1
CHAPTER 4 – CHEMICAL BONDING
 Dalton’s atom story starts even before Mendeleev’s periodic
table
 Frankland (1852) “each element has a fixed valence that
determines its bonding capacity”
 Kekulé (1858) – illustrating a bond by using a dash between
bonding atoms
 van’t Hoff & Le Bel (1874) extended these structures to 3-D
 all this without knowing about the nucleus and bonding
electrons, etc.
 Abegg (1904) suggested “that the stability of the inert or noble
gases was due to the number of outermost electrons in the
atom”
 Theorized that a Cl atom had 1 less electron than Ar so if Cl
would gain 1 electron it would become more stable, also if Na
having 1 electron in its outer shell would loss it to become
stable
i.e. Na + Cl -- Na+ + Cl Lewis (1916) produced the 1st clear understanding of chemical
bonding:
1. atoms are stable if they have a noble gas-like
electron structure; i.e. stable octet of electrons
2. electrons are most stable when paired
3. atoms form chemical bonds to achieve a stable
octet of electrons
4. a stable octet may be achieved by an exchange of
electrons between metal & non-metal atoms
5. a stable octet of electrons may be achieved by the
sharing of electrons between non-metal atoms
6. the sharing of electrons results in the formation of
covalent bonds
 to show these bonds Lewis devised Lewis Structures
2
 example of Lewis Structures:
Cl2
··
··
·· ··
·· ··
··Cl· +
·Cl·· --- ··Cl··Cl·· ------- ··Cl-Cl··
··
··
·· ··
·· ··
NaCl
··
··
+
Na· + ·Cl·· ------[Na] + [··Cl··]··
··
Homework:
Nelson 12 Review Page 225 Sample Problem
Lewis Structures and Quantum Mechanics
 putting Lewis structures and quantum mechanics together a
connection between the 2 theories becomes apparent
 in a Lewis structure the 4 sides of the rectangle of electrons
around the core represent the s & p energy sublevels
 Summerfeld added these energy sublevels in 1915 & Lewis
proposed his theory in 1916
 Combining the 2 theories we can see for Mg
6s
5p 5p
5p
4d
4d 4d
4d
4d
3d
3d 3d
3d
3d
5s
4s

3s

2s

1s
4p 4p
4p
3p 3p
 
2p 2p
3p

2p
which corresponds to Mg, valence of 2, ·Mg· , 1s2 2s2 2p6 3s2
3
 For N, valence 3, 1s2 2s2 2p3
6s
5s
5p 5p
4p 4p
5p
4d
4d 4d
4d
4d
3d
3d 3d
3d
3d
4p
4s
3s

