Bonding and
Nomenclature
Four Types of Naming
 Binary
compounds
 Ternary
compounds
 Coordination
 Organic
compounds
compounds
Contain only two types of elements
Contain more than two types of elements
These will not be covered
We will cover these in a separate unit
Chemical Bonding
 BOND
= Forces of attraction between the
nucleus of one atom and the electrons of
another atom
 ONLY Valence electrons participate in
bonding
 Bonds are formed as a result of a
CHEMICAL REACTION
Energy & bonding
 Energy
associated with bonding:
Exothermic (spontaneous)
&
Endothermic (not spontaneous)
Exothermic
 Spontaneous
bond formation = energy is
released
∞ go from HIGH energy (unhappy atoms) to
Lower energy (happy atoms)
∞ Creating a bond creates STABILITY
 Energy is RELEASED as a product
A + B → C + energy
Endothermic
 Breaking
Bonds (not spontaneous) =
Energy is consumed
 Go from LOW energy (happy atoms) to
HGHER energy (unhappy atoms)
 Ripping two atoms apart takes energy
 Energy is CONSUMED or needed as an
ingredient to fuel the process
A + energy → B + C
Keeping Track of Electrons
 The
electrons responsible for the chemical
properties of atoms are those in the outer
energy level.
 Valence electrons - The s and p electrons
in the outer energy level.
 Core electrons -those in the energy levels
below.
Keeping Track of Electrons






Atoms in the same column
Have the same outer electron configuration.
Have the same valence electrons.
Easily found by looking up the group number
on the periodic table.
Group 2A - Be, Mg, Ca, etc.2 valence electrons
Electron Dot diagrams
A
way of keeping track of
valence electrons.
 How to write them
 Write the symbol.
 Put one dot for each valence
electron
 Don’t pair up until they have to
X
The Electron Dot diagram
for Nitrogen
Nitrogen has 5 valence
electrons.
 First we write the symbol.
Then add 1 electron at a
time to each side.
Until they are forced to pair up.

N
Write the electron dot diagram for







Na
Mg
C
O
F
Ne
He
1s22s22p63s1
Na
Mg
1s22s22p63s2
C
1s22s22p2
O
1s22s22p4
F
1s22s22p5
1s22s22p6
1s2
Ne
He
Electron Configurations for Cations
 Metals
lose electrons to attain noble gas
configuration.
 They make positive ions.
 If we look at electron configuration it makes sense.
 Na 1s22s22p63s1 - 1 valence electron
 Na+ 1s22s22p6 -noble gas configuration
Electron Dots For Cations
 Metals
will have few valence electrons
 These will come off
 Forming positive ions
20
Ca
40.078
1s22s22p63s2
+2
Ca
Electron Configurations for Anions
 Nonmetals
gain electrons to attain noble
gas configuration.
(anions)
 They make negative ions.
 If we look at electron configuration it
makes sense.
 S 1s22s22p63s23p4 - 6 valence electrons
 S-2 1s22s22p63s23p6 -noble gas
configuration.
Electron Dots For Anions
 Nonmetals
will have many valence electrons.
 They will gain electrons to fill outer shell.
P
P
-3
Stable Electron Configurations
 All
atoms react to achieve noble gas
configuration.
 Noble gases have two s and six p electrons.
 Eight valence electrons .
 Also called the octet rule.
Ar
Types of Bonding
Ionic
&
Covalent
Guiding Questions?
What is that?
How do we figure out what the chemical formula is?
What does it mean to be "free of chemicals"?
Ionic Bonding
Formed from:
Metal
&
Nonmetal
Name cation first followed by anion
Ionic Bonding
 Anions
and cations are held together by opposite
charges. (Neutral)
 Ionic compounds are called salts.
 Simplest ratio is called the formula unit.
 The bond is formed through the transfer of
electrons.
 Electrons are transferred to achieve noble gas
configuration.
Ionic Bonding

