Electronegativity
+
–
0
H
Cl
H
0
H
The basic units: ionic vs. covalent
• Ionic compounds form repeating units.
• Covalent compounds form distinct molecules.
• Consider adding to NaCl(s) vs. H2O(s):
Na Cl Na Cl
Cl Na Cl Na
H
O
H
O
H
H
• NaCl: atoms of Cl and Na can add individually
forming a compound with million of atoms.
• H2O: O and H cannot add individually, instead
molecules of H2O form the basic unit.
Holding it together
Q: Consider a glass of water. Why do
molecules of water stay together?
A: there must be attractive forces.
Intramolecular
forces are
much stronger
Intramolecular forces
Intermolecular forces
occur between atoms occur between molecules
• We do not consider intermolecular forces in
ionic bonding because there are no molecules.
• We will see that the type of intramolecular bond
determines the type of intermolecular force.
I’m not stealing, I’m sharing unequally
• We described ionic bonds as stealing electrons
• In fact, all bonds share – equally or unequally.
• Note how bonding electrons spend their time:
H2
0
H H
0
covalent
(non-polar)
HCl
+
H Cl
–
polar covalent
LiCl [Li]+[
+
Cl ]–
–
ionic
• Point: the bonding electrons are shared in each
compound, but are not always shared equally.
• The greek symbol  indicates “partial charge”.
Electronegativity
• Recall that electronegativity is “a number that
describes the relative ability of an atom, when
bonded, to attract electrons”.
• The periodic table has electronegativity values.
• We can determine the nature of a bond based
on EN (electronegativity difference).
• EN = higher EN – lower EN
NBr3: EN = 3.0 – 2.8 = 0.2 (for all 3 bonds).
• Basically: a EN below 0.5 = covalent,
0.5 - 1.7 = polar covalent, above 1.7 = ionic
• Determine the EN and bond type for these:
HCl, CrO, Br2, H2O, CH4, KCl
Electronegativity Answers
HCl: 3.0 – 2.1 = 0.9
polar covalent
CrO: 3.5 – 1.6 = 1.9
ionic
Br2:
2.8 – 2.8 = 0
covalent
H2O: 3.5 – 2.1 = 1.4
polar covalent
CH4: 2.5 – 2.1 = 0.4
covalent
KCl: 3.0 – 0.8 = 2.2
ionic
Electronegativity & physical properties
a lower
other
factors
2 would have can
•CaCl
Electronegativity
help toNote:
explain
properties
point:
as atomic size
ofmelting/boiling
compounds like
those insuch
the lab.
CaCl2 = 3.0 – 1.0 = 2.0
within+molecules
–
+ –
• Lets
at
HCl:
partial
charges
CaFlook
=
4.0
–
1.0
=
3.0

also affects melting
2
keep
molecules
LiBr would
have together.
a lower
and boiling points.
melting/boiling point:
EN is an important
• The
KClsituation
= 3.0 – is
0.8similar
= 2.2 in NaCl,
factor but
– not
+ the
but
the=attraction
even greater
LiBr
2.8 – 1.0 is
= 1.8
only factor. It is
+ –
(EN
=
2.1
vs.
0.9
for
HCl).
most useful when
H2S would have a lower
comparing atoms
melting/boiling
point:
• Which
would have
a higher melting/boiling
point?
andEN.
molecules of
H2NaCl
O= 3.5
– 2.1 =of1.4
because
its greater
size.
H2S
= 2.5pick
– 2.1
0.4 with thesimilar
• For
each,
the=one
lower boiling
point a) CaCl2, CaF2 b) KCl, LiBr c) H2O, H2S
CaCl2 would have a lower Note: other factors
melting/boiling point:
such as atomic size
CaCl2 = 3.0 – 1.0 = 2.0
within molecules
CaF2 = 4.0 – 1.0 = 3.0
also affects melting
LiBr would have a lower
and boiling points.
melting/boiling point:
EN is an important
KCl = 3.0 – 0.8 = 2.2
factor but not the
LiBr = 2.8 – 1.0 = 1.8
only factor. It is
most useful when
H2S would have a lower
comparing atoms
melting/boiling point:
and molecules of
H2O= 3.5 – 2.1 = 1.4
similar size.
H2S = 2.5 – 2.1 = 0.4
Why oil and water don’t mix
• Lets take a look at why oil and water don’t
mix (oil is non-polar, water is polar)
+ – +

+ – +

+ – +

The partial charges on
water attract, pushing
the oil (with no partial
charge) out of the way.
+
–
+ – +

+ – +

+
+ – +

+ – +
