To: 2006-2007 AP Chemistry students
From: Past students of Dr.Kumar
SUBJECT: Hints / strategies to survive AP Chemistry.
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Think categorically: know acid from base, strong from weak, metal from
nonmetal, ionic from covalent, etc.
Know your nomenclature, you will need it for everything.
When you are taught something, learn it: many concepts are reused within
other concepts so if you didn’t learn something in the first place it will hurt you
later.
Know your solubility rules: you will be surprised how much you will actually
use these rules to answer questions in other topics, like electrochemistry, for
example.
Know all of the predicting product templates: Learn them immediately after
they are taught to you. These can get you a lot of points on tests and they are just
memorizing simple templates and being able to categorize reactants.
Learn the details: How does temperature, pressure, etc. affect a certain system?
What does it mean if something is solid, liquid or gas? What are certain numbers
dependent upon?
Knowing the following concepts when entering AP Chem will be helpful in the first
quarter and throughout the year.
NOMENCLATURE
nitrate
nitrite
perchlorate
chlorate
chlorite
hypochlorite
acetate
hydroxide
permanganate
azide
chromate
dichromate
sulfate
sulfite
hydrogen sulfate
hydrogen sulfite
thiosulfate
thiocyanate
cyanide
oxalate
NO3 -1
NO2 -1
ClO4 -1
ClO3 -1
ClO2 -1
ClO -1
C2H3O2 -1 or
CH3COO-1
OH -1
MnO4 -1
N3 -1
CrO4 -2
Cr2O7 -2
SO4 -2
SO3 -2
HSO4 -1
HSO3 -1
S2O3 -2
SCN -1
CN -1
C2O4 -2
carbonate
hydrogen carbonate or
bicarbonate
silicate
phosphate
phosphite
hydrogen phosphate
dihydrogen phosphate
arsenate
arsenite
borate
peroxide
ammonium
hydronium
CO3 -2
HCO3 -1
SiO3 -2
PO4 -3
PO3 -3
HPO4 -2
H2PO4 -1
AsO4 -3
AsO3 -3
BO3 -3
O2 -2
NH4 +1
H3O+
Knowing your nomenclature is essential to doing well in AP Chem.
OXIDATION NUMBERS
The oxidation number of an element indicates the number of electrons lost, gained, or
shared as a result of chemical bonding.
When determining oxidation numbers remember these rules:
 The oxidation number of an atom in the elemental state is zero
Ex. I2 and Al both have an oxidation number of zero
 The sum of the oxidation numbers of all atoms in a molecule add up to zero.
 The sum of the oxidation numbers of all the atoms in a polyatomic ion must add
up to the charge of the ion
The following elements usually have fixed oxidation numbers:
 Alkali metals: +1
 Alkaline earth metals: +2
 Group 3A metals: +3
 Oxygen: -2, ( O -2 )
 Hydrogen: +1, ( H + )
 Zinc and Cadmium: +2, (Zn+2 Cd +2)
 Silver: +1 (Ag +)
Mercury: Hg2+2 Hg+2
Copper: Cu+1 Cu+2
Iron:
Fe+2 Fe+3
Tin:
Sn+2 Sn+4
Lead:
Pb+2 Pb+4
Examples:
1. What is the oxidation number of carbon in the compound carbon dioxide ( CO2 )?
What you know: The oxidation numbers of the elements in the compound must add up to
zero and the oxidation number of oxygen is known as -2.
You can solve for the oxidation number of carbon in an equation with carbon as X.
1. X + (-2)2 = 0
2. X – 4 =0
3. X = +4 … the oxidation number of carbon is +4 in CO2
2. What is the oxidation number of sulfur in the polyatomic ion sulfate (SO4 -2 )?
What you know: The oxidation numbers of the elements in the polyatomic ion must add
up to the charge of the ion (-2) and the oxidation number of oxygen is -2.
You can solve for the oxidation number of sulfur in an equation with sulfur as X.