a.
b.
c.
d.
Compounds formed by the TRANSFER of
electrons from one atom to another
Metal loses e− to a Nonmetal
𝑁𝑎+
𝐶𝑙 −
Metal combines with a Polyatomic ion
𝑁𝑎+ 𝑁𝑂3
Polyatomic ion combines with a Nonmetal
𝑁𝐻4 𝐶𝑙 −
Polyatomic ion combines with another
Polyatomic ion
𝑁𝐻4 𝑁𝑂3
Formula writing for Ionic
Compounds
 IUPAC
= International Union of Pure and
Applied Chemistry
 Compounds have a common name and
a chemical name
 There is a systematic method for naming
ionic compounds
 Need 2 types of ions and the CHARGE of
each ion.
Criss-cross rule
 Name
the Cation first (Metal
 Name the anion second (Nonmetal)
Replace the ending of the anion with
“IDE”
Ex: Flourine – Flouride
oxygen - oxide
Criss-Cross Rule
Example: Aluminum Chloride
Step 1:
Aluminum
Chloride
write out name with space
Step 2:
Al
3+
Cl
1-
write symbols & charge of elements
Step 3:
Al 1
Cl 3
criss-cross charges as subsrcipts
Step 4:
combine as formula unit
(“1” is never shown)
AlCl3
Criss-Cross Rule
Example: Aluminum Oxide
Step 1:
Aluminum
Oxide
Step 2:
Al3+
O2-
Step 3:
Al 2
O3
Step 4:
Al2O3
Ionic Bonding
transfer of electron
+
Na Cl
NaCl
-
Ionic Bonding
+2
Ca
+2
Ca
+2
Ca
P
P
-3
-3
 All the electrons must be accounted for!
Ionic Bonding
Bonding Activity
Properties of Ionic Compounds
 Crystalline
structure.
 A regular repeating arrangement of ions
in the solid.
 Ions are strongly bonded.
 Structure is rigid.
 High melting points- because of strong
forces between ions.
Formula writing for Ionic
Compounds
 Polyatomic
ions are named exactly as
they are seen on Table
 Parentheses are REQUIRED only when you
need more than one “bunch” of a
particular Polyatomic ion.
 Examples:
NaOH – sodium hydroxide
KNO3 –
Potassium nitrate Ammonium hydroxide –
NH4OH
Calcium Phosphate – CaP
𝑀𝑔+2 and 𝑁𝑂2− - Mg(NO2)2
and CLO3 - - (NH4)3 N
𝑁𝐻4
Polyatomic Ion:
a group of atoms that stay together and have a single, overall charge.
BrO41-
Perbromate ion
CO42ClO41IO41NO41PO53SO521 more oxygen
BrO31-
BrO1-
Bromate ion
BrO21-
Bromite ion
CO32-
CO22-
CO2-
ClO31-
ClO21-
ClO1-
IO31-
IO21-
IO1-
NO31-
NO21-
NO1-
PO43-
PO33-
PO23-
SO42-
SO32-
SO22-
“normal”
1 less oxygen
Carbonate ion
Chlorate ion
Iodate ion
Nitrate ion
Phosphate ion
Sulfate ion
Hypobromite ion
2 less oxygen
Polyatomic Ions - Memorize
Eight “-ATE’s”
PO43SO4
……………
2- ……………
CO32ClO3
NO3
…………..
1- …………..
1- ………..….
phosphate
phosphATE
Exceptions:
sulfate
sulfATE
carbonate
carbonATE
chlorate
chlorATE
nitrate
nitrATE
NH41+
…………… ammonium
OH1-
……………
hydroxide
CN1-
…………..
cyanide
Naming Ternary Compounds from
Oxyacids
The following table lists the most common families of oxy acids.
one more
oxygen atom
most
“common”
HClO4
perchloric acid
HClO3
chloric acid
one less
oxygen
HClO2
chlorous acid
two less
oxygen
HClO
hypochlorous acid
H2SO4
sulfuric acid
H3PO4
phosphoric acid
HNO3
nitric acid
H2SO3
sulfurous acid
H3PO3
phosphorous acid
HNO2
nitrous acid
H3PO2
hypophosphorous acid
(HNO)2
hyponitrous acid
An acid with a
