H2O
2s
2px
.
..
O . 2py
..
2px
2pz
2py
2F-1 (of 14)
H2O
1s
2s
H
2px
1s
.
..
O . 2py
..
2pz
H
2px
1s
1s
2py
Each bond is formed by combining a H 1s atomic orbital (with 1 e-) with
an O 2p atomic orbital (with 1 e-)
This predicts the shape of a water molecule to be bent with a 90º angle
However, the molecule is bent with a 105º angle
2F-2 (of 14)
1939 LINUS PAULING
Showed that linear combinations of an atom’s valence
atomic orbitals produce another set of equivalent valence
atomic orbitals
HYBRIDIZATION – The combining of 2 or more orbitals of different
sublevels to make an equal number of HYBRID ORBITALS of equivalent
energy (said to be DEGENERATE)
2F-3 (of 14)
2p
hybridization
sp3
2s
E
These 4 hybrid orbitals are called sp3 orbitals
2F-4 (of 14)
sp3
2pz
2py
hybridization
2px
sp3
2s
sp3
sp3
The sp3 hybrid orbitals are arranged in a tetrahedron, with an angle of
109.5º between them
2F-5 (of 14)
sp3
2pz
2py
hybridization
2px
sp3
2s
sp3
sp3
The sp3 hybrid orbitals are arranged in a tetrahedron, with an angle of
109.5º between them
2F-6 (of 14)
sp3
sp3
sp3
sp3
Each bond is formed by combining an O sp3 atomic orbital (with 1 e-) with
a H 1s atomic orbital (with 1 e-)
This forms a bent molecule with a theoretical bond angle of 109.5º
2F-7 (of 14)
Each bond in H2O is completely symmetrical around its internuclear axis
SIGMA BOND (σ) – A bond that is completely symmetrical around its
internuclear axis
Each bond is named: σ(sp3+1s)
2F-8 (of 14)
NH3
2py
2s
2px
.
..
N . 2py
.
2px
2pz
2pz
2F-9 (of 14)
NH3
1s
2py
2s
H
2px
.
..
N . 2py
.
H
2px
2pz
.
H
1s
1s
2pz
This predicts the shape of an ammonia molecule to be trigonal pyramidal
with a 90º angle
However, the molecule is trigonal pyramidal with a 107º angle
2F-10 (of 14)
2p
hybridization
2s
E
2F-11 (of 14)
sp3
sp3
sp3
sp3
sp3
This forms a trigonal pyramidal molecule with a bond angle of 109.5º
Each sigma bond in NH3 is formed by combining a N sp3 atomic orbital
(with 1 e-) with a H 1s atomic orbital (with 1 e-)
Each bond is named: σ(sp3+1s)
2F-12 (of 14)
CH4
sp3
2s
2px
.
..
C
.
sp3
2py
2p
hybridization
E
2F-13 (of 14)
.
C . sp3
.
sp3
2pz
2s
.
sp3
sp3
sp3
sp3
sp3
This forms a tetrahedral molecule with a theoretical bond angle of 109.5º
Each bond is named: σ(sp3+1s)
Central atoms with SN = 4 always undergo sp3 hybridization when bonding
2F-14 (of 14)
BH3
H
H
B
H
sp3
SN = 3
Trigonal Planar
sp3
H
sp3
B
H
sp3
H
If the B undergoes sp3 hybridization:
Trigonal pyramidal
Central atoms with SN = 3 only need to hybridize 3 valance atomic orbitals
2G-1 (of 19)
BH3
2p
2p
hybridization
sp2
2s
E
sp2
2s
2px
.
..
B
2pz
2G-2 (of 19)
2py
sp2
.
.