1. X + 4(-2) = -2
2. X – 8 = -2
3. X = + 6…. The oxidation number of sulfur + 6 in sulfate.
Diatomic Molecules: H2, N2, O2, F2, Cl2, Br2, I2
Categorizing is very important in AP Chem. You must be able to look at a
compound and classify it. Is it ionic or covalent? Is it soluble? Is it an acid or a base?
If so is it weak or strong?
SOLUBILITY RULES
A compound is soluble if it ionizes completely in water.
You must know your solubility rules to do well in AP Chem!
 Salts of one of the following are always soluble:
nitrates (NO3 -1), acetates (C2H3O2 -1), chlorates (ClO3 -1), perchlorates (ClO4 -1)
Group 1A Alkali metal ions (Na+, K+, etc), ammonium (NH4 + )
 Salts of the following are usually soluble but have exceptions:
chlorides (Cl -1), bromides (Br -1), and iodides (I -1), are always soluble except for
those of Ag+, Pb+2 , and Hg2+2
sulfates (SO4-2) are always soluble except for those of Ag+, Pb+2 , Hg2+2, Ba+2, Ca+2
and Sr+2
All other compounds can be considered insoluble.
Considering these solubility rules we can predict the products of precipitation reactions.
An insoluble, solid ionic compound is known as a precipitate.
What would happen if we added a solution of NaCl to a solution of AgNO3?
According to the solubility rules both NaCl and AgNO3 are soluble, meaning that they
completely ionize in solution. When we add these two solutions together we are really
adding the following ions together:
Na+ + Cl- + Ag++ NO3-
The only possible new compounds that could form from the addition of these ions would
be AgCl or NaNO3. According to solubility rules, AgCl is insoluble while NaNO3 is
soluble.
The addition of these two solutions can therefore be written as:
Na+ + Cl- + Ag++ NO3-  AgCl + Na+ + NO3The remaining ions which do not precipitate but remain on both sides of the reaction can
be considered spectator ions and are left out of the final equation.
The spectator ions here are Na+ and NO3- The net ionic equation, or the equation without
spectator ions, is therefore: Ag+ + Cl-  AgCl
PREDICTING PRODUCTS
The following are not balanced.
Decomposition reactions (usually occur as a result of heating):
1. metal carbonate  metal oxide + carbon dioxide (CO2 )
ex. CaCO3 CaO + CO2
2. metal chlorate  metal chloride + oxygen (O2)
ex. KClO3 KCl + O2
3. metal oxide metal + oxygen (O2)
ex. MgO  Mg + O2
ACIDS AND BASES
3 different definitions of acids and bases:
Acid
H+ donor
Arrhenius
proton donor
Bronsted-Lowry
e- pair receiver
Lewis
Base
OH -1 donor
proton acceptor
e- pair receiver
Strong acids and bases completely ionize in water while weak acids and bases only
partially ionize in water.
Seven Strong Acids
HCl
HBr
HI
HClO3
HClO4
HNO3
H2SO4
If an acid is not one of these seven you know that it is not strong!
Strong Bases
Alkali metal hydroxides, Ca(OH)2, Sr(OH)2, Ba(OH)2
Weak Bases follow the formula:
NxHy ( ex. NH3 ) or CvHwNxHy ( ex. CH3NH2 )
ELECTROLYTES
Strong electrolytes: Compounds which ionize completely in water: strong acids, strong
bases and soluble salts
Weak Electrolytes: Compounds which ionize only slightly in water: weak acids and
weak bases
Non-electrolytes: Compounds which do not produce ions when dissolved in water:
non-soluble salts, sugars, alcohols.
VSEPR ( valence-shell electron-pair repulsion) is a method for determining the
molecular geometry of a molecule based on the idea that pairs of valence electrons in
bonded atoms repel one another.
Using this theory the following geometries are predicted for varying numbers of valence
electrons in a molecule
Molecular
geometry
Linear
Trigonal
Planar
Angular
Tetrahedral
Trigonal
pyramidal
Angular
Trigonal
bipyramidal
Seesaw
T-Shaped
Linear
Octahedral
Square
Pyramidal
Square
Planar
Number of
Bonded
electron
pairs
2
3
Number of
lone
electron
pairs
0
0
Ideal
Bond
angle
Always
polar?
