chemistry 161 chapter 10
DESCRIPTION
CHEMISTRY 161 Chapter 10. PREDICTING THE GEOMETRY OF MOLECULES. H. O. H. 1. derive Lewis structure of the molecule. 2. discriminate between bonding and non-bonding electron pairs. 3. VALENCE SHELL ELECTRON PAIR REPULSION. V ALENCE S HELL E LECTRON P AIR R EPULSION. VSEPR. - PowerPoint PPT PresentationTRANSCRIPT
PREDICTING THE GEOMETRY OF MOLECULES
1. derive Lewis structure of the molecule
2. discriminate between bonding and non-bonding electron pairs
O HH
3. VALENCE SHELL ELECTRON PAIR REPULSION
VALENCE SHELL ELECTRON PAIR REPULSION
3. valence electron pairs stay as far apart as possible
VSEPR
1. identify in a compound the central atom
2. electrons repel each other
4. non-bonding electrons repel more than bonding electrons
B
120°
120°
120°
F B F
F
THREE ELECTRON PAIRS AROUND THE BORON ATOM
TRIGONAL PLANAR ARRANGEMENT BEST
better arrangement for four electron pairs
109.5°
C
TETRAHEDRAL
4 electron pairs tetrahedral
bigger than 90 ° in square planar
put on the H-atoms
P PFF
F
F
F
Bond angle
900
1200
shape of PF5 is trigonal bipyramidal
two of the F atoms different from the others
Se
VSEPR treats double bonds like a single bondO Se O
THREE ELECTRON PAIRS AROUND SELENIUM
ELECTRON PAIR GEOMETRY
TRIGONAL PLANAR
AB2E2 AB4
H O H four electron pairs around the oxygen atom
O
PUT ON THE 2 H-ATOMS
O
HH
shape of H2O is V-shaped or bent
SF
F
F
F
SFF
F
F
WHERE DOES LONE PAIR GO?
OR
lone pairs occupy the trigonal plane (the “equator”) to minimize the number of 90° repulsions
SF4
1 lone pair
See-saw shaped
ClF3
2 lone pairs
T-shaped
SF
F
F
F
ClF
F
F
lone pairs occupy the trigonal plane (the “equator”) first to minimize the number of 90° repulsions
AB4E AB3E2
F
F
F
Xe
XeF2
3 lone pairs
Linear
AB2E3
Summary of Molecular ShapesTotal valence electron pairs
Electron Pair Geometry
Lone electron pairs
Shape of Molecule
2
3
4
Linear
Trigonal planar
Tetrahedral
0
0
1
0
1
2
Linear
Trigonal planar
V-shaped
Tetrahedral
Trigonal pyramid
V-shaped
Total valence electron pairs
Electron Pair Geometry
Lone electron pairs
Shape of Molecule
5
6
Trigonal bipyramidal
Octahedral
1
2
3
0
1
2
See-saw
T-shaped
Linear
Octahedral
Square pyramid
Square planar
0 Trig. bipyramid.
molecules with no single central atom
we apply our VSEPR rules to each atom in the chain
POLYATOMICS
Example: ETHANOL
ETHANOL
The atoms around the carbons form a.
The atoms around the oxygen form a
OCH H
H
H
C
H
H
tetrahedral arrangement
V-shaped structure.
C2H5OH
simplest moleculeH2
two H-atoms 1s1
two H-atoms approach each other and the electron waves interact
OVERLAP to form a region of increased electron density between the atoms
a covalent bond is formed by an
overlap of two valence atomic
orbitals that share an electron pair
VALENCE BOND THEORY
the better the overlap the stronger the bond
the orbitals need to point along the bonds
C
H
HH
H
What orbitals are used?
hydrogen atoms bond using their 1s orbitals
carbon needs four orbitals to bond with.
