Download Gupta 2014 Credit: Google Images for the pictures Chapter 1

Gupta 2014
Credit: Google Images for the pictures
Chapter 1 Summary Notes
Main Concepts
 Elements: substances that cannot
be decomposed into simpler
substances
 Compounds: substances composed
of two or more elements
 Law of Constant
Composition or law of
definite proportions: the
relative masses of elements
are fixed in a given
chemical substance.
 Law of Multiple
Proportions: Applies
ONLY when two elements
combine to form two or
more compounds. The
masses of one element
which combine with a
fixed mass of the second
element are in a ratio of
whole numbers
 Mixtures: combinations of two or
more substances
 Techniques for separating
mixtures: filtration,
distillation,
chromatography
 Properties:
 Physical vs. chemical: Did
the sample (really) change?
 Intensive vs. extensive:
Does the measurement
depend on quantity of
sample?
Explanations
-Law of Constant Composition: Ex. In pure H2O, H and O
combine in a 1:8 mass ratio. Does law of constant composition
hold good for CuSO4.5H2O? Why or why not?
-Law of Multiple Proportions: Explain the following in terms of
multiple proportions:
-Separation Techniques:
Hand Separation- for mixtures that can be visually differentiated
based on mass, color, shape etc.
Filtration: Fitrate, precipitate, heterogeneous mixtures
Separating Funnel: For immiscible liquids, layers separate
with lesser density layer on top.
Centrifugation: Separates particles of different masses based
on centrifugal force. Heavier particles at the bottom and the
lighter particles on top.
Distillation- uses differences in the boiling points to separate a
homogeneous mixture.
Chromatography-separates homogenous mixtures (mostly inks)
based on the differences in solubility of the mixture in a
solvent. There is a stationary and a mobile phase
Summary of the Chapter and Important things to remember:
Gupta 2014
Credit: Google Images for the pictures
Chapter 2 Summary Notes
Main Concepts
Naming Compounds Review: Before
naming a compound, it is important to
know its type because naming depends
upon the type. For naming purposes, we
classify compounds as ionic compounds,
molecular compounds, and acids.
 Ionic Compounds can be identified
by the presence of a metal in it.
(generally solids) Ex. NaCl, K2SO4,
PbSO4
 Molecular compounds are made up
of all non metals. (generally liquids
and gases) Ex. H2O, N2O5
 Acids begin with H (generally present
as aq solutions or gases) Ex. HCl,
H2SO4, HClO3
Coordination compound: compound that
contains a complex ion or ions.
Ex. [Cu(NH3)4]Cl2
1. Name cation before anion; one or
both may be a complex. (Follow
standard nomenclature for noncomplexes.)
2. Within each complex (neutral or ion),
name all ligands before the metal.
-Name ligands in alphabetical order
-If more than one of the same ligand is
present, use a numerical prefix: di, tri,
tetra, penta, hexa, …
-Ignore numerical prefixes when
alphabetizing.
 In any (uncharged) atom: The #f
protons= atomic number (Z)
# of e= # of p
Mass # (A)= atomic #- # of neutrons
Atomic Symbol: 6C12
 Isotopes are atoms of the same
element containing different numbers
of neutrons and therefore having
different masses.
Explanations
Naming Compounds Flow Chart
Does the formula start with H?
NO
YES
Does it begin with a metal that has more than one
oxidation number? (e.g. Fe, Ni, Cu, Sn, Hg)
NO
YES
Name the first element followed by its
oxidation number (Roman Numeral) or
“old school” –ic or –ous endings.
Does the formula contain a polyatomic ion?
NO
YES
Are both elements nonmetals?
NO
YES
Name the first element,
Then the second element
with an –ide ending.
Name the first element, then
the polyatomic ion. If two
elements are present, name
both, then the polyatomic ion
(e.g. NaHCO3 is sodium
hydrogen carbonate).
Name the first element using the proper prefix (never
mono–). Name the second element with the proper
prefix (including mono–) and –ide ending.
1 = mono– 4 = tetra– 7 = hepta– 10 = deca–
2 = di–
5 = penta– 8 = octa–
3 = tri–
6 = hexa– 9 = nona– (not nano–)
It is an acid (must be aqueous).
Does the acid contain a polyatomic ion?
NO
YES
Does the acid end with a
polyatomic ion?
–ite
–ate
Name the polyatomic
ion, replacing the –ate
ending with –ic. Add
the word acid.
Name the polyatomic ion,
replacing the –ite ending
with –ous. Add the word
acid.
Write the prefix hydro–, then the
name of the second element with –ic
ending. Add the word acid.
Practice: 1. Sodium Sulfide (Na2S), Potassium Nitrate (KNO3),
Ferrous Sulfate Fe (SO4), Ammonium Chloride (NH4Cl),
Phosphoric Acid (H3PO4)
2. What is the O.N. of P in PO43- ion?
3. What is the O.N. of Fe in Fe(NO3)3?
Summary of the Chapter and Important things to remember:
Gupta 2014
Credit: Google Images for the pictures
Chapter 3 Summary Notes
Main Concepts
 Atomic mass units:
1 amu = 1.66054 x 10-24 g
or 1 g = 6.02214 x 1023 amu
1 atom of 12C isotope defined as weighing exactly 12 amu
Explanations
Ex. 12C: 98.892% x 12 amu =11.867
13
C: 1.108% x 13.00335 amu =+ 0.1441
AW = 12.011 amu
Therefore 12.011 g C = 1 mol C

Percent Composition

Moles: Avogadro’s Number = 6.022 x 1023 = atoms in exactly
12.000 g of 12C = 1 mol
-Moles important because they can be used for ratios. Mass
cannot be used for ratios. Converting: grams → moles →
molecules