2s

1s
3p 3p
 
2p 2p
3p

2p
··
· N·
·
Homework:
Nelson 12 Page 227, #’s 1-5
Extending the Lewis Theory of Bonding
 Lewis’ theory explained many simple molecules but some
molecules including polyatomic molecules could not be
explained
 Work by Sidgwick showed Lewis structures can work if we don’t
require that each atom contribute 1 electron to a shared pair in
a covalent bond i.e. one atom could contribute both electron
pairs that are shared (co-ordinate covalent bond)
 Sidgwick also realized that an octet of electrons around an
atom may be desirable but is not necessary in all molecules &
polyatomic ions
4
Method 1: eg. NO31-1) arrange atoms in most symmetrical arrangment
2) find total number of valence electrons including the charge
of the ion e.g. 5+[(3) x 6] + 1 = 24 e3) put pairs between the central atom and each of the
surrounding atoms that uses up 6
4) distribute the rest and you have this;
··
··O··
··
N
··
··
··O·· ··O··
··
··
a. are atoms surrounded by 8 electrons?
5) Because N only has 6 we can move one pair of electrons
from O to help N have a stable octet
··
··O··
··
N
··
·· ··
··O·· ··O··
··
7) draw the Lewis structure with the single and double bonds
and enclosing the polyatomic ion with a square bracket and
including the charge
-
O
N
O
O
Homework:
Nelson 12, Page 230 #’s 1-4
5
Method 2: eg. NO31—
1) identify the central atom as the least electronegative atom exceptions are H and generally O
 central atom is N
2) place the surrounding atoms around the central atom (on
sides)
3) draw single bonds between the surrounding atoms to the
central atom
O
O
N
O
4) determine the number of bonds in the molecule
a) calculate the # valence electrons for all the atoms,
including electrons added or removed due to polyatomic
charge
# valence e-= 23
e added
= 1
# valence e + added e = 24
b) calculate the # electrons if each atom had a filled valence
shell by itself
# e- if valence shell filled for each atom = 32
c) calculate the difference between 4a and 4b
difference = 8
d) each pair of electrons in 4c is a bond
# bonds = 4
5) add any double or triple bonds between the central and
surrounding atoms to achieve the # bonds determined in 4d
- atoms that often form multiple bonds are C, N, O, S
O
O
N
O
6) draw the electron dots to represent the remaining unshared
pair of electrons around the atoms, with charge if needed
1O
O
N
O
6
7) identify whether coordinate bonds are present by
determining which atom “owns” the electrons forming the
bonds
1O
O
N
O
Exceptions to the Octet Rule
1. Expansion of the Valence Level
 elements of the third (3rd) period or greater can form molecules
where the central atom is surrounded by greater than the octet
of valence electrons – due to the presence of the “d” subshell
 eg. Phosphorus has 5 e—that can form as many as 5 bonds
 PF3 is common
F
 PF5 is less common, but does exist
F
—
—
2 e from 3s and 3 e from 3p
F
P
are promoted to 3d subshell,
F
with each 3d orbital having
F
an lone electron for pairing
—
Sulfur has 6 e that can form as many as 6 bonds
 SF2 vs SF6
other examples may include: SF4, XeF2, SeF4, TeF6, XeF5+1
2. Molecules with an Odd Number of Valence Electrons
 the single unpaired electrons are attracted to the atom by the
paramagnetic field
 the strength of attraction is proportional to the # of
unpaired e—
 eg. NO
valence e-= 11
—
total e when valence level filled = 16
5 e—
# of bonds = 2
# e—left over = 1 (unpaired)
7
N=O
N=O
Lewis Diagrams are consistent formulas with NO
NO2
valence e-=
—
total e when valence level filled =
# of bonds =
# e—left over =
3. Molecules with Insufficient Valence Electrons
e—
(unpaired)
 Be and B do not form ionic bonds, but instead form covalent
bonds
 ionization energies are too high for Be and B
 belong to group IIA and IIIB
 eg. BF3
- check electronegativities of B and F
-valence e
= 24
—
total e when valence level filled = 32
8 e—
# of bonds expected = 4
suggests a Lewis structure
F–B=F
F
however, experimental evidence shows the structure has
NO double bond
B is shown to have only 6 valence eF–B–F
i.e. electron deficient
exhibits partial double bond character
F
other examples AlCl3, BeCl2, BCl2, TlCl2+, BeBr2
8
The Nature of the Chemical Bond
 Shroedinger ‘s mathematical equation to describe the standing
wave for an electron works fine for an atom but when a molecule
is considered, with 2 nuclei, extra electrons, etc. quantum
mechanics are still being studied
 Linus Pauling developed the valence bond theory when a covalent
bond is formed 2 orbitals overlap and produce a new combined
orbit with 2 electrons of opposite spin is produced.
 Any 2 half-filled orbitals can overlap in the same way, e.g. HF
i.e. the H’s 1s orbital is thought to overlap the half-filled 2p orbital
 This also works with larger molecules, consider the
O atom

