FINAL CHEMISTRY STUDY GUIDE
Chapter 1: Chemistry – The Study of Matter and the Changes It Undergoes
1.1
= Types of Matter
Matter
Matter
Mixtures
Pure Substances
Heterogeneous
Homogeneous Element
(course) = you
can differentiate
different parts
(visually/with a
microscope)
(Solution) can't
differentiate

1.2




Just a pure,
simple element
such as Gold or
Carbon
Compound
Contains more
than one element.
A compound is 2
or more different
elements
chemically bonded
together, such as
NaCl. Always
contain elements
in same mass
percentages.
To separate mixtures:
o Filtration = to separate heterogeneous mixtures
o Distillation = to separate homogeneous mixtures
o Chromatography = to separate volatile gasses
= Measurements
Metric Prefixes
o 106 = Mega (M)
o 103 = Kilo (k)
o 102 = hecto (h)
o 101 = deka (d)
o 10-1 = Deci (d)
o 10-2 = centi (c)
o 10-3 = milli (m)
o 10-6 = micro ()
o 10-9 = nano (n)
o 10-12 = pico (p)
Precision = basically how many decimal places your measurement has (eg. 4.125 is more precise than 4.1)
Accuracy = how well your measurement agrees with right/defined answer
Significant Digits:
o Non-Zero digits are significant!
o Zeros:
 Leading Zeros = not significant = 0.00025 (only 2 and 5 are significant)
 Captive Zeros = are significant = 1.008 or 1105, in both all numb. are sig
 Trailing Zeros = When number does not have decimal (150) = not significant, but when number
does (1.0500), trailing zeros are significant
o Exact Numbers = have no significant value meaning
 Can be counted = (5 pennies)
 If defined = 12 inches in a foot, 2.54 cm in an inch
o Addition/Subtraction
 Result can only contain as many decimal places as number with least amount of decimal places
o Multiplication/Division
 Result can only have as many significant digits as number with least amount of significant digits.
(501*25 = 12500 but = 1.3x104 because 25 only has 2 sigs)

1.3


Important Conversions To Remember:
o ToF = (9/5*ToC) + 32o
o ToC = (5/9)(ToF-32o)
o T0K = ToC + 273.15
o 2.54cm = 1 inch
= Properties of Substances
Chemical Properties:
o When substance takes part in a chemical reaction, a change that converts it to a new substance.
 Decomposition = Mecury Oxide turns into Mecury and Oxygen when heated 600oC
Physical Properties
o W/o changing chemical identity or make-up
 Melting points
 Boling point
 Color
 Taste
o E.g = Whether Nitrogen freezes or boils its still nitrogen

Density
o D = Mass/Volume
o Doesn’t matter what amount of something, density is still the same

