enginering khalid ahmed

1. 1
Introduction to Chemistry
Chapter 12
Chemical Bonds
Chemical Bonds are the attractive forces that hold groups of two or more atoms together and make them function as a unit.
Chemical bonding occurs with the valence electrons of an atom.
Two types of chemical bonds:
Ionic: Ionic bonding occurs when an atom that loses electrons relatively easily reacts with an atom that gains
electrons relatively easily.
• metals with nonmetals
• salts
• crystals
• held together by electrostatic attraction
E.g. How salts are formed:
Na0 (s) + Cl20
2 Na+ + Cl– + Cl– → NaCl + NaCl
or
2 Na+ + 2Cl– 2 NaCl
Ionic compounds: (formula units)
• are solids with high melting points
• they exist as crystal structures
• they consist of formula units
• most are soluble in water
• molten compounds conduct electricity well because they contain mobile charged ions.
• aqueous solutions (dissolved in water) conduct electricity well because they contain mobile charged ions.
• they are mostly formed between two elements with large differences (∆EN>2) in electronegativity, usually a metal
and a nonmetal.
Covalent bonding: This bonding occurs when electrons are shared by atoms. The electrons are attracted to the
nucleus of the opposing atoms in the bond.
E.g. H2 HCl SiO2
H• •H H··H H—H H2
•• •• ..
H• • Br ‫׃‬ H ‫ ׃‬Br ‫׃‬ H—Br : HBr
•• •• ..
Covalent compounds: (molecules)
• they are solids, liquids, and gases with low melting points
• many are not soluble in water
• liquid and molten compounds do not conduct electricity
• aqueous solutions are usually poor conductors of electricity because most do not contain charged particles (ions).
• they are often formed between two elements with similar electronegativities, usually nonmetals.
To understand chemical bonding, we must first understand the forces between the bonds.
• Electronegativity: is the ability of an atom in a molecule to attract shared electrons to itself.

2. 2
2
0 4
∆ EN
0 = covalent
> 0 = polar covalent
> 2 = ionic
∆EN = H—H 2.1-2.1 = 0 covalent
∆EN = S—H 2.5-2.1 = 0.4 polar covalent
∆EN = NaCl 0.9-3.0 = 2.1 ionic
Lewis Dots:
• Lewis dots represent the valence electrons of an atom.
• The number of valence electrons for Group A elements are the same as the group number.

3. 3
Lewis Structures: The Octet Rule
The theory about why noble gases are basically unreactive and stable was built around the observations:
• Noble gases have completely filled s and p shells
• There are 8 valence electrons for each of the noble gases
• The filled shells of the valence electrons account for the stability of the noble gases
The octet rule states that atoms will share, gain, or lose electrons in order to produce a noble gas electron configuration.
He → 1s2 Li+ → 1s2 Be2+ → 1s2
Ne → 1s22s22p6 Na+ → 1s22s22p6 Mg2+→ 1s22s22p6 Al3+→1s22s22p6
F– → 1s22s22p6 O2– → 1s22s22p6 N3– → 1s22s22p6
Isoelectronic – elements and ions that contain the same number of electrons.
Ionic Bonding
General Formula Table

4. 4
Ionic compounds: Lewis structures
..
NaCl → Na+ + [ :Cl: ]–
¨
..
MgBr2 Mg2+ + 2 [ :Br:]–
¨
..
K2S 2 K+ + [:S: ]2–
¨
..
Al2O3 2 Al3+ + 3 [:O: ]2–
¨
Lewis Structure for Covalent Compounds:
• Share electrons
• Sometimes the octet rule applies
• Carbon, nitrogen, oxygen, and fluorine always obey the octet rule
H : H The dots between the hydrogen atoms represent shared electrons.
H—H The straight line between the atoms represents two electrons shared between the two hydrogen atoms
Rules to determine the number of bonds needed in covalent bonding:

5. 5
Binary compounds:
1. Determine the number of valence electrons for each atom:
H—Cl
H Cl
1 + 7 = 8 electrons
2. Bond the atoms—sharing 2 electrons. Show all valence electrons for each atom.
..
H—Cl:
¨
3. Count all the electrons in the bonds and electrons that are not bonded (unshared, non-bonded, lone pairs), they
should equal the total number of valence electrons.
..
H—Cl:
¨
2 e– (from bond)
+ 6 unshared electrons around the chloride atom
8 total electrons
Polyatomic compounds that obey the octet rule:
1. Determine the total number of valence electrons
2. Determine central atom
3. Draw a bond for each atom that is bonded to the central atom.
4. Give each atom that requires an octet the appropriate number of electrons.
5. Count all electrons (included bonded atoms)
6. The total number of electrons should equal the number of valence electrons available.
CH4,
1. CH4 has 4+4 = 8 valence electrons
2. Carbon is the central atom
3. Add hydrogen bonds
H
‫׀‬
H ̶ C ̶ H
‫׀‬
H
4. Count the total number of electrons
NH3
1. NH3 has 5+3 = 8 valence electrons
2. Nitrogen is the central atom
3. Add hydrogen bonds
4. Add two electrons to nitrogen; it must have an octet.
..
H ̶ N ̶ H
‫׀‬
H
3. Count the total number of electrons