B . sp2
2pz
2pz
2pz
2py
hybridization
sp2
sp2
2px
2s
sp2
The angle between the sp2 hybrid orbitals is 120º
The 2pz orbital is 90º from the plane of the sp2 hybrid orbitals
2G-3 (of 19)
2pz
2pz
2py
hybridization
sp2
sp2
2px
2s
sp2
The angle between the sp2 hybrid orbitals is 120º
The 2pz orbital is 90º from the plane of the sp2 hybrid orbitals
2G-4 (of 19)
2pz
2pz
2py
hybridization
sp2
sp2
2px
2s
sp2
Rotating the top 90º towards you
sp2
sp2
sp2
2G-5 (of 19)
2pz
sp2
sp2
sp2
Rotating the top 90º towards you
2G-6 (of 19)
Rotating the top 90º towards you
2G-7 (of 19)
This forms a trigonal planar molecule with a bond angle of 120º
Each bond is named: σ(sp2+1s)
2G-8 (of 19)
BeH2
H
Be
H
SN = 2
Linear
H
Be
H
If the B undergoes sp3 hybridization: Bent, 109.5º
If the B undergoes sp2 hybridization: Bent, 120º
Central atoms with SN = 2 only need to hybridize 2 valance atomic orbitals
2G-9 (of 19)
BeH2
2p
2p
hybridization
sp
2s
E
2s
sp
..
2px
Be 2py
2pz
2G-10 (of 19)
sp
.
.
Be 2py
2pz
2pz
2pz
2py
2py
hybridization
2px
sp
sp
2s
The sp hybrid orbitals are linear,
with an angle of 180º between them
The 2py and 2pz orbitals are 90º
from the sp hybrid orbitals
2G-11 (of 19)
2pz
2py
sp
sp
This forms a linear molecule with a bond angle of 180º
Each bond is named: σ(sp+1s)
2G-12 (of 19)
PH5
5 + 5(1) = 10 valence e-s
H
H
H
H
P
H
Central atoms with SN = 5 need to hybridize 5 valance atomic orbitals
2G-13 (of 19)
PH5
3d
3p
3d
hybridization
3s
E
These 5 hybrid orbitals are called sp3d orbitals
2G-14 (of 19)
sp3d
PH5
sp3d
sp3d
sp3d
sp3d
sp3d
This forms a trigonal bipyramidal molecule
Each bond is named: σ(sp3d+1s)
2G-15 (of 19)
SH6
6 + 6(1) = 12 valence e-s
H
H
H
S
H
H
H
Central atoms with SN = 6 need to hybridize 6 valance atomic orbitals
2G-16 (of 19)
SH6
3d
3p
3d
hybridization
sp3d2
3s
E
These 6 hybrid orbitals are called sp3d2 orbitals
2G-17 (of 19)
SH6
sp3d2
sp3d2
sp3d2
sp3d2
sp3d2
sp3d2
This forms an octahedral molecule
Each bond is named: σ(sp3d2+1s)
2G-18 (of 19)
SN
2
3
4
Hybrid.
Hybrid Orbital
Geometry
sp
Linear
sp2
sp3
sp
sp2
Trigonal
Planar
Tetrahedral
sp
sp2
sp3
sp3
sp3
sp3d
sp3
5
6
sp3d
sp3d2
sp3d
Trigonal
Bipyramidal
Octahedral
sp3d
sp3d
sp3d2
sp3d2
2G-19 (of 19)
sp2
sp3d2
sp3d2
sp3d2
sp3d2
sp3d
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/hybrv18.swf
HYDROCARBON – Molecules composed of only carbon and hydrogen
SATURATED HYDROCARBON – Hydrocarbons with only single bonds
between the carbon atoms
The prefix tells the number of carbon atoms in the molecule
1
2
3
4
5
methethpropbutpent-
6
7
8
9
10
hexheptoctnondec-
The ending -ane indicates only single bonds between the carbon atoms
2H-1 (of 10)
methane : CH4
sp3
H
H
C
H
H
sp3
sp3
SN of C = 4
 Hybridization = sp3
sp3
Tetrahedral
Nonpolar
Each bond: σ(sp3+1s)
2H-2 (of 10)
H
C
H
H
H
ethane : C2H6
H
H
H
C
C
H
H
sp3
H
sp3
SN of each C = 4
 Hybridization = sp3
sp3
sp3
H
H
C
H
H
C
H
C-C bond: σ(sp3+sp3)
Each C-H bond: σ(sp3+1s)
H
2H-3 (of 10)
UNSATURATED HYDROCARBON – Hydrocarbons with at least one
double or triple bond between carbon atoms
The ending -ene indicates a double bond between carbon atoms