name ending in
A salt with a
name ending in
-ous
forms
-ite
-ic
forms
-ate
Hill, Petrucci, General Chemistry An Integrated Approach 1999, page 60
Oxyacids  Oxysalts
If you replace hydrogen with a metal, you have formed an oxysalt.
A salt is a compound consisting of a metal and a non-metal. If the
salt consists of a metal, a nonmetal, and oxygen it is called an
oxysalt. NaClO4, sodium perchlorate, is an oxysalt.
OXYACID
OXYSALT
HClO4
perchloric acid
NaClO4
sodium perchlorate
HClO3
chloric acid
NaClO3
sodium chlorate
HClO2
chlorous acid
NaClO2
sodium chlorite
HClO
hypochlorous acid
NaClO
sodium hypochlorite
Suffixes have meaning
 “-ide”
binary compound
sodium chloride (NaCl)
 “-ite”
or “-ate”
sulfite (SO32-)
sulfate (SO42-)
 “-ol”
polyatomic compound
“-ate” means one more oxygen
than “-ite”
alcohol
methyl alcohol (methanol)
 “-ose”
sugar
sucrose
 “-ase”
sucrase
enzyme
Ionic bonding
 Transition
metals tend to have more than
one oxidation state.
 Use roman numerals to indicate their
oxidation #.
 Roman numeral appears in ( ) AFTER the
element symbol(stock system)
Covalent bonding
 Formed
from 2 or more Nonmetals
 Atoms SHARE e to get a full valence shell
 GOAL… to have a full outer shell
 Exception: H needs 2 𝑒 −
 Examples: 𝐶𝐻4
& 𝐻2 O
Oxidation States in Formulas and Names
Traditional System
Stock System
(Two non-metals)
+1
dinitrogen monoxide
N2O
+3
dinitrogen trioxide
sulfur dioxide
sulfur trioxide
nitrogen (V) oxide
-2
SO2
+6
nitrogen (III) oxide
-2
N2O5
+4
nitrogen (I) oxide
-2
N2O3
+5
dinitrogen pentoxide
-2
sulfur (IV) oxide
-2
SO3
sulfur (VI) oxide
stock system is NOT preferred for two non-metals
 Fluorine
Covalent bonding
has seven valence electrons
• A second atom also has seven
By sharing electrons
…both end with full orbitals
8 Valence
electrons
F
F
8 Valence
electrons
Single Covalent Bond
A
sharing of two valence electrons.
 Only nonmetals and Hydrogen.
 Different from an ionic bond because
they actually form molecules.
 Two specific atoms are joined.
 In an ionic solid you can’t tell which atom
the electrons moved from or to.
Covalent compounds
 FORGET
Charges
 Use greek prefixes to indicate how many
atoms of each element, but do not use
1 – mono 6 – hexa
mono on first element.
2- di
7–
 Examples:
hepta
Carbon dioxide CO2
3 – tri
8 – octa
Dinitrogen trioxide - N2O3
4 – tetra 9 – nona
5–
penta
10 deca
Rules for writing Covalent
compounds
 Least
electronegative element is written
first
 Most electronegative element is written
last
 Subscripts tell you the prefix of each
element in the formula
How to show how they
formed
 It’s
like a jigsaw puzzle.
 I have to tell you what the final formula is.
 You put the pieces together to end up
with the right formula.
 For example- show how water is formed
with covalent bonds.
H
O
Water
Each hydrogen has 1 valence
electron
Each hydrogen wants 1 more
The oxygen has 6 valence electrons
The oxygen wants 2 more
They share to make each other happy
Water



