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Chapter 8
Basic Concepts of
Chemical Bonding
Dr V Paideya
2022
© 2015 Pearson Education
Learning Outcomes
 Differentiate between polar and nonpolar covalent bonding.
 Describe the significance of relative electronegativity in
bonding.
 Identify each atom in a polar bond as having a partial negative
charge or a partial positive charge.
 Compare the relative polarity of two or more polar bonds.
 Describe the nature of the bonding in molecular compounds.
 Describe the importance of the octet rule.
 From its Lewis symbol, predict the number of covalent bonds
an element typically forms.
 Draw Lewis formulas for the diatomic elements, molecular
compounds, and polyatomic ions.
Basic Concepts
of Chemical
 Describe the number of shared and unshared electrons in
a
Bonding
Lewis
formula.
© 2015 Pearson Education, Inc.
Basic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Polar Covalent Bonds
• The electrons in a covalent bond are not always
shared equally.
• Fluorine pulls harder on the electrons it shares
with hydrogen than hydrogen does.
• Therefore, the fluorine end of the molecule has
more electron density than the hydrogen end.
Polar covalent
bond
Nonpolar covalent
bond
Basic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Electronegativity
• Electronegativity is the ability of an atom in a
molecule to attract electrons to itself.
• On the periodic table, electronegativity generally
increases as you go
– from left to right across a period.
Metals have low electronegativities,
&
nonmetals have high electronegativities
– from the bottom to the top of a group.
Basic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Electronegativity and
Polar Covalent Bonds
• When two atoms share electrons unequally, a
polar covalent bond results.
• Electrons tend to spend more time around the
more electronegative atom. The result is a
partial negative charge (not a complete transfer
of charge). It is represented by δ–.
• The other atom is “more positive,” or δ+.
Basic Concepts
of Chemical
Bonding
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Electronegativity Values
Basic Concepts
of Chemical
Bonding
7
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Electronegativity & Bond type
Classification of chemical compounds:
EN  2

Ionic
EN  2 & > 0 
Polar covalent
EN = 0
Nonpolar covalent

Basic Concepts
of Chemical
Bonding
8
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Polar Covalent Bonds
The greater the
difference in
electronegativity,
the more polar is
the bond.
Basic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Dipole Moments
Symbol 
describes the degree of polarity of a bond
E.g.
HCl:
+
ENH = 2.1 & ENCl = 3.0,  EN = 0.9
-
Indicates direction
of polarity with
arrow head towards
 Negative end
Basic Concepts
of Chemical
Bonding
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Is a Compound Ionic or Covalent?
• Simplest approach: Metal + nonmetal is ionic;
nonmetal + nonmetal is covalent.
• There are many exceptions: It doesn’t take into
account oxidation number of a metal (higher
oxidation numbers can give covalent bonding).
• Electronegativity difference can be used; the table
still doesn’t take into account oxidation number.
• Properties of compounds are often best: Lower
melting points mean covalent bonding, for
example.
Basic Concepts
of Chemical
Bonding
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Comparison of Ionic vs Covalent
Property
NaCl
CCl4
solid
liquid
Melting point (C)
801
-23
Molar heat of fusion
(kJ/mol)
30.2
2.5
Molar heat of vaporization
(kJ/mol)
600
30
State
Electrical conductivity
Good
Concepts
PoorBasic
of Chemical
Bonding
12
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Writing Lewis Structures
The steps for drawing plausible Lewis structures:
1. Sum the valence e-‘s(Ve) from all atoms, taking into
account overall charge. If anion, add one electron for
each negative charge. If cation, subtract one electron
for each positive charge.
2. Use single bonds to join the atoms, (2 e-‘s/bond);
subtract these Bonde from the total Ve;
3. Place Lone pairs(LPe) on the terminal atoms to satisfy
the Octet rule; subtract LPe’s from remaining Ve; place
any remaining Ve on the central atom.
4. If the Octet rule for the central atom is NOT satisfied,
form multiple covalent bonds by converting LP’s on
Basic Concepts
the terminal atoms into bond pairs.
