Gen Chen Chap 1 Notes

10. Chemical Bonding II: Molecular Geometryand 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

P....................................................

....................................................

....................................................

........................

........................

....

....................................................F····

··F ······

F ······

F ······F······

– Typeset by FoilTEX – 2

(2) VSEPR notation: AB5

GChem I 10.2

(3) electron-pair geometry: trigonal bipyramidal

(4) no lone pairs ⇒molecular geometry ≡ electron-pair geome-

try: 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 =⇒ 3 bonds =⇒ S ↗ 8

S....................................................

....................................................

........................

........................

....

....................................................

··F······

F ······

F ······

F······


S has expanded valence

(2) VSEPR notation: AB4E

(3) electron-pair geometry: trigonal bipyramidal

(4) lone pair ⇒ molecular geometry 6≡ electron-pair geometry

molecular geometry: seesaw (sawhorse, distorted tetrahe-

dron)

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

N....................................................

....................................................

........

........

........

........

........

........

....

··

F ···· ··

F ······

F······


(2) VSEPR notation: AB3E

(3) electron-pair geometry: tetrahedral


molecular geometry: trigonal pyramidal

Exercise: Determine the shape of carbon suboxide C3O2

(1) Lewis structure

A = 2×6+3×4 = 24

GChem I 10.4

N = 5×8 = 40

S = N − A = 40−24 = 16 =⇒ 8 bonds

O ........................................................................................................··

··C ....................................................

.................................................... C ........................................................................................................ C ....................................................

.................................................... O····


(2) VSEPR notation for each carbon atom: AB2

(3) electron-pair geometry: linear

(4) no lone pairs ⇒molecular geometry ≡ electron-pair geome-

try

molecular geometry: linear

Exercise: Determine the shape of ozone O3

(1) Lewis structure

A = 3×6 = 18

N = 3×8 = 24

S = N − A = 24−18 = 6 =⇒ 3 bonds

O ........................................................................................................··

··O ....................................................··

O ······

O ....................................................·· ····

O ........................................................................................................

··O····


↔

O ........................................................................................................··

··O ....................................................··

O ······

O ....................................................·· ····

O ........................................................................................................

··O····


(2) VSEPR notation for the central oxygen atom: AB2E

GChem I 10.5

(3) electron-pair geometry: trigonal planar


molecular geometry: bent (angular)

expected bond angle: 120◦; experimental value: 117◦

polar molecules

polar covalent bond

example:

δ+H

δ−Cl

+−−−→d

measure of the charge separation (or polarity) is the dipole mo-

ment

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 2.5and 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

O····

C ............................................................................................................................................................

.................................................... O····


δ− 2δ+ δ−

+� -

µ=0

dipole moment of a molecule depends on the shape of the mol-

ecule

water molecule: O: EN = 3.5, H: EN =2.1

polar bonds exist in H2O

shape of H2O

(1) Lewis structure

A = 2×1+1×6 = 8

N = 2×2+1×8 = 12

S = N − A = 12−8 = 4 =⇒ 2 bonds

GChem I 10.7

H .................................................... O ....................................................····

H


(2) VSEPR notation: AB2E2

(3) electron-pair geometry: tetrahedral

(4) lone pairs ⇒ molecular geometry 6≡ electron-pair geometry

molecular geometry: bent (angular)

bond angle: 104◦

H .................................................... O ....................................................····

H

O

HH


2δ−

δ+δ+ +@@@I

+�� 6

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

CH3Cl AB4 polar

GChem I 10.8

different theories of the chemical bond

Lewis: simple; relies on electron pairs =⇒ problems: O2, odd-

electron species, resonance

VSEPR: predicts molecular shape; relies on electron pairs

Quantum Mechanics

valence-bond theory: covalent bond = overlap of atomic or-

bitals

H2

H2S

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

C H

H

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 bet-

ter approximation:

hybridization

GChem I 10.10

2s +3×2p = 4sp3 hybrid orbitals

ammonia NH3

Lewis

N H

H

H ....................................................

....................................................

....................................................··


VSEPR: AB3E =⇒ electron-pair geometry: tetrahedral (molecular

geometry: trigonal pyramidal) =⇒ hybrid: sp3

2p2s

⟩−→ sp3

N

H

H

H

GChem I 10.11

water H2O

Lewis

H .................................................... O ....................................................····

H


VSEPR: AB2E2 =⇒ electron-pair geometry: tetrahedral (molecu-

lar geometry: bent) =⇒ hybrid: sp3

2p2s

⟩−→ sp3

O

H

H

BF3 AB3

electron-pair geometry: trigonal planar

3 sp2 hybrid orbital + 1 p orbital

BeCl2 AB2

electron-pair geometry: linear

GChem I 10.12

2 sp hybrid orbital + 2 p orbital

ethene (ethylene) C2H4

Lewis

C

H

H

C

H

H

C

H

H

....................................................

........

........

........

........

........

........

....

....................................................

.................................................... C

H

H

....................................................

........

........

........

........

........

........

....


each C: AB3

electron-pair geometry: trigonal planar

=⇒ hybrid: sp2 + 1 p

σ-bond: overlap on the internuclear axis

for ethene: 1s − (1)sp2 and (1)sp2 − (1)sp2

π-bond: overlap off the internuclear axis

weaker than a σ-bond

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) C2H2

Lewis

CH .................................................... ............................................................................................................................................................ 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

H2molecule: 1s + 1s↗ σ∗

1s antibonding

↘ σ1s bonding

GChem I 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 nu-

clei

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 cor-

responding antibonding orbital

GChem I 10.15

3. Aufbau principle (filling of MOs proceeds from low to high en-

ergies)

4. Pauli exclusion principle (at most two electrons per MO, op-

posite 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 1s1 =⇒ H2 two electrons

molecular electron configuration of H2: (σ1s)2

bond order = 12(2−0) = 1

helium molecule He2

He 1s2 =⇒ He2 four electrons

GChem I 10.16

molecular electron configuration of He2: (σ1s)2(σ∗1s)2

bond order = 12(2−2) = 0

the helium dimer does not exist

lithium dimer Li2

Li 1s22s1 =⇒ Li2 six electrons

molecular electron configuration of Li2: (σ1s)2(σ∗1s)2(σ2s)2

bond order = 12(2+2−2) = 1

stable diamagnetic molecule; found in vapor phase

beryllium dimer Be2

Be 1s22s2 =⇒ Be2 eight electrons

molecular electron configuration of Be2: (σ1s)2(σ∗1s)2(σ2s)2(σ∗

2s)2

bond order = 12(2+2−2−2) = 0

beryllium dimer is unstable

boron dimer B2

B 1s22s22p1 =⇒ 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

molecular electron configuration of B2: (σ1s)2(σ∗1s)2(σ2s)2(σ∗

2s)2(π2p)2

(σ1s)2(σ∗1s)2(σ2s)2(σ∗

2s)2(π2py)1(π2pz

)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(σ2px)2(π2py

)2(π2pz)2(π∗

2py)1(π∗

2pz)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

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Gen Chen Chap 1 Notes