Covalent Bonds
Other Bonds
James
Bond
Treasury
Bond
Sen. Kit
Bond
Bonding
• A metal and a
nonmetal form an
ionic bond by the
transfer of e-.
• Nonmetal atoms
form covalent bonds
by sharing e-.
Bonding
• A metal and a
nonmetal form an
ionic bond by the
transfer of e-.
• Nonmetal atoms
form covalent bonds
by sharing e-.
…that’s not the whole story.
Consider electronegativity
Call a bond with electronegativity difference of:
0-.4 -- nonpolar covalent bond
.5-1.8 -- polar covalent bond
1.9 and above -- ionic bond
(We can also define % ionic character)
• What kind of bond forms between chlorine
and phosphorus atoms?
• What kind of bond forms between chlorine
and phosphorus atoms?
Cl P
3.0 - 2.1=.9
• This is a polar covalent bond
• (use an absolute value for the difference)
What kind of bond forms between
these atoms?
H and Cl
Cl and F
Na and O
Mg and Mg
Cl and C
F and F
Mg and N
N and O
The Lewis Diagram
• Bar represents a covalent bond,
2 shared electrons
• Unshared pairs fill out the octets.
• Double and triple bars represent double
and triple bonds.
Lewis diagrams
Step 1: Count the total valence electrons available
--use the columns of the periodic chart
--negative ions have extra electrons,
--positive ions are missing electrons
Step 2: Count the total valence electrons needed
--duet rule for hydrogen, or the
--octet rule for everything else
Step 3: Number of bonds = (electrons neededelectrons available) / 2 electrons per bond
Lewis diagrams
Step 4: Choose the central atom (almost always
the unique one), surround it with the others.
Step 5: Connect with one bond to each outer atom.
(PS: Recheck your formula!)
Step 6: Fill in enough multiple bonds to satisfy
step 3
Step 7: Draw in unshared pairs to fill valence
levels.
Don’t
Don’t try to figure out whose electrons are
whose. Electrons are identical.
Don’t string the atoms along. Put one atom
in the center, unless you have 6 or more
atoms.
Don’t EVER put two bonds or an unshared
pair on H.
Lewis diagrams
• Draw a Lewis diagram of hydrogen
cyanide, HCN
Lewis diagrams
• Draw a Lewis diagram of hydrogen
cyanide, HCN
H C N
Try an ion.
• Draw a Lewis diagram for the nitrite, NO2- ,
ion
Lewis diagrams
• Draw a Lewis diagram for the nitrite, NO2- ,
ion
[O N O
]
Resonance
• Which one is preferable?
[O N O
]
[O N O
]
Resonance
• Each is valid. The multiple bond exists in
both locations. This is called resonance.
[O N O
]
[O N O
]
(the double-headed arrow signifies resonance)
Resonance
• Draw three resonance structures for
carbon dioxide.
Formal Charge
• (#valence e-) - (#bonds/2 + unshared e-)
• Try to minimize each. The more
electronegative atom gets a more negative
formal charge.
Formal Charge
• Draw two valid arrangements each for the
atoms in:
H2CO
H2O2
N2O
• (count bonds, put them in, fill in lone pairs)
Formal Charge
• Which is preferable?
H O C H
H
O C H
Formal Charge
• Which is preferable?
H
O C H
H O C H
0 +1 -1 0
or
0 000
Formal Charge
• Which is preferable?
H
O C H
H O C H
0 +1 -1 0
or
0 000
Choose
this one!
Formal Charge
• Which is preferable?
H O O H
H
OO H
Formal Charge
• Which is preferable?
H
OO H
H O O H
0 0 0 0
or
-1 +1 0 0
Formal Charge
• Which is preferable?
H
OO H
H O O H
0 0 0 0
Choose
this one!
or
-1 +1 0 0
Formal Charge
• Which is preferable?
O N N
N O N
Formal Charge
• Which is preferable?
O N N
N O N
-1 +1 0
or +1 -2 +1
Formal Charge
• Which is preferable?
O N N
N O N
-1 +1 0
or +1 -2 +1
Choose
this one!
Formal Charge
• Which is preferable?
O N N
O N N
Formal Charge
• Which is preferable?
O N N
-1 +1 0
O N N
or
0 +1 -1
Formal Charge
• Which is preferable?
O N N
-1 +1 0
Choose
this one!
O N N
or
0 +1 -1
Coordinate covalent bonds
• How did this
C
O
become this?
C
O
Coordinate covalent bonds
• How did this
C
O
become this?
