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CHEMISTRY 103 – Help Sheet #13
Chapter 9 (Part I) – 9.1-9.6
http://www.chem.wisc.edu/areas/clc (Resource page)
Prepared by Dr. Tony Jacob
Nuggets: VSEPR; Angles; Dipole Moment/Molecular Polarity; Valence Bond Theory; Assign Hybrid Orbitals
(HO); Assign HO that Make a Bond
VALENCE SHELL ELECTRON PAIR REPULSION MODEL (VSEPR): A model used to determine
molecular shape of the molecule. VSEPR shape will be wrong if you have an incorrect Lewis structure.
Method for using VSEPR:
1. Draw correct Lewis Dot Structure
2. Domains = (#lone pairs of e- on central atom) + (#atoms bonded to central atom)
3. Compare to VSEPR table. (memorizing VSEPR table is best)
Electron Domain Geometry (EDG): refers to total = (#lone pairs of e- on + #atoms bonded to) central atom
Molecular Geometry (MG): refers to the actual arrangement of atoms in space
Electron Domain Geometry and Molecular Geometry are often not the same!
Repulsions: lone pair e-–lone pair e- > lone pair e-–bonding pair e- > bonding pair e-–bonding pair e- >
MOLECULAR POLARITY/DIPOLE MOMENT: The sum of all of the bond polarities.
If all the individual bond polarities cancel, the molecule is nonpolar and the dipole moment = 0.
If they do not all cancel, the molecule is polar and the dipole moment ≠ 0.
(Dipole moment – measure of molecular polarity)
Example: a. Is the C–O bond in CO2 polar? b. Is the CO2 molecule polar?
Answer: a. Draw the Lewis dot structure for CO2:
Draw vector for the C–O bond; since C and O have different electronegativities, C–O bond is polar.
b. To see if the CO2 molecule is polar → draw the molecule with the correct shape. Correct shape (linear): vectors cancel:
→ CO2 is nonpolar. If molecule is NOT drawn with the correct shape:
→
→ vectors don’t cancel! (Correct shape comes from VSEPR!)
Polar molecules can absorb microwave radiation and heat up in a microwave oven. Non-polar molecules
do not absorb microwave radiation and do not heat up in a microwave oven.
Example: a. Draw the Lewis dot structure for IF2 . b. Using this structure draw the electron domain geometry (EDG) and the molecular geometry (MG) and name
them. c. In the MG diagram, draw the polarity vectors for each bond. Is this molecule polar?
F
F
I
F
I
F
-
F
F
EDG; on central atom:
MG; on central atom:
2 atoms + 3 lone pairs =
2 atoms + 3 lone pairs = linear;
trigonal bipyramid
b.
c. vectors cancel = nonpolar molecule
FORM A BOND: Shown is the energy change as H2 forms a
covalent bond from two H atoms. Covalent Bond: Two orbitals
overlap and contain two shared electrons of opposite spin in the
overlap region.
Potential Energy (kJ/mol)
Answer: a.
I
4. Nuclei repel one
another as they approach
and energy increases
200
1. Atoms far apart;
not interacting
3. Bond formed; energy at
lowest point; internuclear
distance at smallest value
100
0
-100
Energy released
when bond is
formed
2. Atoms approach one
another; bond starts to
form and energy drops
-200
-300
-436 -400
74
100
200
Internuclear Distance (pm)
VALENCE BOND THEORY/HYBRIDIZATION – Modeling bonds in molecules
Problem: With more complicated molecules such as CH4, VSEPR theory predicts the molecular geometry is
tetrahedral with bond angles of 109.5˚; this implies the orbitals also lie at 109.5˚. The C atomic orbitals in CH4
available to form bonds are 2s and 2p with angles of 90˚ (p orbitals). These angles don’t align with the molecular
angles of 109.5˚ predicted by VSEPR so something must change when the molecular bonds form. Let atomic
orbitals combine to form new orbitals called hybrid orbitals (HO) that then form bonds.
Sigma (σ ) Bonds: The bonds/electrons lie between the nuclei of the two atoms on the internuclear axis. All
bonds (single, double, or triple) contain one σ bond. Sigma (σ) bonds are typically made from the overlap
of two hybrid orbitals, HO–HO, (e.g., sp2–sp3) or 1s–hybrid orbital, 1s–HO (e.g., 1s–sp3). (Some instructors
present sigma bonds made from p-p orbital overlap with halogen atoms.)
