Covalent Bonding Orbitals

1

• Covalent Bonding Orbitals
• Sec 9.1 Hybridization and the Localized Electron
  Model
• In this section we will consider the orbitals
  used for bonding in covalent compounds (or in
  ions containing non-metals). The orbitals
  available for bonding are one s and three p
  orbitals, and for elements in period 3 (and
  beyond) we also have ___ orbitals available for
  bonding.
• sp3 hybridization pg. 414-415
• Consider the molecule methane which has __
  regions of electrons around the central atom (all
  of which are bonding regions). This would mean
  using the s and 3 p orbitals (a total of 4
  orbitals). We know the shape of this molecule is
  tetrahedral with bond angles of ______. This
  fact poses 2 problems for what we know of
  orbitals
• 1. The 3 p orbitals (px, py and pz) are all ___
  to each other which would mean bond angles of
  90o.
• 2. The overlap of an s orbital with the s
  orbital of hydrogen would yield a bond of
  different length and strength than the overlap of
  a p orbital with an s orbital of hydrogen.
  Experimental evidence indicates that all of the
  _________ are identical.
• Explanation the s and 3 p orbitals combine to
  form ___ identical orbitals called sp3 orbitals.
  The bond formed from the overlap of the sp3
  orbital and the s orbital of hydrogen creates a
  bond called a _______ (?) bond (sigma bonds are
  the result of the lobes of orbitals overlapping
  as the point toward each other).

d
4
109.5o
-
bonds
4
sigma
Molecular Geometry Review For sp3 hybridization
there are ___ regions of electrons around the
central atom. a. If all 4 regions are bonding,
the molecular geometry (shape and bond angle)
is b. If 3 regions are bonding and 1 is a lone
pair, the molecular geometry is c. If 2 regions
are bonding and 2 are lone pairs, the molecular
geometry is See sample exercise 9.1 pg. 416
4
Tetrahedral 109.5o
Trigonal pyramidal 107o
Bent 104.5o
NH3 has 4 identical orbitals (3 bonding pairs, 1
lone pair) Therefore it has sp3 hybrid orbitals
2

• sp2 hybridization pg 416-418
• When there are __ regions of electrons around the
  central atom, the atoms orbitals are sp2
  hybridized.
• Here, there will be 3 identical ___ bonds and 1
  free p orbital to overlap with another p orbital
  to form what is called a pi (?) bond (see page
  418). The combination of one ? and one ? bond in
  the same region is called a double bond.
• Note In any double bond there is 1 ? and 1 ?
  bond.
• Molecular Geometry Review
• For sp2 hybridization there are ___ regions of
  electrons around the central atom.
• a. If all 3 regions are bonding, the
  molecular geometry (shape and bond angle) is
• b. If 2 regions are bonding and 1 is a lone
  pair, the molecular geometry is
• sp hybridization pg 418-420
• When there are __ regions around the central atom
  then the atoms orbitals are sp hybridized.
  Here, a triple bond may form, composed of 2 ___
  bonds and 1 ____ bond.
• .
• See Sample Exercise 9.2 pg. 420

s
3
trigonal planar 120o
bent 118o
2
s
p
In N2 there are 2 regions of electrons around
each atom, thereby making each N atom sp
hybridized.
In the triple bond there is 1 s and 2 p bonds
dsp3 hybidization pg 421 When there are ___
regions of electrons around the central atom the
atoms orbitals are dsp3 hybridized (here there
is 1 d, 1 s and 3 p orbitals)
5
3
5

• For dsp3 hybridization there are ___ regions of
  electrons around the central atom.
• If all 5 regions are bonding, the molecular
  geometry (shape and bond angle) is
• If 4 regions are bonding and 1 is a lone pair,
  the molecular geometry is
• If 3 regions are bonding and 2 are lone pairs,
  the molecular geometry is
• If 2 regions are bonding and 3 are lone pair
  regions, the molecular geometry is
• See sample Exercise 9.3 pg. 421
• d2sp3 hybidization pg. 422-423
• When there are ___ regions of electrons around
  the central atom the atoms orbitals are d2sp3
  hybridized (here there is 2 d, 1 s and 3 p
  orbitals)
• For d2sp3 hybridization there are ___ regions of
  electrons around the central atom.
• If all 6 regions are bonding, the molecular
  geometry (shape and bond angle) is
• If 5 regions are bonding and 1 is a lone pair,
  the molecular geometry is
• If 4 regions are bonding and 2 are lone pairs,
  the molecular geometry is
• See Sample Exercise 9.4 pg. 422
• See summary on fig.9.24 pg. 424

Trigonal bipyramidal 120o 90o
seesaw 180o 118o 90o
T-shaped 180o 90o
Linear 180
5 regions of electrons around the central I so it
is dsp3 hybridized
The outer Is have 4 regions of electrons so they
are sp3 hybridized
6
6
octahedral 90o
Square pyramidal 180o 90o
square planar 90o
6 regions of electrons around the Xe so it is
d2sp3 hybridized
CO
Each atom is sp hybridized (linear shape). The
triple bond is the result of 1 sigma and 2 pi
bond
4
-1
..
F
..
.. ..

