Unformatted text preview: Chem 210 / WANG / Chapter 9A
VSEPR Method for predicting shape Chapter 9.1 – 9.3
Molecular Geometry and Polarity
9.1 Molecular Shapes
9.2 The VSEPR Method
9.3 Molecular Shape and Molecular Polarity
http://www.up.ac.za/academic/chem/mol_geom/mol_geometry.htm http://www.chempractice.com/drills/java_G_structure.php ElectronElectron-group repulsions and the five basic
molecular shapes. linear Shell
Repulsion The single molecular shape of the linear electronlinear electrongroup arrangement. The VSEPR Theory
______ ______ _____ ______ _______ • Electron groups around the central atom repel
each other. •
• They end up staying as far apart as possible.
Both bonding and non-bonding pairs do this.
non- To predict shape around a central atom, use the following format: A - central atom
X -surrounding atom AXmEn E -nonbonding
valence electron-group integers Shape of CO2 or O=C=O is linear
Double bond at each side of the C atom
counts as one electron group, as each group
is shared with the same O atom. tetrahedral trigonal planar trigonal bipyramidal The Shape of Molecules
Molecules have specific shapes.
The shape of a molecule affects
properties such as melting point,
density and solubility.
Shape can be predicted by the VSEPR
Valence The molecule is
repulsion among the
two sets of bonding
pairs causes them to
stay as far apart as
CO2, HCN, BeF2 octahedral Note bond angles and names of each basic shape.
The two molecular shapes of the trigonal
planar electron-group arrangement.
electronEx: SO3, BCl3, NO3-, CO32BCl Class Shape
Class Shape Ex:SO2, O3, PbCl2, SnCl2
SnCl Fall 2009 / UM-SJTU JI Trigonal planar, BCl3
Note that all 4 atoms (B and 3 Cl)
are on the same plane.
The molecule is
because the repulsion
among the three
bonding pairs makes
them stay as far apart
as possible. ACTUAL versus IDEAL Bond Angles
Bond angles deviate from theoretical angles when
• Atoms attached to the central atom are different.
• Sets of electron groups are different.
Effect of Double Bonds:
Double bond has
H 1200 real 1200 1220 H
C O C 1160 H H O 1220 Effect of Nonbonding(Lone) Pairs
Lone pairs repel bonding pairs
more strongly than bonding
pairs repel each other. >1200 >1200
Cl Cl <1200
~ 950 1 Chem 210 / WANG / Chapter 9A
VSEPR Method for predicting shape The three molecular shapes of the
tetrahedral electrontetrahedral electron-group arrangement. Lewis structure
2D, does not tell positions of atoms.
Both Lewis structures specify the same molecule!
SO42-, ClO4- Examples:
NH3 , PF3, ClO3, H3O+
PF ClO H C
H NonNon-bonding electron domain
not connected to any atom
Atoms are found at the end of bonding
electron domains only. Examples:
H2O, OF2, SCl2
Molecular Shape ↔ bonding pairs
AX4 How 4 electron domains relate
to various molecular shapes. 4-bonding pairs
Molecular shape AX2E2 H O H H H AX3E
shape Non-bonding pairs influence molecular shape. Tetrahedral shape: CCl4
Note that the 4 Cl atoms are on the
corner of the tetrahedron. The C atom
sits in the center.
The molecule is
there are four
other. Trigonal pyramidal, NH3
Note that the N atom is NOT on
the same plane as the 3 H atoms.
The molecule is
Trigonal pyramidal, as
there is a (invisible)
nonbonding pair of
down the three
(visible) bonding pairs.
bonding But, as “electron clouds”, they are “invisible”. Bent,
Note the O atom is NOT in the
same line as the two H atoms.
The molecule is
bent because there
are two (invisible)
nonbonding pairs of
the two (visible)
bonding pairs. Fall 2009 / UM-SJTU JI VSEPR method Effect of Nonbonding(Lone) Pairs
Explain why the bond angles in H2O and NH3 should differ
the bond angle in CH4. CH4 has 4 equal e groups.
H2O has two lone-pairs.
NH3 has one lone pair. List
List the above three compounds in order of increasing bond
angles. H2O < NH3 < CH4
104.5 < 107 < 109.5 from Molecules with Expanded Valence Shells 5 electron domains
• Axial positions
• Equatorial positions
two 90o repulsions, two
•Lone pairs occupy
equatorial positions to
minimize 90o repulsions. 2 Chem 210 / WANG / Chapter 9A
VSEPR Method for predicting shape The four molecular shapes of the trigonal
bipyramidal electron-group arrangement.
electron- VSEPR method
Molecules with Expanded Valence Shells The three molecular shapes of the octahedral
electronelectron-group arrangement. 6 electron
domains SF6 Octahedral
PF5, AsF5, SOF4 SF4, XeO2F2, IF4+, IO2F2- IOF5 •Axial positions
four 90o repulsions.
four 90o repulsions
•All positions are
equal! BrF5 A summary of common molecular shapes
with two to six electron groups. ICl4- XeOF4 XeF2 , I3- , IF2- ClF3 , BrF3 XeF4 TeF5- Review the angles between bonding
electron domains in each shape. A molecule may have more than one central atoms.
The rules regarding molecular geometry still hold.
around this C
also. H Cl C H H
around this C H 1,2-dichloroethane List geometry for each central atom.
around this C. C H C H H O
around this C H H C H N
H Polarity of molecules
A molecule with ONLY one polar bond
will be polar. Example: HF is a polar
around this N Fall 2009 / UM-SJTU JI H F •Slight positive side
•Smaller electronegativity •Larger electronegativity A molecule with TWO OR MORE polar
bonds may or may not be polar.