C
O
Carbon monoxide really does have the third
bond. The oxygen donates both electrons
to share. This is a coordinate covalent
bond
C
O
Coordinate covalent bonds
• Draw a Lewis diagram of the ozone (O3)
molecule. Count the formal charge for
each atom and mark a coordinate covalent
bond
Coordinate covalent bonds
• Draw a Lewis diagram of the ozone (O3)
molecule. Count the formal charge for
each atom and mark a coordinate covalent
bond
O O
-1 +1
O
0
Exceptions to the octet rule
Draw a Lewis diagram for the triiodide ion, I3-
Exceptions to the octet rule
Draw a Lewis diagram for the triiodide ion, I3-
Gadzooks!
• When you try to find the number of bonds,
(24-22)/2=1 bond.
• That’s not enough to tie the ion together.
Exceptions to the octet rule
• When that happens—go old school. Circle
your electrons
[ I I I
]
Exceptions to the octet rule
• When that happens—go old school. Circle
your electrons
[ I I I
]
Exceptions to the octet rule
• When that happens—go old school. Circle
your electrons
[ I I I
]
Two single bonds will satisfy the outer two
iodine atoms, the middle one breaks the
octet rule (with 10 electrons).
Exceptions to the octet rule
• Draw a Lewis diagram for XeF4
(The point here is to find out how many
unshared pairs are on the central atom)
Exceptions to the octet rule
• Draw a Lewis diagram for XeF4
(The point here is to find out how many
unshared pairs are on the central atom)
F
F
Xe
F
F
Polar bonds
• We use a
symbol to show a polar
covalent bond.
O
H
HH
• The arrow points toward the more
electronegative atom, the (+) end is less
electronegative
Polar bonds
• Or, mark the molecule’s (+) and (-) parts
dO
d+
H
HH
d+
• The d is the small Greek delta. It indicates
a small change. In this case, a partial
charge
Three properties of polar bonds:
• The less electronegative end of a polar
bond:
d+
d-
H
Cl
--is more positive
--cannot attract the electrons as well
--is farther from the shared pair of
electrons
Molecular Shapes
Most molecules have a central atom that
follows the octet rule. This allows the
following shapes.
– Tetrahedral
– Trigonal pyramid
– Bent
– Linear and
– Trigonal planar
(trigonal=having three
corners)
Molecular Shapes
• Four bonds in four directions makes a
tetrahedral shape
Molecular Shapes
• Three bonds and one lone pair in four
directions makes a trigonal pyramid shape
Molecular Shapes
• Two bonds and two lone pairs in four
directions makes a bent shape
Molecular Shapes
• A double bond holds two electron pairs in
the same direction. With no lone pairs,
this makes a trigonal planar molecule
Molecular Shapes
• One lone pair, with a single and a double
bond gives a bent molecule.
Molecular Shapes
• Two double bonds, or a single and a triple
makes a linear molecule
Molecular Shapes
• Two atoms are always in a straight line, a
linear molecule.
If A=central atom, B=atoms bonded
to it, E=e- pairs:
• AB4—tetrahedral
(no double bonds)
• AB3E-trigonal pyramid
(no double bonds)
• AB2E2 –bent
(no double bonds)
• ABE3 –linear
(no double bonds)
• AB3— trigonal planar
(one double bond)
• AB2E –bent
(one double bond)
• ABE2—linear
(one double bond)
• AB2—linear
(2 doubles or 1 triple)
• ABE—linear (“)
Look for double bonds and unshared pairs
• Determine the shape of each molecule
and ion on the lab that has a single central
atom.
Polarity of molecules
• When polar bonds are not cancelled by
symmetry, you get a polar molecule. A
polar molecule has (+) and (-) parts.
• POLARITY is the first property to look for
when analyzing a molecule !
Polarity
• CH4 has no polar bonds. It is symmetric
• PH3 has no polar bonds It is not symmetric
• CO2 has polar bonds. It is symmetric
• H2O has polar bonds. It is not symmetric
•
Polarity
• CH4 has no polar bonds. It is symmetric
• Not polar!
• PH3 has no polar bonds It is not symmetric
• Not polar!
• CO2 has polar bonds. It is symmetric
• Not polar!
• H2O has polar bonds. It is not symmetric
• Polar!
• Mark each molecule on the lab that is
polar.
• For those that are not polar—why not?
