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Chem 20 Final Review Chemical Bonding

Transferring of electrons (ionic)

Sharing of electrons (molecular)

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Neil Bohr’s model of the atom:

Valence Electrons: • 2, 8, 8, 18

These are the electrons in the highest • The number of energy levels equals the
energy level around the nucleus. They are period number
found in the outer most energy level of an
• The number of valence electrons is equal
atom.
to the last digit of the group number

• valence electrons are involved in the

chemical reaction.
• Orbitals may be occupied by 0, 1 or
• They are either going to be exchanged, 2 electrons, but no more than 2
so as to form an ionic bond electrons

• Or they are going to be shared, so as to

form a covalent/molecular bond

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Rules for drawing Lewis Dot Diagrams:

• lone pair: contains 2 electrons • Write out the atomic symbol

• bonding electrons: electrons that • place a dot on the left side of the
are not paired atomic symbol

• Continue placing dots around the

symbol in a clockwise order

Has three
– 3e– – Electronegativity gradually increases
Valence
– 8e– – Electrons as we move from left to right in a period
– 2e– – These are
bonding
13 p+ electrons

Al And the Electronegativity decreases

Al as we move down the group

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magnesium oxide
Ionic Bonds difference between the Mg O
electronegativities are quite large

[Mg]2+ [ O ]2–

calcium fluoride Ionic bonds are very strong bonds and

thus require a large amount of energy
Ca F to break them apart.
F This explains why ionic compounds:
•Are solid at room temperature
[Ca]2+ 2x[ F ]1– •Have high melting and boiling
points

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A Covalent Bond is formed when both • The atom with the most possible
electronegativities are high. bonding electrons becomes the
Central Atom
The two atoms end up “sharing” the
bonding electrons

Let’s look at Cl2

Let’s look at O2 (O has 2 bonding e–)

sharing of sharing of
electrons electrons
forms a single bond forms a double bond
Can also be written as: Cl–Cl Can also be written as: O=O

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Three hydrogen atoms are bonded to one nitrogen VSEPR stands for:
atom. This compound is ammonia.
Valence Shell Electron Pair Repulsion
N 3x(H )We should have 8
electrons in total

H NH We have 8
electrons in total
H

According to VSEPR theory

Double and triple bonds can be
 Only the valence electrons of the treated just as single bonds for
central atom are important in predicting shapes using VSEPR theory
determining the molecule’s shape

 Bonded pairs and lone pairs of

electrons are treated approximately
equally

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A solid line represents a bond in

the plane of the page

A dashed line represents a bond

behind the plane of the page

A wedged or triangle represents

a bond in front of the plane of AX2 AX4
AX3 AX3E AX2E2
the
page

If there is unequal sharing of the

electrons the bond is considered to be
a polar covalent bond δ+ H Cl δ−
2.2 3.2

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Non-Polar Covalent Molecule: carbon dioxide

 CO2 (g) 
dipoles cancel each other out
(net dipole of zero) O C O

methane
H  CH 4 (g)  Molecules that are symmetrical are
2.2

HCH non-polar because their dipole moments

H 2.2
2.6
2.2 balance.
2.2

Here we can see that the dipoles cancel each

other out, so CH4 is considered a non-polar
compound

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Polar molecules: water

The bond dipoles do not cancel out H2O(l)
leaving a non-zero net dipole OH
O
H
Here we can see that the dipoles do not cancel
each other out, so water is considered a polar
compound

ammonia
δ–
NH3(g) N
HNH N 3.0 δ+
H
δ+ When in comes to solubility
remember that like dissolves like
H H
2.2 2.2 H δ+
H 2.2
H
Here the dipoles do not cancel each other out, so
H
ammonia is considered a polar compound

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London Dispersion forces The strength of the London force is:

directly related to the number of
Dipole-Dipole forces electrons in the molecule

Hydrogen bonding

Dipole–Dipole Forces Hydrogen Bond:

These forces are only found in polar For hydrogen bonding to occur, it must
compounds be bonded to a highly electronegative
The strength of the dipole–dipole atom: nitrogen, oxygen or fluorine.
force is dependent on the overall
polarity of the molecule

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hydrogen bond (intermolecular)

Ranking the intermolecular bonds from

covalent bond
+ strongest to weakest we get:
(intramolecular) 1. Hydrogen bonding
– 2. Dipole-dipole bonding
3. London dispersion bonding
covalent bond Remember if it is an ionic compound it will have a higher
boiling point
(intramolecular)

Comparing Kelvin and

Celsius Scales
• To convert degrees Celsius to Kelvin , you add
273.
K = °C + 273
Gases
• To convert Kelvin to degrees Celsius, you subtract
273.
°C = K - 273

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Atmospheric Pressure Gas Laws

• They are based on the temperature, pressure and
volume relationships that all gases have in common
Standard Temperature and Pressure (STP)
= 101.325 kPa and 0°C
1. Boyle’s Law P1V1 = P2V2

Standard Ambient Temperature and Pressure (SATP) 2. Charles’ Law V1 = V2

T1 T2
= 100 kPa and 25°C

3. Combined Gas Law P1V1 =P2V2

T1 T2

Example: Law of Combining Volumes Molar Volume

Use the law of combining volumes to predict the volume of oxygen required for the
complete combustion of 120 mL of butane gas from a lighter. – At STP, molar volume = 22.4 L/mol

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) – At SATP, molar volume = 24.8 L/mol

120 mL V=?