Example
180°
120°
No
No
BeCl2
BF3
2
4
3
1
0
1
120°
109.5°
109.5
Yes
No
Yes
SO2
CH4
NH3
2
5
2
0
Yes
No
H2O
PCl5
4
1
Yes
SF4
3
3
6
5
2
3
0
1
109.5
90°,
120°,180°
90°,
120°,180°
90°, 180°
180°
90°, 180°
90°
Yes
No
No
Yes
ClF3
XeF2
SF6
BrF5
4
2
90°
No
XeF4
Polarity
VSEPR is important for chemistry because it determines the POLARITY of the
molecule. On multiple choice section of the AP Chem exam, it is not uncommon to be
given a list of compounds/molecules and asked to determine its polarity.
A molecule is polar if it contains lone pairs which do not cancel each other out such as in
a square planar molecule, or if atoms surrounding the central atom are different. If an
atom is polar it is said to have a dipole moment.
The polarity of two compounds can be used to determine their solvency in one another.
Generally, the rule “like dissolves like” can be used to to determine whether one
substance will dissolve in another. This means that a polar solute will dissolve in a polar
solvent, but not in a non-polar solvent and vice-versa.
Ex. Ethanol, an alcohol, is a polar substance and therefore dissolves in water which is
also polar. Methane (CH4 ) will not dissolve in water because it is nonpolar.
Polarity is also important in determining intermolecular forces.
SIGNIFICANT FIGURES
Rules:
1. Non-zero digits are always significant: 2.345 has four sig-figs
2. Zeroes within a number are always significant: 10.101 has five sig-figs
3. Zeroes which come after the decimal place but before(to the left of) any non-zero
digit are not significant: 0.00063 has two sig-figs, 6 and 3
4. Zeroes which come after(to the right of) any non-zero digits after a decimal place
are significant: 0.0006300 has four sig-figs, 6, 3, 0, 0
5. Scientific notation
Atlantic/Pacific rule for determining sig figs
This is based on the fact that when looking at a map of the US the Atlantic Ocean is on
the right and the Pacific Ocean is on the left.
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If a decimal point is Absent then start counting digits from the Atlantic side of a
number(the right). Start with the first non-zero digit and count towards the left.
All counted digits are significant.
If a decimal point is Present then start counting digits from the Pacific side of a
number(the left). Start with the first non-zero digit and count all digits towards the
right. All counted digits are significant.
Multiplication/Division of Sig-figs
 When multiplying or dividing two numbers, the answer should contain the
amount of sig-figs of the number with the least amount of sig-figs: 25(2 sig figs)
X 10.25 (4 sig figs)=260(2 sig-figs)
Addition/Subtraction
 When two numbers are added or subtracted the answer will contain no more
decimal places than the number with the lowest amount of decimal places:
25.1(1 decimal place) +25.12(2 decimal places)=50.2(1 decimal place)
PERIODIC TABLE TRENDS
Moving left  Right
 Atomic radius decreases
 Ionization energy increases
 Electronegatvity increases
Moving top bottom
 Atomic radius increases
 Ionization energy decreases
 Electronegativity decreases
STOICHIOMETRY
Chemical calculations are often carried out using stoichiometry which takes into account
the mole ratios in chemical reactions to come up with a series of conversion factors for
solving problems.
Consider the reaction:
C3H8 + 5O2  3CO2 + 4 H2O
Let’s say we wanted to find out how many moles of CO2 five moles of C3H8 would
produce when reacted in excess O2. First we need to look at the mole ratio. The mole
ratio in the reaction is 1:5:3:4. These numbers are the balanced coefficients of the moles
of reactants and products. This means that for every 1 mole of C3H8 burned in 5 moles of
O2 , 3 moles of CO2 and 4 moles of H2O will be produced. This ratio will always remain
the same for this reaction no matter how many moles of each reactant are present. If 5
moles of C3H8 are present, they will require 25 moles of O2 to be completely consumed
because of the1:5 mole ratio between the reactants.
Mole ratio is one of the most important things to remember when doing math in
chemistry. Forgetting it is one of the most common mistakes.