A double bond consists of a SIGMA BOND and a PI BOND
PI BOND () – A bond that is only symmetrical upon a 180º rotation around
its internuclear axis
2H-4 (of 10)
ethene : C2H4
H
H
H
C
C
2pz
sp2
H
sp2
sp2
SN of each C = 3
 Hybridization = sp2
σ(sp2+sp2)
(2p+2p)
Each C-H bond: σ(sp2+1s)
C-C bonds:
2H-5 (of 10)
H
H
C
C
H
H
The ending -yne indicates a triple bond between carbon atoms
A triple bond consists of a 1 SIGMA BOND and 2 PI BONDS
2H-6 (of 10)
ethyne : C2H2
2pz
2pz
H
C
C
H
SN of each C = 2
sp
1s sp
2pz
sp
sp
sp
sp
2py
 Hybridization = sp
σ(sp+sp)
(2pz+2pz)
(2py+2py)
Each C-H bond: σ(sp+1s)
C-C bonds:
2H-7 (of 10)
H
C
C
H
1s
BOND ORBITAL MODELS (BOM’s) – Geometric diagrams of the bonding
in molecules
To draw a BOM for a molecule:
1 – Draw the Lewis structure and predict the hybridizations of each atom
2 – Draw the correct geometries for the hybrid atomic orbitals of each atom,
starting with atoms involved in double or triple bonds
3 – Have atoms bond by overlapping atomic orbitals, and label each bond
2H-8 (of 10)
Draw a BOM for CH3CHCH2
H
H
C
H
C
H
C
H
C1 C2 C3
sp3 sp2 sp2
H
H
H
H
C
C
H
H
σ(sp2+sp2)
(2p+2p)
C2-H & C3-H bonds: σ(sp2+1s)
C2-C3 bonds:
C1-C2 bond:
C1-H bonds:
2H-9 (of 10)
C
H
σ(sp3+sp2)
σ(sp3+1s)
Draw a BOM for CH2CCH2
H
C
H
H
H
C
C
C1 C2 C3
sp2 sp sp2
H
H
H
C
C-C bonds:
C-H bonds:
2H-10 (of 10)
C
C
H
σ(sp2+sp)
(2p+2p)
σ(sp2+1s)
Valence Bond Theory cannot predict
(1) magnetism of molecules
(2) the e- arrangement of molecules with an odd number of e-s
(3) the e- arrangement in molecules with resonance
MOLECULAR ORBITAL THEORY – Electrons in molecules exist in
molecular orbitals that extend over the entire molecule
2I-1 (of 7)
H2
The number of valence MO’s in the H2 molecule equals the number of
valence AO’s in the 2 individual H atoms
2 valence AO’s in the atoms,  2 valence MO’s in the molecule
1s AO
1s AO
This MO is the result of the addition of the 2 AO’s: 1s + 1s
This MO is the result of the subtraction of the 2 AO’s: 1s - 1s
2I-2 (of 7)
MO1
σb
MO2
σ*
SIGMA MOLECULAR ORBITAL – An orbital that is completely symmetrical
around its internuclear axis
BONDING MOLECULAR ORBITAL – An orbital lower in energy than the
original atomic orbitals because most of the e- density is between the 2
nuclei
ANTIBONDING MOLECULAR ORBITAL – An orbital higher in energy than
the original atomic orbitals because most of the e- density is outside the 2
nuclei
2I-3 (of 7)
MO Energy Level Diagram for H2
σ1s*
E
2I-4 (of 7)
1s
σ1sb
1s
atom 1
molecule
atom 2
Bond Order
=
Bonding e-s – Antibonding e-s
_______________________________________
2
Bond Order for H2 =
2–0
______
=
1
2
PARAMAGNETIC – A substance that is magnetic in a magnetic field
Due to unpaired e-s
DIAMAGNETIC –
A substance that is never magnetic
Due to no unpaired e-s
H2 is diamagnetic
Electron configuration for H2 :
2I-5 (of 7)
(σ1sb)2
He2
The valence AO’s of each He atom combine to form MO’s in the He2
molecule
MO Energy Level Diagram for He2
σ1s*
1s
σ1sb
1s
E
Bond Order for He2
2I-6 (of 7)
=
(2 – 2) / 2
=
0
 He2 does not exist
Does He2+ exist?