Determining Empirical Formula
Start with the number of grams of each element, given in the
problem.
-If percentages are given, assume that the total mass is 100
grams so that the mass of each element = the percent given.
-Convert the mass of each element to moles using the molar
mass
-Divide each mole value by the smallest number of moles
calculated.
-Round to the nearest whole number. This is the mole ratio of
the elements and is represented by subscripts in the empirical
formula.
-If the number is too far to round (x.1 ~ x.9), then multiply
each solution by the same factor to get the lowest whole
number multiple. If one solution is 1.5, then multiply each
solution in the problem by 2 to get 3.If one solution is 1.25,
then multiply each solution in the problem by 4 to get 5.
Summary of the page and Important things to remember:
Ex. What is the percent of O in
CuSO4.5H2O? 57.7%
Ex. How many molecules of H2O in 100.0
g H2O? 3.343 x 1024 molecules H2O
Ex. What is the empirical formula of a
compound that is composed of 80.%
Carbon and 20.% Hydrogen? If the molar
mass is found to be 30 g/mol, what is the
molecular formula? CH3, C2H6
Gupta 2014
Credit: Google Images for the pictures
Chapter 3 Summary Notes Contd.
Main Concepts
 EF from Combustion Data: To find out the EF of a
compound, the compound is burned in the air. The
equation for combustion of these compounds is all
follows.
For Hydrocarbons: CxHy + O2 CO2 + H2O
For Compounds Containing CHN:
CxHyNz+ O2 CO2 + H2O+ NO2
For Compounds containing C, H and O
CxHyOz + O2 CO2 + H2O
Explanations
Ex. A 0.6349 g sample of the unknown
produced 1.603 g of CO2 and 0.2810 g of H2O.
Determine the empirical formula of the
compound. Ans. C7H6O2
Ex. When carbon-containing compounds are burned
in a limited amount of air, some CO(g) is produced
as well as CO2 (g). A gaseous product mixture is
35.0 mass % CO and 65.0 mass % CO2. What is the
mass % C in the mixture? 32.7% C
Ex. If 6.0 g hydrogen gas reacts with 40.0 g oxygen
gas, what mass of water will be produced?
OR
2H2 (g) + O2 (g) → 2 H2O (g)

Ex. A mixture of morphine (C17H19NO3) and an inert
solid is analyzed by combustion with O2. The
unbalanced equation for the reaction of morphine
with O2 is
C17H19NO3 + O2 → CO2 + H2O + NO2
The inert solid does not react with O2. If 4.000 g of
the mixture yields 8.72 g of CO2, calculate the
percent morphine by mass in the mixture. 83.1%
Limiting Reactant: The reactant that runs out first. LR can be
determined by comparing the mole ratios of the reactants. LR
determines the amount of the products formed.

% Yield 
Ex. If in the previous example, only 40.0 g
water were formed, what is the percent yield and
percent error?
Actual yield
100
Theoretica l yield
Accepted value - Measured result

% Error 

Determining the Formula of a Hydrate: To determine the
formula of a hydrate, a certain mass of hydrate is heated to
drive off the water. Then the mass of water driven-off is
calculated. Now the mole ratio of water to the compound
is calculated.
Accepted value
100
Summary of the page and Important things to remember:
Ex. When 2.000 gram of Na2CO3.xH2O was
heated, 0.914 gram of anhydrous residue
remained. What is the formula of this
compound? Ans: Na2CO3.7H2O Sodium
carbonate heptahydrate
Gupta 2014
Credit: Google Images for the pictures
Chapter 4 Summary Notes
Main Concepts
Explanations
Reactions: To be able to successfully write reactions, you will need to know the Synthesis Reactions:
following: Solubility Rules , Nomenclature Types of Reactions (explained later in
this worksheet), How to write net ionic equations MUST KNOW For AP
Chemistry reaction prediction:
1. Always write balanced net ionic equations (meaning dissociate soluble
compounds (based on solubility rules),
2. Metal are insoluble and are atomic, written as (s) in these equations. Ex. Mg(s)
3. Molecular compounds such as gases (CO2, H2S Etc.) are written as (g) and will
not dissociate into ions.
4. Water is written as (l) and does not dissociate.
5. Ionic compounds may or may not dissociate depending on solubility rules. Ex.
PbSO4 insoluble and NaNO3 soluble.
6. Even a soluble ionic compound may NOT dissociate if it is in solid form
(meaning no water present to actually dissociate the ions.)
7. Weak acids and bases partly dissociate or ionize and are written with a
reversible arrow.
8. Remember PSHOFBrINCl. Phosphorus occurs as P4, Sulfer as S8 and rest as
diatomic.
9. While we are reviewing, remember the difference between Zn and Zn2+ and Cl2
and 2 Cl10. Strong acids (HCl, HBr, HI, HNO3, H2 SO4 (first dissociation only!), HClO4 and
HClO3) and strong bases (Group 1 alkali metal hydroxide and Ca, Ba, Sr
hydroxides from group 2) dissociate in aq. Solutions. Weak acids and bases are
not dissociated in net ionic equations.
Solubility Rules Always soluble: alkalies, NH +, NO -, C H O 4
3
2
3
Decomposition
Single Replacement
2
Types of Reactions: Double displacement. Precipitation, neutralization, gas forming.
H2CO3 in water = H2O & CO2
Single displacement or redox replacement: (metals displace metals and nonmetals displace
nonmetals)
Combination or synthesis = two reactants result in a single product
• Metal oxide + water  metallic hydroxide (base)
• Nonmetal oxide + water  nonbinary acid
• Metal oxide + nonmetal oxide  salt
Decomposition = one reactant becomes several products
• Metallic hydroxide  metal oxide + water
• Acid nonmetal oxide + water
• Salt  metal oxide + nonmetal oxide
• Metallic chlorates  metallic chlorides + oxygen
• Electrolysis decompose compound into elements (water in dilute acids or solutions of
dilute acids)
• Hydrogen peroxide  water + oxygen
• Metallic carbonates --> metal oxides + carbon dioxide
• Ammonium carbonate  ammonia, water and carbon dioxide.
Hydrolysis = compound reacting with water.
• Watch for soluble salts that contain anions of weak acid the anion is a conjugate base and
cations of weak bases that are conjugate acids.
Reactions of coordinate compounds and complex
Combustion Reactions:
Gupta 2014
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Chapter 5 Summary Notes
• Complex formation by adding excess source of ligand to transitional metal of highly
charged metal ion such
as Al3+ Al =4 ligands others 2X ox #
• Breakup of complex by adding an acid  metal ion and the species formed when
hydrogen from the acid
reacts with the ligand
Redox = change in oxidation state= a reaction between an oxidizer and a reducer.
1. Familiarization with important oxidizers and reducers
2. “added acid” or “acidified”
3. an oxidizer reacts with a reducer of the same element to produce the element at
intermediate oxidation state
Molarity (M) = moles solute
volume of solution
=
Precipitation
mol
L
Titration is a method to determine the molarity of unknown acid or base. In
titration, an acid or base of unknown molarity is titrated against a standard solution
(whose M is known) of acid or base.The end point in a titration is indicated by a
color change by the indicator. Indicators are weak acids or bases and are added in
small quantity (1-3 drops) to indicate the end point. At equivalence point (which
should be close to end point),
moles of H+= moles of OHM1V1= M2 V2 (sometimes used to get moles , M= moles/L , so moles= M XV)
-What other ways can you get the moles- for a solid acid or base? For a gas?
Electrolyte: substance which, in aqueous solution, ionizes and thus conducts
electricity. Ex: salt in water.
Non-electrolyte: substance which, in aqueous solution, does not dissociate and thus
does not conduct electricity
Strong & weak electrolytes: conductivity depends on degree of dissociation and
equilibrium position:
HA (aq) ↔ H+ (aq) + A- (aq)
Strong = nearly completely dissociated
Weak = partially dissociated
Molecular equation: shows complete chemical equation with states of matter,
undissociated
BaCl2 (aq) + Na2SO4 (aq) → 2 NaCl (aq) + BaSO4 (s)
Complete ionic equation: shows complete chemical equation with states of matter,
dissociated if appropriate
Ba2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + SO42-(aq) →
2 Cl-(aq) + 2 Na+(aq) + BaSO4 (s)Spectator ions:
present in reaction but do not “participate”; depend on solubility rules
Cl- (aq) and Na+ (aq)
3. Net ionic equation: shows chemical equation without spectator ions
Ba2+ (aq) + SO42- (aq) → BaSO4 (s)
Summary of the page and Important things to remember:
Ex. How many mL of a 3M
NaOH solution are required to
completely neutralize 20.0 mL of
1.5M H2SO4? (Start by writing a
balanced equation!) Ans. 20.0
mL
Ex. How many g of NaOH is
required to completely react with
100. mL of 1M HCl?
Gupta 2014
Credit: Google Images for the pictures
Main Concepts
 0th and 1st Laws of Thermodynamics:
0. 2 systems are in thermal equilibrium when they are at
the same T
1. Energy can be neither created nor destroyed, or, energy
is conserved
 Internal Energy
 Includes translational, rotational, vibrational energy
 Change (ΔE = Efinal - Einitial) is often measured
o ΔE > 0: Energy of system increases (gained from surr.)
o ΔE < 0: Energy of system decreases (lost to surr.)
o ΔE = q + w
 q = heat added/liberated from system
 q > 0 : heat added to system
 q < 0: heat removed from system
 w = work done on or by the system
 w > 0 : work done to system
 w < 0 : system does work on surr.
 Calorimetry: Measurement of heat flow, experimental
technique used to measure the heat transferred in a physical or
chemical process
o Calorimeter: the apparatus used in this procedure; two
types: constant pressure (coffee cup) and constant
volume (bomb calorimeter)
o Coffee Cup Calorimeter: system in this case is the
“contents” of the calorimeter and the surroundings are
cup and the immediate surroundings
o qrxn + q solution = 0
 qrxn: heat gained/ lost in the chemical reaction
 qsolution: the heat gained/lost by solution