2
1s

2
2s

_
2p
4
_
9
 If this reacted with H, it would seem reasonable that the 1s orbital
would overlap the 2 half-filled 2p orbitals of O with an angle
between the 2 bonds to be 90°.
 In fact, what is seen is that an angle of 105° is observed which
means that there is a problem with the theory or other factors are
involved
Homework:
Nelson 12 page #’s 1-4
Hybrid Orbitals
 Lewis bonding theory could not explain the 4 equal bonds in
compounds like CH4(g), nor its tetrahedral shape, and the existence
of double & triple bonds
 Pauling and others proposed that C bonds form in the shape of a
tetrahedron
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 Consider the energy level diagram for carbon in its ground state:
3s

2s

1s
 
2p 2p
2p
 Initially they proposed that an electron from the 2s orbital was
promoted to the third 2p orbital to create 4 orbitals with single
electrons
3s

2s

1s
 
2p 2p

2p
 Each electron in the orbitals could then pair up with an electron
from another atom, such as hydrogen - it was believed that more
energy would be gained when the bond was formed than for the
electron to be promoted
 Unfortunately this bonding arrangement would create 3 equal
bonds with the 2p orbitals, but one different bond with the 2s
orbital
 Evidence showed that all the bonds were equal in shape & energy
so out went that thought
 Now CH4 is explained by hybridization to 4 identical hybrid sp3
orbitals
carbon – ground state
carbon - hybridized
3s

2s

1s
 
2p 2p
2p
3s


 
3
3
sp sp sp3 sp3

1s
11
 The explanation being that the 2s electrons and the 2p electrons
form 4 identical hybrid sp3 orbitals with one bonding electron in
each, which explains the bonding capacity of 4 in C
 These orbitals are hybridized only when bonding occurs not in an
isolated atom
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 Hybridization is used to explain the bonding of atoms in covalent
molecules ONCE the shape has been determined
 Conditions for bonding:
 occurs along the axis of orbitals involved
 covalent bond formed by the overlap of orbitals sharing
two (2) electrons
 bond is a result of nuclei attraction on electrons from
both atoms
 # of hybrid orbitals formed always equals # of atomic
orbitals used
 Pauling suggested that a whole series of hybridizations could occur
to explain BF3(g) and BeH2(g) , etc (see Table 1 in text page 234)
13
Additional hybrid orbitals include:
Hybrid Orbital
sp
sp2
sp3
sp3d
sp3d2
Geometry
#Orbitals
linear
trigonal planar
tetrahedral
trigonal bipyramidal
octahedral
2
3
4
5
6
Example
Be in BeF2
B in BF3
C in CH4
P in PCl5
S in SF6
 Shape of molecule and therefore hybrid orbital used is determined
by the single bonds to surrounding atoms and any lone pairs of
electrons (see VSEPR Theory)
 questions in homework below do not involve lone pairs of
electrons; as such the number of single bonds will determine
the number of hybrid orbitals
Homework: Nelson 12 page 235 #’s 8,10,11,12 & 14.
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Double & Triple Covalent Bonds
 Difficult to explain double & triple bonds i.e. explaining the
differences between C2H6(g), C2H4(g) and C2H2(g)
 Lewis structures suggests that there must be sharing of 1, 2 or 3
electron pairs in order to obtain a stable octet
 How could electrons in the sp3 orbitals overlap not once but two
or three times with just one atom
 Valence bond theory tells us that two kinds of orbital overlap are
possible:
1. the end to end overlap of s orbitals, p orbitals, hybrid orbitals
or a combination of these orbitals
This type of the overlap produces a sigma “” bond (figure 9)
(a) represents s orbitals overlapping end to end
(b) represents p orbitals overlapping end to end
(c) represents hybrid orbitals overlapping end to end
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2. two orbitals can overlap side to side to form a pi “π” bond
(see Fig. 10)
Double Bonds
Fig. 11 (above)
 C is the most common central atom in molecules with double or
triple bonds
 Orbitals of C atoms can be hybridized to form 4 sp3 orbitals (4
single bonds)
 “New” theory is that we have partial hybridization of available
orbitals to leave 1 or 2 p orbitals with single unpaired electrons
carbon – ground state
3s

