Final Exam: Cumulative! W 1-3 pm

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Final Exam Cumulative! W 1-3 pm

• Review Session. M 3 pm
• Office Hours T 1-3

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A polar bond is characterized by separation of
electrical charge. Polar molecules, therefore,
have nonzero dipole moments. For HCl, we can
represent the charge separation using d and d-
to indicate partial charges. Because Cl is more
electronegative than H, it has the d- charge,
while H has the d charge.
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1. To determine whether a molecule is polar, we need
   to determine the electron-dot formula and the
   molecular geometry. We then use vectors to
   represent the charge separation. They begin at d
   atoms and go to d- atoms. Vectors have both
   magnitude and direction.
2. We then sum the vectors. If the sum of the
   vectors is zero, the dipole moment is zero. If
   there is a net vector, the molecule is polar.

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To illustrate this process, we use arrows with a
on one end of the arrow. Well look at CO2 and
H2O. CO2 is linear, and H2O is bent.
The vectors add to zero (cancel) for CO2. Its
dipole moment is zero.
For H2O, a net vector points up. Water has a
dipole moment.
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The relationship between molecular geometry and
dipole moment is summarized in Table 10.1.
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• Polar molecules experience attractive forces
  between molecules in response, they orient
  themselves in a d to d- manner. This has an
  impact on molecular properties such as boiling
  point. The attractive forces due to the polarity
  lead the molecule to have a higher boiling point.

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We can see this illustrated with two compounds
cis-1,2-dichloroethene
trans-1,2-dichloroethene
There is no net polarity this is a nonpolar
molecule.
The net polarity is down this is a polar
molecule.
Boiling point 60C.
Boiling point 48C.
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• Which of the following molecules would be
  expected to have a zero dipole moment?
• a. GeF4
• b. SF2
• c. XeF2
• d. AsF3

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GeF4 1(4) 4(7) 32 valence electrons. Ge is
the central atom. 8 electrons are bonding 24 are
nonbonding. Tetrahedral molecular
geometry. GeF4 is nonpolar and has a zero
dipole moment.
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SF2 1(6) 2(7) 20 valence electrons. S is the
central atom. 4 electrons are bonding 16 are
nonbonding.Bent molecular geometry. SF2
is polar and has a nonzero dipole moment.
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XeF2 1(8) 2(7) 22 valence electrons. Xe is
the central atom. 4 electrons are bonding 18 are
nonbonding.Linear molecular geometry. XeF2
is nonpolar and has a zero dipole moment.
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AsF3 1(5) 3(7) 26 valence electrons. As is
the central atom. 6 electrons are bonding 20 are
nonbonding.Trigonal pyramidal molecular
geometry. AsF3 is polar and has a nonzero
dipole moment.
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• Which of the following molecules would be
  expected to have a zero dipole moment?
• a. GeF4 tetrahedral molecular geometry zero
  dipole moment
• b. SF2 bent molecular geometry nonzero dipole
  moment
• c. XeF2 linear molecular geometry zero dipole
  moment
• d. AsF3 trigonal pyramidal molecular
  geometry nonzero dipole moment

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1. Valence bond theory is an approximate theory put
   forth to explain the electron pair or covalent
   bond by quantum mechanics.

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• A bond forms when
• An orbital on one atom comes to occupy a
  portion of the same region of space as an orbital
  on the other atom. The two orbitals are said to
  overlap.
• The total number of electrons in both orbitals
  is no more than two.

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1. The greater the orbital overlap, the stronger the
   bond.
2. Orbitals (except s orbitals) bond in the
   direction in which they protrude or point, so as
   to obtain maximum overlap.

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1. To obtain the bonding description about any atom
   in a molecule
2. 1. Write the Lewis electron-dot formula.
3. 2. Use VSEPR to determine the electron
   arrangement about the atom.
4. 3. From the arrangement, deduce the hybrid
   orbitals.
5. 4. Assign the valence electrons to the hybrid
   orbitals one at a time, pairing only when
   necessary.
6. 5. Form bonds by overlapping singly occupied
   hybrid orbitals with singly occupied orbitals of
   another atom.