2s, 2px , 2py, 2pz
[He] 2s22p2
CH4
[He] 2s22p2
1. The electronic configuration of carbon is
the orbital diagram is: [He]
the Lewis dot structure is C ....
necessary to promote one 2s electron
[He] 2s12p3
[He]
PROMOTE AN ELECTRON
[He]
[He] 2s22p2
excited state (valence state)
Lewis dot structure
four unpaired electrons
we can use these to form chemical bonds
C
2. bonds formed with s orbitals will be different to bonds formed with p orbitals
3. three p orbitals are mutually perpendicular, suggesting 90° bond angles
combining the orbitals
1. a covalent bond is formed by an overlap of two
valence atomic orbitals that share an electron pair
Experiment shows that all four bonds are identical
experiment shows that methane has 109.5° bond angles
we need four orbitals pointing to the vertices of a tetrahedron
orbitals are just mathematical functions
HYBRIDIZATION
C
H
HH
Hwe can combine them
COMBINING ORBITALS TO FORM HYBRIDS
HYBRIDIZATION
number of atomic orbitals that are combined
the number of resulting hybrid orbitals
IS EQUAL TO
Combine one s and one p a sp- hybrid
++
s + p
The positive part cancels negative part
2s+ 2p
HYBRIDIZATION
The positive part adds to positive part
CONSTRUCTIVE INTERFERENCE
DESTRUCTIVE INTERFERENCE
What do we get?
Combine one s and one p to give a sp- hybrid
+
s + p
REMEMBER IF WE MIX TWO WE MUST GET TWO BACK
The other combination is s - p
HYBRIDIZATION
2s+ 2p
Where is the nucleus?
HYBRIDIZATIONCombine one s and one p a sp- hybrid
+
+SUBTRACT the
orbitals2s- 2p
SUBTRACTING THE p ORBITAL CHANGES ITS PHASE
HYBRIDIZATIONCombine one s and one p a sp- hybrid
+
+SUBTRACT the
orbitals2s- 2p
SUBTRACTING THE p ORBITAL CHANGES ITS PHASE
HYBRIDIZATIONCombine one s and one p a sp- hybrid
+ +
s - p2s- 2p
The positive part cancels negative part
The positive part adds to positive part
CONSTRUCTIVE INTERFERENCE
DESTRUCTIVE INTERFERENCE
What do we get?
HYBRIDIZATIONCombine one s and one p a sp- hybrid
+
s - p
The positive part cancels negative part
We get two equivalent sp orbitals
ORIENTED AT 1800
2s- 2p
Where is the nucleus?
COMBINE one s-orbital and two p-orbitals
Get three sp2 - orbitals oriented at 1200
s and p orbitals
three sp2-hybrids
directed at 1200
four hybrid orbitals needed to form four bonds
s + px + py + pz
an atom with sp3 hybrid orbitals is said to be
4 sp3 hybrids
The four sp3 hybrid orbitals form a tetrahedral arrangement.
sp3 hybridizationEPG of 4 pairs
METHANE: CH4
sp3 hybridized
What happens to the energies of the orbitals?
C
H
HH
H
E2p
2s
Orbitals in free C atom
E sp3
Hybridized orbitals of C atom in methane
When orbitals are hybridized they have the same energy:
The FOUR sp3 hybrids are DEGENERATE.
HYBRIDIZE
C
Each bond in methane results from the overlap of a hydrogen 1s orbital and a carbon sp3 orbital.
H
HH
H
Hydrogen 1s orbital
Carbon sp3 orbitals
Each hybrid ready to overlap with H 1s orbitals
Form a chemical bond by sharing a pair of electrons.
VALENCE BOND MODEL
Step 1: Draw the Lewis structure(s)
Step 2: Determine the geometry of the electron pairs around each atom using VSEPR OR preferably use the EXPERIMENTAL GEOMETRY
Step 3: Specify the hybrid orbitals needed toaccommodate the electron pairs on each atom
Hybrid orbital model
sp3 hybrids are also employed in …...
all molecules that have a 4 pair EPG….
NH3, H2O, NH4+ , CCl4
OTHER MOLECULES USING sp3 HYBRIDS
AMMONIA…..
AMMONIA: NH3
2s 2pN [He]
NVSEPR
Valence shell has four pairs
EPG is TETRAHEDRALNitrogen electronic configuration
Need sp3 hybrids
HYBRIDIZE
HH
H
E2p
2s
Orbitals in free N atom
E sp3
Hybridized orbitals of N atom in ammonia
When orbitals are hybridized they have the same energy:
sp3 hybridization…….