Solubility
o Process/Extent to which solute dissolves in solvent
 Ex. 487g sugar dissolves completely in 100g water at 100oC, and 247g dissolves in 100g water at
20oC, how much water does it take to dissolve 100g of sugar at 100oC?
 100g sugar x (100g water/ 487g sugar) = 20.5 grams of water
 If cooled to 20oC: 20.5g water x (204g sugar/100g water) = 41.8g of sugar is in the water
and the remaining 58.2 crystalizes
o Saturated = completely dissolved
o Unsaturated = not enough solute
o Supersaturated = too much solute
o Crystallization = mixture tries to correct an imbalance
 Ex. a solution that is saturated at 30g sugar/100g sugar is instead mixed with 35g sugar/100g
water. So, when shaken, some of the mixture crystallizes. To find the amount, (35/100) –
(30/100) = 5g sugar crystallizes.
There are 113 elements, 91 of which occur naturally
Chapter 2
1) Atoms and Atomic Theory
a) Early Schools of Thought
i) Democrtitus = indivisible particles, atoms from atomos
ii) Aristotle = matter is infinitely divisible; 4 elements (earth, wind, fire, water)
iii) Lavosisier = the father of modern analytical chemistry
(1) Law of the Conservation of Matter = In chemical reactions, mass in the beginning and mass after are
same, mass is neither lost nor gained
iv) Proust
(1) Law of Definite Properties = Elements in compound are present in fixed proportion by weight. (E.g.
Water is always H2O)
v) Dalton
(1) Law of Multiple Proportions = Many elements combine in more than one ratio (e.g. H2O water is 2:1,
H2O2 is Hydrogen Peroxide 1:1)
(2) Dalton Atomic Theory = On page 27
(a) Chemical elements are made of atoms
(b) Atoms of an element are the same mass (except for isotopes)
(c) Atoms of different elements have different masses (except for isobars)
(d) Atoms combine in small number ratios (1:1, 1:2)
2) Components of the Atom
a) * Electrons, then protons, then neutrons were discovered
b) * A cathode ray is an electron beam
3) Ionic Compounds = With Metal and Non-metal!
a) Ion = Any charged particle
b) Ionic Bond = Bond formed between 2 oppositely charged particles, from transfer of electrons from one element
to another
i) Charge is caused because atoms strive to be noble gasses
ii) Generally form between metals and nonmetals
iii) Element that loses electron becomes positively charged, called “CATION”, and one that gains electrons
becomes negatively charged and is called a “ANION”
iv) Metals = Lose electrons, positively charged (CATION)
v) Nonmetals = Gain electrons, negatively charged (ANION)
c) Important Notes
i) LEAVE NUCLEUS ALONE, only manipulate electrons
ii) No such thing as an ionic “molecule” because the compound
iii) Ionic compounds do not exist independently, just as crystals
iv) In chemical reaction, reaction is transfer of electrons
d) Naming
i) For higher charge Cation use suffix –ic
ii) For lower charge Cation use suffix -ous
4) Empirical Formula (Formula for Ionic Compounds)
a) It provides information about the (quantity) or ratio or combined elements in a compound written in lowest
terms
b) When writing empirical formula, Cation (+) is listed before Anion (-)
c) In an ionic compound, all electrons must be accounted for, (no leftovers)
d) To solve the formula, recognize what the charges on the elements to become noble gasses (Rb1+) and then use
“Crisscross Method”
i) Examples
(1) Ionic Compound of Rubidium and Oxygen
(a) A compound of Rubidium (Rb) and Oxygen (O) are charged as ions as Rb1+ and O21e-
1e-
(b) Rb
O
Rb = Rb2O = 2 Rubidiums for every oxygen
(c) OR use Crisscross method and simply move integers from charges of opposite atom to the number of
elements: For example move the 2 from the O to the number of Rubidiums and move the 1 from the
Rubidium to the number of oxygen = Rb2O1 = Rb2O
(2) Sr and S
(a) Charges are Sr2+ and S2(b) Use Crisscross and the compound is Sr2S2 and reduce to smallest terms = SrS
e) In Naming Ionic Compounds, keep leading Cation the same name, and change the suffix of the Anion to –ide
i) Example
(1) Na and Cl = Na1+ and Cl1- = NaCl = Sodium Chloride
5) Ionic Compounds W/Polyatomic Ion
a) Polyatomic Ion = a group of atoms such as NO3
b) With polyatomic ions, the charge is over the whole group of atoms. The identity of the polyatomic ion must be
preserved. Use parenthesis if you need more than one, and don’t just multiply through
i) Ex. Ca(NO3)2 not CaN2O6
c) To make compounds, use same crisscross method as a regular ionic compound, while preserving identity of
groups of atoms.
d) To name Ionic Compounds w/ Polyatomic Ions use the suffix -ate
i) Examples
(1) Ca and NO2 = Ca2+ and NO31- = Ca(NO3)2 = Calcium Nitrate
(2) Ga and C2H3O2 = Ga3+ and C2H3O21- = Ga(C2H3O2)3 = Gallium Acetate
(3) Ammonium (NH4+) and Phosphate (PO43-) = (NH4)3PO4 = Ammonium Phosphate
ii) For groups of atoms with variable charges, use appropriate roman numerals after name to clarify or use latin
name if available
iii)
6) Molecule = With Non-metals!
a) Molecule = compound between two non-metals
i) Joined by Covalent Bonds = which mean that elements share electrons instead of exchanging them
7) Binary Molecular Compounds
a) Different than ionic compounds because they have no charge, and empirical formula ratios don’t apply
i) Ex. NO2 and N2O4 are different compounds
b) Naming
i) End in -ide
ii) Prefixes
(1) Mono(2) Di(3) Tri(4) Tetra(5) Pent(6) Hex(7) Hept(8) Oct(9) Non(10)
Deciii) ***Don’t use mono for first occurring element in compound
iv) Examples
(1) SF6 = Sulfur Hexfluoride
(2) Cl2O7 = Dichloride Heptoxide
(3) CF4 = Carbon Tetrafluoride
8) Acids
a) Acid = A compound with the ability to produce H+ ions in a solution
i) Example
(1) Cl
In H2O
H+ ClH
b)
c)
d)
e)
(2) This is an example of Disassociation = come off a molecule
The strength of an acid is its ability to produce H+ ions
To identify a compound as an acid, look for a “leading” hydrogen
Acids are polar covalent so treat acid like ionic with H as a 1+
Naming
i) Suffix changes
(1) –ate goes to –ic
(a) Nitrate = Nitric
(2) –ite goes to –ous
ii) Hydrogen names = If acid does not contain oxygen, hydrogen is called hydro and acts as a prefix to the
second element. If there is an oxygen, hydrogen is not included in the name.
iii) Examples
(1) HBr = Hydrogen, Bromine and no oxygen = Hydrobromic Acid
(2) HNO3 = Hydrogen and Nitrate = Nitric Acid
(3) H2SO4 = Hydrogen and Sulfate = Sulfuric Acid
(4) HNO2 = Hydrogen and Nitrite = Nitrous Acid
(5) HC2H3O2 = Hydrogen and Acetate = Acetic Acid
iv) Acetic Acid
(1) Can be called multiple names:
(a) HC2H3O2
(b) HCH3COO
(c) But most common is CH3COOH
Chapter 3
Atomic Masses:
 Molar Mass = Atomic weight of an element, converted to Grams
o Ex. Molar Mass of 15.994O = 15.994 Grams
 Calculating Atomic Mass from Isotopes w/ abundances
o Multiply each of the Isotope’s weights by their abundances, and add
 Ex. Oxygen exists in 3 isotopes 16.00O (99.76%), 17.00O (.04%), and 18.00O (.2%)
 16(.9976) + 17(.0004) + 18(.002) = 16 amu
 Calculating Isotope Abundance
o Set up algebraic equation:
 Known mass of element = isotope1 weight(x) + isotope weight 2(x-1)
 Ex. Oxygen (atomic mass = 15.9994amu) exists in 3 isotopes with masses 15.9949amu,
16.9993amu, and 17.9992amu occurring .204%
o 15.9994 = 16.9993x + 15.9949(1-x) + 17.9992(.00204)
o 15.9582 = 16.9993x + 15.9949 – 15.9949x
o -.0367 = 1.0044x
o x = 3.65% = 16.9993O
o 15.9949O = 100% - 3.65% - .204% = 99.76%
Converting From Grams – Moles – Units and So on
 Grams -> Moles = Divide by MM
 Moles -> Units = Multiply by Avogadro’s Number (6.022e23)
 Units -> Moles = Divide by Avogadro’s Number (6.022e23)
 Moles-> Grams = Multiply by MM
 Example:
o Convert 1e-8 g Silver (Ag) to moles, then to Atoms of Ag
 (1e-8)/(MM Ag) = 9.271e-11 mol Ag(6.022e23) = 5.58625e13 Atoms Ag
Molar Mass and Mole-Gram Conversions
 To Find Molar Mass of Molecule: Multiply molar mass of each element by number of atoms, then add results
o Examples:
 Molar mass of CF2Cl2 = 1(MM C) + 2(MM F) + 2(MM Cl) = 120.914
 Convert 35g of CF2Cl2 to moles:
 35/MM of CF2Cl2 = 35/120.914 = .289mol CF2Cl2
 To Find number of atoms in a molecule:
o Convert into unit/molecules and multiply by corresponding number of atoms of element in the molecule
Percent Composition From Analysis:
 Take molar mass of element multiplied by its number of atoms in molecule, and divide by molar mass of
molecule
o Example: Percent composition of H in C6H10O2S
 (10(mm of H))/(MM of C6H10O2S) = 6.9%
Simplest Formula From Analysis
 Take percent of each element and use a 100g sample so each percent is converted to grams of an element = 30%
O = 30g O, Then divide by molar mass of element, then divide by lowest common term to find simplest form
o Example: 45.9% C, 2.75%H, 26.2% O, 17.50% S, 7.65% N
 45.9/mm C, 2.75/mm H, 26.2/mm O, 17.5/mm S, 7.65/mm N
 C3.822H2.728O1.638S.546N.546, divide all by .546 because it is the smallest
 C7H5O3SN
 If after you have divided each by smallest term and you are left with non-whole numbers, multiply by an integer
to get round numbers. Ex if you had C2.67, multiply by 3 = C8
Writing and Balancing Equations:
 Be extra sure to write molecules correctly, charge balancing if between metal and non-metal
 Make sure number of each type of element is same on both sides
Mole-Mass Relations in Reactions
 Basic Procedure: If you start with grams, convert to moles by dividing by MM of element. Then multiply
moles by ratio of element you are looking for over the element you know. Then multiply those moles of the
element you were looking for back to moles if asked to.
 Example: Phosphine gas reacts with oxygen according to equation
o 4PH3 + 8O2  P4O10 + 6H2O
o Find mass of Tetraphosphorous decaoxide produced from 12.43 mole of phosphine
o 12.43 mol Phosphine (1mol P4O10/4mol PH3) = 3.1075mol P4O10
o 3.1075mol P4O10 (mm of P4O10) = 882.184 g P4O10
Theoretical Yield Problems:
 Work backwards from theo yield
 Example: Problem 66
Limiting Reagent Problems:
 Obvious because problem gives you masses you have of the reactants (what combines to react)
 Work equation first starting with one reactant to find the amount needed of other reactant to react completely.
If you find you need a greater amount of the other reactant to react completely, the other reactant is the limiting
reagent. Then work the problem with the limiting reagent to find whatever other amount of molecule you react
with.
 Example: # 68
Combustion Reaction:
 Usually involves Carbon and Hydrogen, and you burn it