H
Put the pieces together
The first hydrogen is happy
The oxygen still wants one more
O
 The
Water
second hydrogen attaches
 Every atom has full energy levels
 A pair of electrons is a single bond
HO
H
HO
H
Lewis Structures
1) Count up total number of valence electrons
2) Connect all atoms with single bonds
- “multiple” atoms usually on outside
- “single” atoms usually in center;
C always in center,
H always on outside.
(not H, though)
3) Complete octets on exterior atoms
4) Check
- valence electrons math with Step 1
- all atoms (except H) have an octet;
if not, try multiple bonds
- any extra electrons?Put on central atom
Multiple Bonds
 Sometimes
atoms share more than one
pair of valence electrons.
 A double bond is when atoms share two
pair (4) of electrons.
 A triple bond is when atoms share three
pair (6) of electrons.
C
O
Carbon dioxide
 CO
2
- Carbon is central atom ( I
have to tell you)
 Carbon has 4 valence electrons
 Wants 4 more
 Oxygen has 6 valence electrons
 Wants 2 more
Carbon dioxide
 Attaching
1 oxygen leaves the oxygen 1 short and
the carbon 3 short
CO
Carbon dioxide

Attaching the second oxygen
leaves both oxygen 1 short and the
carbon 2 short
OC O
Carbon
dioxide
 The only solution is to share more
Requires two double bonds
 Each atom gets to count all the
atoms in the bond

8 valence 8 valence
8 valence
electrons electrons
electrons
O CO
How to draw them
 Add
up all the valence electrons.
 Count up the total number of electrons to
make all atoms happy.
 Subtract.
 Divide by 2
 Tells you how many bonds - draw them.
 Fill in the rest of the valence electrons to
fill atoms up.
Examples
N
H
 NH
N
3
- has 5 valence electrons wants 8
 H - has 1 valence electrons wants 2
 NH
3 has 5+3(1) = 8
 NH wants 8+3(2) = 14
3
 (14-8)/2=
3 bonds
 4 atoms with 3 bonds
Examples
 Draw
in the bonds
 All 8 electrons are accounted for
 Everything is full
H
H NH
Examples








HCN C is central atom
N - has 5 valence electrons wants 8
C - has 4 valence electrons wants 8
H - has 1 valence electrons wants 2
HCN has 5+4+1 = 10
HCN wants 8+8+2 = 18
(18-10)/2= 4 bonds
3 atoms with 4 bonds -will require multiple
bonds - not to H
HCN
 Put
in single bonds
 Need 2 more bonds
 Must go between C and N
HCN
HCN
Put in single bonds
 Need 2 more bonds
 Must go between C and N
 Uses 8 electrons - 2 more to add

HC N
HCN
Put in single bonds
 Need 2 more bonds
 Must go between C and N
 Uses 8 electrons - 2 more to add
 Must go on N to fill octet