The FOUR sp3 hybrids are DEGENERATE.
HYBRIDIZE
Overlap of two of oxygen sp3 hybrids with …..H atom 1s orbitals.
WATER.
Lone pairs in two of the sp3 hybrids.
To form two bonds.
Think about H3O+ !!!
O
H
H
Overlap of Oxygen sp3 hybrids containing a lone pair
H+ ion empty 1s orbitals.
HYDRONIUM ION.
O
H
H
H+
ISOELECTRONIC WITH?
NH3
QUESTIONWhich of the following molecules is uses sp3 hybrids in the valence bond description of its bonding?
A CO2
B NF3
C O3
D NO2+
E F2OANSWER…….
1 C and D
2 B and E
3 A and D
4 B and C
5 B and A
QUESTIONWhich of the following molecules is uses sp3 hybrids in the valence bond description of its bonding?
A CO2
B NF3
C O3
D NO2+
E F2O
WHAT ABOUT OTHER EPG’S …….
1 C and D
2 B and E
3 A and D
4 B and C
5 B and AOF F
C OO
O N O[ ]+
NF FF
O O O
A four electron pair EPG uses sp3 hybrids
The three electron pair EPG uses sp2 hybrids
VALENCE BOND THEORY FOR OTHER ELECTRON PAIR GEOMETRIES
The two electron pair EPG uses sp hybrids
X
180°
180°
X
120°
120°
120°
X
109.5°
EPG’s2 3
4
HYBRIDS
sp sp2 sp3
lets look at a molecule that needs sp2
Ethylene: C2H4 C CH
H
H
H
The CARBON is sp2 hybridized
3 effective electron pairs
3 sp2 hybridss + px + py
The 3 sp2 hybrid orbitals form a trigonal planar arrangement.
three hybrid orbitals on each carbon for the trigonal planar EPG.
VSEPR trigonal planar EPG around each C-atom. a HCH angle of 1200.
sp2 hybridization
GROUND STATE C atom
E2p
2s
sp2 hybridized orbitals of C
FORMATION OF sp2 hybrids
E sp2
2p
E2p
2s
VALENCE STATE C atom
HYBRIDIZE
This leaves one p orbital unhybrized…….
x
y
z
The unhybridized p orbital is perpendicular to sp2 plane.
sp2 - hybrid orbitalUNHYBRIDIZEDp- orbital
An sp2 hydridized C atom
Lets put it all together…….
x
y
z
x
y
z
OVERLAP the sp2 hybrids from the two carbons to form a sigma bond between them.
C C
BONDING IN ETHYLENE
bond
PUT THE ELECTRONS IN AND…..
C CH
H
H
H
x
y
z
x
y
z
overlap two sp2 hybrids on each carbon with hydrogen 1s orbitals to form sigma bonds and...
The two unhybridized p orbitals are left over to form…..
H
HH
H
C CH
H
H
H
x
y
z
x
y
z
The two unhybridized p orbitals are left over to form a …..
H
HH
H
pi bond ( bond)
The second part of the carbon-carbon double bond !
C CH
H
H
H
x
y
z
x
y
z
H
HH
H
pi bond ( bond)The second part of the carbon-carbon double bond !
Electrons are shared between the unhybridized p orbitals in an area above and below the line between nuclei.
C CH
H
H
H
x
y
z
x
y
z
H
HH
H
pi bond ( bond)
sp2
sigma bonds ( bond)
THE COMPLETE PICTURE!!!!!!!
C CH
H
H
H
SUMMARY...
C C
H
H
H
H
H(1s)-C(sp2)
:C(sp2)-C(sp2)C(2p)-C(2p)
H(1s)-C(sp2)
Now look at ethyne (acetylene)
bonding
H(1s)-C(sp)
:C(sp)-C(sp)
C(2p)-C(2p) TWO OF THESE!!
H(1s)-C(sp)
BONDING SCHEME IN ETHYNE
C CH H
What does this look like????
x
y
z
x
y
z
OVERLAP the sp hybrids from the two carbons to form a sigma bond between them.
C C
Put in the unhybridized p orbitals
x
y
z
x
y
z
OVERLAP the sp hybrids from the two carbons to form a sigma bond between them.