2s

1s
 
2p 2p
2p
carbon – hybridized for
double bond (3, 1π)
3s




2p
2
2
2
sp sp sp

1s
16
 i.e. promote an electron from a 2s orbital to a 2p orbital and we
will have 3 sp2 orbitals and 1 p orbital with a single electron (still
have 4 orbitals)
 the 3 sp2 orbitals gives the trigonal planar shape from the
central C atom bonded to two H atoms and one C atom
 the single p orbital aligns itself perpendicular to the trigonal
planar shape – the “dumb bell” shape is above and below the
plane
 see Fig 12
 in C2H4, 3 hybrid sp2 orbitals are used to form  bonds between
the C and two H’s and the C and the other C (Fig. a)
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 half-filled p orbitals overlap sideways (Fig. b) to form pi bonds, a
region of electron density above & below the C-C sigma bond
 a pi bond is a combined orbital which has a pair of electrons
having opposite spins
 the extra shared pair of electrons in the π bond makes a greater
attraction between the 2 C nuclei which is why the double
covalent is shorter and stronger than a single bond.
Homework: Nelson 12 Page 238 #’s 18-21
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Triple Bonds
 ethyne (acetylene) is a linear molecule of C2H2 so that a
tetrahedral sp3 hybridization is not likely
 the energy level diagram that permit the formation of the triple
bond of ethyne is:
carbon – ground state
3s