How To Solve: Find mass percentages of Hydrogen and Carbon in resulting molecules, and then multiply each
by corresponding masses of corresponding molecules. Go through usual procedure of Simplest Form, and you
are left with empirical formula of Hydrocarbon
Chapter 4
Molarity:
 M = mol/volume (L)
o Ex. How many moles of Potassium Chloride are there in 20 ml of KCL?
 .15M = mol KCL/.02 L
 .15M(.02)= .003 Mol KCL
 Dilutions: When you have a concentrate with a certain molarity that you add water to to dilute it.
o Use M1V1 = M2V2
o Calc. Volume of .2M stock solution needed to produce 600ml of a desired .05M solution
 .2V1 = .05(.6)
 V1 = .15 L then add water up to 600 ml
Double Displacement Reaction/Precipitations:
 In double displacement reactions, switch cations and anions and balance equations. If a reaction occurs, one
becomes solid and the other is soluble. The ions of the soluble one are eliminated from both sides of the
equation
Solubility Rules:
 Soluble:
o Nitrates, all alkali, Ammonium (NH4+)
o Sulfates: all except Ag, Ca, Sr, Ba, Hg, Pb
o Salts of F- except Mg, Ca, Sr, Ba, Pb
o Salts of Cl, Br, and I (AS IN STRONG ACIDS) except Ag, Hg, Pb
 Insoluble: NOTE: SOULBLE RULES OVER INSOLUBLE
o Carbonates
o Phosphates
o Sulfides
o Hydroxides
Acid – Base Neutralization Reaction:





