3.68 whereisit

CIEG 632
Review of General
Chemistry
C. P. Huang
University of Delaware
1
Content
1.
2.
3.
4.
5.
6.
7.
8.
The periodical table
Fundamental particles
State of matter
Stoichemistry
Dual properties of matter
Atomic orbitals
Chemical bonding
Water structure/properties
2
3
4
Periodic Trends in Electron Configurations
5
Periodic Trends in Oxidation Numbers
6
Periodic Trends in Radii
7
Periodic Trends in Ionization Energy
Na(g) → Na+(g) + e-
8
Periodic Trends in Electron Affinity
Cl(g) + e- → Cl-(g)
9
Periodic Trends of Electronegativity
10
Electronegativity
• Electronegativity is a measure of the tendency of an atom to attract a
bonding pair of electrons.
• The Pauling scale is the most commonly used. Fluorine (the most
electronegative element) is assigned a value of 4.0, and values range down to
caesium and francium which are the least electronegative at 0.7.
• No electronegativity difference between two atoms leads to a pure nonpolar covalent bond.
• A small electronegativity difference leads to a polar covalent bond.
• A large electronegativity difference leads to an ionic bond.
11
Electronegativity
Electronegativity
Difference
Type of Bond Formed
0.0 to 0.2
nonpolar covalent
0.3 to 1.4
polar covalent
> 1.5
ionic
12
Fundamental Particles
A = 65
Z = 30
N = 65-30 = 25
Atomic number (Z) = number of protons
Atomic weight (A) = Z + N (neutrons)
13
Fundamental Particles
•
Relative atomic weight
– Mass of individual atom is very small, the heaviest
one is still less than 5x10-22g
– Unified atomic mass (u) is defined as 1/12 of the
mass of a C12 atom
• C12 = 12u; N23 = 22.9898u;
• Proton = 3.68x10-27u = 1.67x10-24 g;
• electron = 9.11x10-28 g; 6.023x1023 amu = 1 g
– Most chemical reaction do not discriminate
against isotopes.
– In iron ore, meteorites, the iron components are
• Fe54:Fe56:Fe57:Fe58 = 5.82%:91.66%:
2.19%:2.33%;
• average = 3.1426u + 51.3296u + 1.2483u +
0.1914u = 55.912u
14
Fundamental Particles
Particle
Symbol
Mass
Charge
Proton
p
1.00728u
+1
Neutron
n
1.00867u
0
Negative
electron
e-, e, , -
0.0005486u
-1
Positive electron
(positron)
e+, +
0.0005486u
+1
15
State of Mater
• Phase: solid; gas; liquid
http://www.harcourtschool.com/activity/states_of_matter/
http://www.chem.purdue.edu/gchelp/atoms/states.html
16
Microscopic Characteristics
gas
liquid
solid
•
assumes the
shape and
volume of its
container
particles can
move past one
another
•
assumes the shape of
the part of the container
which it occupies
particles can move/slide
past one another
•
retains a fixed
volume and shape
rigid - particles
locked into place
•
compressible
•
lots of free space
between
particles
not easily compressible
little free space between
particles
•
not easily
compressible
little free space
between particles
•
flows easily
particles can
move past one
another
flows easily
particles can move/slide
past one another
•
does not flow
easily
rigid - particles
cannot move/slide
17
past one another
•
Gas State of Matter
• Measurement of gas
– Pressure (P) = force/area = height x density
– Atmosphere =
• 760 mmHg at oC
• 7600 cm x 13.595 g/cm3 = 1033 g/cm2
=1033 cm H2O
• 29.9 in Hg
• 29.9 x 0.491 = 14.7 lb/in2
18
Gas State of Matter
•
•
•
•
•
General gas laws
Boyles law: PV = constant
Charles law: VT = constant
Gay-Lussac law: P/T =
constant
R = PV/nT=
– (1 atm x 22.4 L)/(1 molx
273 oK) = 0.082 (L-atm/oKmol)
– 0.73 ft3-atm/oR-#mol)
– 1544 ft-lb/lb-mole
– 1.99 cal/mol-oK
– 8.31 joule/mol-oK
PV
P2V2
1 1

T1
T2
Ideal gas
PV  nRT
19
Non-Ideal Gas

an 2  V

 P  2   b   RT
V  n


Gas
Helium
Hydrogen
Nitrogen
Oxygen
Benzene
a (L6-bar-mol-2
0.034598
0.24646
1.3661
1.3820
18.876
b (L3-mol-1)
0.023733
0.026665
0.038577
0.031860
0.11974
20
Gas State of Matter
• Dalton law of partial pressure: PT = Pi
• Example: Collection of hydrogen gas over water at
34oC which vapor pressure is 40 mm; atmospheric
pressurize is 753 mm, volume of gas is 425 mL. What
is the volume of hydrogen gas at STP?