C C
OVERLAP the hydrogen 1s orbitals
x
y
z
x
y
z
OVERLAP the sp hybrids from the two carbons to form a sigma bond between them.
C C HH
sigma framework of bonds pi bonding?
VALENCE BOND THEORY
Step 1 Lewis Dot Structure
Step 2 Get Molecular Geometry
VSEPR EXPERIMENTAL
Step 3 Choose hybrids
Step 4 Describe bonds…...
What about molecules with more than an octet around the central atom?
Examples: PCl5, or SF4 or SiF62-
Four pairs needs Four orbitals
Five pairs needs Five orbitals
six pairs needs six orbitals
PCl5
Cl
PClCl
Cl Cl
We ignore the chlorine atoms and just describe central atom.
Need five hybrid orbitals on the phosphorus
d + s + px + py + pz
5 effective electron pairs dsp3 hybridization
5 dsp3 hybrids
to fit the trigonal bipyramidal EPG.
Five equivalent orbitals……..
dsp3 - hybrid orbitals
x
y
zTRIGONAL BIPYRAMID EPG 5 PAIRS
overlap with orbitals on chlorine to form 5 bonds.
1200
900
SIX PAIRS…..
SF6
F
S
F
F
F
F
F
We need six hybrid orbitals on the sulfur to allow for the octahedral EPG and six bonds.
6 d2sp3 hybridsd2sp3 hybridization
d + d + s + px + py + pz
6 effective electron pairs
SIX equivalent orbitals……..
We ignore the chlorine atoms and just describe central atom.
EXAMPLES
Xe FF
EPG 5 pairs
dsp3 hybridsTwo axial bonds at 1800
Three lone pairs in equatorial
hybrids
Describe the molecular structure and bonding in XeF2 and XeF4
Linear
EXAMPLES
Xe FF
EPG 5 pairs
dsp3 hybridsTwo axial bonds at 1800
Three lone pairs in equatorial
hybrids
EPG 6 pairs
d2sp3 hybridsfour bonds at 900 in a plane
Two lone pairs in axial hybrids
Describe the molecular structure and bonding in XeF2 and XeF4
Linear
XeF
F
F
F
Square planar
MOLECULAR ORBITAL THEORY
electrons occupy orbitals each of which spans the entire molecule
molecular orbitals each hold up to two electrons and obey Hund’s rule, just like atomic orbitals
H2 molecule:
1s orbital on Atom A 1s orbital on Atom B
the H2 molecule’s molecular orbitals can be constructed from the two 1s atomic orbitals
1sA + 1sB = MO1
1sA – 1sB = MO2
constructive interference
destructive interference
ADDITION OF ORBITALSbuilds up electron density in overlap region
1sA + 1sB = MO1
combine them by addition
A B
ADDITION OF ORBITALSbuilds up electron density in overlap region.
1sA + 1sB = MO1
A Bwhat do we notice?
electron density between atoms
SUBTRACTION OF ORBITALSresults in low electron density in overlap region..
1sA – 1sB = MO2
A B
subtract
SUBTRACTION OF ORBITALSresults in low electron density in overlap region..
1sA – 1sB = MO2
A Bwhat do we notice?
no electron density between atoms
COMBINATION OF ORBITALS
1sA + 1sB = MO1
builds up electron density between nuclei
1sA – 1sB = MO2results in low electron density between nuclei
BONDING
ANTI-BONDING
EEnergy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A B
COMBINING TWO 1s ORBITALS
EEnergy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A B
1sA-1sB
MO
1sA+1sB
MO
1s
1s*
E1s
1s*
1s
1s
He2: (1s)2(1s*)2
He HeHe2
bonding effect of the (1s)2 is cancelled by the
antibonding effect of (1s*)2
BOND ORDER
net number of bonds existing after the cancellation of bonds by antibonds
the two bonding electrons were cancelled out by the two antibonding electrons
He2
(1s)2(1s*)2
the electronic configuration is….