HC N
Another way of indicating
bonds
 Often
use a line to indicate a bond
 Called a structural formula
 Each line is 2 valence electrons
H O H =H O H
Structural Examples
H C N
H
C O
H
C
has 8 electrons
because each line is
2 electrons
 Ditto for N
 Ditto
for C here
 Ditto for O
Resonance
 When
more than one dot diagram with
the same connections are possible.
 NO
2
-
 Which
one is it?
 Does it go back and forth.
 It is a mixture of both, like a mule.
 NO
3
POLAR VS NONPOLAR
Polar
Nonpolar
Asymmetrical molecules
Symmetrical Molecules
UNUEQUAL sharing of
electrons
Does not pass the “mirror
test”: can not be folded to
reflect itself
EQUAL sharing of electrons
Does pass the “mirror test”:
can be folded to reflect
itself
2 atoms
different
Elements (EN)
2 atoms
same
element (EN)
> 2 atoms
unbonded e > 2 atoms
No
or lone pairs around
unbonded e or lone pairs
central atom
around the central atom
Ex: HCl 𝐻2 O
Ex: 𝐶𝑙2 𝐶𝑂2
𝐶𝐶𝑙4
BEWARE! There are often POLAR bonds inside
NONPOLAR molecules
Polar Bonds
 When
the atoms in a bond are the same,
the electrons are shared equally.
 This is a nonpolar covalent bond.
 When two different atoms are
connected, the atoms may not be shared
equally.
 This is a polar covalent bond.
 How do we measure how strong the
atoms pull on electrons?
Electronegativity
A
measure of how strongly the atoms
attract electrons in a bond.
 The bigger the electronegativity
difference the more polar the bond.
0.0 - 0.5
Covalent nonpolar
0.5 - 1.0
Covalent moderately
polar
1.0 - 2.0
Covalent polar
> 2.0 Ionic
How to show a bond is
polar
 Isn’t
a whole charge just a partial charge
 d+ means a partially positive
 d- means a partially negative
d+
H
 The
d-
Cl
Cl pulls harder on the electrons
 The electrons spend more time near the Cl
Polar Molecules
Molecules with ends
Polar Molecules
 Molecules
with a positive and a negative
end
 Requires two things to be true
 The molecule must contain polar bonds
This can be determined from
differences in electronegativity.
Symmetry can not cancel out the effects
of the polar bonds.
Must determine geometry first.
HF
H2O
NH3
CCl4
CO2
Is it polar?
Bond Dissociation Energy
 The
energy required to break a bond
 C - H + 393 kJ
C+H
 We get the Bond dissociation energy
back when the atoms are put back
together
 If we add up the BDE of the reactants and
subtract the BDE of the products we can
determine the energy of the reaction (DH)
Find the energy change for
the reaction
 CH
4 + 2O2
CO2 + 2H2O
 For the reactants we need to break 4 C-H bonds at
393 kJ/mol and 2 O=O bonds at 495 kJ/mol= 2562
kJ/mol
 For the products we form 2 C=O at 736 kJ/mol and 4
O-H bonds at 464 kJ/mol
 = 3328 kJ/mol
 reactants - products = 2562-3328 = -766kJ
Intermolecular Forces
What holds molecules to
each other
Intermolecular forces
 ONLY
in covalent molecules
 Weak forces that act between molecules
that hold molecules to each other
 Only exist in gaseous & liquid states
 Called weak forces because they are
much weaker than chemical bonds
REMEMBER: IMF’s occur b/t molecules,
whereas
BONDING occurs within molecules
Other types of bonds
 London
dispersion forces (LDF’s)
 Dipole (dipole-dipole)
 Hydrogen Bonds
Intermolecular Forces
 They
are what make solid and liquid
molecular compounds possible.
 The weakest are called van derWaal’s
forces - there are two kinds
 Dispersion forces
 Dipole Interactions



depend on the number of electrons
more electrons stronger forces
Bigger molecules
London dispersion
 Weakest
of all the IMF’s
 Only important for NONPOLAR molecules
 More electrons = Greater LDF’s
 Electron-electron repulsion creates brief
dipoles in atoms and molecules.
Dipole interactions
 Occur
when polar molecules are
attracted to each other.
 Slightly stronger than dispersion forces.
 Opposites attract but not completely
hooked like in ionic solids.
Dipole interactions
 Depend
on the number of electrons
 More electrons stronger forces
 Bigger molecules more electrons
Fluorine
is a gas
Bromine is a liquid
Iodine is a solid
Dipole interactions
 Occur
when polar molecules are
attracted to each other.
 Slightly stronger than dispersion forces.
 Opposites attract but not completely
hooked like in ionic solids.
+
d
d
H F
+
d
d
H F
Dipole-Dipole
 Molecules
such as HCl have both POSITIVE
and a NEGATIVE end or POLES
 Two poles = DIPOLE
 Results from an UNEQUAL/ASYMMETRICAL
sharing of electrons
 Dipole-Dipole = 2 molecules with
permanent dipoles are attracted to one
another temporarily.
Dipole-dipole
 Dipole
moment = measure of the strength
of the dipole within a molecule (Polarity)
 The greater the difference in
electronegativity between atoms, the
greater the polarity/dipole moment.
 The higher the dipole moment, the
stronger the intermolecular forces.
 The stronger the IMF’s the higher the
melting and boiling point.
Dipole Interactions
+
d
d+
d-
d
Hydrogen bonding
 Specific
type of DIPOLE interaction
 In a Polar bond, hydrogen is basicallly
reduced to a bare proton with almost no
atomic radius.
 Strongest of all IMF’s by far
 ONLY occur in molecules containing
hydrogen and fluorine, Oxygen, or
Nitrogen.
Hydrogen bonding
 Are
the attractive force caused by
hydrogen bonded to F, O, or N.
 F, O, and N are very electronegative so it
is a very strong dipole.
 The hydrogen partially share with the lone
pair in the molecule next to it.
 The strongest of the intermolecular forces.
Hydrogen Bonding
d+ dH O
+
H d
Hydrogen bonding
H O
H
Metallic Bonds
 In
metals, valence shells of atoms overlap
and are free to travel between atoms
through material.
 Atoms of a metal DO NOT bond with
other metal atoms.
 Metals share a sea of MOBILE valence
electrons
 This allows metals to conduct electric
current.
Empirical and Molecular
Formulas
A pure compound always consists of the same
elements combined in the same proportions by
weight.
Therefore, we can express molecular
composition as PERCENT BY WEIGHT.
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O
Empirical Formula
Quantitative analysis shows that a compound contains 32.38% sodium,
22.65% sulfur, and 44.99% oxygen.
Find the empirical formula of this compound.
 1mol Na 
 =
 23 g Na 
sodium sulfate
1.408 mol Na / 0.708 mol = 2 Na
32.38% Na
32.38 g Na 
22.65% S
22.65 g S
 1mol S 