Tetrahedral Arrangement – sigma (σ ) bonds
z
z
+
y
sp3
sp3
sp3
y
x
px
+
x
py
+
p
+
x
s
p
z
y
x
p
z
y
pz
+
+
sp3
sp 3
s
sp 3
sp 3
sp 3
sp 3
Trigonal Planar Arrangement – sigma (σ ) bonds
z
z
y
+
x
p
p
sp2
+
x
px
p
sp2
y
x
s
p
z
y
+
pz
+
sp2
s
sp 2
sp 2
sp 2
sp 2
Linear Arrangement – sigma (σ ) bonds
z
z
y
y
x
+
x
s
p
p
p
p
sp
s
px
p
+
sp
sp
sp
sp
Pi (π ) Bonds: Bonds/electrons do not lie on internuclear axis and occur with double or triple bonds. In the
trigonal planar (sp2) and linear (sp) arrangements, there are unused p-orbitals; The pi (π) bonds are made from
overlap of these two unused p-atomic orbitals, p–p, as shown below.
2 pz-orbitals before overlapping
z
2 pz-orbitals overlapping to form π bond
z
y
y
x
+
pz
y y
x
x
pz
π bond
Make bonds: σ: HO–HO, (e.g., sp2–sp3) or 1s–HO (1s–hybrid orbital); π : p–p (two p-atomic orbitals)
Count Bonds: Single bond = σ bond; Double bond = σ bond + π bond; Triple bond = σ bond + 2π bonds
Example: a. Draw the Lewis dot structure for CF2=CH2. b. Identify the hybridization for each atom. c. Indentify what orbitals make up each bond.
4 domains = sp3
3 domains =
sp2
Single bond = ! bond
! bond = HO–HO = sp2(C)–sp3(F)
F
Answer: See image:
H
C
F
4 domains = sp3
3 domains =
sp2
1s
C
Single bond = ! bond
! bond = HO–1s = sp2(C)–1s(H)
H
1s
Double bond = ! + " bonds
! = HO–HO = sp2(C)–sp2(C)
" = p–p = p(C)–p(C)
1. Predict the electron domain geometry and the molecular geometry of the following molecules.
a. IF4+ b. SF5+ c. SeH2 d. NF3 e. NH2- f. BrF3 g. BrF5 h. SCl4 i. PF6- j. IF42. According to the valence shell electron pair repulsion theory the electron domain geometry around sulfur in the
molecule SF4 will be best described as
a. tetrahedral
b. octahedral
c. square planar
d. trigonal planar
e. trigonal bipyramidal
3. Which molecule(s) below would be nonpolar? There may be more than one molecule.
a. H2S
b. PF3
c. TeF6
4. For each molecule, indicate the polarity of each bond and the overall polarity of the molecule (net dipole) by
drawing arrows representing the polarity of each bond and then an arrow representing the sum of the bond
polarities. For each molecule, state whether that molecule is polar.
a. CF4
b. NH3
c. COCl2 (ENCl = 3.2; ENO = 3.5)
d. NO35. For each molecule below, determined the electron domain geometry and hybridization around the central
atom.
a. CO3-2 b. SO2 c. SO3-2 d. BCl3 e. XeF4 f. PCl5 g. I36. I. a. For the molecule, C2H4, using valence bond theory (hybrid orbital theory), what orbitals are used to make
the C=C bonds? (Hint: Draw the Lewis dot structure first.)
b. What orbitals are used to make the C–H bonds?
II. a. For the molecule, Cl2CO, using valence bond theory (hybrid orbital theory), what orbitals are used to make
the C=O bonds? (Hint: Draw the Lewis dot structure first.)
b. What orbitals are used to make the C–Cl bonds?
a
H
H
H
7. a. Draw in the lone pairs in the structure that are missing.
H
C
C
C
C
C
H
b. What is the bond order for the carbon-carbon bond labeled “a”?
C
C
c. What is the hybridization on the N labeled “b”?
d
d. What is the angle for the H–C–C labeled “c”?
C
C
c
H
C
O
H
e. What is the electron domain geometry for the O atom labeled “f”?
f. What orbitals overlap to form the bond labeled “e”?
f
N
b
g. How many σ and π bonds are in the molecule?
H
H
h. What is the molecular geometry around the N labeled “b”?
i. What orbitals overlap to form the σ bond labeled “d”? What orbitals are used to form the π bond labeled “d”?
8. Use the structure to answer the questions below about the structure and bonding in this molecule.
a. What is the angle between the H-C-H labeled “c”?
H
d
b. What is the electron domain geometry around the carbon atom labeled “b”?
H
C
c. What is the hybridization of the carbon atom labeled “a”?
H
C
C
C
d. What bonds make up the triple bond between carbon atoms labeled “d”?
C
e. What orbitals overlap to make the bonds between the C atoms labeled “d”?
