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TRANSCRIPT
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Ch. 8 Bonding and Molecular Structure
Ch. 9 Orbital Hybridization
The content of Ch. 8 – 9 will be taught through both this PPT and
LM 8-9.1
Text references: 8.1, 8.2, 8.4-8.9, 9.1-9.2
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The rest of this PPT…• Focuses on substances with covalent bonds
– Molecules – Polyatomic ions
• For any molecule or PAion, you will learn to…– draw the Lewis Dot Structure (LDS) – identify the 3D molecular geometry– determine the polarity– Identify hybridization and sigma/pi bonding– (These are all components of LM 8-9.1)
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Chapter 8
LM 8-9.1 Part I
Lewis Dot Structures
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How to draw LDS for molecules and ions:1. Write formula for compound.2. Write symbol for least electronegative atom in center
and write symbols for all other atoms around it.3. Add valence e- from all atoms to find out total # of e- you
must assign to the compound.4. Draw one line (single bond) from central atom to all
outer atoms. (1 line = 2 e-) Subtract 2e- for each line you’ve drawn from the total # valence e-.
5. Give all OUTER atoms their complete “octet.” Subtract this value from total.
6. If there are ANY remaining e-, assign them in pairs around the central atom.
7. If there are no more e- but the central atoms still needs more to complete the octet, remove a lone pair from an outer atom and turn this into a double bond. Repeat if you need to form a triple bond. Ex. #1 Ex. #2
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Example #1
• Draw LDS for nitrogen trifluoride:
Formula = NF3
NF F
F
Total valence e- = 5 + 3(7) = 26e-
Remaining e- = 26 – 6 = 20e-
••••• •
••• •••
••••
• •
Remaining e- = 20 – 18 = 2e-
••
Click this link and go to Lewis Dot Structures for more info.
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Example #2
• Draw LDS for carbon dioxide:
Formula = CO2
O = C = O
Since C did not have enough e- to complete the octet, each oxygen must donate a lone pair of e- to make a double bond.
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Electron Deficient Atoms
• Elements from Groups I, II, & III will not hold 8 e- when bonded
• Group I has 2 e- in bonding
• Group II has 4 e- in bonding
• Group III has 6 e- in bonding
1 valence e-
2 valence e-
3 valence e-
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Expanded Octet
• Some atoms (usually group V and higher) can POTENTIALLY take on more than 8 e- in bonding– Group IV – VIII must always have at least
8 e-, however some will take more
• Ex. AsF5
• As has 5 fluorine atomsbonded to it and thus issharing 10 total e- with the outer atoms
F
F As F
F F
Click on this link and go to Expanded Octet for more info.
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Resonance Structures• When there are equivalent LDS for a molecule
or ion we say that the structure experiences resonance.
• Applies to a symmetric molecule that contains at least one double bond
• The double bond could be in multiple places and still have the same structure.
Click this link and go to Resonance for more info.
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Why Resonance Structures?
• Electrons in molecules w/double bonds are often delocalized between two or more atoms.
• Electrons in a single Lewis structure are assigned to specific atoms-a single Lewis structure is insufficient to show electron delocalization.
• Composite of resonance forms more accurately depicts electron distribution. Molecule or ion doesn’t actually switch between resonance structures. The structure behaves as a blend or composite of all possible structures
• Proof of resonance?– Bond lengths in resonance structures shorter than single
bonds/longer than double bonds– Bonds are stronger than single bonds/weaker than double
bonds
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Resonance Structures
• Draw all resonance structures for ozone (O3)
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Getting a 3D perspective of molecules…For a preview of what’s to come, click on
this link and click on VSEPR Model.
Molecular Geometry
Chapter 8
LM 8-9.1 Part II
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Applying VSEPR to Balloons
• Valence Shell Electron Pair Repulsion = model for predicting molecular geometry (3D arrangement of BP and LP)
• Watch what your instructor can do w/balloons…
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VSEPR• Valence Shell Electron Pair Repulsion model for
predicting molecular geometry• Model describes the 3D arrangement of atoms in
molecule based on # of lone pairs (LP) and bond pairs (BP) of e- on a central atom.
• A molecule adopts the geometry that minimizes the repulsive force among the e- pairs on central atom
• By minimizing repulsion around central atom you get the most stable geometry of a molecule (minimize repulsion to maximize stability)
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For the next slide, draw LDS for:
• BF3, O3
• CH4, NH3, H2O
• PCl5, SF4, ClF3, XeF2
• SF6, BrF5, XeF4What does each grouping have in common? What changes as you go from one grouping to the next? How
might this effect the placement of e- or “geometry” according to VSEPR theory?