Heat Capacity, C: Amount of heat required to raise T of an
object by 1 K
o q = CΔT
Specific Heat (or Specific Heat Capacity), c: heat capacity of
1 g of substance
o q = mcΔT
Explanations
Ex. Octane and Oxygen gases combust
within a closed cylinder in an engine. The
cylinder gives off 1150J of heat and a
piston is pushed down by 480J during the
reaction. What is the change in internal
energy of the system? (Ans: ΔE = -1630J)
Ex. How much energy is required to heat
40.0 g of iron (c = 0.45J/gK) from 0.0oC to
100.0oC? (Ans: q = 1800J)
Ex. 0.500g of Mg chips are placed in a
coffee-cup calorimeter and 100.0mL of
1.00M HCl is added to it. The reaction is:
Mg(s) + 2HCl(aq)  H2(g) + MgCl2(aq)
The temp. of the solution increases from
22.2C to 44.8C. What’s the enthalpy
change for the reaction, per mole of Mg?
Assume specific heat capacity of solution is
4.20J/gK and density of the HCl solution is
1.00 g/mL. (Ans: ΔH= -4.64*105 J/molMg)
Gupta 2014
Credit: Google Images for the pictures
Chapter 5 Summary Notes Contd.
Main Concepts
 Enthalpy, H: change in heat content of a reaction at
constant P
o H = E + PV  ΔH = ΔE + PΔV  ΔH = (qp+w) +
(-w)  ΔH = qp
 qp = heat content
 ΔH > 0: heat gained from surr. + ΔH in
endothermic reaction
 ΔH < 0: heat released to surr. + ΔH in
exothermic reaction

Enthalpy of Reaction, ΔHrxn: heat of reaction, extensive
property, depends on states of reactions and products
o ΔHrxn = -ΔHreverse rxn

Hess’s Law: If a rxn is carried out in a series of steps:
ΔHrxn = Σ(ΔHsteps) = ΔH1 + ΔH2 + ΔH3 + · · ·

Enthalpy of Formation, ΔHf:
heat needed to form substance
from its elements
o Standard Enthalpy of
Formation, ΔHfo: forms
1 mole of compound
from its elements in their
standard state (at 298K)
o ΔHf of pure element (C,
O2, H2, etc) is 0.
ΔHrxno = Σ [n*ΔHfo(products)] - Σ[n*ΔHfo(reactants)]

