Transcript VSEPR Power Point

Chapter 10
Molecular Geometry and Chemical Bonding
Theory
8–1
John A. Schreifels
Chemistry 211
Chapter 10-1
Overview
• Geometry and Directional Bonding
–
–
–
–
Valence-Shell Electron-Pair Repulsion Theory
Dipole Moment and Molecular Geometry
Valence Bond Theory
Description of Multiple Bonding
• Molecular Orbital Theory
– Principles of Molecular Orbital Theory
– Electron Configurations of Diatomic Molecules of the
Second-Period Elements
– Molecular Orbitals and Delocalized Bonding
8–2
John A. Schreifels
Chemistry 211
Chapter 10-2
Molecular Geometry and Directional
Bonding
• Atoms oriented in very well defined relative positions
in the molecule.
• Molecular Geometry = general shape of the
molecule as determined by the relative positions of
the atomic nuclei.
• Theories Describing the structure and bonding of
molecules are:
– VSEPR = considers mostly electrostatics in determining the
geometry of the molecule.
– Valence Bond Theory = considers quantum mechanics and
hybridization of atomic orbitals.
– Molecular Orbital Theory = claims that upon bond formation
new orbitals that are linear combinations of the atomic
orbitals are formed.
John A. Schreifels
Chemistry 211
Chapter 10-3
8–3
Valence Share Electron Pair Repulsion
(VSEPR) Theory
• Valence Share Electron-Pair Repulsion (VSEPR)
model allows us to predict the molecular shape by
assuming that the repulsive forces of electron pairs
cause them to be as far apart as possible from each
other.
– Only the valence electron pairs are considered in
determining the geometry.
– Review: valence electrons are the additional electrons in an
atom beyond the inert gas core; for the neutral atom the
number is equal to the group number, i.e. C has 4, N has 5,
O has 6, and F has 7.
8–4
John A. Schreifels
Chemistry 211
Chapter 10-4
Effect of the number of electron pairs
around the central atom
2 charge clouds,
linear
3 charge clouds,
trigonal planar
4 charge clouds,
tetrahedral
8–5
John A. Schreifels
Chemistry 211
5 charge clouds,
trigonal bipyramidal
6 charge clouds,
Octahedral
Chapter 10-5
PREDICTING EXPECTED GEOMETRY
ACCORDING TO VSEPR THEORY.
•
•
•
Lewis dot structure determines the total # of electrons around the
central atom. Multiple bonds (double and triple) count as one.
The number of bonding and nonbonding electron pairs determines the
geometry of electron pairs and the molecular geometry.
E.g. Determine the geometry of the following:
–
–
–
–
BeCl2.and CO2 linear
BF3, COCl2, O3, SO2
CH4, PCl3, H2O
PCl5, SF4, ClF3, XeF2 (lone pair in axial position for a trigonal bipyramidal
structure).
– SF6, IF5, XeF4
•
•
•
Lone e Pairs affect geometry more than bonding pairs.
E.g. NH3 has one lone pair of electrons. e pairs repulsion from reduces
angle from 109° to 107. H2O with two lone pairs and the angle
between the H's and the O is only 105.
8–6
Multiple bonds have larger affect on geometry than single bonds:
H2C=O (116° instead of 120°); H2C=CH2 (117° instead of 120°).
John A. Schreifels
Chemistry 211
Chapter 10-6
Figure 9.2 Molecular Shapes 2,3,4
electron pairs.
8–7
John A. Schreifels
Chemistry 211
Chapter 10-7
Figure 9.3 Molecular shapes 5, 6 electron
pairs
8–8
John A. Schreifels
Chemistry 211
Chapter 10-8
Polarity of Molecules
• Bond dipole a positive charge next to a negative charge.
• Dipole moment,  the magnitude of the net bond dipole of a
molecule  = Qxr Q = the net charge separation; r = the
separation distance. Units: debyes (D) where 1 D = 33.36x1030
Cm.
• A polar bond forms when two atoms of between two atoms
involved in a bond have significantly different electronegativities.
– Most electronegative substance will have a slight negative charge
(represented as )
– The positive (electron poor) side of the bond is represented as +
or
–  points in direction of the negative charge.
• Net polarity (dipole moment) of a molecule is obtained using the
vector sum of polarities of individual bonds.
• E.g. determine if NH3, H2O, CO2 have dipole moments.
• E.g.2 determine if either the cis- or trans- isomer of C2H2Cl2
John A. Schreifels
Chemistry 211
Chapter 10-9
8–9
Dipole Moments of Molecules
•
•
•
Dipole moments are easily measured in the laboratory and allow the
determination of the net ionization of the molecule.
Complete ionization gives the charge of an electron (1.60x1019 C )
multiplied by r.
the net charge on each of the atoms in a polar bond can be obtained
from the measured dipole moment.
19
1.60 x10
C
1D
  Qxr  n 