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1. Lets look at the methane molecule, CH4. Simply
   using the atomic orbital diagram, it is difficult
   to explain its four identical CH bonds.
2. The valence bond theory allows us to explain this
   in two steps promotion and hybridization.

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1. Hybrid orbitals are orbitals used to describe the
   bonding that is obtained by taking combinations
   of the atomic orbitals of the isolated atoms.
2. The number of hybrid orbitals formed always
   equals the number of atomic orbitals used.

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• Hybrid orbitals are named by using the atomic
  orbitals that combined
• one s orbital one p orbital gives two sp
  orbitals
• one s orbital two p orbitals gives three sp2
  orbitals
• one s orbital three p orbitals gives four
  sp3 orbitals
• one s orbital three p orbitals one d
  orbital gives five sp3d orbitals
• one s orbital three p orbitals two d
  orbitals gives six sp3d2 orbitals

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Hybrid orbitals have definite directional
characteristics, as described in Table 10.2.
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First, the paired 2s electron is promoted to the
unfilled orbital. Now each orbital has one
electron. Second, these orbitals are hybridized,
giving four sp3 hybrid orbitals.
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• Use valence bond theory to describe the bonding
  about an N atom in N2H4.

The Lewis electron-dot structure shows three
bonds and one lone pair around each N atom. They
have a tetrahedral arrangement. A tetrahedral
arrangement has sp3 hybrid orbitals.
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• The orbital diagram of the ground-state N atom is

The sp3 hybridized N atom is
Consider one N in N2F4 the two NF bonds are
formed by the overlap of a half-filled sp3
orbital with a half-filled 2p orbital on F. The
NN bond forms from the overlap of a half-filled
sp3 orbital on each. The lone pair occupies one
sp3 orbital.
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1. One hybrid orbital is required for each bond
   (whether a single or a multiple bond) and for
   each lone pair.
2. Multiple bonding involves the overlap of one
   hybrid orbital and one (for a double bond) or two
   (for a triple bond) nonhybridized p orbitals.

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1. To describe a multiple bond, we need to
   distinguish between two kinds of bonds.
2. A s bond (sigma) has a cylindrical shape about
   the bond axis. It is formed either when two s
   orbitals overlap or with directional orbitals (p
   or hybrid), when they overlap along their axis.
3. A p bond (pi) has an electron distribution above
   and below the bond axis. It is formed by the
   sideways overlap of two parallel p orbitals. This
   overlap occurs when two parallel half-filled p
   orbitals are available after s bonds have formed.

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Figure A illustrates the s bonds in C2H4. The
top of Figure B shows the p orbital on each
carbon at 90 to each other, with no overlap.
The bottom of Figure B shows parallel p orbitals
overlapping to form a p bond.
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In acetylene, C2H2, each C has two s bonds and
two p bonds. The s bonds form using the sp
hybrid orbital on C. This is shown in part A. The
two p bonds form from the overlap of two sets of
parallel p orbitals. This is illustrated in
Figure B.
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The description of a p bond helps to explain the
cis-trans isomers of 1,2-dichloroethene. The
overlap of the parallel p orbitals restricts the
rotation around the CC bond. This fixes the
geometric positions of Cl either on the same
side (cis) or on different sides (trans) of the
CC bond.
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One single bond and one triple bond requires two
hybrid orbitals and two sets of two parallel p
orbitals. That requires sp hybridization.
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1. Describe the bonding about the C atom in
   formaldehyde, CH2O, using valence bond theory.

The electron arrangement is trigonal pyramidal
using sp2 hybrid orbitals. The ground-state
orbital diagram for C is
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1. After promotion, the orbital diagram is

1. After hybridization, the orbital diagram is

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1. The CH s bonds are formed from the overlap of
   two C sp2 hybrid orbitals with the 1s orbital on
   the H atoms.
2. The CO s bond is formed from the overlap of one
   sp2 hybrid orbital and one O half-filled p
   orbital.
3. The CO p bond is formed from the sideways
   overlap of the C 2p orbital and an O 2p orbital.