2s

1s
 
2p 2p
2p
carbon – hybridized for
triple bond (2, 2π)
3s

sp

1s

sp

2p

2p
 the linear shape is achieved by the overlap of one sp orbital from
the C atom with the s orbital of the H atom and the overlap of the
other sp orbital from the C atom with the sp orbital of the second
C atom (see Fig. 16a) - 2 identical sp hybrid orbitals orient at 180°
resulting in a linear molecule
 according to the valence bond theory, the unpaired electrons in
the 2 p orbitals of adjacent C atoms form 2 π bonds (see Fig. 16b)
 In the C-C triple bond, the C’s are bonded with 1 sigma bond and
2 π bonds (see Fig. 16c)
19
Homework: Nelson 12 Page 239 #’s 25-27
As indicated before the section on Double and Triple Bonds:
 the shape of the molecule and therefore hybrid orbital used is
determined by the single bonds to surrounding atoms and any
lone pairs of electrons (see VSEPR Theory)
 in the case of molecules with double and triple bonds the shape
surrounding each “central” atom (involved with a double or triple
bond) is determined by the single bonds and lone electron pairs the double or triple bond is treated as a single bond in
determining the shape
 identifying the shape allows one to identify the hybrid orbital
involved in the single  bonds and the  bond of the double or
triple bond
 the double or triple π bond will result from the overlap of one or
two p orbitals of the atoms involved in the double or triple bond
 as you will see in the next section on VSEPR Theory the starting
point for the molecular shape and therefore the hybrid
arrangement is the Lewis Diagram (or modified structure)
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VSEPR Theory
“Valence Shell Electron Pair Repulsion Theory”
 theory based on the electrical repulsion of bonded and unbonded
electron pairs in a molecule or polyatomic ion
 like charges repel, unlike charges attract
 the number of electron pairs are determined by adding the
number of bonded atoms plus the number of lone pairs of
electrons
 the 3-D shape can be determined once this number is known
by arranging all the pairs of electrons as far apart as possible
 Basic Shapes: (see below)
1. Linear
 2 electron pairs on central atom
 180° angle between pairs
2. Trigonal Planar
 3 electron pairs on central atom
 120° angles between pairs
3. Tetrahedral
 4 electron pairs on central atom
 109.5° angles between pairs
4. Trigonal Bipyramidal
 5 electron pairs on central atom
5. Octahedral
 6 electron pairs on central atom
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Trigonal Bipyramidal
Octahedral
 All of the shapes of molecules are derived from one of the above
five basic shapes
 the molecular shape that results is dictated by the electron
pairs that form bonds to other atoms versus lone electron pairs
22
23
24
Steps to predict the shape of a molecule are:
1. Draw the Lewis diagram for the molecule, including the
electron pairs around the central atom
2. Count the total number of bonding pairs (bonded atoms) and
lone pairs of electrons around the central atom. Count a
multiple bond as one pair of electrons.
3. Arrange the pairs of electrons according to the appropriate
diagram – use basic shapes
4. Obtain the molecular geometry from the directions of bonding
pairs
 Refer to Table 1 (p 245) / Table 2 (p 247) or Hand-out (from
class) and draw the shape of the molecule using a structural
formula as per text drawings
 Students are responsible for knowing the shapes (and their
modification based on lone pairs of electrons) from Table 1 (p
245) and the basic shapes trigonal pyramidal and octahedral –
Table 2 (p 247)
Hybridization and VSEPR Theory
 As mentioned during the discussion on Hybridization the shape of
the molecule and therefore hybrid orbital used is determined by
the single bonds to surrounding atoms and any lone pairs of
electrons
 The steps to follow when determining the hybrid orbital used are:
1. Draw the Lewis electron-dot diagram for the molecule,
including the electron pairs around the central atom
2. Count the total number of bonding pairs (bonded atoms –
remember that multiple bonds are treated as one bonding pair)
and lone pairs of electrons around the central atom
3. Use the VSEPR model to obtain the arrangement of electron
pairs about the central atom
4. From the geometric arrangement of the electron pairs, deduce
the type of hybrid orbitals on the central atom required for the
bonding description (see Table C3 above)
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5. Assign electrons to the hybrid orbitals of the central atom one
at a time, pairing them only when necessary
6. Form  bonds to this central atom by overlapping single
occupied orbitals of other atoms with the singly occupied hybrid
orbitals of the central atom
7. If a double or triple bond exists, then pair up the p orbitals
from the two atoms that are involved to form a π bond
26
Polar Molecules
Electronegativity & Polarity of Bonds
 Pauling realized that electron pairs could be shared evenly or
unevenly
 Created “electronegativity” to explain & predict the polarity of
molecules based on
 F has highest electronegativity (EN) at 4.0
 When you compare the electronegativities between 2 atoms, the
greater the ∆EN (difference in electronegativity) the greater the
polarity of the chemical bond
 The smaller the ∆EN the more non-polar the bond
 A very polar bond is an ionic bond while a non-polar bond is a
covalent bond