• F - B F

..
Each atom is sp3 hybridized (tetrahedral
shape) Each bond is a sigma bond

F
..
Each F atom is sp3 hybridized

.. .. .. .. .. .. ..

The Xe atom is dsp3 hybridized (linear
shape)
F-Xe F
Each bond is a sigma bond

5

• Sec 9.5 Combining the Localized Electron and
  Molecular Orbital Models
• The main problem with the localized electron
  model is the assumption that electrons are
  _______________. This problem becomes most
  apparent for molecules that have Lewis structures
  that can be drawn in different ways, such as an
  octet or lowest ________ charge.
• 1. Only experimental evidence can truly show the
  correct structure in terms of _______ length and
  strength.
• 2. ______________ structures help to eliminate
  the problem although they fail as well.
• 3. Examine ozone and the nitrate ion on page 438
  fig. 9.45 and notice that lowest formal charge is
  not used for these _____________. Further,
  notice that the double bond, which changes
  positions, is composed of one ____ and one ____
  bond. The ? bond results from an overlap of ___
  orbitals where the ? bond is a result of overlap
  of ___ orbitals (a ? bond in a triple bond would
  result from an overlap of ___ orbitals). Here,
  the p orbitals are ___ to the sp2 hybrid orbitals
  and since all the atoms have these p orbitals
  which are __ to each other the electrons can
  oscillate between the different p orbitals. Note
  the ___ bond is always localized between the
  atoms.
• 4. Therefore, ___ bonds are localized and ___
  bonds can oscillate positions (or are called
  delocalized).
• 5. Consider the benzene molecule ( )
  on pg. 439 fig. 9.46 where alternating double
  bonds can oscillate positions. Actually, the
  electrons are oscillating between the mutually __
  p orbitals (this is why we say that they are
  delocalized.
• See figs. 9.48, 9.49 pg. 439

localized
formal
bond
Resonance
bonds
p
s
p
sp2
sp
-

s
pi
s
C6H6

6

• Liquids and Solids
• The arrangements of molecules are very different
  for different ________ of matter (solid, liquid,
  gas). In chapter 5 we saw that gases have much
  larger distances between molecules than liquids
  and liquid molecules slip past each other while
  solids are in _________ positions relative to
  each other. The difference in the distance
  between the molecules lies in their difference in
  ____ forces. Solids have relatively _______ IM
  interactions compared to liquids and gases.
• Look at the enthalpy of fusion (energy required
  to ____ 1 mol of a substance) compared to the
  enthalpy of vaporization (energy required to
  __________ 1 mol of a substance) of water on
  page 449. The _______ energy required for water
  to change from solid to liquid compared to change
  from liquid to gas indicates that solids and
  liquids have a much more similar arrangement of
  molecules than gases. See fig. 10.1 pg. 450
• Sec 10.1 Intermolecular Forces
• 1. Ionic compounds (salts whose bonding is a
  result of a __________ of electrons) have high
  melting and boiling points due to the strong
  attraction of oppositely charged ions.
• Covalent molecules (bonding a result of
  __________ of electrons between non-metals) also
  have attractions for each other but much weaker
  than ionic attractions. Attractions between
  molecules are called intermolecular forces, while
  attractions between atoms ________ the molecule
  are called intramolecular forces (also known as
  bonds, covalent bonds).
• Dipole-Dipole Forces
• 1. Molecules that have an overall dipole moment
  are said to be _______. For polar molecules,
  there is an attraction for ____________ charged
  ends of the molecules. This attraction is
  referred to as a ___________________ force (see
  fig. 10.2 pg. 451).
• 2. Dipole-dipole forces are about _____ as
  strong as covalent or ionic bonds and these
  attractions become rapidly diminished with
  distance.
• 3. Polar molecules that have hydrogen bonded to
  very electronegative elements (N,O,F) exhibit
  particularly strong dipole-dipole forces called
  _____________ bonds (see fig. 10.3 on pg. 451 and
  graph on pg. 452, note the effect of H-bonding on
  strength of intermolecular forces).