6. 6
H2O
1. H2O has 5+3 = 8 valence electrons
2. Oxygen is the central atom
3. Add hydrogen bonds
4. Add electrons to oxygen, two at a time, until it has an octet ( it must have an octet).
..
H ̶ O:
‫׀‬
H
5. Count the total number of electrons
More examples will be given in class.
Polyatomic compounds that do not obey the octet rule:
BeBr2,
• BeBr ,
2
– has a total of 16 valence electrons
– Be is the central atom
– Give each bromine an octet
.. ..
: Br ̶ Be ̶ Br :
¨ ¨
Count the electrons; all the valence electrons have been used.
BCl3
– has a total of 24 valence electrons
– B is the central atom
– Give each chlorine an octet
..
: Cl :
‫׀‬
B
.. ..
: Cl Cl:
¨ ¨
– Count the electrons—all the valence electrons have been used.
Double bonds and triple bonds:
O2
– has a total of 12 valence electrons
– either O is the central atom
– each oxygen must have an octet
– There are not enough valence electrons to give each nitrogen and octet; therefore, we must form double
bonds.
.. ..
:O=O:

7. 7
N2
– has a total of 10 valence electrons
– either N is the central atom
– each nitrogen must have an octet
– There are not enough valence electrons to give each nitrogen and octet; therefore, we must form triple
bonds.
:N≡N:
CO2
– has a total of 16 valence electrons
– either C is the central atom
– form a bond with each oxygen
– each oxygen must have an octet
– There are not enough valence electrons to give each oxygen an octet; therefore, we must form double
bonds.
.. ..
: O =C ̶ O :
¨ ¨
Resonance structures: Delocalized electrons determined by bond strength. Double bonds are shorter and stronger
than single bonds. (Bond strength , the amount of energy required to break a bond)
Write Lewis dot and dash formulas for sulfur trioxide, SO .
3
– 4 x 6 = 24 valence electrons
– sulfur is central atom
– give all oxygen an octet , a double bond will have to be formed between one oxygen and the sulfur
central atom.
·· ··
· O · S· · O ·
· or ·O S O·
· ·· · ·
·· ··
·
·· ·· ··
· O · ·O·
· ·· ·
· ·
··
There are three possible structures for SO3.
– The double bond can be placed in one of three places.
·· ·· ·· ··
·O S O ·
· ·
·O
· S O· · ·O S O ·
· ·
·· ·· ·· ·· ·· ··
·O ·
· ·· · ·O· ·O ·
· ·· ·
· ·
When two or more Lewis formulas are necessary to show the bonding in a molecule, we must use equivalent resonance
structures to show the molecule’s structure. Double-headed arrows are used to indicate resonance formulas.
Resonance is a flawed method of representing molecules.
There are no single or double bonds in SO .
– 3
In fact, all of the bonds in SO3 are equivalent.
The best Lewis formula of SO that can be drawn is:
3
O S O
We will do these in class.
SO2 O

8. 8
NO3
Polyatomic ions: The charges on the ion must be either added or substracted from the total valence electrons.
NH4+ (Must substract an electron because the molecule has a + 1 charge.)
5 + 4 – 1 = 8 valence electrons.
We will do these in class:
SO42-
NO2-
CO32-
Salts with polyatomic ions
Shapes and Polarity of Molecules
(VSEPR—valence shell electron pair repulsion theory)
Electronic shape—counting regions of high electron density
1. Each bond is counted as one region of high electron density.
2. Each double bond and each triple bond is counted as one region of high electron density.
3. Each pair of unshared electrons is counted as one region of high electron density.

9. 9
Rules:
1. Determine the central atom
2. Determine the number of high electron densities around the central atom.
3. Determine the electronic structure based on geometric structure of molecule.
4. Polarity of a molecule is determined by the symmetry of the electron density around the central atom. (The
Electronegativity must be equal and opposite in magnitude)
Number of high Electronic
Molecule Central Atom Polarity
electron densities structure
BeCl2 Be 2 linear nonpolar
BCl3 B 3 trigonal planar nonpolar
CH4 C 4 tetrahedral nonpolar
NH3 N 4 tetrahedral polar
H2O O 4 tetrahedral polar
trigonal
PCl5 P 5 nonpolar
bipyramidal
SF6 S 6 octrahedral nonpolar