PT = PH2O + PH2
PH2 = PT – PH2O = 753- 40 = 713 mm
V2 = V1 (T2/T1)(P1/P2) = 425 (273/(273+34)) (713/760) =
355 mL
21
(Gas state) Concentrated nitric acid acts on copper to give
nitrogen oxide and dissolved copper ions according to the
following reaction
Cu(s) + 4H+(aq) + 2NO3- (aq)  2 NO2(g) + Cu2+ + 2 H2O (l)
Suppose that 6.8 g of copper is consumed from the mass of
the know reactant of product (in this case a pressure of 0.97
atm and a temperature of 45oC. Calculate the volume of NO2
product.
The mole ratio of NO2 / Cu = 2/1 =2
Mole of NO2 =2* 6.80 (g of Cu) / 63.55 (g/mol) = 0.214
(mol)
PV=nRT
V = 0.214(mol)*0.08206(L atm /mol K)*(273.15+45)(K) /
0.97 (atm)
= 5.76 (L)
22
Solid State of Matter
triclinic
monoclinic
hexagonal
orthorhombic
rhombohedra
tetragonal
cubic
7 Crystal Systems
23
Solid State of Matter
•
•
•
•
•
•
•
•
•
•
•
•
There are seven systems of crystals. The system refers to the
shape of the undecorated unit cell. They are:
Triclinic: a >< b >< c,  >< ><  >< 90 deg
Monoclinic: a >< b >< c,  =  = 90 deg,  >< 90 deg
Orthorhombic: a >< b >< c,  = 90 deg
Tetragonal: a = b >< c,  = 90 deg
Hexagonal: a = b >< c,  = 90 deg, = 120 deg
Rhombohedral: a = b = c,  < 120 deg >< 90 deg
Cubic: a = b = c,  = 90 deg

Covalent solids: sugars, camphors, iodine, ice
Ionic solids: table salts
Atomic solids: iron, gold, silver
Van der Walls solids: graphite
c



a
b
24
Bravais lattices
• There are fourteen Bravais lattices.
• The Bravais lattices are constructed from the simplest
translational symmetries applied to the seven crystal
systems.
• A P lattice has decoration only at the corners of the unit
cell.
• An I lattice has decoration at the body center of the cell
as well as at the corners.
• An F lattice has decoration at the face centers as well
as at the corners.
• A C lattice has decoration at the center of the (001)
face as well as at the corners.
• Likewise P and I lattices have decoration at the centers
of the (100) and (010) faces respectively. R lattices are
a special type in the trigonal system which possess
25
rhombohedral symmetry.
Bravais lattices
26
(Solid state) The composition of a sample of wustite is
Fe0.930O1.00. What is the percentage of the iron is in the
form of iron(III)?
Assume x mole of iron is +3, then 0.930-x mole of iron is
+2, the total positive charge is equal to the total negative
charge. So,
3x+2(0.930-x) =2
x = 0.140
% of Fe(III) = 0.140/0.930 = 0.151 = 15.1 %
27
Crystal Coordination
Bragg equation: nλ= 2 d Sinθ
Example:
With λ=0.709Å, n =2,
θ =20.2/2=10.1
2*(0.709) = 2 d Sin(10.1),
d= 4.04 Å = a
28
(Solid state) Metallic aluminum has a face-centered
crystal structure. Calculate the fraction of the v
volume of metal aluminum that is occupied by its
atoms. Given the atomic radius of Al atoms, r1, is
related to the dimension of the cubical unit cell, “a”,
by the following expression: 4r1 = a√2.
4r1= a*20.5
r1= a*20.5/4
fraction of the v volume = 4* (4/3)π(r1)3 / a3
=4* (4/3)π(a*20.5/4)3 / a3 = 0.740
29
(Solid state) A diffraction pattern of metallic aluminum is
obtained by using x-rays which wave length l = 0.709 Å.