In order to find out how many moles of CO2 five moles of C3H8 would produce when
burned in excess O2 we can look at the mole ratio and see that for everyone 1 mole of
C3H8, 3 moles of CO2 will be produced.
In stoichiometry we multiply the number we are given by this conversion factor to
determine our answer.
5 moles C3H8
3 moles CO2
1 mole C3H8
=
15 moles CO2
If we want to know how many grams of CO2 would be produced when 5 moles of C3H8 are
burned in excess O2, we can just multiply by the conversion factor 1 mole of CO2 is 44
grams to arrive at an answer.
5 moles C3H8
3 moles C3H8 44 grams CO2 = 660 grams CO2
1 mole C3H8
1mole CO2
In stoichometric calculations all units should cancel out so that the unit that you want your
answer to be in is all that remains. In the above problem moles of C3H8 and moles of
CO2 both cancel out across the division line, leaving grams of CO2 as the only
remaining unit, which is what our answer should be in .
Limiting Reagent
The limiting reactant or limiting reagent in a reaction is the reactant which is
completely consumed, thus limiting the formation of more products.
Consider the combustion reaction: 2 C2H6 + 7 O2  4 CO2 + 6 H2O
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
If two moles of C2H6 were allowed to react with seven moles of – then ideally
both reactants would be completely consumed because they are in proportion with
each other for the amount needed for the reaction.
If two moles of C2H6 were allowed to burn in excess ---, once all the – was
consumed the reaction would stop, yet – would still remain. --- would therefore be
the limiting reactant.
A limiting reagent can be understood in terms of everyday things.
Lets say 2 wheels and one handle bar are needed to complete a bicycle.
You have 10 wheels and seven handle bars, how many bicycles can you make?
Which is the limiting reagent, the wheels or the bicycle? How much of the other one will
remain in excess?
We can use factor-labeling:
10 wheels 1 bicycle = 5 bicycles
2 wheels
7 hand. bars
1 bicycle
1 hand.bar
=
7 bicycles
The amount of product made will always be the smaller number. Once five bicycles are
made all the wheels have been used up, and therefore no more bicycles can be made. The
wheels are the limiting reagent because they limit bicycle production to five bicycles.
Five handle bars would be used in the five bicycles and two handle bars would therefore
be in excess.
As you can see, the limiting reagent is not always the reactant which you have less of.
There were more wheels present, but more were needed than handles bars to make the
bike so they were the limiting reagent.
Limiting reagent problems can also be calculated when grams of each substance are
given.
2 C2H6 + 7 O2  4 CO2 + 6 H2O
How many grams of CO2 can be made through the burning of 15 grams C2H6 in 27 grams
O2?
15 g C2H6
1 mole C2H6
30.1 g C2H6
4 mole CO2
2 mole C2H6
44 g CO2
1 mole CO2
= 44 g CO2
27 g O2
1 mole O2
32 g O2
4 mole CO2
7 mole O2
44 g CO2
1 mole CO2
= 21 g CO2
Only 21 grams of CO2 can be made with the amounts of reactants available. O2 is the
limiting reagent because once 21 grams of CO2 are made all of the O2 will have been used
up.
How many grams of C2H6 are in excess?
There are many ways to determine this, but one is just using factor labeling.
21 g CO2
1 mole CO2
2 mole C2H6 30.1 g C2H6 = 7.2 g C2H6
44 g CO2
4 mole CO2
1 mole C2H6
7.2 g C2H6 is the number of grams of C2H6 which has been consumed in the production
of CO2. If we subtract this number from our original amount of C2H6 we will obtain the
amount of C2H6 left over.
15 grams C2H6 - 7.2 grams C2H6 = 7.8 grams C2H6 in excess.
Percent Yield
In ideal conditions, burning of 15 grams C2H6 in 27 grams O2 would produce 21 grams of
– as the math showed, representing the theoretical yield of the reaction. Yet in actual
conditions somewhat less – would be produced. The measured mass of – produced in the
reaction would be the actual yield of the --. In chemistry yields are often expressed as
percentages. The percent yield of a reaction is the actual yield divided by the theoretical
yield times 100, or {(actual yield)/(theoretical yield)}x 100= percent yield.