H
b
f. How many σ and π bonds are there in the molecule?
c
a
H
H
4. a. CF4 - nonpolar
b. NH3 - polar
c. COCl2 - polar
d. NO3- is nonpolar
5. a. triangular planar, sp2 b. trigonal planar, sp2 c. tetrahedral, sp3 d. triangular planar, sp2
e. octahedral; d2sp3 f. trigonal bipyramidal, dsp3 g. trigonal bipyramidal, dsp3
6. I. a. σ bond = sp2(C)–sp2(C); π bond = p(C)–p(C)
b. σ bond: sp2(C)–1s(H)
II. a. σ bond = sp2(C)–sp2(C); π bond = p(C)–p(C)
b. σ bond = sp2(C)–sp3(Cl)
C
C
H
C
C
C
H
C
H
H
ANSWERS
1. a. electron domain geometry: trigonal bipyramidal, molecular geometry: Seesaw
b. electron domain geometry: trigonal bipyramidal, molecular geometry: Trigonal bipyramidal
c. electron domain geometry: tetrahedral, molecular geometry: Bent
d. electron domain geometry: tetrahedral, molecular geometry: Trigonal pyramidal
e. electron domain geometry: tetrahedral, molecular geometry: Bent
f. electron domain geometry: trigonal bipyramidal, molecular geometry: T-shaped
g. electron domain geometry: octahedral, molecular geometry: Square pyramidal
h. electron domain geometry: trigonal bipyramidal, molecular geometry: Seesaw
i. electron domain geometry: octahedral, molecular geometry: Octahedral
j. electron domain geometry: octahedral, molecular geometry: Square planar
2. e
3. TeF6
e
7. a. See figure right.
b. 1.5
a
H
H
H
c. sp3
H
C
C
C
C
C
H
d. 120˚
C
C
e
d
e. tetrahedral
C
C
c
f. 1s(H)-sp(C)
H
C
O
H
g. 21σ bonds; 6π bonds
f
N
b
h. trigonal pyramid
H
H
i. σ: sp2(C)–sp2(C); π: p(C)–p(C)
8. a. 109.5˚ b. linear c. sp2 d. 1σ + 2π
e. σ bond: sp(C)–sp(C); π bond: px(C)–px(C); π bond: py(C)–py(C) assumes z-axis is internuclear axis
f. 22σ, 5π
VSEPR GEOMETRIES (memorize) Total =
A+B
#atoms bonded
to central atom
A
# lone pairs on
central atom
B
Electron
domain
geometry
2
2
0
Linear
3
3
0
2
4
5
6
Angles
Ex
Polar
HO
Linear
180o
BeCl2
NP*
sp
Trigonal
planar
Trigonal
planar
120o
BF3
NP*
1
Trigonal
planar
Bent
115o or
<120o
AlF2-
P
4
0
Tetrahedral
Tetrahedral
109.5o
CH4
NP*
3
1
Tetrahedral
Trigonal
pyramid
107.5o
or
<109.5o
NH3
P
2
2
Tetrahedral
Bent
104.5o
or
<109.5o
H2O
P
5
0
Trigonal
bipyramid
Trigonal
bipyramid
90o,
120o
PCl5
NP**
4
1
Trigonal
bipyramid
See-saw
90o,
115o
SF4
P
3
2
Trigonal
bipyramid
T-shape
90o
ClF3
P
2
3
Trigonal
bipyramid
Linear
180o
I3-
NP*
90o
SF6
NP**
6
0
5
1
4
2
Picture
Molecular
geometry
Picture
Octahedral
Octahedral
Octahedral
Square
pyramid
90o
IF5
P
90o
XeF4
NP**
Octahedral
Square
planar
sp2
sp2
sp3
sp3
sp3
dsp3
dsp3
dsp3
dsp3
d2sp3
d2sp3
d2sp3
P = polar molecule; NP = nonpolar molecule
* = The nonpolar (NP) molecules assume the atoms around the central atom are identical; if not, the molecule will be polar
** = In the square planar or octahedral nonpolar molecules, the atoms around the central atom can be different atoms as
long as the atoms that are 180˚ from each other are the same. In the case of a trigonal bipyramid molecule the atoms
that are 120˚ apart or the atoms that are 180˚ apart need to be the same for the molecule to be nonpolar (e.g., PCl2F3
can be polar or nonpolar depending on the positions of the Cl and F atoms).
Note: Some lectures only cover sp, sp2, and sp3, and do not include the expanded octet hybridizations, dsp3 and d2sp3.
Images: Public Domain from Wikipedia.org (http://en.wikipedia.org/wiki/VSEPR_theory)