The second-order Bragg diffraction from the parallel faces
of the unit cell is observed at an angel 2 = 20.2o.
Calculate the lattice parameter, a, and the nearest
neighbor distance in metallic aluminum. Estimate the
atomic radius of aluminum.
(1) Bragg equation: nλ= 2 d Sinθ
With λ=0.709Å, n =2, θ =20.2/2=10.1
2*(0.709) = 2 d Sin(10.1),
d = 4.04 Å = a
(2) The nearest neighbor distance in FCC = a/20.5 = 4.04 Å / 20.5
= 2.86 Å
(3) 4r1= a*20.5 = 4.04 Å * 20.5
r1= 1.43 Å
30
Defects
31
Dislocation
32
Graphene
one-atom-thick planar sheets
of sp2-bonded carbon atoms;
densely packed in a
honeycomb crystal lattice.
carbon-carbon bond length :
0.142 nm
interplanar spacing: 0.335 nm
a stack of three million sheets:
one millimeter thick.
The Nobel Prize in Physics for
2010 was awarded to Andre
Geim and Konstantin
Novoselov "for
groundbreaking experiments
regarding the two-dimensional
material graphene".
http://graphenetimes.com/
33
Liquid State of Matter
• General properties
–
–
–
–
–
–
–
Shape and volume
Compressibility
Diffusion
Evaporation and vapor pressure
Surface tension
Viscosity
Density
34
Solution
• Concentration
– % wt: 5% NaCl = 5% NaCl + 95% H2O
– % vol. 12% alcohol= 12 mL alcohol + 100 mL wine
( Prove = 0.5% alcohol)
– Molar solution (M): mol solute per liter solution
– Molal solution (m): mol solute per liter solvent
– Normal (N) equivalent
1 M HCl = ½ M H2SO4 = 1/3 M H3PO4
Fe3+ +e = Fe2+; eq. wt = 1 mol wt
MnO4- + 8H+ + 5e = Mn2+ + 4H2O; eq wt =1/5 mol wt
Cr2O72- + 14 H+ + 6e = 2Cr3+ +7H2O; eq. wt = 1/6 mol wt
1 m NaCl :
58.5 g NaCl + 1000 g H2O
1 M NaCl
Density = 1.02
Wt = 1020 g
NaCl = 58.5 g
H2O = 1020 – 58.5 = 916.5 g
35
Stoichiometry
aA( g )  bB( s )  cC (l )  dD ( aq)
Haber Process
N 2 ( g )  3H 2 ( g )      2 NH 3 ( g )
Fe2O3 ,1000o C ;500 atm
1 mol N2 + 3 mol H2 produces 2 mol NH3
1 metric tons N2 + 6 metric tons H2 produces 34
metric tons NH3
36
(Stoichiometry) The sulfide ore of zinc (ZnS) is reduced to
elemental zinc by “roasting” it (heating it in air) to give ZnO
and then heating ZnO with carbon monoxide. The two
reactions cab be written as
ZnS +3/2 O2  ZnO + SO2
ZnO + CO  Zn +CO2
Suppose 5.32 kg of ZnS is treated in this way and 3.3 kg of
pure zinc is obtained. Calculate the theoretical yield of Zn
and its actual yield.
(1) Theoretical Zn mass:
{5.32*1000(g) / 97.46(g/mol)} * 65.39(g/mol) = 3.57*103 (g)
(2) Actual percentage yield:
3300 / 3570 = 0.924 = 92.4%
37
Mass Balance on Reacting Systems
38
Example
• Ethylene is oxidized to produce ethylene oxide in
accordance with the following reaction:
2C2H4 + O2  2C2H4O.
The feed to the reactor contains 100 kmol/h of
ethylene and 200 kmol/h of molecular oxygen. The
reaction proceeds to a fraction conversion of limiting
reactants of 40%.
Determine the molar flow rate of all the product gas
constituents if the reaction proceeds under steadystate conditions.
39
Example
n10 = 100 -2x40 = 20 kmol/h
n20 = 200 -40 =160 kmol/h
n30 = 2x40 =80 kmol/h
Limiting reactant = C2H4
ER = 0.4 x 100 (kmol/h) = 40 kmol/h
40
Example
• The reaction:
2CO2(g) = 2CO(g) + O2(g) takes place at 2000 K.