• (PS—don’t even look at the ions. If it has
a whole charge, ignore the partial charges)
Hybridization
• Atomic orbitals combine to form hybrid
orbitals before bonding
(Hydrogen is the only exception)
Before bonding
• The first step is a hybridization of the
valence level
p orbitals
C
H
H
H
H
s orbitals
forms…
Hybridization
• The first step is a hybridization of the
valence level
sp3 orbitals
C
H
H
H
H
The s and p orbitals hybridize to form sp3
orbitals. The sp3 designation shows one s
orbital and 3 p orbitals make the new ones
Hybridization
• The first step is a hybridization of the
valence level
sp3 orbitals
C
H
H
H
H
The number of orbitals is preserved
(4 in  4 out)
Hybridization
C
H
H
H
H
All four bonds are identical. Methane is a
symmetrical molecule.
sp2 Hybridization
• When one p orbital is left out of the
hybridization, it is used to make a double
bond
p orbitals
• C
• H
H
O
s orbitals
…forms….
sp2 Hybridization
• When one p orbital is left out of the
hybridization, it is used to make a double
bond
sp2 orbitals
• C
• H
H
O
Unused p orbitals—will
form the second bond
between C and O
sp2 Hybridization
• When one p orbital is left out of the
hybridization, it is used to make a double
bond
Makes the double bond!
• C
• H
H
O
sp2 Hybridization
Carbon & oxygen share electrons in unused p orbitals
H
C
O::
H
Carbon shares electrons in sp2 orbitals
sp2 Hybridization
Carbon & oxygen share electrons in unused p orbitals
p bond
H
C
H
O::
s bonds
Carbon shares electrons in sp2 orbitals
sp Hybridization
• When two p orbitals are left out of the
hybridization, it is used to make two
double bonds, or a triple bond
• C
• O
O
…forms….
sp Hybridization
• When two p orbitals are left out of the
hybridization, it is used to make two
double bonds, or a triple bond
sp orbitals
• C
• O
Unused p orbitals
O
sp2 orbitals
sp Hybridization
• When two p orbitals are left out of the
hybridization, it is used to make two
double bonds, or a triple bond
• C
• O
O
p bonds
sp Hybridization
• When two p orbitals are left out of the
hybridization, it is used to make two
double bonds, or a triple bond
• C
• O
O
sp Hybridization
Carbon & oxygen share electrons in unused p orbitals
::O
C
O::
Carbon shares electrons in sp orbitals
Look for multiple bonds!
#of Multiples Bonding patterns Hybridization
None
AB4, AB3E,
sp3
AB2E2, ABE3
One
AB3, AB2E
sp2
ABE2
Two
AB2, ABE
sp
Look for multiple bonds!
#of Multiples Bonding patterns Hybridization
None
sp3
One
sp2
Two
sp
What is the hybridization of the
carbon atoms in…
• CCl4
• C2H2
• H2CO
• CO
• C2H6
• CH3OH
• C2H4
• HCOOH
Molecular Orbital Theory (MOT)
• Overlapping s orbitals, or hybridized
orbitals makes a s (sigma) bond
• The electron density is on the SAME line
as the nuclei
s*
s
s
s
Molecular orbitals
• Overlapping p orbitals, makes a p (pi)
bond
• The electron density is on a PARALLEL
line to the line of the nuclei p
p
p
+
+
p*
Molecular orbitals
• A single bond is a s bond
• A double bond is a s bond, and a p bond
above and below the s
• A triple bond is a s bond, with two p
bonds– above/below and front/back
For every bonding molecular orbital (s or p)
an antibonding orbital is formed (s* or p*)
A bond is formed when there are more
bonding than antibonding electrons
VSEPR
• Valence Shell Electron Pair Repulsion
Theory
VSEPR
• Valence Shell Electron Pair Repulsion
Theory
--pronounced “Vesper”
Electron pairs repel each other. Just as it
says.
• VSEPR is used to predict bond angles.
The pairs will space themselves out as far
as possible.
• A lone pair will take as much room as a
bond AND MORE!
• Consider sp3 hybridization
• AB4—like methane. Tetrahedral 109.5o
• AB3E—like ammonia. Pyramidal 107o
• AB2E2—like water. Bent 104.5o angles
--the unshared pairs force the bonds closer
together—bond angles decrease
With sp2 hybridization:
• AB3—like carbonate. Trigonal planar: 120o
• AB2E—like nitrite. Bent: less than 120o
• ABE2—like O2(2 atoms, has to be linear)
With sp hybridization:
• AB2—like carbon dioxide. Linear: 180o
• ABE—like carbon monoxide. Linear: 180o
…but that’s just if you always follow
the rules…
• – like the octet rule.
With dsp3 hybridization:
• AB5—trigonal bipyramid
• AB4E—seesaw
• AB3E2—t-shaped
• AB2E3—linear
• ABE4—linear
With d2sp3 hybridization:
• AB6— octahedral
• AB5E—square pyramid
• AB4E2—square planar
• AB3E3—t-shaped
• AB2E4 , ABE5—linear
What is the shape of…
• All of the molecules and ions on the lab?