Lets say in this reaction that 16 grams of – was the actual yield, the percent yield for the
reaction would therefore be:
16 grams X 100= 76%
21 grams
Below is the previous AP CHEM YEAR IN REVIEW papers:
AP CHEM: Things to Know Before You Start the Year in AP Chem
Solubility Rules
Soluble
Group IA (Na+…)
NH4+
NO3-, ClO3-, ClO4-,
Cl-, Br-, Imost SO42-
Insoluble
Exceptions
none
none
none
+
Ag , Hg22+, Pb2+
Ba2+, Ca2+, Sr2+, Ag1+
Hg2+, Pb2+
most OHGroup IA, NH4+ and
Ba2+
most CO32-, PO43-, S2- Group IA, NH4+
Mostly everything
Group IA, NH4+
else not mentioned
here
Determine whether the following are soluble.
YES
NO
1. CaS
_____
_____
2. CaCO3
_____
_____
3. Pb(NO3) 2 _____
_____
YES
NO
4. NaCl
_____
_____
5. AgNO2
_____
_____
Common Polyatomic Ions
Ammonium
Acetate
NH4+
Chlorate
CH3COO- Perchlorate
Nitrate
NO3-
Nitrite
NO2-
Carbonate
Hydroxide
OH-
Sulfate
ClO3-
Peroxide
O2-2
ClO4- Chromate CrO4-2
Permanganate MnO4- Dichromate Cr2O7-2
CO3-2
Azide
SO4-2 Phosphate
N3PO4-3
Hypochlorite
ClO-
Sulfite
SO3-2
Arsenate
AsO4-3
Chlorite
ClO2-
Thiosulfate
S2O3-2
Arsenite
AsO3-3
Cyanide
CN-
Thiocyanate
SCN-
Borate
BO3-3
Bicarbonate
HCO3-
Bisulfate
HSO4-
Bisulfite
HSO3-
Sample Problems:
1. How many significant figures are there in the following numbers:
A. 209.4
B. 1.74 x 10-23
C. 0.00220
D. 0.087686
VSEPR : Valence Shell Electron Pair Repulsion
* VSEPR is used to determine the molecular shape of a given compound based on the
number of electron pairs around a given central atom.
** To draw the molecule you must first calculate the number of electrons. For
example, in the case of PCl3 Phosphorous has 5 electrons and Chlorine has 7. Since
there are 3 moles of Chlorine you must multiply the number of electrons by 3 (7 x 3 =
21). Add the two to get the total number of electrons (21 + 5 = 26). Therefore you
must distribute 26 electrons. Any unpaired electrons are considered lone pairs. (PCl3
has 1 lone pair).
e- pairs
2
3
4
5
6
e- pairs
3
4
4
5
5
5
6
VSEPR shape
Linear
Trigonal Planar
Tetrahedral
Trigonal Bipyramidal
Octahedral
Lone Pairs
1
1
2
1
2
3
1
Bond angle
180º
120º
109.5º
120º, 90º
90º
e- pair geometry
Bent
Trigonal Pyramidal
Bent
Seesaw
T-shape
Linear
Square pyramidal
hybridization
Sp
Sp2
Sp3
Sp3d
Sp3d2
Bond angle
<120º
<109.5º
<<109.5º
<120º, <90º
<90º
180º *Non-polar
<90º
6
2
Square planar
90º * Non-polar
-- If the molecular shape contains lone pairs the shape is polar with the exception of Square
Pyramidal and Square Planar which are both non-polar.
Give the shape, bond angle, hybrid and polarity of the following compounds.
Sample Questions:
1. CCl4
Shape: _______________________
Bond Angle: __________________
Hybridization: _________________
Polarity: ______________________
2. H2O
Shape: _______________________
Bond Angle: __________________
Hybridization: _________________
Polarity: ______________________
3.
NH3
Shape: _______________________
Bond Angle: __________________
Hybridization: _________________
Polarity: ______________________
ACIDS and BASES

The following are three different definitions of acids and bases.