All constituents are gases. If 100 kmol/h of CO2 enter a
system where this reaction take place essentially to
equilibrium at P = 1 atm and under steady state condition,
determine the equilibrium molar flow rate of all the
constituents in the effluent gas. Given the equilibrium constant
at 2000 K, K = 1.73x10-6.
41
(Stoichiometry) Sulfuric acid (H2SO4) forms in the following
chemical reaction: 2 SO2 + O2 + 2H2O  2 H2SO4. Suppose
400 g SO2, 175 g O2 and 125 g H2O are mixed and the
reaction proceeds until one of the reagents is used up. Which
is the limiting reactant? How many grams H2SO4 are formed
and how many grams of the other reactants remain?
2SO2 + O2 + 2H2O  2H2SO4
SO2= 400(g) / 64.06(g/mo) = 6.24(mol)
O2 = 175(g) / 32.00(g/mol) = 5.47(mol)
H2O = 125(g) / 18.02(g/mol) = 6.94(mol)
So, SO2 is the limiting reactant.
6.24 mol of H2SO4 was produced from the 6.24 mol of SO2
H2SO4 = 6.24(mol) * 98.07(g/mol) = 612 (g)
Mole of O2 remaining = 5.47- 6.24/2 = 2.35(mol)
Mass of O2 remaining = 2.35 * 32.00 = 75.2 (g)
Mole of H2O remaining = 6.94 - 6.24 = 0.70(mol)
Mass of H2O remaining = 0.7 * 18.02 = 12.7 (g)
42
Example
2CO2(g) = 2CO(g) + O2(g)
or
Y10 = (100-ER)/(100 + 0.5ER)
Y20 = ER)/(100 + 0.5ER)
Y30 = 0.5ER/(100 + 0.5ER)
43
ER
ln K
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
3 Ln ER
‐13.27
‐13.27
‐13.27
‐13.27
‐13.27
‐13.27
‐13.27
‐13.27
‐13.27
‐13.27
Ln 0.5 2 ln(100‐ER) Ln(100+0.5ER) SUM D
0.00
‐0.69
9.19
4.61 ‐14.49
2.08
‐0.69
9.17
4.62 ‐12.40
3.30
‐0.69
9.15
4.62 ‐11.17
4.16
‐0.69
9.13
4.62 ‐10.29
4.83
‐0.69
9.11
4.63 ‐9.60
5.38
‐0.69
9.09
4.63 ‐9.04
5.84
‐0.69
9.07
4.64 ‐8.56
6.24
‐0.69
9.04
4.64 ‐8.14
6.59
‐0.69
9.02
4.65 ‐7.77
6.91
‐0.69
9.00
4.65 ‐7.44
1.23
‐0.87
‐2.10
‐2.98
‐3.66
‐4.23
‐4.71
‐5.12
‐5.50
‐5.83
ER = 1.5 kmol/h
n10 = 100 - R = 98.5 kmol/h
n20 = 0.5ER = 0.75 kmol/h
n30 = 0.5ER = 0.75 kmol/h
44
Dual Properties of Matter
Particle with wave-like
character, energy and mass
are interchangeable
M = E/C2
E = hv
v =C
C = 299,792,458 m/s
= 2.998 x108 m/s
h = Plaks constant =6.625x10-27 erg-s
1 erg = 6.24x1011 eV
45
Electromagnetic Spectrum
46
Electromagnetic Spectrum
47
(Chemical
bonding) The bond dissociation energy of a
typical C-F bond in a chlrorofluorocarbon is approximately
440 kJ mol-1. Calculate the maximum wavelength of light
that cam photodissociate a molecule of CCl2F2, breaking
such a C-F bond.
E = hC/λ = 6.626*10-34(J s) *2.998*108(m/s) /λ
= 440*103(J/mol)/6.02*1023(1/mol)
 = 2.72*10-7 m
48
Atomic Orbitals
• Heisenberg Uncertainty Principle: it is impossible to
define what time and where an electron is and where is it
going next.
• Since it is impossible to know where an electron is at a
certain time, a series of calculations are used to
approximate the volume and time in which the electron
can be located. These regions are called Atomic Orbitals.
These are also known as the quantum states of the
electrons.
• Only two electrons can occupy one orbital and they must
have different spin states, ½ spin and – ½ spin (easily
visualized as opposite spin states).
• Orbitals are grouped into subshells.
• This field of study is called quantum mechanics.