If yes, give bond order, magnetism, and electron configuration notation
MO Energy Level Diagram for He2+
σ1s*
1s
σ1sb
1s
E
 He2+ does exist
Bond Order for He2+ =
(2 – 1) / 2
He2+ is paramagnetic
Electron configuration for He2+ : (σ1sb)2 (σ1s*)1
2I-7 (of 7)
= ½
Diatomic Homonuclear Molecules of the 2nd Period
Valence Orbitals
2p
2p
?
2s
E
2J-1 (of 15)
2s
atom 1
molecule
atom 2
The 2 2s AO’s combine the same way as 2 1s AO’s, forming
a s2sb MO and a s2s* MO
2pz
2pz
2py
2px
2py
2px
Each pair of p orbitals will combine to form a bonding MO and an
antibonding MO
2J-2 (of 15)
2px
2px
s2pxb
2px
2px
s2px*
2J-3 (of 15)
2pz
2pz
2pzb
PI MOLECULAR ORBITAL – An orbital that is symmetrical only upon a 180º
rotation around the internuclear axis
2pz
2pz
2pz*
2J-4 (of 15)
The 2 2py AO’s combine the same way as 2 2pz AO’s, forming
a 2pyb MO and a 2py* MO
The 2pzb MO and the 2pyb MO are equal in energy
The 2pz* MO and the 2py* MO are equal in energy
2J-5 (of 15)
Experimental data has determined the energies of the 8 MOs
s2px*
2pz*
2py*
2p
2p
s2pxb
2pzb
2pyb
s2s*
2s
2s
E
s2sb
atom 1
2J-6 (of 15)
molecule
atom 2
C2
s2px*
2pz*
2py*
2p
2p
s2pxb
2pzb
2pyb
s2s*
2s
2s
E
Bond Order for C2
s2sb
=
(6 – 2) / 2
= 2
Electron configuration for C2 : (s2sb)2 (s2s*)2 (2pb)4
2J-7 (of 15)
C2 is diamagnetic
F2
s2px*
2pz*
2py*
2p
2p
s2pxb
2pzb
2pyb
s2s*
2s
2s
E
Bond Order for F2
s2sb
=
(8 – 6) / 2
= 1
F2 is diamagnetic
Electron configuration for F2 : (s2sb)2 (s2s*)2 (2pb)4 (s2pb)2 (2p*)4
2J-8 (of 15)
C2
Bond Order = 2
F2
Highest Bond Energy:
C2
Longest Bond Length:
F2
2J-9 (of 15)
Bond Order = 1
Diatomic Heteronuclear Molecules of the 2nd Period
N’s greater nuclear charge makes its AO energies lower that C’s
CN
s2px*
2pz*
2py*
2p
2p
s2pxb
2pzb
2pyb
s2s*
2s
E
Bond Order for CN =
2s
s2sb
(7 – 2) / 2
= 2½
CN is paramagnetic
Electron configuration for C2 : (s2sb)2 (s2s*)2 (2pb)4 (s2pb)1
2J-10 (of 15)
Molecular Orbital Theory can also explain the e- arrangement in molecules
with resonance
Benzene, C6H6
H
C
H
H
H
H
H
C
C
C
C
C
C
C
C
C
H
H
H
Each C is sp2, and all atoms are planar
Each C has a p orbital perpendicular to the plane
2J-11 (of 15)
C
H
C
H
H
Resonance structures assume that these p orbitals make distinct pi bonds
Molecular Orbital Theory predicts the 6 2p AO’s will combine to make 6 MO’s
2J-12 (of 15)
Resonance structures assume that these p orbitals make distinct pi bonds
Molecular Orbital Theory predicts the 6 2p AO’s will combine to make 6 MO’s
2J-13 (of 15)
DELOCALIZED PI SYSTEM – A group of pi molecular orbitals spread out
over more than 2 atoms
The e-s in the delocalized pi system strengthen the bonds in the ring
Molecules possessing resonance always bond with delocalized pi systems,
which are formed from PARALLEL P ORBITALS
2J-14 (of 15)
NO3-
O
N
-
↔
O
O
O
N
-
↔
O
O
O
N
O
O
All atoms that can form a double bond in at least 1 resonance structure must
be sp2 to have an unhybridized p orbital to form the delocalized pi system
O
N
O
2J-15 (of 15)
O
O
N
O
O
EMPIRICAL FORMULA CALCULATIONS
EMPIRICAL FORMULA – The simplest whole-number ratio of the atoms of
different elements in a compound
2K-1 (of 8)
Molecular Formula:
C6H6
C6H12O6
H2O
Empirical Formula:
C1H1
C1H2O1
H2O
Find the empirical formula of a compound that is 75.0% carbon and
25.0% hydrogen by mass.