Switch cations and anions. Form water when possible
Strong Acids:
o HClo HBr
o HI
o Sulfiric Acid (Sulfate)
o Nitric Acid (Nitrate)
o Perchloric Acid (HCLO4)
Strong Bases
o All alkali metal hydroxides
o Calcium Hydroxide
o Strontium Hydroxide
o Barium Hydroxide
In an acid – base equation, strong vs weak is based on the ability to dissociate
Net Equations:
o Strong acid/Strong Base:
 End up with water
o Weak Acid/Strong Base:
 End up with water and other part of acid. Because base is strong, it will become a spectator and
will be crossed out
o Strong Acid/Weak base
 End up with base plus an H+ and the second part of the acid will be crossed out
Titration


When you are given a solution of precisely known content and use it to find another solution.
o Steps:
 Write a balanced acid-base equation
 Convert molarity to moles
 Compare mole ratios
 Convert moles back to molarity
Example:
o 25ml of .02 Strontium Hydroxide is required to titrate 35ml of Nitric Acid to its endpoint. What is the
molarity of Nitric Acid.
 Sr(OH)2 + 2HNO3  2H2O + Sr(NO3)2
 Str base str acid
 Sr(OH)2 mols = (.02M)(.025L)= .0005 Mol Sr(OH)2 (2HNO3/1Sr(OH)2) = .001 Mol HNO3/.035L
= .029M
Redox Reactions




A redox reaction is when there is a transfer of electrons from both sides to the other sides
When you reduce the charge of the left side by adding electrons = reduction
When you increase the charge of the left side by adding electrons to the right side = oxidation
o When electrons show up on right = Reduction
o When electrons show up on left = Oxidation
Steps for doing a redox problem:
o Break the equation into 2 half reactions
o Balance elements except for oxygen and hydrogen
o Balance oxygen by adding water
o Balance hydrogen by adding H+ ions
o Balance charge by adding e-. Mult by each half reaction by the LCF of the two electron numbers.
o The two electrons should be on opposite sides of the two half reactions. Cancel them out
o Add the two sides of the half reactions together (left + left  right + right)
o Simplify water and hydrogen ions by adding and subtracting from sides
o IF IN BASE SOLUTION: Add an OH- for every H+ ion. Simplify by creating waters. Simplify waters,
H, and OH
Chapter 6
OVERALL EQUATIONS