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Applying VSEPR Theory• Lowest energy arrangement when there are:
– 2BP, 0LP = – 3BP, 0LP = – 4BP, 0LP = – 5BP, 0LP = – 6BP, 0LP =
• Bond angle = angle formed when 3 atoms bond together
Linear 180o
Trigonal Planar 120o
Tetrahedral 109.5o
Trigonal bipyramidal 120o, 90o
Octahedral 90o
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Linear Shapes
Examples;
BeH2
CO2
HCN
Examples;
HF
HCl
H2
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
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Trigonal Planer Geometry
Examples;
BF3
CO32-
COCl2
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
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Tetrahedral Geometry
Examples;
CH4
CCl4
SO42-
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
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Trigonal Bipyramidal Geometry
Examples;
PF5
PCl5
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
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Octahedral Geometry
Examples;
SF6
SiF63-
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
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Applying VSEPR Theory (cont’d)• Lowest energy arrangement when there are lone
pairs??? Leads to molecular geometry• LP’s push harder on surrounding BP’s
– LP can spread out more in nonbonding orbital than BP can (BP more localized/constrained in space)
– LP’s make bond angles slightly smaller than expected– The bond angle tends to decrease as # of LP
increases• Steric # (total # of e- pairs) determines e- pair
arrangement to minimize repulsion; #BP vs. LP determines name of molecular geometry (as opposed to electronic geometry); LP influence bond angles
• See how this works…try link above
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Steric # = 3 (3 Total Pair of E-)
• 3BP, 0LP
• 2BP, 1LP
Examples;NO2
-
SO2
O3
Trigonal Planar
Bent
<120o
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Steric # = 4 (4 Total Pair of E-) • 4BP, 0LP
• 3BP, 1LP
• 2BP, 2LP
Examples;NH3
NF3
PCl3
Examples;H2OClO2
OF2
Tetrahedral
Trigonal pyramidal
Bent
<109.5o
<<109.5o
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Steric # = 5 (5 Total Pair of E-)• 5BP, 0LP
• 4BP, 1LP
Examples;
SF4
BrF4+
Trigonal bipyramidal
Seesaw
<90o
<120o
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Steric # = 5 (contd)
• 3BP, 2LP
• 2BP, 3LP
Examples;ClF3
BrF3
SeO32-
Examples;XeF2
ICl2-
I3-
T-Shaped
Linear
<90o
180o
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Steric # = 6 (6 Total Pair of E-) • 6BP, 0LP
• 5BP, 1LP 4BP, 2LPExamples;
BrF5
IF5
Examples;XeF4
ICl4-
Octahedral
Square Pyramidal
Square Planar
<90o 90o
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What if there’s >1 central atom?
• Number each atom in the chain (this one has 4 central atoms)
• Count the number of BP and LP on each central atom
• Identify the geometry of each central atom (there’s no geometry for the whole structure)
1 2 3 4
• C1: 4BP, 0LP tetrahedral• C2, 3, 4: 3BP, 0LP trigonal planar (each one)
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Chapter 8Lab 8-9.1 Part III
Molecular Polarity
To be polar or not to be polar, that is the question!
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Bond Polarity vs. Molecular Polarity• Bond polarity – refers to
distribution of e- cloud around 2 atoms in a bond– Polar if unequal distribution
(dipole-one end of bond slightly pos and the other slightly neg) (has a dipole moment)
– Nonpolar if equal distribution (has no dipole moment)
• Determined by difference in EN values of atoms
More info on polarity at this link and select Partial Charges and Bond Dipoles
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Chemical Bond Formation• Chemical bond – results when a chem
rxn occurs between 2 or more atoms and the valence e- reorganize so that a net attractive force occurs between them– Ionic Bonds
• Transfer of valence e-• Electrostatic attractn of opp. charged ions• Usu. btwn M + NM, M + PAion or PAion + NM
– Covalent Bonds• Sharing of valence e-• Electrostatic attractn of e- of one atom to nucleus
of another• Usu. btwn NM + NM
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Covalent Bond Formation
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Electronegativity (EN)• Definition: a value that represents the
relative ability of an atom to attract e- towards itself in a bond– EN is NOT an energy– Scale ranges from 0 - 4
• What this means: larger EN = atom has greater attraction for e- in a bond
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Back
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Electronegativity (EN)• Trend:
– Down Group = decrease– L to R in Period = increase
• Explain Trend:– Down Group = inc shielding e- = decrease pull
on own valence e- = decrease pull on any e- in bond
– L to R in Period = inc Z = increase pull on own valence e- = increase pull on any e- in bond
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Electronegativity (EN)• Deviations:
Group 8: Most Grp 8 elements tend to have EN = 0 b/c they have full outer en levels. This means they tend not to form bonds like other elements. Some Grp 8 elements do have EN values, though, b/c they are large enough to form bonds.