VO2: 120 ml C4H10 x ( 13 mL O2) = 780 mL

2 mL C4H10

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IDEAL GAS LAW

Calculate the volume occupied by 0.024
mol of carbon dioxide at SATP – IDEAL GAS – does not really exist
Follows all gas laws perfectly under all conditions
• Does not condense when cooled
• Assumes that the particles have no volume and are not attracted
to each other

– REAL GAS – does not follow gas laws exactly, it deviates at

low temperatures and high pressures
• Condenses to liquid or sometimes solid when cooled or under pressure
• Particles are attracted to each other and have volume
• Behaves like an ideal gas at higher temperatures and lower pressures

Using the Ideal Gas Law

Example One: What mass of neon gas should be
introduced into an evacuated 0.88L tube to
produce a pressure of 90 kPa at 30°C?

PV = nRT
Solutions
n = PV
RT

n = (90kPa)(0.88L)
(8.314 L•kPa/mol•K)(303K)

n = 0.0314 mol x 20.18 g = 0.63 g

1 mol

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Dissociation vs. Ionization SUMMARY

• What is the difference between dissociation and Substance Process General Equation
ionization? Molecular Disperse as individual, XY (s/l/g)  XY (aq)
neutral molecules
• Both produce (aq) ions... Ionic Dissociate into individual MX (s)  M+ (aq) + X-(aq)
ions
• Dissociation, however, is the separation of ions Base (ionic hydroxide) Dissociate into positive ions MOH (s)  M+ (aq) + OH-(aq)
that already exist before dissolving in water and hydroxide ions
HX (s/l/g)  H+ (aq) + X-(aq)
M+X- (s)  M+ (aq) + X-(aq) Acid Ionize to form hydrogen
ions and anions
Reference pg. 201
• Ionization involves the production of new ions,
specifically hydrogen ions
HX0 (aq)  H+ (aq) + X-(aq)

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• Amount Concentration (aka Molar

Concentration) (mol/L) • Al2(SO4)3 (aq)  2 Al 3+(aq) + 3 SO42- (aq)
C = n solute (mol) • c = 0.30 mol/L
Vsolution (L)
• [Al 3+(aq)] = 0.30 mol/L x (2) = 0.60
mol/L 1

• [SO42- (aq) ] = 0.30 mol/L x (3) = 0.90

mol/L 1

Determining the volume of Stock Solution for

a Standard Solution
dilution formula:
C1V1 = C2V2
•Example: How would you prepare 100 mL of 0.40 mol/L MgSO4(aq) from
a solution of 2.0 mol/l MgSO4(aq)

C 1 V1 = C 2 V2
(2.0mol/L) (V1) = (0.40mol/L) (100mL)
V1 = 20 mL

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Empirical Definitions
Acid – a substance which dissolves in water to produce a solution that:
– Tastes sour
– Turns blue litmus red
– Conducts electricity
–
Acids and Bases –
Reacts with active metals to produce H2(g)
Neutralizes Bases
Base – a substance which dissolves in water to produce a solution that:
– Tastes bitter; feels slippery
– Turns red litmus blue
– Conducts electricity
– Neutralizes acids

Theoretical Definitions Theoretical Definitions

a) Arrhenius: b) Modified Definition:

Acid – a substance that forms an acidic solution by dissolving in Acid – a species that forms an acidic solution by reacting with water
water to produce free hydrogen ions (H+(aq)) in solution to produce hydronium ions (H3O+(aq))

– Example: HCl (aq)  H+ (aq) + Cl- (aq) – Example: HCl (aq) + H2O(aq) H3O+(aq) + Cl- (aq)

Base – a substance that forms a basic solution by dissolving in Base – a species that forms a basic solution by reacting with water
water to produce free hydroxide ions (OH-(aq)) in solution to produce hydroxide ions (OH-(aq))

– Example: NaOH(aq)  Na+ (aq) + OH– (aq) – Example: NH3 (aq) + H2O(aq) NH4+(aq) + OH–(aq)

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Summary Acid – Base Indicators

Common name Color of pH range of colour Color of In-
pH = -log [H3O+(aq)] [H3O+(aq)] =10 –pH HIn(aq) change (aq)
Bromothymol blue Yellow 6.0-7.6 Blue
pOH = -log [OH -(aq)] [OH -(aq)] =10 –pOH Phenolphthalein Colourless 8.2-10.0 Pink

• The number of digits following the decimal point in a pH or pOH

value is equal to the number of significant digits in the
corresponding hydronium or hydroxide concentration.

