10. Chemical Bonding II: Molecular Geometry
and Hybridization of Atomic Orbitals
molecular shape
most molecules are 3-D objects
bond lengths and bond angles
Lewis structures convey no 3-D information
Valence Shell Electron Pair Repulsion model: VSEPR (“vesper”)
electron pairs repel each other =⇒ electron pairs tend to remain
as far apart as possible
lone pair – lone pair À lone pair – bond pair > bond pair – bond
pair
electrons are anchored to central atom, lie on a sphere: valence
shell
◦
two pairs −−→ W 180 −−→ linear
◦
three pairs −−→ W 120 −−→ trigonal planar
◦
four pairs −−→ W 109.5 −−→ tetrahedral
GChem I
10.1
electron-pair geometry
does not have to be equal to
molecular geometry
molecular geometry = geometry of the atomic nuclei bound to
the central atom: bond pairs
VSEPR notation: A = central atom, B = atom bonded to central
atom, E = lone pair on the central atom
Exercise: Determine the shape of PF5
procedure: (1) obtain Lewis structure; (2) write down VSEPR
notation; (3) determine electron-pair geometry; (4) determine
molecular geometry
(1) we have obtained the Lewis structure of PF5 previously
·· ··
F
··
·· ··
F
··
.....
.....
.....
.....
.....
.
.
.....
.....
.....
.....
.....
P
...
...
...
...
...
...
....
.....
.....
.....
.
.
.
....
.....
.....
.....
.....
.....
.
·· F ··
··
·· ·
F
·· ·
·· ·
F
·· ·
(2) VSEPR notation: AB5
GChem I
10.2
(3) electron-pair geometry: trigonal bipyramidal
(4) no lone pairs ⇒ molecular geometry ≡ electron-pair geometry: trigonal bipyramidal
Exercise: Determine the shape of SF4
(1) Lewis structure
A = 1 × 6 + 4 × 7 = 34
N = 5 × 8 = 40
S = N − A = 40 − 34 = 6
·· ··
F
··
·· ··
F
··
.....
.....
.....
.....
.....
.
..
.....
.....
.....
.....
.
.
.
.
.
.....
.....
.....
....
.
.
.
..
.....
.....
.....
.....
.....
.
·· S
=⇒
3 bonds =⇒
S %8
·· ·
F
·· ·
·· ·
F
·· ·
S has expanded valence
(2) VSEPR notation: AB4 E
(3) electron-pair geometry: trigonal bipyramidal
(4) lone pair ⇒ molecular geometry 6≡ electron-pair geometry
molecular geometry: seesaw (sawhorse, distorted tetrahedron)
GChem I
10.3
Exercise: Determine the shape of NF3
(1) Lewis structure
A = 1 × 5 + 3 × 7 = 26
N = 4 × 8 = 32
S = N − A = 32 − 26 = 6
=⇒
3 bonds
·· ··
F ··
....
...
..
...
...
·· ··
F
··
.
.....
.....
.....
.....
.....
N
··
.....
.....
.....
.....
.....
.
·· ·
F
·· ·
(2) VSEPR notation: AB3 E
(3) electron-pair geometry: tetrahedral
(4) lone pair ⇒ molecular geometry 6≡ electron-pair geometry
molecular geometry: trigonal pyramidal
Exercise: Determine the shape of carbon suboxide C3 O2
(1) Lewis structure
A = 2 × 6 + 3 × 4 = 24
GChem I
10.4
N = 5 × 8 = 40
S = N − A = 40 − 24 = 16
··
O
··
..........................
..........................
C
..........................
..........................
C
C
..........................
..........................
8 bonds
=⇒
..........................
..........................
··
O
··
(2) VSEPR notation for each carbon atom: AB2
(3) electron-pair geometry: linear
(4) no lone pairs ⇒ molecular geometry ≡ electron-pair geometry
molecular geometry: linear
··
··
··
O
O
O ··
Exercise: Determine the shape
· · of ozone O3· ·
..........................
..........................
..........................
(1) Lewis structure
A = 3 × 6 = 18
N = 3 × 8 = 24
S = N − A = 24 − 18 = 6
··
O
··
..........................
..........................
··
O
..........................
··
O ··
··
↔
=⇒
··
·· O
··
3 bonds
..........................
··
O
..........................
..........................
