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Final Exam F 12/11 9 am We have the room until 12 pm. Review Session R 12/10 2 pm SL 110. These arrangements are illustrated below with balloons and models of molecules for each.

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Final exam f 12 11 9 am we have the room until 12 pm l.jpg

Final ExamF 12/119 am We have the room until 12 pm.

Review Session

R 12/10 2 pm SL 110



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Predicting Molecular Geometry Using VSEPR positions of the atoms, not the lone pairs. The direction in space of the bonding pairs gives the molecular geometry.

  • Write the electron-dot formula from the formula.

  • Based on the electron-dot formula, determine the number of electron pairs around the central atom (including bonding and nonbonding pairs).

  • Determine the arrangement of the electron pairs about the central atom (Figure 10.3).

  • Obtain the molecular geometry from the directions of the bonding pairs for this arrangement (Figure 10.4).


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The electron-pair arrangement is tetrahedral. positions of the atoms, not the lone pairs. The direction in space of the bonding pairs gives the molecular geometry.

Any three pairs are arranged as a trigonal pyramid.

When one pair of the four is a lone pair, the geometry is trigonal pyramidal.



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ICl species:3 has 1(7) + 3(7) = 28 valence electrons. I is the central atom. The electron-dot formula is

There are five regions: three bonding and two lone pairs.

The electron-pair arrangement is trigonal pyramidal.

The geometry is T-shaped.


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ICl species:4- has 1(7) + 4(7) + 1 = 36 valence electrons. I is the central element. The electron-dot formula is

There are six regions around I: four bonding and two lone pairs.

The electron-pair arrangement is octahedral.

The geometry is square planar.


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Predicting Bond Angles species:

  • The angles 180°, 120°, 109.5°, and so on are the bond angles when the central atom has no lone pair and all bonds are with the same other atom.

  • When this is not the case, the bond angles deviate from these values in sometimes predictable ways.

  • Because a lone pair tends to require more space than a bonding pair, it tends to reduce the bond angles.


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The impact of lone pair(s) on bond angle for tetrahedral electron-pair arrangements has been experimentally determined.


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Multiple bonds require more space than single bonds and, therefore, constrict the bond angle. This situation is illustrated below, again with experimentally determined bond angles.


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Dipole Moment therefore, constrict the bond angle. This situation is illustrated below, again with experimentally determined bond angles.

A quantitative measure of the degree of charge separation in a molecule


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Measurements are based on the fact that polar molecules are oriented by an electric field. This orientation affects the capacitance of the charged plates that create the electric field.


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In part A, there is no electric field; molecules are oriented randomly.

In part B, there is an electric field; molecules align themselves against the field.


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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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  • 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.

  • 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. We’ll 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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  • 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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  • The formula AX molecules; in response, they orient themselves in a 3 can have the following molecular geometries and dipole moments:

  • Trigonal planar (zero)

  • Trigonal pyramidal (can be nonzero)

  • T-shaped (can be nonzero)

  • Molecule Y is likely to be trigonal planar, but might be trigonal pyramidal or T-shaped.

  • Molecule Z must be either trigonal pyramidal or T-shaped.


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GeF zero dipole moment?4: 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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SF zero dipole moment?2: 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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XeF zero dipole moment?2: 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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AsF zero dipole moment?3: 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 geometryzero dipole moment

  • b. SF2 bent molecular geometry nonzero dipole moment

  • c. XeF2 linear molecular geometryzero dipole moment

  • d. AsF3 trigonal pyramidal molecular geometry nonzero dipole moment


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  • Valence bond theory zero dipole moment? 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 zero dipole moment?

  • • 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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  • Hybrid orbitals zero dipole moment? are orbitals used to describe the bonding that is obtained by taking combinations of the atomic orbitals of the isolated atoms.

  • 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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  • To obtain the bonding description about any atom in a molecule:

  • 1. Write the Lewis electron-dot formula.

  • 2. Use VSEPR to determine the electron arrangement about the atom.

  • 3. From the arrangement, deduce the hybrid orbitals.

  • 4. Assign the valence electrons to the hybrid orbitals one at a time, pairing only when necessary.

  • 5. Form bonds by overlapping singly occupied hybrid orbitals with singly occupied orbitals of another atom.


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  • Let’s look at the methane molecule, CH molecule:4. Simply using the atomic orbital diagram, it is difficult to explain its four identical C—H bonds.

  • The valence bond theory allows us to explain this in two steps: promotion and hybridization.


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First, the paired 2 molecule:s 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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The sp molecule:3 Hybrid Orbitals in NH3 and H2O


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The sp molecule:3d Hybrid Orbitals in PCl5


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The sp molecule:3d2 Hybrid Orbitals in SF6

Sulfur Hexafluoride -- SF6


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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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1 atom in Ns

1s

2s

2p

sp3

  • The orbital diagram of the ground-state N atom is

  • The sp3 hybridized N atom is

  • Consider one N in N2F4: the two N—F bonds are formed by the overlap of a half-filled sp3 orbital with a half-filled 2p orbital on F. The N—N bond forms from the overlap of a half-filled sp3 orbital on each. The lone pair occupies one sp3 orbital.


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3 atom in Ns

3s

3p

3p

3d

3d

  • Use valence bond theory to describe the bonding in the ClF2- ion.

  • The valence orbital diagram for the Cl- ion is

  • After the promotion to get two half-filled orbitals, the orbital diagram is


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sp atom in N3d

3d

  • The sp3d hybridized orbital diagram is

  • Two Cl—F bonds are formed from the overlap of two half-filled sp3d orbitals with half-filled 2p orbitals on the F atom. These use the axial positions of the trigonal bipyramid.

  • Three lone pairs occupy three sp3d orbitals. These are in the equatorial position of the trigonal bipyramid.


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  • To describe a multiple bond, we need to distinguish between two kinds of bonds.

  • 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.

  • 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 two kinds of bonds.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, C two kinds of bonds.2H2, 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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Restricted Rotation of two kinds of bonds.-Bonded Molecules

A) Cis - 1,2 dichloroethylene B) trans - 1,2 dichloroethylene


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cis two kinds of bonds.

trans

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 C=C bond. This fixes the geometric positions of Cl: either on the same side (cis) or on different sides (trans) of the C=C bond.


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We can see this illustrated with two compounds: two kinds of bonds.

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 60°C.

Boiling point 48°C.


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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 orbitals and two sets of two parallel s

2s

2p

  • 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 orbitals and two sets of two parallel s

1s

2s

2p

sp2

2p

  • After promotion, the orbital diagram is

  • After hybridization, the orbital diagram is


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  • The C—H orbitals and two sets of two parallel s bonds are formed from the overlap of two C sp2 hybrid orbitals with the 1s orbital on the H atoms.

  • The C—O s bond is formed from the overlap of one sp2 hybrid orbital and one O half-filled p orbital.

  • The C—O p bond is formed from the sideways overlap of the C 2p orbital and an O 2p orbital.


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That's all, folks!! orbitals and two sets of two parallel