Arrhenius:
Bronsted-Lowry:
Lewis:
ACID
H+ donor
proton donor
e- pair receiver
BASE
OH- donor
proton acceptor
e- pair donor
*Strong acids and bases completely ionize when placed in water while weak acids and
bases only ionize slightly.
Strong acids: HCl HBr HI HNO3 H2SO4 HClO4 HClO3
Strong bases: any group 1 or 2 element (other than Mg or Be) and hydroxide (OH-).
Ex: NaOH, Ba(OH)2
Weak bases are compounds that fit into one of the following patterns: NxHy or CxHyNbHa
Strong/ Weak Electrolytes:
Strong Electrolytes- contain strong acids and bases are because they ionize completely in
water.
Weak Electrolytes- contain weak acids and bases because they only ionize slightly in water.
Nonelectrolytes- are compounds that do not ionize in solution, though they may be soluble.
Questions:
State whether each of the following are Strong, Weak, or Nonelectrolytes.
1. Na2SO4
2. CCl4
3. NH3
S
S
S
W
W
W
N
N
N
4. HClO2
5. C6H6
6. HC2H3O2
Diatomic Molecules
Hydrogen H2
Nitrogen N2
Oxygen
O2
Fluorine
F2
Chlorine Cl2
S
S
S
W
W
W
N
N
N
Bromine
Br2
Iodine
I2
PERIODIC TABLE TRENDS
Moving Left --> Right



Atomic Radius Decreases
Ionization Energy Increases
Electronegativity Increases
Moving Top --> Bottom



Atomic Radius Increases
Ionization Energy Decreases
Electronegativity Decreases
STP, Standard Temperature and Pressure
273 K (0 degrees Celsius) and 1 atm pressure. STP is often used for measuring gas density and
volume.
FACTOR LABELING
In factor labeling use tic-tac-toe lines to go from one unit to another, with conversion
factors in between.
Some common ratios used are:
1 mole=6.02 x 10E23 molecules
Mole=grams/gmw
Moles=liters/(22.4L/mol)
Practice:
a) ZnSO3  ZnO + SO2
What is the STP volume of SO2 gas produced by the above reaction when the grams of
ZnSO3 are consumed?
B) Propane was completely burned in air at STP. The reaction was C3H8 + O23CO2
+4H20. If 67 liters were produced and all of the carbon in the CO2 came from the
propane, what the mass of the propane sample?
C)CaCO3  CaO + CO2. A sample of pure CaCO3 was heated and decomposed
according to the reaction given above. If 28 g of CaO were produced by the reaction,
what was the initial mass of CaCO3?
OXIDATION NUMBERS
The oxidation state (or oxidation number) of an atom is the number of electrons that it
gains or looses when it forms a bond.
Follow these rules:
 Elements are zero
 The sum of the oxidation numbers of all atoms in a molecule add up to zero
 The sum of the oxidation numbers of all the atoms in a polyatomic ion must add
up to the charge on the ion
The following elements usually have this oxidation number:
 Alkali metals: +1
 Alkaline earths: +2
 Group 3A: +3
 Oxygen: -2
 Hydrogen: +1
Example:
a) K+
+1 (due to the charge of this ion)
b) K2O:
set up an algebraic equation: 2(+1) +x = 0 solve and get x=-2
-2
c) CrO4
set up an algebraic equation: x + 4(-2) =-2 solve and get x=+6
Practice problems:
a) MnO2
b) ClO4
c) Mn
d) Ca+2
MOLECULAR/EMPIRICAL FORMULA
Determine the empirical formula for a compound with the following elemental
composition:
40.00% C, 6.72% H, 53.29% O.