Bond Enthalpy: Amount of energy required to break a
particular bond between two elements in gaseous state
o Indicates the “strength” of a bond
ΔHrxn ≈ Σ [ΔHbonds broken] - Σ[ΔHbonds formed]
o NOTE: this is the “opposite” of Hess’s Law
Summary of the page and Important things to remember:
Explanations
Ex. What is the ΔH of combustion of 100g CH4
if ΔHrxno = -890kJ? (Ans: -5550kJ)
Ex. What is ΔHrxno of the combustion of
propane?
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
Givens:
3C(s) + 4H2(g)  C3H8(g)
ΔH1 = -103.85kJ
C(s) + O2(g)  CO2(g)
ΔH2 = -393.5kJ
H2(g) + ½O2(g)  H2O(l)
ΔH3 = -285.8kJ
o
(Ans: ΔHrxn = -2219.8kJ)
Ex. What is ΔHrxno for the combustion of liquid
benzene?
C6H6(l) + 15/2O2(g)  6CO2(g) + 3H2O(l)
Givens:
ΔHfo(C6H6(l)) = +49 kJ/mol
ΔHfo(CO2(g)) = -394 kJ/mol
ΔHfo(H2O(l)) = -286 kJ/mol
(Ans: ΔHrxno = -3268 kJ/mol)
Ex. What is ΔHrxno for the following reaction?
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
Given:
ΔH/mol of C-H bond: +413 kJ/mol
ΔH/mol of H-Cl bond: +431 kJ/mol
ΔH/mol of C-C bond: +348 kJ/mol
ΔH/mol of Cl-Cl bond: +242 kJ/mol
ΔH/mol of C-Cl bond: +328 kJ/mol
ΔH/mol of C=C bond: +614 kJ/mol
(Ans: ΔHrxno ≈ -104 kJ/mol)
Gupta 2014
Credit: Google Images for the pictures
Chapter 6 Summary Notes
Main Concepts
 Electromagnetic Spectrum: radiant energy can travel without
matter
λν
c = speed of light = 3.0 x 108 m/s
λ = wavelength (m)
ν = frequency (Hz)

Planck’s Theory: Blackbody radiation can be explained if
energy can be released or absorbed in packets of a standard
size called quanta
h c h = Planck’s constant = 6.63 x 10-34 J-s
E  h 

 Photoelectric Effect: As first explained by Einstein in 1905,
the photoelectric effect is the spontaneous emission of an
electron from metal struck by light if the energy is sufficient

Atomic Emission Spectra: spectrum for specific wavelengths
of light emitted from pure substances

Bohr’s Model of the H Atom: Bohr applied idea of
quantization of energy transfer to atomic model, theorizing that
electrons travel in certain “orbits” around the nucleus
Allowed orbital energies are defined by:
R
 2.178 10 18
n = principal quantum number = 1, 2, 3 ...
En  2 H 
n
n2




Explanations
Atomic Emission Spectra for
H
Line
Series
Transition down
to (emitted)
or up from
(absorbed)…
Type of EMR
Lyman
1
UV
Balmer
2
Visible
Paschen
3
IR
Brackett
4
Far IR
RH = Rydberg’s constant = 2.178 x 10-18 J
Line Series: transitions from one level to another
Heisenberg’s uncertainty principle: The position and
momentum of a particle cannot be simultaneously measured
with accuracy.
Schrödinger’s wave function: Relates probability (  2) of
predicting position of e- to its energy.
h2 d 2
d
E
 U  ih
2
2m dx
dt
Matter as a Wave:
m = h / c Particles (with mass) have an associated wavelength
h / mc
Waves (with a wavelength) have an associated mass
and velocity
Probability Plots for 1s, 2s, and 3s
Orbitals
Gupta 2014
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Credit: Google Images for the pictures
Pauli Exclusion Principle: no two charges in an atom
can have the same set of four quantum numbers n, l, m1,
ms.
Effective Nuclear Charge: the net positive charge acting
on the outermost electron.
Shielding Effect: inner electrons shielding the outer
electron from the full charge of the nucleus.
Electron Configuration: the way the electrons are
distributed among the various orbitals of an atom.
 The most stable, or ground, electron configuration is
one in which the electrons are in the lowest possible
energy states.
Hund’s Rule: for degenerate orbitals (orbitals with the
same energy), the lowest energy is attained when the
number of electrons with the same spin is maximized.
The periodic table is your best guide to the order in
which orbitals are filled.
 s-block and p-block contain the representative (main
group) elements.
 The ten columns in the middle that contain transition
metals, elements in which d-orbitals are being filled.
 f-block metals are the ones in which the f-orbitals are
being filled.
Diamagnetic: paired electrons
Paramagnetic: unpaired electrons
Mass Spectroscopy: Helps identify # and abundance of
isotopes and structures of different compounds. Chlorine
has two isotopes, 35Cl and 37Cl, in the approximate ratio
of 3 atoms of 35Cl to 1 atom of 37Cl. You might suppose
that the mass spectrum would look like this but that is
not the case because Chlorine consists molecules that
Ex: Give ground state electron configurations for
the following: Ni2+ and Ni3+
Ans: Ni2+ = [Ar]3d8, Ni3+= [Ar]3d7
E
x: Is Cupric ion dia or paramagnetic? Why?
Paramagnetic, due to an unpaired d e.
Mass Spectrograph for Chlorine (Cl2)
fragment
(chemguide.co.uk)
You could have the following mass fragments--35 + 35 = 70,
35 + 37 = 72, 37 + 37 = 74. So the actual mass spectrograph
will look like the one on the right.
(chemguide.co.uk)
Summary of the page and Important things to remember:
Gupta 2014