states
fixed
IM
strong
melt
vaporize
small
transfer
sharing
within
polar
oppositely
Dipole-dipole
1
hydrogen
7

• London Dispersion Forces
• 1. ____ molecules (molecules are covalent
  compounds), particularly non-polar molecules,
  exhibit forces of attraction referred to as
  London dispersion forces.
• 2. London dispersion forces arise due to the
  fact that all molecules contain _____________that
  are not uniformly distributed about the nucleus.
  An asymmetric distribution of electrons results
  in formation of a temporary ___________ which can
  induce a temporary dipole in another molecule.
  There is an attraction between the temporary and
  induced dipole. (see fig 10.5 pg 453)
• 3. The strength of the London dispersion forces
  depends upon the _______________ of the molecule
  (polarizability is the ease to which the
  electrons can become asymetrically distributed)
  which is directly related to the ___________ of
  electrons within the molecule.
• Large molecules or elements have more electrons
  than smaller molecules or elements therefore have
  ______________ London dispersion forces (and
  higher melting and boiling points).Table 10.2 pg.
  453
• Sec 10.2 The Liquid State
• 1. Liquids are not very compressible compared to
  gases (Cartesian divers would be an example) and
  have __________ densities than gases.
• 2. When a polar liquid is poured onto a nonpolar
  surface (such as wax), the water will ________
• This is due to the fact that water is ________
  and is not soluble in a nonpolar solvent (like
  dissolves in like in terms of polarity). So the
  water repels the wax and sticks to itself to make
  the smallest volume possible..
• 3. When water is placed in a glass buret, the
  water is attracted to glass and rises up the
  walls to form a concave _________________. The
  glass is polar. ____________ molecules will form
  a convex meniscus when placed in a glass buret
  (fig. 10.7 pg. 454).
• a. When water is placed in a tube of small
  diameter they _______ up the tube to exhibit the
  property of capillary action. The types of forces
  responsible for this property is the difference
  in cohesion (forces of attractions between the
  molecules) and adhesion (forces of attraction for
  the container).

All
electrons
dipole
polarizability
number
stronger
greater
bead
polar
meniscus
Non-polar
rise
8

• b. In a _________ diameter tube there are much
  stronger adhesion forces than cohesion.
• 4. Surface Tension- the _______________ a liquid
  has to increase its surface area.
• a. Surface tension arises due to an ___________
  attraction of molecules at the suface of the
  liquid for other molecules within the liquid.
  (fig. 10.6 pg. 454).
• b. Molecules with ________ IM forces have high
  surface tensions ( the surface tension of water
  allows for bugs to walk on water and for needles
  to float on water).
• 5. Another property which tells of a liquids
  nature is ___________ (a measure of a liquids
  resistance to flow). Liquids with large IM
  forces have ____________ viscosities than those
  with weak IM forces (example glycerol, small
  molecule but much H-bonding).
• Homework pg. 499-501 6 12 14 16 35 37
  38b,c,e,f 40 41

small
resistance
uneven
strong
viscocity
larger
9

• Sec 10.8 Vapor Pressure
• 1. Vaporization (or evaporation) is the process
  of molecules of _____ kinetic energy escaping
  from the surface of a liquid to behave as gas
  particles.
• 2. Molecules within a liquid at a given
  temperature have a _______ range of kinetic
  energy (or velocity). Some have enough KE to
  escape from the liquid into the gas phase (as T
  increases this number increases). fig. 10.41 pg.
  485
• 3. The rate of evaporation depends upon the
  _________ of the IM force. The weaker the IM
  force, the _________ the substance will
  evaporate.
• 4. The process of evaporation takes energy from
  the surroundings (an _________________ process).
  This energy is called the enthalpy of
  vaporization ( ).
• 5. When a liquid evaporates the loss of high KE
  particles _____________ the temperature of the
  liquid.
• 6. When a liquid is placed into a container and
  is sealed a state of dynamic ______________
  between molecules escaping and condensing (at
  equilibrium the s evaporating and condensing
  are equal). The molecules in the gas phase exert
  a pressure called the ___________ pressure which
  can be measured by a barometer. See fig. 10.38
  pg. 484.
• Pvapor Patm Pcolumn
• 7. Substances which evaporate quickly are said
  to be _____________ (the weaker the IM forces,
  the more volatile the substance).
• 8. Vapor pressure will _____________ with
  increasing temperature. Table 10.8 pg. 485
• The relationship between vapor pressure and
  temperature is not _____________. fig. 10.42 pg.
  486

high
broad
strength
faster
endothermic
?Hvap
lowers
equilibrium
vapor
volatile
increase
linear
Pvap ke-?Hvap / RT or ln P ln k - ?Hvap /
(RT) A plot of ln Pvap vs. 1 / T (K-1) will yield
a straight line with slope _________.
-?Hvap/R
10
The slopes of the lines are equal to -?Hvap / R