49
Atomic Subshells
• These are some examples of atomic orbitals:
– s subshell: (Spherical shape) There is one orbital
in an s subshell. The electrons can be located
anywhere within the sphere centered at the atom’s
nucleus.
http://www.chm.davidson.edu/ronutt/che115/AO.htm

p Orbitals: (Shaped like two balloons tied together)
There are 3 orbitals in a p subshell that are denoted as
px, py, and pz orbitals. These are higher in energy than
the corresponding s orbitals.
50
http://www.chm.davidson.edu/ronutt/che115/AO.htm
Atomic Subshells
– d Orbitals: The d subshell is divided into 5 orbitals (dxy,
dxz, dyz, dz2 and dx2-y2). These orbitals have a very
complex shape and are higher in energy than the s
and p orbitals.
http://www.chm.davidson.edu/ronutt/che115/AO.htm
51
Electronic Configuration
• All chemistry is done at the electronic level (that is why
electrons are very important).
• Electronic configuration is the arrangement of electrons in
an atom. These electrons fill the atomic orbitals.
• Atomic orbitals are arrange by energy level (n), subshells
(l), orbital (ml) and spin (ms) - in order:
52
Lithium Electronic Configuration
• The arrows indicate the
value of the magnetic
spin (ms) quantum
number (up for +1/2 and
down for -1/2)
• The occupation of the
orbitals would be written
in the following way:
1s22s1
or, "1s two, 2s one".
http://wine1.sb.fsu.edu/chm1045/notes/Struct/EConfig/Struct08.htm
53
Electronic Configuration
•
short hand which references the last completed orbital
shell (i.e. all orbitals with the same principle quantum
number 'n' have been filled)
– Na: [Ne]3s1
– Li: [He]2s1
•
The electrons in the stable (Noble gas) configuration
are termed the core electrons
The electrons in the outer shell (beyond the stable
core) are called the valence electrons
•
54
Electronic Configurations –
Box Diagram
• The two electrons in Helium represent the complete
filling of the first electronic shell. Thus, the electrons in
He are in a very stable configuration
• For Boron (5 electrons) the 5th electron must be placed
in a 2p orbital because the 2s orbital is filled. Because
the 2p orbitals are equal energy, it doesn't matter which 55
2p orbital is filled.
http://wine1.sb.fsu.edu/chm1045/notes/Struct/EConfig/Struct08.htm
The Order of Filling Orbitals
http://en.wikipedia.org/wiki/Image:Electron_orbitals.svg
56
Valence Electrons
• The valence electrons are the electrons in the last shell
or energy level of an atom.
The lowest level (K), can
contain 2 electrons.
The next level (L) can contain
8 electrons.
The next level (M) can
contain 8 electrons.
www.uoregon.edu
Carbon - 1s22s22p2
four valence electrons
57
Examples of Electronic
Configuration
•
•
•
•
•
Ne  1s2 2s2 2p6
F  1s2 2s2 2p5
F-  1s2 2s2 2p6
Mg  1s2 2s2 2p6 3s2
Mg2+  1s2 2s2 2p6
(10 electrons)
(9 electrons)
(10 electrons)
(12 electrons)
(10 electrons)
• Notice – different elements can have the same number
of electrons
58
Atomic orbitals
• http://www.orbitals.com/orb/
•
ATOMIC STRUCTURE AND BONDING
MENU:http://www.chemguide.co.uk/atommenu.html#top
59
Molecular Orbitals
The goal of molecular orbitals is to describe molecules in
a similar way to how we describe atoms, that is, in terms
of orbitals, orbital diagrams, and electron configurations.
• The molecular orbital volume encompasses the whole
molecule.
• The electrons fill the molecular orbitals of molecules like
electrons fill atomic orbitals in atoms.
Paramagnetic: compounds with unfilled atomic orbitals
Diamagnetic: compounds with fully occupied atomic orbitals
60
Molecular Orbitals
• Electrons go into the lowest energy orbital available to
form lowest potential energy for the molecule.
• The maximum number of electrons in each molecular
orbital is two. (Pauli exclusion principle)
• One electron goes into orbitals of equal energy, with
parallel spin, before they begin to pair up. (Hund's Rule.)
61
Atomic and Molecular Orbitals
•
Orbital Mixing
– When atoms share electrons to form a bond, their
atomic orbitals mix to form molecular bonds. In order
for these orbitals to mix they must:
• Have similar energy levels.