BOND ORDER = 0
BOND ORDER
=
measure of bond strength and molecular stability
If # of bonding electrons > # of antibonding electrons
Bondorder
the molecule is predicted to be stable
BOND ORDER
= {
high bond order indicates high bond energy and short bond length
# of bonding electrons(nb)
# of antibonding electrons (na)
– 1/2 }
measure of bond strength and molecular stabilityIf # of bonding electrons > # of antibonding electrons
Bondorder
the molecule is predicted to be stable
H2+,H2,He2
+
= 1/2 (nb - na)
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
Para-
½
225
106
E
He2+ He2H2
Dia-
1
436
74
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
Para-
½
225
106
E
He2+
Para-
½
251
108
He2H2
Dia-
1
436
74
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
First row diatomic molecules and ionsH2
+
Para-
½
225
106
E
He2+
Para-
½
251
108
He2
—
0
—
—
H2
Dia-
1
436
74
HOMONUCLEAR DIATOMICS
Li2 Li : 1s22s1
both the 1s and 2s overlap to produce bonding and anti-bonding orbitals
second period
E
1s 1s
1s
Electron configuration for DILITHIUM
2s
2s*
2s
2s
(1s)2(1s*)2(2s)2
Li2
nb = 4 na = 2
Bond Order = 1
single bond.
E
1s 1s
1s
Electron configuration for DILITHIUM
2s
2s*
2s
2s
(1s)2(1s*)2(2s)2
the 1s and 1s* orbitals can be ignored when
both are FILLED!
Li2
omit the inner shell
E2s
2s*
2s
2s
Li LiLi2
The complete configuration is: (1s)2(1s*)2 (2s)2
Li2 (2s)2 only valence orbitals contribute to molecular bonding
E2s
2s*
2s
2s
Be2Be BeBe2
Electron configuration for DIBERYLLIUM
Configuration: (2s)2(2s*)2 Bond order = 0
E2s
2s*
2s
2s
(2s)2(2s*)2Be BeBe2
Be2
Electron configuration for DIBERYLLIUM
nb = 2
na = 2
Bond Order = 1/2(nb - na) = 1/2(2 - 2) =0No bond!!! The molecule is not stable! Now B2...
B2
the Boron atomic configuration is
1s22s22p1
form molecular orbitals
we expect B to use 2p orbitals to
addition and subtraction
E
MODIFIED ENERGY LEVEL DIAGRAM
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*
Notice that the 2p and 2p
have changed places!!!!
E
2s
2s*
2s
2s
Electron configuration for B2
2p
2p*
2p2p
2p
2p*
Place electrons from 2s into 2s and 2s*
B is [He] 2s22p1
E
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*(2s)2(2s*)2(2p)2
Abbreviated configuration
Complete configuration
(1s)2(1s*)2(2s)2(2s*)2(2p)2
ELECTRONS ARE UNPAIRED
E
2s
2s*
2s
2s
Electron configuration for B2:
Bond order2p
2p*
2p2p
2p
2p*(2s)2(2s*)2(2p)2
Molecule is predicted to be stable and paramagnetic.
na = 2nb = 4
1/2(nb - na)= 1/2(4 - 2) =1
2p*2p*2p
2p
2s*2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2 C2 N2 O2 F2
E
2p*2p*2p
2p
2s*2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2 N2 O2 F2
E
2p*2p*2p
2p
2s*2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2 O2 F2
E
2p*2p*2p
2p
2s*2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2 F2
E
2p*2p*2p
2p
2s*2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2
Para-
2
495
121
F2
E
NOTE SWITCH OF LABELS
2p*2p*2p
2p
2s*2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2
Para-
2
495
121
F2
Dia-
1
154
143
E
NOTE SWITCH OF LABELS
2p*2p*2p
2p
2s*2s
E
O2 O2+ O2
– O22-
O2 : B.O. = (8 - 4)/2 = 2O2
+ : B.O. = (8 - 3)/2 = 2.5O2
– : B.O. = (8 - 5)/2 = 1.5O2
2- : B.O. = (8 - 6)/2 = 1
2p*2p*2p
2p
2s*2s
E
O2 O2+ O2
– O22-
O2 : B.O. = 2O2
+ : B.O. = 2.5O2
– : B.O. = 1.5O2
2- : B.O. = 1
O2+ >O2 >O2
– > O22-
BOND ENERGY ORDER