VSEPR model

VSEPR model
connects Lewis structures to molecular structure
Shape determines many properties of molecules
Example:
Name
Methane
ethane
propane
butane
Pentane
ALKANES
Molecular
formula
CH4
C2H6
C3H8
C4H10
C5H12
In general: CnH2n+2
Condensed
molecular formula
CH3-CH3
CH3CH2CH3
CH3CH2CH2CH3
CH3CH2CH2CH2CH3
SHAPES OF MOLECULES
Molecular shape: spatial arrangement of atoms
Valence-Shell Electron-Pair Repulsion
VSEPR model
Predict the shape of the molecule, from Lewis
structures.
Concept 1
Electron pairs around a central atom repel each
other. They must stay near the nucleus, but
otherwise, they will stay as far away from each
other as possible.
# of electron
domains
2
3
4
5
6
arrangement of
Example
electron pairs
linear
O=C=O
trigonal planar
BF3
tetrahedral
CH4
trigonal bipyramidal PF5
octahedral
SF6
Describes: electron pair geometry
(See Table 9.1)
Determining Molecular Shape
1. Draw the Lewis structure of the molecule
2. Determine the electron-pair geometry
(count electron domains: double and
triple bonds count as one domain!).
3. Focus on bonded-electron pairs ONLY to
determine the molecular geometry.
Example: Molecules with 4 electron domains
Electron-pair geometry will be tetrahedral
(angles near 109.5o)
# of bonded
atoms
4
3
2
1
molecular
geometry
tetrahedral
trigonal pyramid
bent
linear
example
CH4
NH3
H 2O
HCl
Refinement to Concept 1
VSEPR Concept 2
Æ
Unshared electron-pairs – lone pairs
need more space than shared pairs.
The space they occupy is less
constrained than for bonded pairs.
Æ
Multiple bonds repel like lone pairs.
Consequences of VSEPR Concept 2
H
C
H
109.5
o
H
H
tetrahedral
N
H
107
o
H
O
H
H
104.5o
trigonal pyramidal
Bond angles decrease
lone pairs occupy more space
H
bent
Example 1
Molecules with 2 electron domains
Molecular Geometry Æ LINEAR (bond anlges 180o)
180
o
180
Cl Be Cl
o
O C O
Example 2
Molecules with 3 electron domains
Molec. Geom. Æ Trigonal Planar for 3
bonded atoms
Bent for 2 bonded atoms
120o
F
F
B
F
124o
O
Cl
C
111
Cl
o
O
O
O
117o
Double bonds and lone-pair domains need more space!
Example 3
Molecules with 5 electron domains
Electron domain geometry will be trigonal
bipyramidal (angles near 180o, 120o, or 90o)
# of bonded
atoms
5
4
3
2
molecular
geometry
example
trigonal pyramid
seesaw
T-shaped
linear
2 types of sites:
axial
PCl5
SF4
BrF3
XeF2
90o (to equatorial)
180o (to axial)
120o (to equatorial)
equatorial
Equatorial positions less crowded:
Lone pairs prefer these positions
F
116o
F
S
F
F
F
186o
Br F
F
89o
Example 4
Molecules with 6 electron domains
Electron domain geometry will be octahedral
(angles near 180o, 120o, or 90o)
# of bonded
atoms
molecular
geometry
6
5
4
example
octahedral
square pyramidal
square planar
SF6
IF5
XeF4
5 bonded atoms: where does lone pair go?
all sites are equivalent.
4 bonded atoms: where do the two lone pairs go?
lone pairs will be as far away from each
other as possible: the second pair goes on
opposite side from first pair.
82o
F
F
F
Xe
F
F
square planar
F
F
I
F
F
square pyramidal
Molecular Shapes: Summary
Number of Number of
electron
bonded
Molecular geometry
domains
atoms
Example
2
2
linear
CO2
3
3
trigonal planar
BF3
3
2
bent
O3
4
4
tetrahedral
CH4
4
3
trigonal
pyramidal
NH3
4
2
bent
H 2O
5
5
trigonal
bipyramidal
PCl5
5
4
seesaw
SF4
5
3
T-shaped
BrF3
5
2
linear
XeF2
6
6
octahedral
SF6
6
5
square pyramidal
BrF5
6
4
square planar
XeF4
VSEPR rules can be applied to
other structures
Molecules without a central atom
(organic molecules)
Molecules with multiple bonds
tetrahedral (bent)
122
H H
H
H
H H
H
trigonal planar
o
O
H
117o
H
H
H3 C
tetrahedral
F
O S F
F
F
trigonal pyramidal
O
CH3
tetrahedral
Example
Determine the approximate bond angles
as indicated.
