 degree lone pairlone pair repulsions). Lowest energy is most stable.
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 %
10.
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.
11.
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
triangular
planar
3
sp2
triangular
planar
120
CH3Cl
4
tetrahedral
4
sp3
tetrahedral
109.5
NF3
4
tetrahedral
4
sp3
tetrahedral
109.5
SF4
5
triangular
bipyramidal
5
sp3d
triangular
bipyramidal
90,180
120
XeF2
5
triangular
bipyramidal
5
sp3d
triangular
bipyramidal
90,180
120
XeF4
6
octahedral
6
sp3d2
octahedral
90,180
H2CO
3
triangular
planar
3
sp2
triangular
planar
120
HCN
2
linear
2
sp
linear
180
12.
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.
13.
π M.O.
σ M.O.
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.
14.
In normal π bonding, only two 2pz orbitals overlap to form π M.O.’s while in delocalized
π bonding, there is continuous overlap of 2pZ orbitals from an unbroken chain of 3 or
more atoms forming an equal number of delocalized π M.O.’s.
15.
H2!
He2+
O2
N2+
C22!
Ne2+
Na2
Mg2
P2
CO
NO
BN
16.
VSEPR shape: triangular planar about each carbon giving benzene its flat hexagonal
shape, all C’s sp2 hybridized. Each carbon has three unpaired sp2 electrons for σ
bonding and one 2p electron for π bonding. Three delocalized π bonding MO’s and
three delocalized π antibonding MO’s formed from six 2p AO’s.
17.
NO3!, O3, and C4H6
18.
Non-bonding delocalized Π 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 Π molecular orbital is formed.
(σ1s)2(σ1s*)1 bond order = ½ paramagnetic
(σ1s)2(σ1s*)1 bond order = ½ paramagnetic
(σ2s)2(σ2s*)2(σ2p)2 (π2p)2(π2p)2 (π2p*)1(π2p*)1 bond order=2 paramagnetic
(σ2s)2(σ2s*)2 (π2p)2(π2p)2(σ2p)1 bond order = 2.5 paramagnetic
(σ2s)2(σ2s*)2 (π2p)2(π2p)2(σ2p)2 bond order = 3.0 diamagnetic
(σ2s)2(σ2s*)2 (σ2p)2 (π2p)2(π2p)2 (π2p*)2(π2p*)2(σ2p*)1 bond order = ½ paramagnetic
(σ3s)2 bond order = 1 diamagnetic
(σ3s)2(σ3s*)2 bond order = 0 diamagnetic and unstable
(σ3s)2(σ3s*)2 (π3p)2(π3p)2(σ3p)2 bond order = 3 diamagnetic
(σ2s)2(σ2s*)2 (π2p)2(π2p)2(σ2p)2 bond order = 3.0 diamagnetic
(σ2s)2(σ2s*)2 (π2p)2(π2p)2(σ2p)2 (π2p*)1 bond order = 2.5 paramagnetic
(σ2s)2(σ2s*)2 (π2p)2(π2p)2 bond order = 2.0 diamagnetic
Developed by: Dr. Chris Flinn