Credit: Google Images for the pictures
Photoemission Spectroscopy (PES)
In a photoelectron spectroscopy experiment any electron can
be ionized when the atom is excited. Unlike the first
ionization, in this experiment any electron can be removed,
not just the electron that requires the least amount of energy.
PES gives insight into the structure of atom. Each peak in PES
indicates the number of electrons and the position of the peak
indicates the amount of energy required to remove those
electrons. Note that s electrons will require more energy than
p electrons due to higher ENC hence s electrons will be farther
out on the energy axis.
http://www.chem.arizona.edu/chemt/Flash/photoelectron.
html
Summary of the page and Important things to remember:
Gupta 2014
Credit: Google Images for the pictures
Chapter 7 Summary Notes
Main Concepts
Explanations
Periodic Trends Key Words
 Principal Energy Level: more energy
levels means bigger atoms
 Nuclear Charge (# of p in an atom):
attraction of electrons to nucleus;
increased nuclear charge causes
atomic radius to decrease
 Shielding Effect: inner electrons
increase atomic size by reducing the
attractive force on outermost electrons
Atomic Radius
 Effective Nuclear Charge: force of
attraction felt valence e from nucleus;
a high ENC means smaller ionic
radius (greater attraction to outermost
electrons)
Atomic Radius: half distance between
two covalently-bonded atoms
Periodic Trends
 Atomic Radius: increase going down,
decrease going right (more energy
levels and shielding going down,
higher ENC going right)
 Size of Ions: cations smaller than
anions (fewer e-, less e- repulsion),
bigger going down,smaller going right
 Isoelectronic series have same # of
electrons, different # of protons
 Ionization Energy: the amount of
energy required to remove an electron
from the ground state of a gaseous
atom or ion.
 A(g)  A+ + e Bigger going right, smaller going
down
 Exceptions: between Group 2&13,
Group 5&6
Isoelectronic Series
Ionization Energy
Zeff = Z − S
Zeff = ENC
Z = atomic number
S = screening constant (#
of inner electrons)
Gupta 2014
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Chapter 7 Summary Notes
Main Concepts
Periodic Trends (cont’d)
 Electron Affinity: amount of energy
required to add an electron to a
gaseous atom:
 Cl(g) + e−  Cl−
 Larger going right, smaller going
down
 Exceptions: between Group 1&2,
Group 14&15
 Electronegativity: tendency to attract
electrons in a covalent bond
 Increases going up and right
Metals and Nonmetals
 Metals: tend to form cations, metal
oxides are basic
 Nonmetals: tend to form anions,
nonmetal oxides are acidic, poor
conductors of electricity
 Metallic character increases down a
group, decreases across a period
Explanations
Electron Affinity
Electronegativity
Alkali metals (1A)—The most reactive metal
family, these must be stored under oil
because they react violently with water!
They dissolve and create an alkaline, or
basic, solution, hence their name.
Alkaline earth metals (2A)—These also are
reactive metals, but they don’t explode in
water; pastes of these are used in batteries.
Halogens (7A)—Known as the “salt
formers,” they are used in modern lighting
and always exist as diatomic molecules in
their elemental form.
Noble gases (8A)—Known for their
extremely slow reactivity, these were once
thought to never react; neon, one of the
noble gases, is used to make bright signs.
http://www.sparknotes.com/testprep/books/sat2/chemistry/chapt
er4section6.rhtml
Summary of the Chapter and Important things to remember:
Gupta 2014
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Chapter 8 Summary Notes
Main Concepts
 Intramolecular Bonding
o Ionic: electrostatic attrction between oppositely charged
ions. Generally solids. Ex: NaCl, K2SO4
o Covalent: sharing of e- between two atoms (typically
nonmetals). Generally gases/liquids. Ex: CO2, SO2
o Metallic: “sea of e-”; bonding e- relatively free to move
throughout 3D structure. Generally solid. Ex: Fe, Al
o Covalent Network: large number of atoms/molecules
bonded in network through covalent bonding. Ex: SiO2,
Si, Ge, Diamond, Graphite
 Ionic Bonding
o Results as atoms lose or gain electrons to achieve a
nobel gas electron configuration. Typically exothermic.
 Bonded state lower in energy (more stable)
 Opposite charges create electrostatic attraction,
which determines the strength of the ionic bond
o Occurs when difference in electronegativy is > 1.7
o Use brackets when writing Lewis symbols of ions.
 Lattice Energy: Measurement of the strength of the ionic bond
o Hlattice = energy required to completely separate 1
mole of solid ionic compound into its gaseous ions
o Electrostatic attraction (and thus lattice energy)
increases as ionic charges increase and as ionic radii
decrease.
DH lattice µ