··
O
··
(2) VSEPR notation for the central oxygen atom: AB2 E
– Typeset by FoilTEX –
GChem I
1
10.5
(3) electron-pair geometry: trigonal planar
(4) lone pair ⇒ molecular geometry 6≡ electron-pair geometry
molecular geometry: bent (angular)
◦
expected bond angle: 120 ; experimental value: 117
◦
polar molecules
polar covalent bond
example:
δ+
δ−
H Cl
+−−−→
d
measure of the charge separation (or polarity) is the dipole moment
µ = δ·d
unit of dipole moment [µ] = Cm
too large, use the unit debye ( D) for molecules
1 D = 3.34 × 10−30 Cm
experiments show that CO2 is nonpolar; however the molecule
has polar covalent bonds since the electronegativity of C is 2.5
and that of O is 3.5
GChem I
10.6
why is carbon dioxide nonpolar??
consider the shape of the molecule
the Lewis structure is the same as that of CS2 (Set 9)
··
O
··
..........................
..........................
C
..........................
..........................
··
O
··
VSEPR notation: AB2 =⇒ molecular geometry = linear
δ−
··
O
··
δ−
2δ+
..........................
..........................
C
..........................
..........................
+
µ=0
··
O
··
-
dipole moment of a molecule depends on the shape of the molecule
water molecule: O: EN = 3.5, H: EN =2.1
polar bonds exist in H2 O
shape of H2 O
(1) Lewis structure
A = 2×1+1×6 = 8
N = 2 × 2 + 1 × 8 = 12
S = N − A = 12 − 8 = 4
GChem I
– Typeset by FoilTEX –
=⇒
2 bonds
10.7
1
H
..........................
··
O
··
H
..........................
··
··
(2) VSEPR
AB
H notation:
O
H 2 E2
..........................
..........................
(3) electron-pair geometry: tetrahedral
(4) lone pairs ⇒ molecular geometry 6≡ electron-pair geometry
molecular geometry: bent (angular)
bond angle: 104
◦
2δ−
O
δ+ H+
6@
I
@
H δ+
@+
µ6=0
net dipole moment µ = 1.94 D
molecules of the type ABn are NOT polar if all terminal atoms B
are the same.
Examples:
CH4
AB4
nonpolar
– Typeset by FoilTEX –
CH3 Cl
AB4
1
polar
– Typeset by FoilTEX –
GChem I
1
10.8
different theories of the chemical bond
Lewis: simple; relies on electron pairs =⇒ problems: O2 , oddelectron species, resonance
VSEPR: predicts molecular shape; relies on electron pairs
Quantum Mechanics
valence-bond theory: covalent bond = overlap of atomic orbitals
H2
H2 S
provides information on bond energies: 1s -1s overlap is stronger
than 1s -3p overlap
C
[He]
2s
2p
=⇒ simplest hydrocarbon CH2 with bond angle of 90◦
! CH2 does not exist!
simplest hydrocarbon: methane CH4
???
excited-state
GChem I
10.9
C
[He]
2s
2p
promotion: 2s → 2p
problem: 2s nondirectional, 2p at right angles
Lewis structure of methane
H
...
...
..
...
...
.
H
..........................
C
..........................
H
...
...
...
...
...
...
H
VSEPR: AB4 =⇒ tetrahedral (electron-pair and molecular geom◦
◦
etry) =⇒ bond angle 109.5 6= 90
???
we cannot solve the Schrödinger equation to obtain an exact,
analytical expression for the molecular wave function ψ
atomic orbitals −→ approximation of ψ for multielectron atoms
this approximation does not work well for molecules; find a better approximation:
hybridization
GChem I
10.10
2s + 3 × 2p = 4sp 3 hybrid orbitals
ammonia NH3
Lewis
H
..........................
··
N
..........................
H
...
...
...
...
...
...
H
VSEPR: AB3 E =⇒ electron-pair geometry: tetrahedral (molecular
3
geometry: trigonal pyramidal) =⇒ hybrid: sp
+
2p
2s
−→
sp 3
N
H
H
H
GChem I
– Typeset by FoilTEX –
10.11
1
water H2 O
Lewis
H
..........................
··
O
··
..........................
H
VSEPR: AB2 E2 =⇒ electron-pair geometry: tetrahedral (molecu3
lar geometry: bent) =⇒ hybrid: sp
+
2p
2s
−→
sp 3
H
O
H
BF3
AB3
electron-pair geometry: trigonal planar
2
3 sp hybrid orbital + 1 p orbital
BeCl2
AB2
electron-pair geometry: linear
– Typeset by FoilTEX –
GChem I
1
10.12
C
C
H
H
2 sp hybrid orbital + 2 p orbital
ethene (ethylene) C2 H4
Lewis
H
H
....
...
..
...
....
....
...
..
...
....
C
..........................
..........................
C
...
...
...
...
...
...
...
...
...
...
...
...