• Overlap well.
• Be close together.
This is and example of orbital
mixing. The two atoms share
one electron each from there
outer shell. In this case both 1s
orbitals overlap and share their
valence electrons.
62
http://library.thinkquest.org/27819/ch2_2.shtml
Energy Diagram of Bond Formation by
Orbital Overlap
63
Examples of  Bond Formation
64
• The Aufbau principle, lowest energy MOs fill
first, or fill the orbitals from below, if there
are large orbital energy differences.
• The Pauli exclusion principle, a maximum of
two electrons per orbitals and these must be
of opposite spin
• Hund's rule, when there are equal energy or
"degenerate" orbitals, these fill one electron
at a time before pairing begins. Or, if two
orbitals are not very different in energy,
near-degenerate, or even exactly degenerate,
then two electrons will fill both orbitals with
parallel spin.
65
•
•
•
•
•
•
•
First, we have the 1s + 1s -> 
bonding MO, and then above this
the σ* antibonding MO.
Then, there are the pair 2s + 2s
bonding σ (sigma) and
antibonding σ* MOs.
Next we come to the p + p
interactions. These are a little
more complicated because p
orbitals are orientated in space
along the x, y and z dimensions.
The 2px + 2px orbital interaction
gives a σ bonding MO.
The 2py + 2py and 2pz +
2pz interactions are equivalent,
they only differ in spatial
orientation. Both give rise
to π bonding MOs.
Next come a pair of π*
antibonding MOs.
Finally, we have the 2px + 2px σ*
(sigma star) antibonding MO.
66
67
Atomic and Molecular Orbitals
• In atoms, electrons occupy atomic orbitals, but in
molecules they occupy similar molecular orbitals which
surround the molecule.
• The two 1s atomic orbitals combine to form two
molecular orbitals, one bonding () and one antibonding
(*).
• This is an illustration of
molecular orbital
diagram of H2.
• Notice that one electron
from each atom is being
“shared” to form a
covalent bond. This is an
example of orbital
mixing.
68
http://www.ch.ic.ac.uk/vchemlib/course/mo_theory/main.html
Molecular Orbital Diagram (H2)
http://www.ch.ic.ac.uk/vchemlib/course/mo_theory/main.html
69
MO Diagram for O2
http://www.chem.uncc.edu/faculty/murphy/1251/slides/C19b/sld027.htm
70
Molecular Orbital Diagram (H2O)
71
Molecular Orbital Diagram (HF)
http://www.ch.ic.ac.uk/vchemlib/course/mo_theory/main.html
72
Molecular Orbital Diagram (CH4)
So far, we have only look at molecules with two atoms.
MO diagrams can also be used for larger molecules.
http://www.ch.ic.ac.uk/vchemlib/course/mo_theory/main.html
73
74
75
76
Bond Order
The number of bonds between a pair of atoms is called the bond order
Oxygen, for example, has a bond order of two.
We can calculate the bond order in the O2 molecule by noting that there are
eight valence electrons in bonding molecular orbitals and four valence
electrons in antibonding molecular orbitals in the electron configuration of
this molecule. Thus, the bond order is two.
When there is more than one Lewis structure for a molecule, the bond order
is an average of these structures. The bond order in sulfur dioxide, for
example, is 1.5 the average of an S-O single bond in one Lewis structure
and an S=O double bond in the other.
77
Conclusions:
• Bonding electrons are localized between atoms (or are
lone pairs).
• Atomic orbitals overlap to form bonds.
• Two electrons of opposite spin can occupy the
overlapping orbitals.
• Bonding increases the probability of finding electrons in
between atoms.
• It is also possible for atoms to form ionic and metallic
bonds.
78
References
• http://www.chemguide.co.uk/atoms/properties/atomorbs.
html
•
http://www.ch.ic.ac.uk/vchemlib/course/mo_theory/main.html
• http://en.wikipedia.org/wiki/Molecular_orbital_theory
• http://library.thinkquest.org/27819/ch2_2.shtml
79
Lewis Structure
Lewis symbols of the main group elements
H•
Li •
Na
K
Rb
Cs
He:
• Be
•
Mg
Ca
Sr
Ba
•
•B•
Al
Ga
In
Tl
•
•
•
•
Si
Ge
Sn
Pb
P
As
Sb
Bi
•C•
:N•
•
..
S
Se
Te
Po
Cl
Br
I
At
:O• :F•
''
''
..