Q+Qr+ + r-
Covalent Bonding: Atoms share electrons to achieve noble gas
configuration that is lower in energy (therefore more stable)
o Occurs when difference of electronegativity is ≤ 1.7
 Polar covalent: 0.3 < diff in EN ≤ 1.7
 Nonpolar covalent: 0 ≤ diff in EN ≤ 0.3
 Coordinate covalent: shared pair contributed by
only one of the two sharing species. Ex: Lewis
acids and bases
Explanations
Ex. Draw the Lewis symbol for fluoride.
Ex. In the following pairs, which has a
greater lattice energy and why?
NaCl or KCl
NaCl or MgS
Covalent bond strength is measured by
bond energy. Bond energy is calculated by
= Energy of reactant bonds- Energy of
product bonds
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Metallic Bonding: Forms between metal atoms because of
the movement of valence electrons from atom to atom to
atom in a “sea of electrons”. The metal consists of cations
held together by negatively charged electron “glue”.
o Results in excellent thermal and electrical
conductivity, ductility, and malleability
o A combination of 2 metals is called an alloy
The Octet Rule: Atoms tend to gain, lose, or share
electrons until they are surrounded by 8 electrons in their
outermost energy level (filled s&p shells), and are thus
energetically stable
Lewis symbols (electron-dot symbol)
o Shows a dot for valence electrons of an atom/ion
o Places dots at top, bottom, right, and left sides and
in pairs only when necessary (Hund’s rule).
o Primarily used for representative elements only
(Groups 1A – 8A)
o Transition metals typically form +1, +2, & +3 ions
 Transition metal atoms first lose both “s”
electrons, even though it is a higher energy
subshell. Ex. Cr2+, Cr3+
 Most lose electrons to end up with a filled
or half-filled subshell. Ex. Cu+ ion
Lewis structures: used to depict bonding pairs and lone
Ex. Draw the Lewis structures for the following
pairs of electrons in the molecule
1. Total the # of valence electrons in the system. Add using steps. Show work!
a) Cl2
the total negative charge if you have an ion,
b) CH2Cl2
subtract the charge if you have a cation.
c) NH3
2. Number the electrons if each atom is to be “happy”
d) NaCl
(8 electrons for octet rule, or 2 for hydrogen)
e) HCN(SO4)
3. Calculate the number of bonds in the system.
f) H2O2CNS
Covalent bonds are made by the sharing of
electrons. # of bonds = (electrons in step 2 –
electrons in step 1) / 2
4. Draw the structure. The central atom is usually the
atom with the least electronegativity.
5. Double check your answer by counting total
number of electrons
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Exceptions to the Octet Rule
o Odd-electron molecules. Ex. NO or NO2
o Incomplete octet. Ex. H2, He, BeF2, BF3
o Expanded octet (occurs in molecules when the
central atom is beyond the third period. The empty
3d subshell is used in hybridization). Ex. PCl5, SF6
Formal Charge: the numerical difference between # of
valence electrons in the isolated atom and # of electrons
assigned to that atom in the Lewis structure. Doesn’t
represent real charges, just a useful tool for selecting most
stable Lewis structure
1. Assign unshared electrons (usually in pairs) to the
atom on which they are found
2. Assign one electron from each bonding pair to each
atom in the bond (split the electrons in a bond)
3. Subtract the electrons assigned from the original
number of valence electrons
o Used to select the most stable (and therefore most
likely) structure when more than one structure is
reasonable according to rules
o Most stable:
 Has FC on all atoms closest to zero
 Has all negative FC on most EN atoms
Resonance Structures: Equivalent Lewis structures that
describe a molecule with more than one likely
arrangement of electrons
o Notation: use double-headed arrow between all
resonance structures
o “Real” electron structure of the molecule is an
“average” of all resonance structures
Bond Order: Indication of bond strength and bond length
Bond Enthalpy: Amount of energy required to break a
particular bond between two elements in gaseous state.
Indicates “strength” of a bond.
Hrxn ≈  (Hbonds broken) -  (Hbonds formed)
Summary of the page and Important things to remember:
# Valence electrons in free atom
– # Non-bonding electrons
– ½ (# Bonding electrons)
Formal Charge
Ex. Draw at least 2 Lewis structures for each,
then calculate the formal charge for each atom
in each structure.
a)
SCNb)
N2OBF3
Ex. Draw the Lewis structure and determine the
bond S-O, C-C, and C-H bond orders.
c) SO3
d) C6H6
Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g)
DHrxn = ?
Bond Ave DH/mol Bond Ave DH/mol
C-H
413
Cl-Cl
242
H-Cl
431
C-Cl
328
C-C
348
C=C
614
Ans: Hrxn ≈ -104 kJ/mol
Gupta 2014
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Chapter 9 Summary Notes
Main Concepts
 Valence Shell Electron Pair Repulsion Theory:
helps construct molecular 3-D shape from 2-D
Lewis Structures


Electron Domains: areas of valence electron
density around the central atom

Includes bonding electron pairs and lone
electron pairs

A single, double, or triple bond counts as one
domain
Basis for VSEPR: each group of valence electrons
(electron domains) around a central atom tend to be
as far as possible from each other to minimize
repulsion and this determines molecular geometry
of molecule


lonepair-lonepair repulsion > lonepair-bondpair
repulsion > bondpair-bondpair repulsion
Molecular Dipole Moment: molecules with polar
covalent bonds might have a net dipole moment
depending on the compound's 3-D geometry and
symmetry

Check if individual bonds are polar and if
individual dipole moments cancel out due to
symmetry

Polarity: polar substances are soluble in water and
non-polar substances are soluble in non-polar
solvents like benzene and oil

Valence Bond Theory: predicts bond strengths
based upon orbital overlap for covalent bond
formation

Basis for VB: covalent bond forms when orbitals of
two atoms overlap

An orbital can have a max of two electrons with
opposing signs

The bond strength depends on the attraction of
nuclei for the shared electron, so greater the
Explanations
Gupta 2014
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overlap, the stronger the bond


Sigma Bond: end to end overlap of orbital
which allows free rotation of parts of molecule
(single bonds)
Draw hybrid orbital diagram for C2H4.
Pi Bond: side to side overlap of orbital which
restricts rotation (2nd and 3rd bonds in double
and triple bonds)
Summary of the page and Important things to remember:
Gupta 2014
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
Hybrid Orbital Theory: an extension of VB theory, where
atomic orbitals "hybridize" to form new hybrid orbitals;
explains the bonding in terms of quantum mechanical
model of atom (s, p, d, f orbitals)

Basis for HO: valence atomic orbitals in the molecule are
very different from those in isolated atoms




The number of hybrid orbitals obtained equals the
number of atomic orbitals mixed
 The type of hybrid orbitals obtained varies with types
of atomic orbitals mixed
 ns and np give two sp hybrids
 ns and two np give three sp2 hybrid orbitals
 ns and three np give four sp3 hybrid orbitals
Delocalized Bonds are present in compounds with
resonance structures
Molecular Orbital Theory: is based on the wave nature of
the electrons and is a better model to explain
paramagnetism of oxygen
Basis for MO: Molecular orbitals form through the
combination of atomic orbitals





Bonding MO: stable orbital that forms between nuclei
Antibonding MO: less stable orbital that forms behind
nuclei
 Sigma MO: orbital forming from a combination of
two 1s or 2s orbitals form different atoms or two 2pz
orbitals from different atoms
 Pi MO: orbital forming from a combination of two 2px
or 2py from different atoms (do not appear until B2)
Diamagnetism: all electrons paired; no magnetic
properties
Paramagnetic: at least 1 unpaired electron; drawn into
exterior magnetic field since spins of atoms become
aligned; unlikely to retain alignment when field is
removed
Ferromagnetism: occurs primarily in Fe, Co, and Ni;
drawn into exterior magnetic field since spins of atoms
become aligned; likely to retain alignment when field is
removed
Summary of the page and Important things to remember:
Gupta 2014
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Chapter 10 Summary Notes
Main Concepts
 Characteristics of Gases
 Particles in a gas are very far apart, and
have almost no interaction.
 Gases expand spontaneously to fill their
container (have indefinite volume and
shape.)
 Pressure
 Pressure = a force that acts on a given
area
F
P
A
 Atmospheric pressure: the result of the
bombardment of air molecules upon all
surfaces
 1 atm = 760 mm Hg = 760 torr = 101.3
kPa = 14.7 PSI