A.
B.
C.
D.
E.
Angle #1
Angle #2
109o
180o
120o
109o
120o
109o
120o
90o
120o
109o
Structure of Alkanes
Lewis structures
H H
H
H C H
H C
H
H
H H H
C H
H C
H H
H
H
H
H
C
C
C
C
H
H
H
H
H
C
C
H H H
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
All C atoms have 4 bonded pairs,
no lone pairs
Electron domain geometry:
tetrahedral
Molecular geometry:
tetrahedral
Bond angle = 109.5o
H
H
Alkenes
Contain one or more double bond
H
H
C
H
H
C
C
H
H
ethene
C
H
CH 3
propene
Carbons on either side of the double bond
have three electron domains
elec. domain geom.:
molec. Geometry:
trigonal planar
trigonal planar
The C=C–H or C=C–C angle is 120o
Alkynes
Contain one or more triple bond
H–C≡C–H
ethyne
H–C≡C–CH3
propyne
Carbons on either side of the triple bond
have two electron domains
elec. domain geom.:
molec. Geometry:
linear
linear
The C=C–H or C=C–C angle is 180o
MOLECULAR POLARITY
molecular shape and
electronegativity
Determines many properties of
molecules, hence matter
For example:
Why does ice float?
Why don’t Oil and water mix
But alcohols and water do?
MOLECULAR POLARITY
Purest covalent bond is between
two identical atoms
Cl2 N2 O2
When different atoms are bonded, the
bond is polarized — differing electron
attacting ability, electronegativity
electronegativity
polar bond
2.1
4.0
H–F
δ+
δ-
electron density
the polarity of individual bonds is
determined by electronegativity
difference between atoms
PERIODIC TRENDS IN
ELECTRONEGATIVITY
Fig. 8.6
MOLECULAR POLARITY
Polarity of entire molecule
vector quantity
DIPOLE MOMENT
μ=Qr
distance
charge
Q = 1.60 x 10-19 C
(1 electron)
Unit = Debye 3.34 x 10-30 C-m
For dipole moment of a molecule,
two factors are involved:
(1) polarity of bonds in the molecule
(2) geometric arrangement of atoms and
bonds in the molecule
Dipole moment in Debyes
μ=Qr
distance
charge
Q = 1.60 x 10-19 C
(1 electron)
Unit = Debye 3.34 x 10-30 C-m
How big is μ (in Debye) for Q = 1 e–
and separation r = 100 pm?
μ=
(1.60 x 10–19 C) (100 x 10–12 m)
(3.34 x 10–30 C-m)
μ = 4.8 Debye
H
Cl
d = 1.36 A
0.136 nm
If ionic, full + on H and full − on Cl
μ = Q r = (1.60x10−19 C)(1.36x10−10 m)
μ = 2.18 x 10−29 C•m
2.18x10−29 C•m
= 6.54 D
−30
3.33x10 C•m/D
But 1.03 is experimental value
Different because HCl is covalent
Much less than full (+) on H and (–) on Cl
1.03 x 100% = 15.7%
6.54
MOLECULAR GEOMETRY
& DIPOLE MOMENT
Formula
AX
AX2
AX3
AX4
AX5
AX6
Molecular
Geometry
linear
linear
bent
trigonal planar
trigonal pyramidal
t-shaped
tetrahedral
square planar
seesaw
trigonal bipyramidal
square pyramidal
octahedral
Dipole
Moment
?
0
?
0
?
?
0
0
?
0
?
0
HCl
H2O
BF3
NH3
CH4
DIPOLE MOMENTS
CO2
O=C=O
2.5
3.5
no net dipole moment
CCl4
Cl
C
Cl
tetrahedral
Cl
Cl
each bond
C – Cl
2.5
3.0
no net dipole due to symmetry
DIPOLE MOMENTS
3.4
O
H2O
2.2
bent
H
H
2.2
:
net dipole moment
μ = 1.85 D
NH3
2.2
N 3.0
H
H
H
net dipole moment
μ = 1.47 D
MICROWAVE COOKING
- Works because water is polar
- Water molecules found in food
absorb microwave radiation. They
rotate so their molecular dipoles
align with the crests and troughs
of the microwave radiation
Freq = 2.45 GHz
= 2.45 x 109 sec−1
Consider CH3Cl
3.2
Cl
C
H
tetrahedral
2.5
H
2.2
H
all 4 pointing upwards
net dipole moment
μ = 1.87 D
COMPARE CIS- & TRANSDICHLOROETHYLENE
Cl
Cl
Cl
C=C
H
H
C=C
H
H
cis-
Cl
trans-
Which is polar ? Why ?