• For both pH and pOH, an inverse relationship exist between the

ion concentration and the pH or pOH. The greater the hydronium
ion concentration, the lower the pH is.

More Practice.... Try Three More...

• HI(aq) – explain acidic properties • H3PO4(aq) – explain acidic properties
– HI(aq) + H2O(l)  H3O+(aq) + I- (aq) – H3PO4(aq) + H2O(l)  H3O+(aq) + H2PO4- (aq)

• NaCH3COO(aq) – explain basic properties • Na2SO4 (aq) – explain basic properties

– NaCH3COO(aq)  Na+ (aq) + CH3COO-(aq) Simple Dissociation – Na2SO4 (aq)  2 Na+ (aq) + SO42-(aq) Simple Dissociation

– CH3COO-(aq) + H2O(l)  CH3COOH (aq) + OH-(aq) – SO42-(aq) + H2O(l)  HSO4 -

(aq) + OH-(aq)

• HOCl(aq) – explain acidic properties • Sr(OH)2(aq) – explain basic properties

– HOCl + H2O(l)  H3O+(aq) + OCl- (aq) – Sr(OH)2(aq)  Sr2+(aq) + 2 OH- (aq)

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The Difference: Using the The Difference: Using the

Modified Arrhenius Theory Modified Arrhenius Theory
• Weak Acids: have low conductivity, a lower rate of reaction
w/ active metals and carbonates and a relatively high pH
• Strong Acids: have high conductivity, high rate of reaction w/
metals and carbonates and a relatively low pH
• Based on this evidence, a weak acid reacts incompletely
(<50%) with water to form relatively few hydronium ions
>99%
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq) <50%
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)

Strong and Weak Bases

• Strong Bases – all soluble ionic hydroxides that
dissociate completely (>99%) to release
hydroxide ions
NaOH(s)  Na+(aq) + OH-(aq)
Stoichiometry
• Weak Bases – an ionic or molecular substance
that reacts partially (<50%) with water to
produce relatively few hydroxide ions
<50%

NH3 (aq) + H2O(aq) OH–(aq) + NH4+(aq)

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What do you remember? Using the solubility table:

In Science 10 you learned about five reaction types, can you match them up

 Composition (Formation)  CH4(g) + O2(g)  CO2(g) + H2O(g)

 Decomposition  Mg(s) + O2(g)  MgO(s)

 Combustion  Cu(s) + AgNO3(aq)  Ag(s) + Cu(NO3)2(g)

 Single Replacement  CaCl2(aq) + Na2CO3(aq)  CaCO3(s) +

NaCl(aq)
 Double Replacement
 H2O(l)  O2(g) + H2(g)

Practice Limiting and Excess Reagents

• Write the net ionic equation for the reaction of aqueous barium chloride and • When no further changes appear to be occurring, we assume that all of the AgNO3(aq)
aqueous sodium sulfate. (Refer to the solubility table)
that was initially present has now been completely reacted.

• 1) BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)

• A limiting reagent is the reactant whose entities are completely consumed in a
reaction, meaning the reaction stops.
• 2) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq)  BaSO4(s) + 2Na+(aq) + 2Cl-(aq)
– In order to make sure this happens, more of the other reactant must be present
(Complete ionic equation)
than is required
• 3) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq)  BaSO4(s) + 2Na+(aq) + 2Cl-(aq)
• An excess reagent is the reactant whose entities are present in surplus amounts

• 4) Ba2+ (aq)) + SO42-(aq)  BaSO4(s) (Net ionic equation)

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Stoichiometry Calculations Practice #3 (Mass Stoichiometry)

What mass of iron (III) oxide is required to
produce 100.0 g of iron?
(Measured quantity)
solids/liquids m  n
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
mole
m=? m = 100.0g
ratio
M = 159.70g/mol M = 55.85 g/mol
solids/liquids m  n
(Required quantity)
m Fe2O3(s): 100.0 g x 1 mol x 1 mol x 159.70 g = 143.0 g Fe2O3
55.85 g 2 mol 1 mol

Percent Yield for Reactions Percent Yield Example #1

 Example: In a chemical analysis, 3.00 g of silver nitrate in
– Percent yield = actual yield x 100
predicted yield solution was reacted with excess sodium chromate to produced
2.81 g of filtered, dried precipitate.