: Ne :
''
Ar
Kr
At
Rn
80
Lewis Structure
CF4, H2O and CO2 dot structures
..
:F:
.. .. ..
:F:C:F:
. .
. .
:F:
''
. .
..
..
..
O
/ \
H H
O::C::O
..
..
..
or
..
..
: O=C=O :
81
Formal Charge
The formal charge on any atom in a Lewis structure is a number
assigned to it according to the number of valence electrons of the atom
and the number of electrons around it.
The charge it will have if the bonding were perfectly covalent.
The formal charge of an atom is equal to the number of valence
electrons, Nv.e. subtract the number of unshared electrons, Nus.e. and
subtract half of the bonding electrons (shared), ½ Nb.e..
Formal charge = Nv.e. - Nus.e. - ½ Nb.e.
Formal charge rules
Formulas with the lowest magnitude of formal charges are more stable.
More electonegative atoms should have negative formal charges.
Adjacent atoms should have opposite formal charges.
82
Draw Lewis dot structure for SO2
Put down number of valence electrons:
..
:O:
..
:S:
..
:O:
Put all atoms together to make a molecule and check to see if it satisfy
the octet rule.
..
..
..
:O::S::O:
0
0
0
<= octet rule not satisfied
formal charge
Adjust bonding electrons so that octet rules apply to all the atoms
.. .. ..
:O:S::O:
''
<- octet rule satisfied
-1 +1
formal charge
0
O1 =6-6-(2)/2 = -1
S =6-2-(6)/2 = 1
O2 = 6 -4 –(4)/2 = 0
83
Resonance
..
S
/
\
:O: :O:
''
''
1
«
..
S
//
\
:O: :O:
''
2
..
S
«
/
\\
:O:
:O:
''
3
..
S
«
//
\\
:O:
:O:
4
In structure 1, the formal charges are +2 for S, and -1 for both O
atoms.
In structures 2 and 3, the formal charges are +1 for S, and -1 for
the oxygen atom with a single bond to S.
The low formal charges of S make structures 2 and 3 more stable
or more important contributors.
The formal charges for all atoms are zero for structure 4, given
earlier.
This is also a possible resonance structure, although the octet rule ..
is not satisfied.
S
Combining resonance structures 2 and 3 results in the following /.' '' '.\
structure:
:O: :O
84
Exceptions to the octet rule
NO, NO2, BF3 (AlCl3), and BeCl2 do not satisfy the octet
rule
.N:::O:
compare
with
:C:::O:
.
N
//
\
:O:
:O:
''
..
:F:
|
B
/
\
:F:
:F:
''
''
..
..
:Cl : Be : Cl:
''
''
85
(Chemical bonding) Draw the Lewis electron-dot structure
for the following species, indicating formal charges and
resonance structures where applicable
CO32SCNH2CNN
NO2-
86
(Molecular orbital) Elemental bromine is a brownish-red liquid
that was first isolated in 1826. The current method of
production is to oxidize bromide ion in natural brines with
element chlorine.
a. What is the ground-state electron configuration of the
valence electrons of bromine molecules (Br2)?
b. Is bromine paramagnetic or diamagnetic?
c. What is the electro configuration of Br2+ molecular ion?
d. Will Br2+be stronger or weaker than that of Br2? What is its
bond order?
(a) (σ4s)2(σ*4s)2(σ4pz)2 (π4p)4 (π*4p)4
(b) Diamagnetic
(c) (σ4s)2(σ*4s)2(σ4pz)2 (π4p)4 (π*4p)3
(d) Stronger, the bond order of Br2+ is (8-5)/2=3/2 and Br2
is (8-6)/2=1
87
Chemical Bonding
•
•
•
•
•
•
•
•
•
Ionic bonding
Covalent bonding
Co-ordinate (dative covalent) bonding
Electro negativity . . .
Shapes of simple molecules and ions
Metallic bonding
van der Waals forces
Hydrogen bonding
Bonding in organic compounds
http://www.chemguide.co.uk/atoms/bondingmenu.html#top
http://www.chemguide.co.uk/index.html#top
88
Forming a Covalent Bond
• Molecules can form bonds by sharing electron
– Two shared electrons form a single bond
• Atoms can share one, two or three pairs of electrons
– forming single, double and triple bonds
• Other types of bonds are formed by charged atoms
(ionic) and metal atoms (metallic).