bonds
C—H
C — Cl
2.5
2.5
2.1
Δ=0.4
3.0
Δ=0.5
symmetry
NET
0
RANK BY POLARITY
Cl
Cl
A
B
Cl
Cl
F
Cl
C
D
F
F
F
F
F
E
F
F
F
F
MOLECULAR SHAPE
& HYBRID ORBITALS
We have previously discussed orbitals
in atoms
Now, we will discuss orbitals in molecules
VSEPR model predicts molecular shape,
but it does not explain covalent bonds
Now we have covered atomic orbitals and
molecular shape – put the two together
Given a knowledge of the molecular shape
(atom geometry) …..
Now we will explain it in terms of
MOLECULAR ORBITALS
HYBRID ORBITALS
Consider CH4
Facts we know: C atom 1s22s22p2
1s2
2s2
2p2
valence electrons
C
geometry of p orbitals
H
BUT…. CH4 is tetrahedral
109.5°
C
H
H
H
the atomic orbitals do not explain molecular
geometry
FORMATION OF
HYBRID ORBITALS
THREE STEPS:
1 - promotion
2 - hybridization/mixing
3 - bond formation
In CH4 formation example
4 atomic orbitals: 2s 2px2py2pz
hybridized
4 hybrid orbitals 4 sp3
number of
orbitals
name of type of
hybrid orbitals
HYBRID ORBITALS
IN CH4
step
1
1s2
2s2
2p2
1s2
2s
2p
electron promotion
requires energy
Now have 4 unpaired electrons
HYBRID ORBITALS
IN CH4
step
2
4 orbitals which are half-filled
mix to form a new set of
half-filled orbitals
C atom
hydridized
1s2
sp3
hybrid orbitals
The hydrid orbitals are directed
tetrahedrally: they repulse each other
No energy change
HYBRID ORBITALS
IN CH4
step
3
C–H bonds are formed when
the 4 hybrid orbitals overlap with
the 1s orbitals of H atoms
4 shared electron pairs
sigma (σ) bonds
good overlap
energy is released
Covalent molecule CH4 results
H
109.5°
C
H
H
H
:
HYBRID ORBITALS
H: N : H
H
:
:
NH3
Lewis
N
H
trigonal pyramidal
H
H
sp3
H : O: H
:
Lewis
:
H2O
:
σ bonds
lone pair in one hybrid orbital
:O
H
H
bent
sp3
σ bonds
2 lone pairs in hybrid orbitals
HYBRID ORBITALS
F
Consider BF3
F
1s
1s
2s
shape = trigonal planar
B
2p
sp2
F
hybridization of one s
and two p orbitals gives
three trigonal planar
hybrids, and one p is
left unshared and empty
and is perpendicular
to plane of hybrids
unshared p orbital
HYBRID ORBITALS
Consider BeCl2
Cl–Be–Cl
shape = linear
Be
1s
2s
2p
promote & hybridize
1s
2 sp hybrid
orbitals
sp
2 p atomic orbitals
not used
(perpendicular to
hybrid orbitals)
SUMMARY OF
HYBRIDIZATION TYPES
No. of
e− pairs
atomic
orbitals
used
2
s,p
3
s,p,p
4
s,p,p,p
5
s,p,p,p,d
6
hybrid
type
e− pair
formed geometry
examples
two sp
linear
BeF2
three sp2
trigonal
planar
BF3,SO3
four
sp3
five sp3d
tetrahedral
CH4,NH3,
H2O
trigonal
bipyramidal
PF5,SF4,
BrF3
s,p,p,p,d,d six sp3d2 octahedral
SF6, ClF5,
XeF4
π BONDING ORBITALS
H
Consider ethylene
H
C=C
H
has 120° bond angles
so sp2 hybrid orbitals are involved
1s
1s
2s
H
2p
sp2
2p
unused for hybridization
sigma bonds (σ) : overlap of hybrid sp2 orbitals with H
pi bonds (π) : side-to-side overlap of p orbitals
of two carbons
sigma bonds (σ) : straight-ahead overlap of
hybrid orbitals with each other or with H
pi bonds (π) : side-to-side overlap of p orbitals
of two carbons
ethylene
H
C’s are sp2 hybridized
H
C=C
H
H
5 σ bonds
1 π bond
acetylene
C’s are sp hybridized
3 σ bonds
2 π bonds
H–C≡C–H