Example: CO2 is nonpolar, H2O is polar Polarity of molecules
Is CO a polar molecule?
Answer: CO is a polar molecule.
δ+ C •Slight positive side
•Smaller electronegativity O δ- •Slight negative
•Larger electronegativity 3 Chem 210 / WANG / Chapter 9A
VSEPR Method for predicting shape Orientation of polar molecules in an electric field. Geometry and polarity of molecules Polarity of molecules For a molecule to be polar
1. It must have polar bonds
2. It must have the proper
geometry Is CO2 a polar molecule?
CO2 is not a polar molecule.
δ- δ+ O C •slight negative
Electric field OFF Electric field ON Polar C C H H δ- δ+ F
H H H δ+ CH3F is polar. H H H ENH = 2.1 N H
H H N
H H molecular dipole The dipoles reinforce each
other, so the overall
molecule is definitely polar. F CH2F2 is polar. non-polar
non-polar Polarity of molecules depends
F F ends are more δthan H ends. H Non-Polar C
.. δ+ H Is this one polar or non-polar? δ- C C-F bond is polar, but
all ends are equally δ-.
CF4 is nonpolar.
nonpolar. F F
Polar H end is more δ+
than the F ends.
CHF3 is polar. NH3 is polar.
N-H bond is polar.
H ends are more δ+ than the non-bonding e pair.
more ENN = 3.0 δ- F Non-polar H N Molecular Polarity depends on
Geometry H C-F bond is polar.
F end is δCH3 end is δ+. •slight negative •Slight positive side Molecular Polarity depends on
All C-H bonds
CF4 δ- O F F Water
Water molecule is polar because of its:
1. polar bonds
2. bent geometry. N
H H H YES
Is this a polar molecule? YES
charged. H H C H If so, which part is δ+?
Which part is δ-? H
charged δ- If there is a way to approach the molecule
from two ends, of which one is δ+, the
other δ-, the molecule is POLAR! Fall 2009 / UM-SJTU JI 4 Chem 210 / WANG / Chapter 9A
VSEPR Method for predicting shape Using VSEPR to predict
molecular geometry Is this a polar molecule? YES
H H C More
charged. H FIRST determine electron domain geometry:
electron H N H
charged δ- A second way to approach
this molecule to find δ+, vs. δ-.
It is POLAR! Lewis Structure and Shape
• Draw Lewis structure of HC2H and use
VSEPR theory to predict the shape
around C. • Draw the Lewis structure
• Count the total number of electron pairs
around the central atom (multiple bonds
count as one bonding pair).
• Arrange electron domains to minimize
THEN determine molecular geometry:
• Lone pairs (in equatorial positions if
applicable) are “invisible”! SAMPLE PROBLEM 1: ethane and ethanol.
Predict the various bond angles in both molecules. Lewis Structure and Shape
• Draw Lewis structure of H2CO and use
VSEPR theory to predict the shape around C. H
δ+ HCO H C
δ- •Is this molecule polar? YES
•If so, indicate the + and – ends of the molecule. Solution
(a) BF3 has 24 valence e- and all electrons around
the B will be involved in bonds. The shape is
AX3, trigonal planar. F HCCH B F0
CH3CH3 H C C ethanol
CH3CH2OH H linear Nonpolar molecule – no polar bonds at all!! Solution continued The steps in determining a molecular shape. (b) COS is linear.
1. C and S have the same EN (2.0)
2. O has higher EN (3.5).
C=O bond is quite polar.
3. The molecule is polar overall.
polar S C O Molecular
formula Step 2 Electron-group
arrangement e- Count all groups around
central atom (A)
angles P Step 4 Count bonding &
nonbonding egroups separately. (c) For PF3 - there are 26 valence electrons,
1 nonbonding pair F The shape is based upon the
tetrahedral arrangement. F Note lone pairs &
double bonds Molecular
(AXm En) Fall 2009 / UM-SJTU JI PROBLEM 2
Answer F Step 1
structure 1. F (EN 4.0) is more electronegative
than B (EN 2.0).
2. All dipoles directed from B to F.
3. Because all are at the same angle
and of the same magnitude, the
molecule is nonpolar. The F-P-F bond angles should be
<109.50 due to the repulsion of the
nonbonding electron pair. P
F F F <109.50 The final shape is trigonal pyramidal. The type of shape is
AX3E 5 Chem 210 / WANG / Chapter 9A
VSEPR Method for predicting shape PROBLEM 2
Answer Cl (d) For COCl2, C has the lowest EN
and will be the center atom.
Cl O There are 24 valence e-,
3 atoms attached to the center atom. C does not have an octet; a pair of nonbonding electrons
will move in from the O to make a double bond.
The shape for an atom with three
atom attachments and no
nonbonding pairs on the central
atom is trigonal planar.
The Cl-C-Cl bond angle will be less
0 due to the electron
density of the C=O.
O Fall 2009 / UM-SJTU JI SAMPLE
PROBLEM 4 F
F F F Sb
F Determine the shape around each of the
central atoms in acetone, (CH3)2C=O. F
tetrahedral (a) SbF5
• 40 valence e• all electrons in bonding pairs;
• shape is AX5 - trigonal bipyramidal. O
C H (b) BrF5
• 42 valence e• 5 bonding pairs 1 nonbonding pair
on central atom.
• Shape is AX5E, square pyramidal.
square C HH H H
H O tetrahedral H CCH
H trigonal planar F
F Br F >1200 F
<1200 6 ...
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