H
H
each C: AB3
electron-pair geometry: trigonal planar
=⇒ hybrid: sp 2 + 1 p
σ-bond: overlap on the internuclear axis
2
2
for ethene: 1s − (1)sp and (1)sp − (1)sp
2
π-bond: overlap off the internuclear axis
– Typeset by
FoilTE
– -bond
weaker
than
aX σ
1
for ethene: p − p
double bond = σ-bond + π-bond
GChem I
10.13
stronger than a single bond (σ-bond), but not twice as strong
shape of a molecule: determined by σ-bond framework
rotation about a double bond is severely restricted
ethyne (acetylene) C2 H2
Lewis
H
..........................
C
..........................
..........................
..........................
C
..........................
H
each C: AB2
electron-pair geometry: linear
=⇒ hybrid: sp + 2 p
triple bond: 1 σ-bond + 2 π-bonds
still no explanation why O2 is paramagnetic
fourth description of chemical bond:
molecular-orbital theory (MO)
use wave functions or orbitals that belong to the entire molecule
H2 molecule: 1s + 1s
GChem I
%
σ∗1s
antibonding
&
σ1s
bonding
10.14
a bonding molecular orbital has lower energy than the atomic
orbitals from which it was formed
lower energy =⇒ greater stability
bonding orbital: electron density greatest between the two nuclei
formation of bonding orbital: constructive interference
an antibonding molecular orbital has higher energy than the
atomic orbitals from which it was formed
higher energy =⇒ lower stability
antibonding orbital: electron density goes to zero between the
two nuclei
formation of bonding orbital: destructive interference
sigma molecular orbital: electron density is concentrated on the
internuclear axis, cylindrical symmetry
Molecular Electron Configuration (molecular-orbital diagram)
1. The number of molecular orbitals formed = number of atomic
orbitals combined
2. The more stable the bonding orbital, the less stable the corresponding antibonding orbital
GChem I
10.15
3. Aufbau principle (filling of MOs proceeds from low to high energies)
4. Pauli exclusion principle (at most two electrons per MO, opposite spins)
5. Hund’s rule (when MOs of identical energy are available,
electrons occupy those orbitals singly if possible and have
the same spin orientation)
6. Number of electrons in MOs = total number of electrons on
the bonding atoms
bond order = 12 (# of electrons in bonding MOs − # of electrons
in antibonding MOs)
a molecule is stable if the bond order is greater than zero
First and Second Period Homonuclear Diatomic Molecules
hydrogen molecule H2
atomic hydrogen H 1s
1
=⇒ H2 two electrons
2
molecular electron configuration of H2 : (σ1s )
bond order = 21 (2 − 0) = 1
helium molecule He2
He 1s 2 =⇒ He2 four electrons
GChem I
10.16
2
∗ 2
molecular electron configuration of He2 : (σ1s ) (σ1s )
bond order = 12 (2 − 2) = 0
the helium dimer does not exist
lithium dimer Li2
Li 1s 2 2s 1 =⇒ Li2 six electrons
2
∗ 2
2
molecular electron configuration of Li2 : (σ1s ) (σ1s ) (σ2s )
bond order = 12 (2 + 2 − 2) = 1
stable diamagnetic molecule; found in vapor phase
beryllium dimer Be2
Be 1s 2 2s 2 =⇒ Be2 eight electrons
2
∗ 2
2
∗ 2
molecular electron configuration of Be2 : (σ1s ) (σ1s ) (σ2s ) (σ2s )
bond order = 12 (2 + 2 − 2 − 2) = 0
beryllium dimer is unstable
boron dimer B2
B 1s 2 2s 2 2p 1 =⇒ B2 ten electrons
p -orbitals come into play
GChem I
10.17
σ2p , σ∗2p
1×
π2p , π∗2p
2×
molecular-orbital energy level diagram
2
∗ 2
2
∗ 2
2
molecular electron configuration of B2 : (σ1s ) (σ1s ) (σ2s ) (σ2s ) (π2p )
(σ1s )2 (σ∗1s )2 (σ2s )2 (σ∗2s )2 (π2p y )1 (π2p z )1
bond order = 12 (2 + 2 + 2 − 2 − 2) = 1
stable paramagnetic molecule
molecular-orbital diagrams for 2nd period homonuclear diatomic
molecules
oxygen molecule
O2 : (σ1s )2 (σ∗1s )2 (σ2s )2 (σ∗2s )2 (σ2p x )2 (π2p y )2 (π2p z )2 (π∗2p y )1 (π∗2p z )1
bond order = 12 (2 + 2 + 2 + 2 + 2 − 2 − 2 − 2) = 2
stable paramagnetic molecule
molecular orbital theory explains the magnetic properties of
the oxygen molecule and other diatomic molecules and ions
GChem I
10.18