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Is bond ionic or polar?• Difference in EN values between two atoms will
determine if bond is Ionic or Covalent (found in Ch. 8.7)
• On a continuum of EN:
• Can also use distance on PT as indicator of bond type: the closer two atoms are, the more covalent the bond; the farther apart they are, the more ionic the bond
0.4 1.7 40
Nonpolarcovalent
Polarcovalent
Ionic
100% covalent 0% covalent
0% ionic 100% ionic
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Is bond ionic or polar?
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• Molecular polarity – related to distribution of e- cloud around entire molecule – Polar molecule – experiences unequal
distribution of e- cloud (aka: dipole – one end of molecule slightly + and other end slightly -) (has a dipole moment)
– Nonpolar molecule – experiences equal distribution of e- cloud (no dipole moment)
• Determined by symmetry of molecule (geometry and arrangement of LP)
• LP tend to cause asymmetry in structure which results in a shift in the e- cloud
Bond Polarity vs. Molecular Polarity
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What is a dipole moment?• A substance possesses a dipole moment if its centers of
positive and negative charge
do not coincide and cancel each other out.
= e x d
(expressed in Debye units)
(you won’t need to use this)
• The larger the dipole moment, the more polar is the bond or
molecule (due to a greater asymmetry in the distribution of
electrons around the bond or molecule)
• The greater the EN, the larger the dipole moment for a bond.
——++
polarpolar
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Why is water polar?• H—O bonds (polar bonds) are not
“balanced” around the structure (the pull of each bond does not cancel the other)
• Lone pairs cause asymmetry (imbalance)
• E- cloud of molecule is shiftedsuch that there is greaterneg chargenear O and greater pos charge by H
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General Rule for Molecular Polarity• If all outer atoms are same and structure is
symmetrical (dipole moments of bonds “cancel” each other out b/c same magnitude but opposite direction) molecule is NP (no overall dipole moment)– If 0 LP then NP (if all outer atoms same)
• If structure is asymmetrically arranged (has LP or different outer atoms) molecule is Polar (has dipole moment)– If has LP* then polar– *Exception: Square planar (4 BP, 2LP) and Linear (2 BP,
3 LP) are NP even though they have LP– LP’s cancel each other out
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Attractive forces in & btwn molecules
• Intramolecular forces = attractive forces within the molecule (bonds)
• Intermolecular forces (IMF) = attractive forces between molecules (sticky factors)
• Properties of covalent compounds are attributed to their IMF– Ex. Polar cmpds tend to have higher MP and
BP due to stronger IMF
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HybridizationThe “real story” behind bonding…
For a preview of what’s to come, click on this link and select
Hybridization.
Chapter 9Lab 8-9.1 Part IV
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Draw the LDS of CH4:
What is the e- configuration for C? for H?
Carbon = 2s2 2p2 H = 1s1
s px py pz s s
s s
Is there really room for each H to share e- with Carbon?
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Carbon = 2s2 2p2 H = 1s1
s px py pz s s
s s
Is there really room for Hydrogen to share 4 e- with Carbon?
No! Carbon must make room for 2 more e- so it promotes an e- (moves it to a p orbital)
s px py pz
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Here’s what it would look like so far:
p
p
p
s
Carbon = 2s2 2p2H = 1s1
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And then if Hydrogen bonds with Carbon:
Remember that each p orbital is on an X, Y or Z axis at 90o angles from one another.
Also remember that the s orbital is smaller in radius than the p orbitals.
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Evidence that something “different” is happening…
• Each C-H bond in methane has the same length (109 pm)
• Each H-C-H bond angle is the same (109.5o)• How is this possible if :
– the s and p orbitals are different sizes from one another (this would lead to bonds having different lengths)?