 Predicted value: 2AgNO3(aq) + Na2CrO7(aq) Ag2CrO7(s) +2NaNO3(aq)

m = 3.00 g m=
M = 169.88g/mol M = 331.74 g/mol

3.00g x 1 mol x 1 mol x 331. 74 g = 2. 93 g

169.88g 2 mol 1 mol

 Percent yield = actual yield x 100% = 2.81g x 100% = 95.9%

predicted yield 2.93g

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Gas Stoichiometry Gas Stoichiometry

• If 300g of propane burns in a gas barbecue, • What volume of ammonia at 450kPa and 80oC can be obtained from the
what volume of oxygen measured at SATP is complete reaction of 7.5kg of hydrogen with nitrogen?

required for the reaction? 2N2(g) + 3H2(g)  2NH3(g)

• Remember: 24.8L/mol for SATP m = 7500g V=?

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) M = 2.02 g/mol P = 450kPA

T = 353.13K
m = 300g V=?
44.11g/mol 24.8L/mol 7500 g x 1 mol x 2 = 2475.2475 mol NH3(g)
2.02 g 3
300 g x 1 mol x 5 mol x 24.8 L = 843 L O2(g)
44.11 g 1 mol 1 mol
PV = nRT  V = nRT = (2475.2475 mol)(8.314kpa•L/mol•K)(353.15K)
P (450kPa)

= 16150.10L  1.6 x 104 L of NH3(g)

Solution Stoichiometry
 In an experiment, a 10.00 mL sample of sulfuric acid solution reacts
completely with 15.9 mL of 0.150 mol/L potassium hydroxide. Calculate
the amount concentration of the sulfuric acid.

 H2SO4(aq) + 2KOH(aq)
V = 10.00mL V = 15.9 mL
 2H2O(l) + K2SO4(aq)
Chemical Analysis
c=? 0.150 mol/L

15.9mL x 0.150 mol x 1mol x 1 = 0.119 mol/L

1L 2 mol 10.0 mL

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2) If 10.0g of copper is placed in solution of 20.0g of silver

nitrate, which reagent will be the limiting reagent?
Chemical Analysis by Titration
 All reactants must be converted to moles, then using the mole
ratio, determine which reactant will run out first. • Titration – is a common
experimental design used to
Cu(s) + 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq) determine the amount concentration
of substances in solution.
10.0g 20.0g – The solution of known concentration
given: 0.157 mol 0.0589 mol may be either the titrant or the
sample; it makes no difference to the
analysis
• Titration breakdown:
n Cu(s): 10.0g x 1 mol = 0.157 mol x 2 = 0.315 mol
63.55 g 1 – Carefully adding a solution (titrant)
from a burette into a measured, fixed
n AgNO3: 20.0g x 1 mol = 0.118 mol x 1 = 0.0589 mol volume of another solution (sample) in
169.88 g 2 an Erlenmeyer flask until the reaction
is judged to be complete

Chemical Analysis by Titration Chemical Analysis by Titration

• Burette – precisely marked glass cylinder with a
stopcock at one end. Allows precise, accurate • An initial reading of the
measurement and control of the volume of burette is made before any
reacting solution. titrant is added to the sample.
• Then the titrant is added until
• When doing a titration, there will be a point at the reaction is complete;
which the reaction is complete; when chemically when a final drop of titrant
equivalent amounts of reactants have combined. permanently changes the
This is called the equivalence point: colour of the sample.
– Equivalence point – the point during a • The final burette reading is
titration at which the exact theoretical then taken.
chemical amount of titrant has been added to • The difference between the
the sample. (QUANTITATIVE) readings is the volume of
titrant added.
• To measure this equivalence point experimentally,
we look for a sudden change in an observable
property, such as color, pH, or conductivity. This
is called the endpoint. (QUALITATIVE)
Near the endpoint, continuous gentle swirling of the solution is important

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Sample Problem
• Determine the concentration of hydrochloric acid in a commercial solution.
• A 1.59g mass of sodium carbonate, Na2CO3(s), was dissolved to make 100.0mL of
solution. Samples (10.00mL) of this standard solution were then taken and
titrated with the hydrochloric acid solution.

• The titration evidence collected is below. Methyl orange indicator was used.

Trial 1 2 3 4
Final burette reading (mL) 13.3 26.0 38.8 13.4
Initial burette reading (mL) 0.2 13.3 26.0 0.6
Volume of HCl(aq) added 13.1 12.7 12.8 12.8
Indicator colour Red Orange Orange Orange

TIP: In titration analysis, the first trial is typically done very quickly. It is just for practice, to learn
what the endpoint looks like and to learn the approximate volume of titrant needed to get to the
endpoint. Greater care is taken with subsequent trials

General
Rule
Strong Acid to
Weak Base: pH
at equivalence
point is always
lower than 7

Strong Base to
Weak Acid: pH
at equivalence
point is always
higher than 7

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