Download If electronegativity difference is greater than 2, then the bond is ionic

Recall electronegativity (Linus Pauling Caltech)­­
A measure of how well an atom competes for electrons when bonded to another atom. F is the most (4.0). O is the next most electronegative (3.5). N and Cl are next.
If electronegativity difference is greater than 2, then the bond is ionic. Remember, the greater the electronegativity difference the more ionic character. It is simplistic and erroneous to state a bond as either 100% ionic or 100% covalent.
If the electronegativity difference is between 0.4 and 2.0, then the bond is polar covalent. Electrons are shared between the two atoms but they are not shared equally.
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How do I tell if an atom has a nonzero formal charge?
Remember that the central atoms in the structures below have zero formal charge. You should calculate formal charge if any atom in your Kekule structure does not have an octet, or look like one of the atoms below.
Group 14
Group 15
Group 16
Group 17
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These atoms all have +1 formal charges and an octet. Note they have one more bond and 1 less lone pair than when neutral with an octet. You will encounter all of these situations in this course except C with 5 bonds. Carbon CANNOT form 5 bonds, do not draw C with 5 bonds.
Group 14
Group 15
Group 16
4 bonds
0 lone pais
3 bonds
1 lone pair
5 bonds
0 lone pairs
NO!!!
Group 17
2 bonds
2 lone pairs
Can carbon ever be +1? Yes if it has an incomplete octet. These are called carbocations. They are not stable, but they are common intermediates in chemical reactions. N, O, and the halogens do NOT have an incomplete octet. In general, if an electronegative atom has a negative charge, then it has an octet.
Examples of carbocations. Remember all are unstable, and there are 3 subclasses of the most common type. We will examine these further later in the course.
Most common carbocation is sp2 with 3 bonds to separate atoms. We subdivide this type of carbocation into four different types: Primary, secondary, tertiary, and benzylic. More on this later in the course.
This is a vinyl or an sp carbocation. It is very unstable and it is not common even as an intermediate in reactions
NOT GOOD
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These atoms all have ­1 formal charges and an octet. Note they have one less bond and 1 more lone pair than when neutral with an octet. You will encounter all of these situations in this course. The carbanions (Group 14) are the least stable, because C is the least electronegative, and therefore less able to bear a negative charge.
Group 14
3 bonds
1 lone pairs
Group 15
Group 16
2 bonds
2 lone pais
1 bonds
3 lone pair
Group 17
0 bonds
4 lone pairs
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Draw the Lewis structure of the nitrate ion. Label all atoms that have a formal charge.
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Draw the Lewis structure of the folowing ions. Label all atoms that have a formal charge.
• sulfate ion
• carbonate ion
• sulfite ion
• chlorate ion
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Draw the Lewis structure of the folowing ions. Label all atoms that have a formal charge.
• sulfate ion
• carbonate ion
• sulfite ion
• chlorate ion
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VSEPR Theory Steps in predicting molecular geometries.
1. Draw Lewis Structure
2. Count the number of bonded atoms + lone pairs
3. Repulsion of these areas of negative charge determines shape. Maximize distance between areas of (­) charge to determine geometry of the orbitals.
4. Although lone pairs affect the shape, we disregard them when naming the shape.
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Draw Lewis Structures of the following
PCl5
TeCl4
ClF3
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I3
BrF5
SF6
XeF4
C2H4
AlCl3
BeCl2
SO42­
NO3­
C2H2
CH2O
ICl4­
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SnCl5
CO32­
CH4
CO2
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Draw Lewis Structures of the following
PCl5
TeCl4
ClF3
I 3­
BrF5
SF6
XeF4
C2H4
SO42­
NO3­
C2H2
CH2O
ICl4­
SnCl5­
CO32­
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Bonded hybridization
atoms +lone bonding pairs
arrangement
example
bond
angles
shape
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When atoms bond, atomic orbitals become hybrid orbitals.
Atomic orbitals cannot explain how carbon is bonded in methane. Bond angles would be 90 degrees.
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To predict whether a molecule is polar.
1. Draw Lewis structure
2. Draw shape.
3. Draw in the individual dipoles.
4. Decide if they cancel.
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Is carbon tetrachloride polar?
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Bond line formulas.
Organic Molecules are sometimes large and we need a more convenient way to represent them. Bond line formulas do not include C atoms or H atoms that are bonded to C
There is a C wherever there is an END or a BEND. There is the correct number of H­atoms on the C so that it has four bonds.
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Energy
Atomic Orbitals
Hybrid orbitals
2p
They hybridize
4 sp3 hybrid orbitals that are equal in energy. They are also at 109.5 degrees.
2s
Covalent bond is overlap between sp3 on C and a 1s on H.
C is 1s22s22p2
1s
sp3
sp3
1s
sp3
sp3
1s
1s
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Energy
Atomic Orbitals
Hybrid orbitals
One of the p­orbitals is not hybridized. It is perpendicular to the sp2 orbitals.
2p
2p
sp2
They hybridize
2s
3 sp2 hybrid orbitals that are equal in energy. They are at 120 degrees.
C is 1s22s22p2
Overhead view
side view
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Energy
Atomic Orbitals
Hybrid orbitals
2p
2p
They hybridize
sp2
Two of the p­orbitals are not hybridized. They are perpendicular to each other and the two sp orbitals.
2s
2 sp hybrid orbitals that are equal in energy. They are at 180 degrees.
C is 1s22s22p2
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Resonance hybrids are multiple Lewis Structures used to describe a molecule or ion. Electrons in resonance structures are delocalized across multiple atoms.
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Rules:
1. atoms must be bonded in the same arrangement.
2. Total number of electrons must be the same in all resonance structures.
3. Total charge must be the same.
4. Follow octet rule if at all possible.
5. Most important resonance structures contain negative charges on electronegative atoms.
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Draw resonance structures of 1. nitrate ion
2. CH2N2
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Calculating enthalpy of reaction using bond dissociation energies. Bond dissociation energy (BDE)­­the energy required to pull two bonded atoms apart, giving each atom 1/2 of the shared electrons.
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Using the table on page 356, please find ∆H for the following reactions.
CH3COOH + CH3OH 2 NO2
N2O4
H2 + I2
2 HI
CH3COOCH3 + H2O
C2H2 + 2 H2
C2H6
2 C3H6 + 9 O2
6 CO2 + 6 H2O
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Using the table on page 356, please find ∆H for the following reactions.
CH3COOH + CH3OH 2 NO2
N2O4
H2 + I2
2 HI
CH3COOCH3 + H2O
C2H2 + 2 H2
C2H6
2 C3H6 + 9 O2
6 CO2 + 6 H2O
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