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Exercise 4
Question:
Write a plausible Lewis structure for phosphorus
trichloride, PCl3.
Solution:
Step 1. Count the total number of valence electrons:
P – group 15 .: 5 valence electrons
Cl – group 17 .: 7 valence electrons
Total Ve
= (3 x Cl) + (1 x P)
= (3 x 7) + (1 x 5)
= 26 eStep 2. Identify the central atom:
ENP = 3.0, and ENCl = 4.0
.: P is the central atom
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Basic Concepts
of Chemical
Bonding
Exercise cont…
Step 3.
Draw a skeletal structure:
Step 4.
Subtract 2 electrons for each bond formed
from the total Ve:
Remaining electrons
= 26 – (3 bonds x 2 e-/bond)
= 26 – 6
= 20 eBasic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Exercise cont…
Step 5.
Complete the octets for the terminal atoms,
by placing lone pairs on the atoms:
Step 6.
Subtract the Lone pair electrons from the
Remaining electrons
Rem. e-’s
= 20 – 3 x (3 Lone pairs x 2 e-/LP)
= 20 – 18
= 2 e-
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Basic Concepts
of Chemical
Bonding
Exercise cont…
Step 7. Place any remaining electrons on the
central atom, and check that the Octet rule is
satisfied for ALL atoms
Rem. e-’s
= 2 e-
Octet rule for all atoms is satisfied!
© 2015 Pearson Education, Inc.
Basic Concepts
of Chemical
Bonding
Exercise 5
Question: Attempt this question on your own
Write a plausible Lewis structure for cyanogen, C2N2, a
poisonous gas used as a fumigant and rocket
propellant.
Solution:
Step 1. Count the total number of valence electrons:
C – group 14 .: 4 valence electrons
N – group 15 .: 5 valence electrons
Step 2. Identify the central atom:
Basic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Example 5 cont…
Step 3.
Draw a skeletal structure:
Step 4.
Subtract 2 electrons for each bond formed
from the total Ve:
Step 5.
Complete the octets for the terminal atoms,
by placing lone pairs on the atoms:
Basic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Example 5 cont…
Step 6.
Subtract the Lone pair electrons from the
Remaining electrons
Step 7.
The central atoms, only have the 4 Be, so
need another 4 each, .: move LPe to form
multiple bonds
Basic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Exercise
Question: Attempt this question on your own
Write the Lewis structure for the nitronium ion, NO2+.
Solution:
Step 1. Count the total number of valence electrons:
N – group 15 .: 5 valence electrons
O – group 16 .: 6 valence electrons
Basic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Example 6
Question:
Write the Lewis structure for the nitronium ion, NO2+.
Solution:
Step 1. Count the total number of valence electrons:
N – group 15 .: 5 valence electrons
O – group 16 .: 6 valence electrons
Total Ve
= 1 x N + 2 x O – 1e (charge)
= (1 x 5) + (2 x 6) – 1e
= 16 e-
Step 2. Identify the central atom:
ENN = 3.0, and ENO = 3.5
.: N is the central atom
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Basic Concepts
of Chemical
Bonding
Example 6 cont…
Step 3.
Draw a skeletal structure:
O–N–O
Step 4.
Subtract 2 electrons for each bond formed
from the total Ve:
Remaining electrons
= 16 – (2 bonds x 2 e-/bond)
= 16 – 4
= 12 eBasic Concepts
of Chemical
Bonding
© 2015 Pearson Education, Inc.
Example 6 cont…
Step 5.
Complete the octets for the terminal atoms,
by placing lone pairs on the atoms:
O N O
Step 6.
Subtract the Lone pair electrons from the
Remaining electrons
Rem. e-’s
= 12 – 2 x (3 Lone pairs x 2 e-/LP)
= 12 – 12
= 0 eBasic Concepts
of Chemical
Bonding
24
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Example 6 cont…
Step 7.
The central atom, N, has only the 4 Be, so
need another 4, .: move 2 LP’s to form
multiple bonds.
There are 2 possible ways:
O N O
OR
O N O
Which result in 2 plausible products:
O N O
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OR
O N O
Basic Concepts
of Chemical
Bonding