– The s and p orbitals are at 90o not at 109.5o angles
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Then answer is….Hybridization
• All of the s, p atomic orbitals mix to create new hybrid orbitals
• # of new hyrbrid orbitals = # of atomic orbitals that were mixedEX. If s and p atomic orbitals mix get 2 new hybrid orbitals, each called sp hybrid
• Each hybrid orbital is exactly like the other and the geometry is based on these (not atomic orbitals)
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More on Hybridization…• Identify the hybridization if the following atomic
orbitals are mixed:– s + p + p– s + p + p + p + d
• What geometry could you expect from the above hybridization?
sp2 hybridization
sp3d hybridization
3 hybrids (3BP or 2BP, 1LP)
5 hybrids (5BP or 4BP, 1LP, or 3BP, 2 LP or 2BP, 3LP)
Hybrids can contain LP or BP. # BP & LP determines geometry. Even if you know hybridization, must know what each orbital contains to determine geometry.
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http://www.mikeblaber.org/oldwine/chm1045/notes/Geometry/Hybrid/Geom05.htm
BeF2
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BH3
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NH3
H2O
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Sulfur difluoride1. Draw the LDS.
2. Identify the geometry.
3. Is the molecule polar?
4. What is the hybridization of the central atom?
S
FF
•• ••
Bent
Yes, it is polar
2BP+2LP = 4 hybrids = sp3 hybridization
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Xenon tetrafluoride1. Draw the LDS.
2. Identify the geometry.
3. Is the molecule polar?
4. What is the hybridization of the central atom?
XeFF
•
Squar Planar
No, it is not polar
4BP+2LP = 6 hybrids = sp3d2 hybridization
F F•
••
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Sigma and Pi Bonding• Draw LDS and Label all sigma and pi bonds
in each structure:
• Ethane (C2H6)
• Ethene (Ethylene) (C2H4)
• Ethyne (Acetylene) (C2H2)
Go to this link for more info about this section.
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Sigma and Pi Bonding• Ethane (C2H6)
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Sigma and Pi Bonding• Ethene (Ethylene) (C2H4)
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Sigma and Pi Bonding• Ethyne (Acetylene) (C2H2)
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Sigma and Pi Bonding• Sigma Bonding () – head to head overlap
of hybrid or unhybridized orbitals
• Pi Bonding () – side to side overlap of unhybridized p bonds (contains 2 e- total)
*All single bonds consist of bonds.*One of the double bonds is a bond.
*The second and third bond of a double/triple bond are bonds.*Can only occur when central atom is sp or sp2 hybridized
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Benzene (C6H6)
b/c of resonance we write…
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Covalent Bond Characteristics
• Bond Order: The number of bonds between two atoms
• C—C• C==C • C C
• Bond Energy: Energy a bond must absorb to break (sign?)
• Bond Length: Avg distance between the centers of two nuclei (Next slide…)
• How are they all related?
What’s the BO in a resonance structure? BO = 1
BO = 2
BO = 3
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Determining Bond Length
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Factors Affecting Bond Length
Larger atoms = longer bonds
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How are they all related?___BL = ___BE
___BO = ___BE
___BO = ___BL
indirect
direct
indirect
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Using Bond Energies to find Hrxn
• Draw the dot structures of the reactants and the products
• Determine the energy needed to BREAK all the bonds in the reactants (Endothermic, positive value)
• Determine the energy change to MAKE all the bonds in the products (Exothermic, negative value)
• Add them together to get an approximate value for the H of the reaction
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Using Bond Energies to find Hrxn
• Determine the H for the combustion of methane gas. (use BE values on next slide)
• What is the Hrxn when you use Hf values?
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CH4 + 2O2 CO2 + 2H2O
BEreactants = 4(C-H) + 2(O=O) = 4(413kJ) + 2(498 kJ)
BEproducts = 2(C=O) + 4(H-O) = 2(-745 kJ) + 4(-463 kJ)
BEreactants = 2648 kJ
Notice these are pos energies b/c bonds are
breaking!!!
BEproducts = -3342 kJ
H Rxn = BEproducts + BEreactants = -694 kJ
Notice these are neg energies b/c bonds are
being made!!!
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More links
• Click this link for a site that provides an overview of this information
• The first 2min 15sec of this video gives a great visual representation of the big picture (where this unit fits into all of chemistry)
• A song about ionic and covalent bonds