• Sample Exercise 10.5 pg .487
• Once ?Hvap is known. If you know the vapor
  pressure at 1 temperature (like normal boiling
  point where the pressure is 1 atm.), the vapor
  pressure may be determined at any other
  temperature.
• ln P1 ?Hvap / (RT1 ) ln k
• ln P2 ?Hvap / (RT2 ) ln k
• so ln P1 ?Hvap / (RT1 ) ln P2 ?Hvap /
  (RT2 )
• ln (P1 / P2) ? Hvap / R (1 / T2
  1/T1) (Clasius-Clapeyron equation)
• Sample Exercise 10.6 pg. 487

To determine ?Hvap multiple the slope by R and
change the sign
Since water has the most negative slope, it will
have the most positive ?Hvap
ln P1 ? Hvap 1 1 P2 R
T2 T1
(0.0030960 K-1 - 0.0033557 K-1 )
P2 23.8 torr /
5279.9 K
( )
e
P2 23.8 torr /
(-0.0002597K-1 )
P1 ? Hvap 1 1 P2 e R
T2 T1
5279.9 K
( )
( )
e
? Hvap 1 1 R T2 T1
P2 94 torr
( )
( )
P2 P1 /
e
43,9000 J/mol 1 1 8.3145 J/molK
323K 298K
( )
( )
P2 23.8 torr /
e
11

• Changes of State
• Terms
• Boiling point the temperature where the vapor
  pressure of the liquid is equal to the
  atmospheric pressure.
• Boiling point will lower with increasing
  altitude table 10.10 pg. 495
• Melting point where the vapor pressures of the
  solid and liquid states are equal. Fig. 10.45 pg.
  490
• Supercooling Cooling a liquid quickly so that
  it may remain a liquid below the melting point at
  atmospheric pressure.
• Superheating Heating a liquid quickly so that
  its temperature is above the boiling point at
  atmospheric pressure (causes bumping which can
  be reduced by boiling chips).
• Enthalpy of fusion energy required to melt
  1mole substance. (6.02 kJ/mol for water)
• Enthalpy of vaporization energy required to
  vaporize 1 mole of substance
• (40.7 kJ/mol for water)
• Specific heat the amount of heat required to
  raise the temperature of 1 g of substance 1oC.
• sice 2.1 J/goC swater 4.2 J/goC ssteam
  2.0 J/goC
• s Cp
• Read cases 1-3 pg. 490 491
• Discuss fig. 10.44 pg. 489

12
q1
q3
q2

• Problem 1 How much heat is required to raise
  the temperature of 20.0 g of H2O at -20.0oC to
  water at 20.0oC?

q1mCp?t
q1(20.0g)(2.1J/goC)(0 (-20.0oC))
qtot q1 q2 q3
q1(20.0g)(2.1J/goC)(20.0oC)
qtot 840.0 J 6683.3 J 1680 J
q1840.0 J
mol 18.015g
q2 20.0 g
6683.3 J
qtot 9203.39 J
6.02.103 J mol
qtot 9200 J
q3mCp?t
q3(20.0g)(4.2J/goC)(20.0oC- 0oC)
q3(20.0g)(4.2J/goC)(20.0oC)
q3 1680 J
13

• 5.00 kJ of heat is added to an unknown mass of
  ice at 0.0oC. If the ice attains a temperature
  of 10.0oC. Determine the mass of the ice.

qtotal q1 q2
5.00.103 J m . mol . 6.02.103 J m (4.2
J/goC) (10.0oC) 18.015 g
mol
5.00.103 J 334.17 m 42.00 J/g m
5.00.103 J 376.17 J /g m
m 5.00.103 J / 376.17 J / g
m 13.3 g
14

• Critical Temperature Temperature above which,
  no matter how great the pressure, a substance
  cannot be liquefied.
• Critical Pressure-Pressure required to liquefy a
  gas at the critical temperature.
• Triple point Pressure and temperature where a
  gas may exist as a solid liquid and gas.
• Sublimation change of state from solid to gas.
• Regelation-application of pressure to liquefy ice
  (application ice skating) due to ice taking up
  more space as a solid than a liquid near
  freezing.
• See differences in phase diagrams for water (fig.
  10.51 pg. 495 and CO2 fig. 105.2 pg. 497)