• I3-, SF6, XeF4, PCl5, IF5?
• Count the s and p bonds in the following
molecule. Label each bond as s or p
H
H C C
H
C
C
H
O
C H
H
11
3
• Count the s and p bonds in the following
molecule. Label each bond as s or p
H
H C C
H
C
C
H
O
C H
H
• Determine the hybridization of the carbons
and the oxygen atom
H
H C C
H
C
C
H
O
C H
H
• Determine the hybridization of the carbons
and the oxygen atom
sp sp sp3 sp2 sp3
H
H C C
H
C
C
H
O
sp2
C H
H
The molecular aufbau diagram
The molecular aufbau order
s1s2s*1s2s2s2s*2s2s2px2p2py,z4p*2py,z4s*2px2….
• For example:
• O2 has 16 electrons. Its electron
configuration is:
O2 s1s2s*1s2s2s2s*2s2s2px2p2py,z4p*2py,z2
The molecular aufbau order
What is the electron configuration of…
N2
NO
Ne2
Remember: we
couldn’t do a Lewis
diagram with an odd
number of electrons!
The molecular aufbau diagram
The molecular aufbau diagram
These two can
switch places—
no effect on
bonding, but it
causes magnetic
effects we can
measure
The molecular aufbau diagram
These two can
switch places—
no effect on
bonding, but it
causes magnetic
effects we can
measure
Bond order
• The order of a bond in a diatomic molecule
is half the number of shared electrons not
cancelled by antibonding electrons.
• Or:
• (number of bonding e- in the atomsantibonding e-)/2
Bond order
What is the bond order of…
N2
NO
Ne2
Remember: we
couldn’t do a Lewis
diagram with an odd
number of electrons!
You will be responsible for:
• Writing the molecular orbital electron
configuration and
• Calculating the bond order…
• …of any pair of atoms from the second
period as they attempt to form a diatomic
molecule.
Bond Energies
The energy it takes to break a bond is the
amount of energy released as the bond is
formed.
• --measured in kJ/mol
• --can be used to estimate DHrxn
• --can be absorbed or emitted as light.
What is the DHf of NH3?
What is the DHf of NH3?
• Write the reaction
N2 + 3H2 2NH3
What is the DHf of NH3?
• Count the bonds made and broken
N2 + 3H2 2NH3
1 NN triple bond, 3 HH single bonds broken
6 NH single bonds made
What is the DHf of NH3?
• Look up bond energies, and find a total
N2 + 3H2 2NH3
1 molx941kJ/mol+3 molx436kJ/mol= 2249kJ used
6 molx393 kJ/mol=2358 kJ released
What is the DHf of NH3?
• Find the difference, express as kJ/mol
N2 + 3H2 2NH3
2358 kJ-2249kJ=109 kJ more is released, as 2mol
NH3 is produced, DHf=-109kJ/2mol=-55kJ/mol
• It’s an estimate.
• My book claims -46 kJ/mole.
What is the heat of reaction for:
CH4 + O2 H2O +CO2
What is the heat of reaction for:
CH4 +2O2 2H2O +CO2
What is the heat of reaction for:
CH4 +2O2 2H2O +CO2
Break 2 O=O and 4 C-H
Form 4 H-O and 2 C=O
What is the heat of reaction for:
CH4 +2O2 2H2O +CO2
Break 2 O=O and 4 C-H
=2@500 kJ + 4@393kJ=2572kJ
Form 4 H-O and 2 C=O
=4@464kJ + 2@799kJ= 3454 kJ
What is the heat of reaction for:
CH4 +2O2 2H2O +CO2
Break 2 O=O and 4 C-H
=2@500 kJ + 4@393kJ=2572kJ
Form 4 H-O and 2 C=O
=4@464kJ + 2@799kJ= 3454 kJ
-882kJ/mol
What is the heat of reaction for:
2H2O2 2H2O +O2
Tasks
On your test, you will be asked to:
• Describe how and why ionization of metals
and non-metals occurs
• Write EC’s for atoms and ions
• Show formation of ionic and covalent
bonds by electron dot diagrams
• Describe metallic bonding
• Define and identify electrolytes
On your test, you will be asked to:
• Identify particles and types of substances
by bonding
• Draw Lewis diagrams
• Identify shapes of molecules and ions
• Identify types of bonds between atoms
• Describe polarity
On your test, you will be asked to:
•
•
•
•
•
Identify polar and nonpolar molecules
Identify hybridizations
Describe single and double bonds by MOT
Estimate DHrxn by bond energies
Use VSEPR to predict molecular shapes
and bond angles.
• Calculate and justify bond orders for
diatomics from the second period