32
g
S


= 0.708 mol S / 0.708 mol
=1S
44.99% O
44.99 g O
 1 mol O 


16
g
O


= 2.812 mol O / 0.708 mol
=4O
Step 1) %  g
Step 2) g  mol
Step 3) mol
mol
Na2SO4
Empirical Formula
A sample weighing 250.0 g is analyzed and found to contain the following:
27.38 g
27.38%
1.19%
1.19
g
14.29%
14.29
g
57.14%
57.14
g
Na
sodium
H
hydrogen
C
carbon
O
oxygen
Assume sample is 100 g.
Determine the empirical formula of this compound.
Step 1) convert %  gram
Step 2) gram  moles
  1.1904 mol Na
x mol Na  27.38 g Na1mol Na
/ 1.19 mol = 1 Na
23 g Na 

  1.19 mol H / 1.19 mol = 1 H
x mol H  1.19 g H1mol H

1
g
H


  1.1908 mol C / 1.19 mol = 1 C
x mol C  14.29 g C1mol C

12
g
C



  3.5712 mol O / 1.19 mol = 3 O
x mol O  57.14 g O1mol O

16
g
O


Step 3) mol / mol
NaHCO3
Empirical
&
Molecular
Formula
(contains only hydrogen + carbon)
(~17% hydrogen)
A 175 g hydrocarbon sample is analyzed and found to contain ~83% carbon.
The molar mass of the sample is determined to be 58 g/mol.
Determine the empirical and molecular formula for this sample.
Determine the empirical formula of this compound.
Step 1) convert %  gram
Assume sample is 100 g.
Then, 83 g carbon and 17 g hydrogen.
Step 2) gram  moles
  6.917 mol C
x mol C  83 g C1mol C