89
Ionic (Electrovalent) Bonding
90
Coordination Number
Rr
<0.155
0.155-0.225
0.225 – 0.414
0.414 – 0.732
CN
1 or 2
3
4
4 or 6
0.732 – 1.00
>1
8
12
Linear or bent molecules
Triagonal planar of A
Tetrahedral of A
Square planar of A(4)
Octahedral of A (6)
Body centered cube
Close-packed structure of
metals
Rr = rc/ra
91
Covalent BondingSingle Bonds
92
Co--Ordinate
Dative Covalent
• A covalent bond is formed by
two atoms sharing a pair of
electrons.
• In the formation of a simple
covalent bond, each atom
supplies one electron to the
bond - but that doesn't have to
be the case.
• A co-ordinate bond (also called
a dative covalent bond) is a
covalent bond (a shared pair of
electrons) in which both
electrons come from the same
atom.
93
Metallic Bonding
94
Intermolecular Bonding- Vander
Walls Forces
• Intermolecular attractions are attractions between
one molecule and a neighbouring molecule.
95
Intermolecular Bonding- Hydrogen
Bonds
96
Water Structure
97
Water Properties
Maximum density at 4 oC - 1,000 kg/m3, 1.940 slugs/ft3
Specific Weight at 4 oC - 9.807 kN/m3, 62.43 Lbs./Cu.Ft, 8.33
Lbs./Gal., 0.1337 Cu.Ft./Gal.
Freezing temperature - 0 oC (Official Ice at 0 oC)
Boiling temperature - 100 oC
Latent heat of melting - 334 kJ/kg
Latent heat of evaporation - 2,270 kJ/kg
Critical temperature - 380 oC - 386 oC
Critical pressure - 221.2 bar, 22.1 MPa (MN/m2)
Specific heat water - 4.187 kJ/kgK
Specific heat ice - 2.108 kJ/kgK
Specific heat water vapor - 1.996 kJ/kgK
Thermal expansion from 4 oC to 100 oC - 4.2x10-2
Bulk modulus elasticity - 2.15 x 109 (Pa, N/m2)
98
Phase Transition
 Heat released
Heat absorbed 
Phase transition in water
facilitates heat transfer in the
environment.
99
Temperature factor
100
Water Properties
101
Surface Tension
Surface tensin (dyne/cm)
 = 76.229-0.633T – 0.0119T2
100
Kelvin equation
80
60
40
20
0
0
10
20
30
40
 p 
2σ v
ln   
rRT
 po 
T (oC)
102
Viscosity
 = 1.805 - 0.0494 T – 0.0005 T2
Darcy equation
2
q = (-k/)P
Viscosity (CP)
1.5
1
0.5
0
0
5
10
15
20
25
30
T (0C)
103
Vapor Pressure
es = 4.7017 + 0.225T + 0.0213 T2
Vapor Pressure (mm Hg)
60
50
40
30
vapor pressure deficit
20
Humidity
10
0
0
10
20
30
40
T (oC)
104
Density
= 1.000 +2x10-5 T – 5x10-6 T2
1
0.998
1.00005
0.997
1
0.996
density
Density (g/cm
0.999
0.995
0.994
0.99995
0.9999
0.99985
0.993
0.9998
0.992
0
10
20
30
40
50
Temperature (oC)
0.99975
0.9997
0
2
4
6
8
10
temperature (oC)
Stratification
Epilimnion
Thermocline
Hypolimnion
105
Water Composition
Abundances (% or half-life) of hydrogen and oxygen
isotopes
2D
3T
H
99.985% 0.015% 12.33 y
14O
15O
16O
17O
18O
70.6 s
122 s
99.762% 0.038% 0.200%
Relative abundance of isotopic water
H216O
99.78
%
18
H218O H217O HD16O
0.014
0.20% 0.03%
9%
20
19
19
D216O HT16O
0.022
trace
ppm
20
20 amu
106
Water Structure
107
Water Structure
108
“threaded” water
(Inorg. Chem. 2003, 44(4) pp 816 - 818)
a suitable molecular backbone (above) can cause water molecules to
form a "thread" that can snake its way though the more open space of
the larger molecules.
water can have highly organized local structures when it interacts with
molecules capable of imposing these structures on the water.
109
Physical Properties of water
• http://www.thermexcel.com/english/tables/
eau_atm.htm
110