Barometer: measures atmospheric P
compared to a vacuum
* Invented by Torricelli in 1643
* Liquid Hg is pushed up the closed
glass tube by air pressure

Manometers: measure P of a gas
*Closed-end: difference in Hg levels
(Dh) shows P of gas in container
compared to a vacuum
* Open-end: Difference in Hg levels
(Dh) shows P of gas in container
compared to Patm
Gas Laws

Boyle’s Law: the volume (V) of a fixed
quantity (n) of a gas is inversely
proportional to the pressure at constant
temperature (T).
Explanations
Closed
V  constant 
1
P
P1V1  P2V2
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Main Concepts
Charles’ Law: V of a fixed
quantity of a gas is directly
proportional to its absolute T at
constant P.
Gay-Lussac’s law: P of a fixed
quantity of a gas is directly
proportional to its absolute T at
constant V.
Seen as derivative of C’s and B’s
laws
Avogadro’s hypothesis: Equal
volumes of gases at the same T &
P contain equal numbers of
molecules
Combined Gas Law


Explanations
V  constant T
V1 V2

T1 T2
P  constant  T
P1 P2

T1 T2
PV
 constant
T
Summary of the Chapter and Important things to remember:
P1V1 P2V2

T1
T2
 Chapter 11 Summary Notes
2014
Credit: Google Images Explanations
for the pictures
MainGupta
Concepts
 INTRAmolecular Forces: the forces holding
atoms together to form moleculs
 INTERmolecular Forces: Forces between
molecules between ions, or between molecules
and ions
Intermolecular Forces (IMF)
-IMF < intramolecular forces (covalent,
metallic, ionic bonds)
-IMF strength: solids>liquids>gases
-Types of IMFs Ion-Ion Forces, Ion-Dipole
Forces, Dipole-Dipole Forces, H-bonds
extreme dipole-dipole, LDFs
Types of IMF
 Electrostatic Forces: act over larger distances
in accordance with Coulomb’s Law
 Ion-Dipole: between an ion and a dipole (a
neutral, polar molecule has separate partial
Boiling points and melting points are good indicators of relative
charges)
IMF strength
-Increasing with increasing polarity of
molecule and increasing ion charge
 Ion-Permanent Dipole
-Water is highly polar and can interact with
positive ions to give hydrated ions in water
QQ
-Attraction between ions and dipole depends
F  2
d
on ion charge and ion-dipole distance
 Dipole-Dipole
-Weakest electrostatic force (not all IMFs,
LDFs weaker than dipole-dipole); exist
between neutral polar molecules
-Increase with increasing polarity (dipole
moment) of molecule
 Hydrogen Bonds (H-bonds)
-H is unique among elements because it has a
single e- that is also a valence e-When e- is “hogged” by a highly
electronegative atom (very polar covalent
bond), the H nucleus is partially exposed and
becomes attracted to e- rich atom nearby
-Explains why ice floats on water, has latticelike structure, explains why molecules with Hbonds have higher boiling points, H-bonding
in water, O -H bond is very polar
 H-bonding in biology
-DNA bases bind to each other due to specific
hydrogen bonding between Lewis Bases
Summary of the page:
Gupta 2014
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Inductive Forces arise from induced distortion of
e- cloud
 London Dispersion: between polar or
nonpolar molecules or atoms, but is generally
mentioned for non polar molecules when other
forces are absent. Very weak, motion of ecreates an instantaneous dipole moment which
induces a dipole in an adjacent atom
 Nonpolar molecules can dissolve in water due
to LDFs. Water induces a dipole in electron
cloud. Solubility increases with mass of gas
due to greater distortion.
 When induced forces between molecules are
very weak, the solid will sublime (solid to gas)
Liquids
 Molecules are in constant motion, molecules
close together
 Liquids are almost incompressible
Evaporation
-To evaporate, molecules must have sufficient
energy to break IMFs
-Condensation is reverse (remove energy and
make IM bonds)
 Vapor Pressure
 Heat of Vaporization heat required (at
constant P) to vaporize the liquid
 Equilibrium vapor pressure & the
Clausius-Clapeyron Equation
o Used to find ∆vapH˚
o Logarithm of vapor pressure P is
proportional to ∆vapH˚ and to 1/T
o lnP = -(∆vapH˚/RT) + C
o Surface Tension leads to spherical
liquid droplets
 Properties resulting from IMFs
 Viscosity: resistance of a liquid to flow
 Surface Tension: energy required to increase
the surface area of a liquid
 Intermolecular forces lead to capillary action
and concave meniscus for a water column
-Capillary Action: movement of water up a
piece of paper depends on the H-bonds
between H2O and the OH groups of the
cellulose in the paper
Summary of the page:
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Cohesion: attraction of molecules for other
molecules of the same compound
 Adhesion: attraction of molecules for a
surface
 Meniscus: curved upper surface of a liquid in
a container; a relative measure of adhesive and
cohesive forces
 London Dispersion Forces
-Increase with increasing molar weight,
increasing # of e-, increasing # of atoms
-“Longer” shapes (more likely to interact with
other molecules)
 Phase Changes
-Endothermic: melting, vaporization,
sublimation
-Exothermic: condensation, freezing,
deposition
Structures of solids
 Amorphous: without orderly structure (ex.
Rubber, glass)
 Crystalline: repeating structure; have many
different stacking patterns based on chemical
Molecular
formula, atomic or ionic sizes, and bonding
Types of Crystalline Solids:
-Atomic: Properties: poor conductors, low
melting point
-Molecular: Properties: poor conductors, low
to moderate melting point
-Ionic: Properties:hard and brittle, high
melting point, poor conductors, some
solubility in H2O
-Covalent (a.k.a. covalent network):
Properties: very hard, very high melting point,
Ionic
Covalent Network
Metallic
generally insoluble, variable conductivity
Credits: Google Images
-Metallic: Properties: excellent conductors,
malleable, ductile, high but wide range of
melting points
Summary of the page
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Chapter 13 Summary Notes
Main Concepts
Vocabulary
 Molarity: measure of concentration in
solutions, mol/L M  mol solute
Liters solution
 Solvation: dissolving; the interactions
between solute and solvent
 Crystallization: process by which
solute particles leave solvent
Explanations
Describe the steps with the lab equipment needed to make 0.1 M
100 ml NaOH? (Hint: volumetric flask)
How are supersaturated solutions created?
Dissolve solute with heat, then cool solution slowly. Solution is
“tricked” into appearing unsaturated
Saturation
 Saturation: solution that is in
equilibrium with undissolved solute
 Unsaturated solution contains less
solute than saturated solution
 Supersaturated solution contains
more solute than saturated solution
but appears unsaturated
Factors Affecting Solubility
 Miscible liquids mix; both are polar or
both are nonpolar
 Covalent network solids do not
dissolve in polar or nonpolar
solvents
 Increasing temperature increases
solubility for most solids, but decreases
solubility for gases
Summary of the Chapter and Important things to remember:
At 40oC, the solubility of KNO3
in 100g of water is 64 g.
What is the solubility of KCl at
10oC?
30 g
At which points would an
unsaturated solution appear?
Supersaturated solution?
How much KClO3 needs to be
added to 10g of KClO3 at 60oC
to make a saturated solution?
20 g
What is normal
boiling point of
ethanoic acid?
117 oC
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Chapter 14 Summary Notes
Main Concepts
Reaction rate: A measure of the (average)
speed of a reaction
Reaction rate is affected by:
1. Concentration of reactants
2. Temperature of the reaction
3. Presence/absence of a catalyst
4. Surface area of solid or liquid
reactants and/or catalysts
-Average Rate: Rate of a reaction over a
given period of time
-Instantaneous Rate: Rate of reaction at
ONE given point of time.
-Initial Rate: Rate of reaction at t=0. (its
instantaneous rate at t=0)
Reaction order: the exponents in a rate
law (can be fractions)
Rate law: shows how the rate of reaction
depends on the concentration of
reactant(s); determined experimentally.
Cannot be determined by the coefficients
of a balanced reaction (unless in an
elementary step)
Rate = k[A]m[B]n
*Units of k change w/order of the rxn
To Find Rate Laws:
1. Using Initial Rates
2. Integration Method: Determining Rate
Law by Determining the Change in
Concentration of reactants over time
 gives rate law either graphically or
by calculations.
Explanations
Rate with regards to other
1 [C ]
1 [ A]
1 [ B ]
Rate