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 Pauling’s polar bond resulted when 2 different kinds of atoms
(non-metals) formed a bond
 If the bond is polar it meant that the electrons were “spending
more time closer to one of the atoms than to the other
 The end where the electrons stayed, was slightly more negative,
or partially negative and represented as  - , and the end where
the electron stayed away from was  +
 General rule is that when the ∆EN is greater than 1.7, the percent
ionic character exceeds 50%
Examples: H – H
2.1- 2.1=0, non-polar bond
P – Cl
2.1-3.0=0.9, indicates a polar covalent bond
Na - Br
0.9 – 2.8=1.9, indicates an ionic bond
Homework: Nelson 12 Page 253, #’s 1-3
28
Polar Molecules
 Polar bonds in a molecule do not mean that you have a polar
molecule
 CO2 is considered to be a non-polar molecule even though the
C=O bonds are considered to be polar
 According to Lewis structures & VSEPR theory, CO2 is a linear
molecule
 The bond dipole – proportional to the ∆EN – is shown by an arrow
pointing from the lower EN ( + ) to the higher EN ( - )
 The bond dipole is a vector quantity where direction is important –
two dipoles of equal magnitude, but opposite direction will cancel
each other out
 shown in the CO2 molecule, i.e. the CO2 molecule exhibits no
polarity and the molecule is said to non-polar
 If we look at H2O, water is polar and the Lewis structure & VSEPR
theory predict that the molecule would be V-shaped (based on the
tetrahedral basic shape)
 The dipoles do not cancel and instead the vertical components
add together to produce a non-zero molecular dipole (red
resultant arrow in middle)
 The molecule has a definite overall polarity and is said to be a
polar molecule
 i.e. it is partially negative at the O end and partially positive at
the H end
29
 In this case the shape helps produce the 2 oppositely charged
ends on the molecule
 From these 2 examples one can see why shape is as important as
the polarity of the molecule
 “Both the shape of the molecule and the polarity of the bond are
necessary to determine if a molecule is polar or non-polar”
 check out CH4
 The outer part of the molecule is positive on all sides and none
of the ends are charged differently
 This is because the CH4 molecule is symmetrical
 “In all symmetrical molecules, the sum of the bond dipoles is 0
and the molecule is non-polar”
 other examples of non-polar molecules are CCl4 and BF3 ,etc.
30
Homework: Nelson 12 Pages 255-256, #’s 6-8 & 9
31
Intermolecular Forces
 Intermolecular forces are the forces of interaction that may exist
between molecules.
 Intermolecular forces are much weaker than covalent bonds: if
covalent bonds are assigned a strength of about 100, then
intermolecular forces are generally 0.001 to 15
 types of intermolecular forces can be classified as: dipole-dipole
forces, London forces, hydrogen bonding
Dipole-Dipole Forces
 molecules may be classified as polar or non-polar: a polar
molecule occurs when the net result of the bond dipoles
(proportional to the difference in electronegativities of the two
atoms involved in the bond) in a molecule are NOT zero
 the dipole-dipole force is an attractive intermolecular force
resulting from the tendency of polar molecules to aligh themselves
such that the positive end of one molecule is near the negative
end of another
 the strength of the dipole-dipole force is dependent on the
polarity of the molecule
32
London (Dispersion) Forces
 Fritz London (1930) accounted for the weak attraction between
any two molecules (specifically non-polar molecules) by
recognizing that electrons orbiting the nucleus of an atom may be
on any one side at any point in time
 as a result there may be a small, instantaneous dipole, with
one side having a partial negative charge and the other side
having a partial positive charge
 if another atom is present nearby, the partial negative charge
of the first atom will repel the electrons of the second atom
creating a partial positive charge of the second atom; the
partial negative charge and the partial positive charge of each
atom will result in an attractive force between the two atoms
 this weak attraction occurs instantaneously because the
electrons are in constant motion; nonetheless the motion of the
electrons in one atom will influence the motion of the electrons
of the other atom
33
 London forces (or dispersion forces) are the weak attractive forces
between molecules resulting from the small, instantaneous dipoles
that occur because of the varying positions of the electrons during
their motion about the nuclei
 London forces increase as the molecular weight increases due to:
 the increased number of electrons in motion
 the increased size of the molecule, which permits electrons to
move further from the nucleus – more polarizable
Van der Waals Forces
 Van der Waals forces is a general term for those intermolecular
forces that include dipole-dipole and London forces
 these forces will affect the ease or difficulty with which a molecule
leaves a liquid
 can observe the effect of intermolecular forces by looking at
molecular weight - increase in MW generally results in increase in
London forces
 can also observe effect of intermolecular forces by looking at
boiling points of liquids
 surface tension is also dependent on intermolecular forces
34
 surface tension is the energy needed to increase the surface
area of the liquid
 to increase the surface area, it is necessary to pull molecules