The first step will be to assume exactly 100 g of this substance. This means in 100 g of
this compound, 40.00 g will be due to carbon, 6.72 g will be due to hydrogen, and 53.29
g will be due to oxygen. We will need to compare these elements to each other
stoichiometrically. In order to compare these quantities, they must be expressed in terms
of moles. So the next task will be to convert each of these masses to moles, using their
atomic weights:
 This leads, using factor labeling, 3.331 mol C, 6.667 mol H, and 3.331
mol O
Then divide through each of the mole quantities by which ever mole quantity is the
smallest number of moles. In this example, the smallest mole quantity is either the moles
of carbon or moles of oxygen (3.331 mol)
 You now achieve 1 mol of C, 2.001 mol H, and 1 mol of O
The ratio of C:H:O has been found to be 1:2:1, thus the empirical formula is: CH2O
Note- to find the molecular formula you would first find the empirical formula and its
mass. You would set up an equation that would be the
 (empirical formula mass)(x)= new mass wanted
 this way you can see what number you need to multiply your empirical
formula by to achieve the mass desired, when x is found you multiply that
number by all the subscripts in your empirical formula
Practice:
a) A hydrocarbon contain 75% carbon my mass. What is the empirical formula for the
compound?
b) A hydrocarbon was found to be 20% hydrogen by weight. If 1 mole of the
hydrocarbon has a mass of 30 grams, what is its molecular formula?
c) When solid potassium iodide is combined with fluorine gas, a compound is formed
that is 27% chlorine and 73% fluorine. What is the empirical formula of the compound?
ELECTRONS
Quantum Numbers:
 Shells: n=1,2,3… It is the average distance from nucleus and its energy. The
higher values are farther away and have more energy and less stability than
electrons in shells with lower values.
 Subshells: l= 0.1,2… describes the shape of the orbital
o n=1, the 1st subshell, has 1 subshell: 2, or l=0
o n=2, the 2nd subshell, has 2 subshells s (l=0) and p (l=1)
o n=3, the 3rd subshell has 3 subshells s (l=0) p(l=1) d (l=2)
o note- s is spherical and p is dumbbell shaped
 Orbitals: ml= -1, 0,+1… It decribes the orientation of the orbital in space.
o s subshell has 1 orbital ml =0
o p subshell has 3 orbitals ml =-1, ml=0 ml=+1
o d orbital has 5 orbitals ml =-2 ml =-1 ml =0 ml =+1 ml =+2
 Spin: ms= +1/2, -1/2
Pauli Exclusion Principle- No two electrons can have the same set of quantum numbers.
So each electron in any atom has its own distinct set of four quantum numbers.
Hand's Rule- When an electron is added to a subshell, it will always occupy an empty
orbital if one is available. Electrons always occupy orbitals singly if possible, and only
pair up if no empty orbitals are available.
To name the electron configuration use the above charts as a tool while using all the
information, from left to right, until you get to the element you are naming.
Example: Cl: 1s22s22p63s23p5
Transition metals are same way. Look at this example of the electron configuration for
the element Niobium1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d3
Practice:
a) Provide the electron configuration for Neon
b) Provide the electron configuration for Carbon
c) What is the quantum numbers for the valence electron in a potassium atom in ground
state?
d) Which of the following is an impossible set of quantum numbers?
4,0,0,1/2
4,0,1,1/2
4,1,0,1/2
4,1,1,1/2
4,2,1,1/2
e) What is the electron configuration for Sc?
Answers
Solubility
1. YES
2. NO
3. YES
4. YES
5. NO
Sig-Figs
A: 4, because the number neither begins nor ends with zeros.
B: 3, All are non-zero digits
C: 3, The first three zeros just mark the decimal place, but the zero after the 2 is significant since
it indicates certainty
D: 5, the zeros serve only to mark the place
VSEPR
1. tetrahedral, 109.5º, sp3, nonpolar
2. bent, <<109.5º, sp3, polar
3. trigonal pyramidal, <109.5º, sp3, polar
Strong/Weak Electrolytes
1. S
2. N
4. W
5. N
Molecular/Empirical Formula Practice
a. CH4
b. C2H6
c. CIF3
Factor Labeling Practice
a. 23 L
b. 44 g
c. 50 g
Oxidation Number Practice Problems
a. +4
b. +7
c. 0
d. +2
Electron Configuration Practice
a. 1s2 2s2 2p6
b. 1s2 2s2 2p2
c. 4,0,0,1/2
d. 1,0,1,1/2
e. 1s2 2s2 2p6 3s2 3p64s23d104p4
3. W
6. W