/ 6.917 mol = 1 C
12
g
C


  17 mol H / 6.917 mol = 2.5 H
x mol H  17 g H1mol H

1
g
H


(2.4577 H)
2 C @ 12 g = 24 g
5H@ 1g = 5g
29 g
MMempirical = 29 g/mol
CH2.5
C2H5
MMmolecular = 58 g/mol
Step 3) mol / mol
58/29 = 2
Therefore 2(C2H5) = C4H10
butane
Common Mistakes when
Calculating
Formula
Given:
Compound consists Empirical
of 36.3 g Zn and 17.8
g S.
Find: empirical formula
36.3 g Zn
17.8
= 2 Zn
Zn2S
17.8 g S
36.3 g Zn
17.8
= 1S
1 mol Zn
65.4 g Zn
17.8 g S
1 mol S
32.1 g S
1
1
= 0.555 mol Zn
0.555 mol
= 0.555 mol S
0.555 mol
Chemical formula
indicates MOLE ratio,
not GRAM ratio
Zn
ZnS
S
zinc sulfide
VSEPR
Valence Shell Electron Pair Repulsion.
 Predicts
three dimensional geometry of
molecules.
 Name tells you the theory.
 Valence shell - outside electrons.
 Electron Pair repulsion - electron pairs try
to get as far away as possible.
 Can determine the angles of bonds.
VSEPR
 Based
on the number of pairs of valence
electrons both bonded and unbonded.
 Unbonded pair are called lone pair.
 CH
4 - draw the structural formula
 Has
4 + 4(1) = 8
 wants 8 + 4(2) = 16
 (16-8)/2 = 4 bonds
H
H C H
H
VSEPR
 Single
bonds fill all
atoms.
 There are 4 pairs of
electrons pushing
away.
 The furthest they can
get away is 109.5º.
Vsepr theory and molecular
molecule
geometry
Type of
Formula
example
0
𝐴𝐵2
Be𝐹2
3
0
𝐴𝐵3
B𝐹3
Bent or
angular
2
1
𝐴𝐵2 E
ONF
Tertrahedr
al
4
0
𝐴𝐵4
C𝐻4
Trigonalpyramidal
3
1
𝐴𝐵3 E
𝑁𝐻3
Bent or
angular
2
2
𝐴𝐵2 𝐸2
𝐻2 O
Trigonalbipyramid
al
5
0
𝐴𝐵5
𝑃𝐶𝑙5
Molecular
shape
Atoms
bonded to
central
atom
Lone pairs
of electrons
Linear
2
Trigonalplanar
 Hybridization
– mixing of 2 or more atomic
orbitals of similar energies on the same
atom to produce new hybrid atomic
orbitals of equal energies
Atomic Orbitals
Type of
hybridization
Number of
hybrid orbitals
S, p
sp
2
S, p, p
𝑠𝑝2
3
S, p, p, p
𝑠𝑝3
4
The Shapes of Some Simple ABn Molecules
The VSEPR Model
SO2
..
O
N
S
O
C
O
O
Linear
O
Bent
F
S
O
F
F
O
Trigonal
planar
Trigonal
pyramidal
AB6
F
F
F
Cl
F
F
T-shaped
F
F
F
Square
planar
Brown, LeMay, Bursten, Chemistry The Central Science, 2000, page 305
F
F
P
Xe
F
F
F
S
F
F
F
F
F
Trigonal
bipyramidal
Octahedral
(by mass...not atoms)
Percentage Composition
24.305
35.453
Mg
Cl
12
17
magnesium
chlorine
partg
24
% Mg
% == whole
100
95 g xx 100
25.52% Mg
Mg2+ Cl1MgCl2
It is not 33% Mg and 66% Cl
74.48% Cl
1 Mg @ 24.305 amu = 24.305 amu
2 Cl @ 35.453 amu = 70.906 amu
95.211 amu
Find the molar mass and percentage composition of zinc acetate
Zn2+ CH3COO1acetate = CH3COO1-
Zn(CH3COO)2
1 Zn @ 65.4 g/mol = 65.4 g
/ 183.4 g x 100% = 35.6 % Zn
4 C @ 12 g/mol
= 48 g
/ 183.4 g x 100% = 26.2 % C
6 H @ 1 g/mol
=
/ 183.4 g x 100% = 3.3 % H
4 O @ 16 g/mol
= 64 g
Zn(CH3COO)2
6g
183.4 g
/ 183.4 g x 100% = 34.9 % O
A compound is found to be 45.5% Y and 54.5% Cl.
Its molar mass (molecular mass) is 590 g.
Assume a 100 g sample size
a) Find its empirical formula
45.5 g Y
1 mol Y
= 0.5118 mol Y
/ 0.5118 mol
88.9 g Y
=1Y
YCl3
54.5 g Cl
1 mol Cl
= 1.535 mol Cl
35.5 g Cl
/ 0.5118 mol
= 3 Cl
1 Y @ 88.9 g/mol = 88.9g
b) Find its molecular formula
590 / 195.4
3 Cl @ 35.5 g/mol = 106.5 g
=3
3 (YCl3)
YCl3
Y3Cl9
195.4 g