apart against the intermolecular forces of attraction
 viscosity of a liquid depends in part on the intermolecular forces
 increasing attractive forces between molecules increases the
resistance to flow
 viscosity is also dependent on other factors, such as the
possibility of molecules tangling together (long molecules)
35
Hydrogen Bonding
 comparison of fluoromethane (CH3F) and methanol (CH3OH) will
show that their boiling points are significantly differently
 fluoromethane: -78°C (gas under normal conditions)
 methanol: 65°C (liquid under normal conditions)
 both molecules have the same molecular weight and polarity
(dipole)
 suggests that another intermolecular force is at work, not just
Van der Waals forces
 chemical structure shows that fluoromethane has a C-F bond
and methanol has a C-OH bond (similar to water H-OH) – the
–OH group creates additional attractive forces between
molecules than only the Van der Waals forces
 Hydrogen Bonding is a weak to moderate attractive force that
exists between a hydrogen atom covalently bonded to a very
electronegative atom, X, and a lone pair of electrons on another
small, electronegative atom, Y
X
H- - -Y
 generally X and Y are the atoms F, O or N
 compare boiling points of hydrides with Group 16 (VIA) elements:
H2O (100°C), H2S (-60°C), H2Se (-40°C), H2Te (0°C)
 if London forces were the intermolecular forces present one
would expect the boiling points to increase from H2O to H2Te,
based on molecular weight – obviously not the only factor as
H2O has a much higher boiling point
 consistent with the view that hydrogen bonding exists in H2O,
but is virtually non-existent in the others
 observe hydrogen bonding in H2O – relates to partial charges
on O and H
36
 can also see the same pattern when looking at other hydrides of:
 group 17: HF, HCl, HBr, Hi
 group 15: NH3, PH3, AsH3, SbH3
 however, pattern is not evident with group 14 hydrides: CH4,
SiH4, GeH4, SnH4
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The Structure and Properties of Solids
 all solids have a definite shape & volume, are incompressible and
do not flow readily
 hardness, melting point, mechanical characteristics and
conductivity can vary…….why?
 due to forces between the particles in elements and compounds
Ionic Crystals
 arrangement of ions forms a crystal lattice
 basically a 3D arrangement of ions in a crystal structure
 wide variety of shapes means that there is a wide variety of
structures
 ionic compounds are hard but brittle solids at SATP, conduct
electricity in liquid form but not in solid state, form conducting
solutions in water and have high melting points
 interpreted to meant that ionic bonds are strong and directional
 in general, ionic bonding is stronger than all intermolecular forces
e.g. Ca3(PO4)2(s) in tooth enamel (ionic bond) is stronger than ice
H2O(s) (hydrogen bonding).
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Metallic Crystals
 metals are shiny, silvery, flexible solids with good thermal &
electrical conductivity
 X-ray diffraction shows that all metals have a continuous and very
compact crystalline structure
 all metals have a closely packed structure (with few exceptions)
 current theory to explain this is the electron sea model where
the fixed positive nuclei are bonded to loosely held, mobile
electrons
 ideas in this model are;
1. low I.E. of metal atoms to explain loosely held electrons
2. empty valence orbitals to explain electron mobility
3. electrostatic attractions of +ve centers and –ve electron
“sea” to explain the empirical properties of metals
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Molecular Crystals
 explain molecular solids which are not hard, have relatively low
melting points, and are non-conductors (as liquids or solids)
 X-ray analysis shows a crystal-like structure with the molecules
packed very tightly together and more complex than in ionic
compounds
Covalent Network Crystals
 include diamonds, quartz (used in making emery sandpapers)
 are brittle, very hard, high melting points, insoluble and nonconductors of electricity
 are usually much harder and have much higher melting points
than ionic and molecular crystals
 the shape & x-ray diffraction analysis shows that diamonds have
carbon atoms in a large tetrahedral network with each C bonded
to 4 other C atoms
 each diamond is a crystal and can be described as a single
macromolecule with a chemical formula of C(s)
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 the network of covalent bonds leads to the common name for
these crystals as covalent network
 properties of hardness & high melting point provide evidence that
the overall bonding is very strong
 the interlocking structure gives it its strength, as the individual C-C
bonds are not strong on their own
 that’s why individual atoms are not easily displaced and the
sample is so hard
 to break a covalent network crystal, many bonds must be broken,
which would need lots of energy which is why the melting points
are so high
 electrons are not free to move so these substances are nonconductors of electricity
Other Covalent Networks
 carbon is extremely versatile and can bond to itself to form many
pure carbon substances
 3-D structures (diamonds), layers of sheets (graphite), large
spherical molecules (buckyballs) and long thin tubes (carbon
nanotubes)
Read semiconductors on your own and copy the Summary into your
notes.