r

 30
e
3.336 x10
Cm
10
4.796 x10 D
 nr 

e m
where n = net charge and r is the radius (in m).
E.g. determine the net charge on HCl if the dipole moment is 1.03 D and
the bond distance is 127 pm; determine the % ionization.
E.g.2 determine the expected dipole moment of NaCl if the bond were
completely ionic and then determine the % ionic character. The ionic
radii of Na+ and Cl are 102 pm and 181 pm, respectively.
Experimental dipole moment is 9.0 D.
John A. Schreifels
Chemistry 211
8–10
Chapter 10-10
MOLECULAR SHAPES:VALENCE
BOND THEORY (VBT)
• Valence Bond Theory: a quantum mechanical description of
bonding that pictures covalent bond formation as the overlap of
two singly occupied atomic orbitals.
• VSEPR effective but ignores the orbital concepts discussed in
quantum mechanics.
• H2 forms due to overlap of two 1s orbitals.
• Electron densities from p-subshell electrons overlap to produce
a bond in F2.
• CH4:The 1s orbital of hydrogen must overlap with the 2s and 2p
orbitals of carbon.
• Presence of electrons from hydrogen adds new waves that are
in contact with each other and undergo constructive interference
– new waves result.
• The s and p orbitals around an atom such as carbon become
equivalent and the orbitals become a hybrid (sp3) of the original
8–11
orbitals.
• Hybrid orbitals are as far apart as possible.
John A. Schreifels
Chemistry 211
Chapter 10-11
Other Kinds of Hybrid Orbitals
• Hybridization varies from sp, sp2, up to sp3d2 depending upon
the number of orbitals involved in the bonding.
• Each of these has a characteristic shape see table in book.
• Hybridization determined by using VSEPR to establish the
geometry, i.e., the number of electron clouds around the central
atom. The number of electron clouds = the number of hybrid
orbitals.
E.g.: Determine the hybridization of B in BF3.
• The  bond formed between an s orbital and a p orbital or even
between two p orbitals.
• E.g. CH3CH2OH, has all  bonds - even though there are C-C
bonds and C-O bonds which each involve the interaction of sp3
orbitals to form the  bonds.
• SF6: sp3d2.
John A. Schreifels
Chemistry 211
8–12
Chapter 10-12
VBT: Multiple bonds
•
C2H4 planar with a trigonal geometry = sp2
hybridization for each of the carbon atoms and
they form  bonds with hydrogen.
•
Each carbon has 4 orbitals in its valence shell.
This means one of the p-orbitals for each C is not
hybridized.
•
Proximity to each other results in overlap to give
a charge distribution resembling a cloud which is
above and below the plane of the molecule and
called a  –bond .
•
Overlap above and below makes rotation of
carbon atoms difficult.
•
E.g. C2H2: sp (linear) hybridized. Leads to the
existence of a  bond as well as two  bonds.
•
Summarizing a
•
single bond is a  bond,
•
double bond is a  bond and a  bond,
•
triple bond is a  bond and 2  bonds.
E.g. Dinitrogen difluoride exists as cis and trans
isomers( a compound having the same formula
with a different arrangement of atoms).
Investigate the bonding.
John A. Schreifels
Chemistry 211
8–13
Chapter 10-13
MO Theory of Bonding
• Molecular Orbital Theory extends quantum theory and states
that electrons spread throughout the molecule in molecular
orbitals = region in a molecule in which an electron is likely to
be which is similar to the concept discussed in quantum theory.
Molecular orbitals are considered to be the result of the
combination of atomic orbitals.
• Hydrogen: when two atomic orbitals from hydrogen approach
each other they form 2 molecular orbitals,  and *, bonding
orbital and antibonding orbital respectively.
– The energy of the bonding orbital is lower than the original atomic
orbital.
– The energy of the antibonding orbital is higher than the original
atomic orbitals and thus destabilizes the molecule.
*
• The electron distribution of H2 would be: 1s  , 1s
. An
excited state of this molecule would be 1s  , *
 .
 2s
John A. Schreifels
Chemistry 211
8–14
Chapter 10-14
•
•
Molecular Orbital Theory of Other
Diatomic Molecules
2
He :  2 *  no net stabilization (or
1s
bonding).1s
1
  2 *   a net of one bonding electron.
He 2
1s
1s
2
Bond order: BO = 1/2(nb  na) where nb is the
number of bonding electrons and na is the number
of antibonding electrons.
E.g. For He2 BO(He2) = 1/2(2 2) = 0.
E.g.2 H2 on the other hand would have a BO(H2) =
1/2(2  0) = 1 or there is a single bond between
the two atoms.
• Li2:
2 * 1
1s  1s 2s 2  BO=1/2(4  2) = 1.
•
•
•
 
Molecule of lithium
should2be stable.
2
2
2
*
O : 1s  1s  2s 2 *2s  2p 4 2p 2 *2p  ;
2
BO= 1/2(10  6) = 2. Last two filled orbitals are
antibonding  one elctron in each orbital (Hund’s
rule) or two unpaired electrons  O2 a
paramagnetic molecule.
John A. Schreifels
Chemistry 211
Fig. 10.34 MO Diagram of
8–15
N2
Chapter 10-15
MO Levels of 2nd Row Elements
Large 2s-2p interaction
*2p
B2
C2
N2
*2p
2p

Small 2s-2p interaction
O2
F2
Ne2

 








2p
 
*2s






2s






1
P
2
D
3
D
2
P
1
D
0
Bond Order
Magnetic behavior
John A. Schreifels
Chemistry 211
 
P = Paramagnetic; D = Diamagnetic
8–16
Chapter 10-16
Delocalized Bonding
•
•
•
•
•
Molecular orbital theory handles delocalization quite nicely since molecular
orbitals can be said to be spread over the entire.
Metals and energy bands formed by them.
Solidification of metal atoms forms large “molecules” with extensive
delocalization of electrons.
Molecular orbitals for all metals are very similar and a continuous band is formed.
They can conduct electricity when the atoms are excited so that an electron
occupies an excited state. The energy separation between the occupied and
unoccupied orbitals is small so that little energy is required to cause this.
8–17
John A. Schreifels
Chemistry 211
Chapter 10-17