1)
Assume you have 100 g of the compound  75.0 g C and 25.0 g H
2)
Calculate the moles of atoms of each element
75.0 g C
x
1 mol C
_____________________
= 6.245 mol C
12.01 g C
25.0 g H
x 1 mol H
____________________
= 24.75 mol H
1.01 g H
3)
Divide each number of moles by the smallest number of moles
6.245 mol C
_______________________
6.245
4)
= 1.00 mol C
24.75 mol H = 3.96 mol H
_______________________
6.245
The integer mole ratio must be the atom ratio:
2K-2 (of 8)
CH4
Find the empirical formula of a compound that is 90.0% carbon and
10.0% hydrogen by mass.
90.0 g C
x
1 mol C
_____________________
= 7.494 mol C
12.01 g C
10.0 g H
x 1 mol H
___________________
= 9.901 mol H
1.01 g H
7.494 mol C = 1.000 mol C  2
3
_______________________
7.494
9.901 mol H
_______________________
= 1.321 mol H  2
3
7.494
= 2.00
3.00 mol C
= 3.96
2.64 mol H
Empirical formula: C3H4
If the moles of all elements are not within 0.1 moles of an integer, they
must all be multiplied by an integer until they are integers
2K-3 (of 8)
MOLAR MASSES OF COMPOUNDS
MOLAR MASS – The mass of one mole of molecules of a molecular
substance, or one mole of formula units of an ionic substance
Al2(SO4)3
2 mol Al (26.98 g/mol) = 53.96 g
3 mol S (32.07 g/mol) = 96.21 g
12 mol O (16.00 g/mol) = 192.00 g
342.17 g
The mass necessary to have 1 mole of Al2(SO4)3 formula units
2K-4 (of 8)
Calculate the molar mass of carbon tetrachloride
CCl4
1 mol C (12.01 g/mol)
4 mol Cl (35.45 g/mol)
= 12.01 g
= 141.80 g
153.81 g 
153.81 g CCl4 = 1 mol CCl4
Calculate the number of chlorine atoms in 1.00 g carbon tetrachloride.
1.00 g CCl4 x
1 mol CCl4
__________________
153.81 g CCl4
= 1.57 x 1022 atoms Cl
2K-5 (of 8)
x
4 mol Cl
______________
1 mol CCl4
x
6.022 x 1023 atoms Cl
_____________________________
1 mol Cl
MOLECULAR FORMULA CALCULATIONS
MOLECULAR FORMULA – The actual number of the atoms of different
elements in a molecule
Empirical Formula:
C1H1
Molecular Formula:
C1H1 or C2H2 or C3H3 or C4H4 etc.
2K-6 (of 8)
Find the molecular formula of a compound that is 43.7% phosphorus
and 56.3% oxygen by mass, and has a molar mass of about 280 g/mol.
43.7 g P
x
1 mol P
_____________________
= 1.411 mol P
30.97 g P
56.3 g O
x
1 mol O
_____________________
= 3.519 mol O
16.00 g O
1.411 mol P = 1.00 mol P  2
_______________________
1.411
3.519 mol O = 2.49 mol O
_______________________
1.411
= 2.00 mol P
Empirical formula: P2O5
Find the molar mass of the empirical formula
2K-7 (of 8)
= 4.98 mol O
2
P2O5
2 mol P (30.97 g/mol)
5 mol O (16.00 g/mol)
=
=
61.94 g
80.00 g
141.94 g
Divide the compound’s actual molar mass by the empirical formula’s
molar mass – it should be very close to an integer
280 g/mol
____________________________
≈ 2
141.94 g/mol
 The molecular formula is 2 times the empirical formula
Molecular formula: P4O10
2K-8 (of 8)