reactants and products
c t
a t
b t
2N2O5 (g) → 4 NO2 (g)
+ O2(g)
If D[O2]/Dt = 5.0 M/s, what is D[N2O5]/Dt?-10.0 M/s
Orders of Reactions & Related Equations:
Using Initial Rates Method: data given
Expe [A]
[B] Initial Rate of Formation of
rimen
C in M
t
1
0.60 0.15 6.3´10-3
2
0.20 0.60 2.8´10-3
3
0.20
0.15 7.0´10-4
Integration Method (Using graphs or equations)
Zero Order
Rate  k[ A] 0 Rate  
[A]t = -kt + [A]0
(k) = M/s
First Order
ln[A]t = -kt + ln[A]0 or log[A]t = -kt / 2.303 + log[A]0
(k) = s-1
Summary of the Chapter and Important things to remember:
A
t
Gupta 2014
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The Collision Model
-Reactants must collide, and with the right
Second Order
orientation and energy for an effective
collision
-Elementary steps: a single event or
step (reaction) in a multi-step reaction
-Molecularity: the # of molecules
participating as reactants in an
elementary step
-Catalyst: Substance that changes the
rate of a reaction without undergoing a
permanent chemical change itself
1 = kt + 1
[A]
[A]0
Check for Permissible Rxn. Mechanism
1. Balanced eq.
2. Rate Determining Step (RDS) is
the slow one.
-Radioactive decay follows a first order
kinetics. Half life (t1/2) can be calculated,
if rate constant is known or vice versa.
-Activation energy ( Ea ): minimum
energy required to initiate a chemical
reaction
Half Life
ln 2 0.693
k

t1/2
t1/2
Activation Energy
For a graph between ln k
and 1/T, the slope can be
used to calculate Ea—
activation energy.
Remember to use 8.314
j/mol. K for R.
Slope 
Summary of the Chapter and Important things to remember:
 Ea
R
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Chapter 15 Summary Notes
Explanations
Main Concepts
Chemical Equilibrium
 Occurs when rate of forward reaction = rate of reverse
reaction. Ex. Vapor pressure: rate of vaporization = rate
of condensation, Saturated solution: rate of dissociation =
rate of crystallization. Conc. of reactants and products
does not have to be equal at the equilibrium, only the
rates of forward and reverse rxn become equal.
 Express concentration in Partial Pressure for gases and
molarity for solutes in liquids
 Rate = kforward [A]
Rate = kreverse [B]

at equilibrium


[ A] 
PA
RT
[B] k f

[A] k r
[ B] 
or
PB
RT
PA
RT
P
Rate  k r B
RT
Rate  k f
PB k f

 constant  K eq
PA k r
-If Kc > 1, then more products at equilibrium
-If Kc < 1, then more reactants at equilibrium
-If Kc = 1, then almost equal concentrations of products
and reactants
There is a spontaneous tendency towards equilibrium.
(spontaneous ≠ quickly, spontaneous = always moving
towards equilibrium)
It is possible to force equilibrium one way or the other
temporarily by altering the reaction conditions, but once
this “stress” is removed, the system will return to its
original equilibrium.
The equilibrium expression is:

Law of Mass Action :
aA+bB↔cC+dD

Concentrations of pure solids and pure liquids are not
included in Keq
c
d
(P ) (P )
[C]c [D]d
Kp  C a D b
Kc 
or
(PA ) (PB )
[A]a [B] b
For a heterogeneous equilibrium:
CaCO3 (s) ↔ CaO (s) + CO2 (g)
K eq 
Summary of the page and Important things to remember:
[CaO] (PCO 2 )
[CaCO3 ]
K p  PCO 2
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