V= or C=
o C= Speed of Light (3x108 m/s)
E=h
E=hC/
o H = Planks Constant 6.626x10-34
^E = -Rh(1/n2hi – 1/n2lo)


o Rh= 2.180x10-18J
d = h/mv
2r=nd
Waves




Frequency () = How many waves pass by a point in a certain amount of time
o Normally measured in Hertz (Hz) = 1/sec
Wavelength () = Distance between two consecutive identical points on a wave
o Measured in nanometers (nm) = 10-9m
EQUATIONS
o V= or C=
 Velocity=(wavelength )(frequency )
 Speed of Light (3x108 m/s) = (wavelength )(frequency )
Wave Spectrum
o
Frequency decreasing 
Gamma
Visible
Long Radio
Wavelength Increasing 

o Visible Light 400nm V I B G Y O R
Photoelectric Effect
o Frequency carries the energy of the wave
o EQUATIONS
 E=h
 Energy(J) = (Planks Constant 6.626x10-34)(Frequency)
 E=hC/
 Energy(J) = (Planks Constant 6.626x10-34)(Speed of light)/(wavelength)
o Light is a massless particle known as a photon
o Threshold Frequency = Minimum frequency required to knock off an electron off metal. Beyond the
threshold frequency, the intensity of the light (amplitude) determined how many electrons were ejected
per unit of time.
o Amplitude = refers to brightness (Photon/time)
o EQUATIONS
 ^E = -Rh(1/n2hi – 1/n2lo)
 Rh= 2.180x10-18J
 d = h/mv
 DeBroglie Wavelength = (Planks constant)/(mass of an electron 9.11x10-31)(velocity
 2r=nd
Quantum Mechanics
 Quantum Numbers and Fields
o Psi ()2 give the 90% probability region in 3D space
o Principle Quantum Number (n) = determines energy level and average distance from nucleus
o Angular Momentum Quantum Number (l) = determines shape of sublevel
 Possible Values: l = 0  (n-1)
 Field Letters: 0=S (Sphere), 1=P(3 dumbbells), 2=D (5), 3=F (7), 4=G (Some People Don’t Fart
Gas)
o Magnetic Quantum Number = determines spatial orientation of sublevel orbital
 Possible Values: -l  0 +l
o Spin Orbital Number (ms) = determines spin orientation of electron in an orbital

1
2
3
4
5
6
7
 Possible Values: 1/2
o *Each orbital can contain a maximum of 2 electrons, so S can have 2, P can have 6, D can have 10
Electron Filling in An Atom
o Aufbau Principle = Electrons fill energy positions from lowest to highest
o Hund’s Rule = When filling orbitals within an energy sublevel, electrons fill orbitals singly before
pairing
o Pauli Exclusion Rule = No two electrons in a given atom can possess the same set of four quantum
numbers
S-Sublevel
1s
2s
3s
4s
5s
6s
7s
D-Sublevel (1 Down from Period)
P-Sublevel
1s
3d
4d
5d
6d
2p
3p
4p
5p
6p
7p
F-Sublevel (2 Down From Period)
4f
5f

Spectroscopic Notation
o Ex. H = 1s1
 1 = (n) Principal Quantum Number
 s= sublevel
 1 = electron in that sublevel
o Ex. Br
 1s22s22p63s23p64s23d104p5
o Condensed: Go back to the previous noble gas and add notations until you reach the element
 Ex. Li = 1s22s1 or [He] 2s1
o *Any atom with filled outer S and P sublevels is considered Valence
 Orbital Box Diagrams
o Related to spectroscopic notation. Fill boxes in according to electrons in energy levels
o Ex. Fluorine 1s22s22p5
0
0
-1
0
+1


  
1s
2s
2p 2p 2p
 Exceptions To Aufbau’s Principle: Want to fill a whole energy level or at least half. Removes an electron from
previous full electron level to supplement
o Element: Chromium (Cr)
 Expected: [Ar] 4s23d4
 Actual: [Ar] 4s13d5
o Element: Copper (Cu)
 Expected: [Ar] 4s23d9
 Actual: [Ar] 4s13d10
 Periodic Table Trends
o Atomic Radius: Radius of an atom from nucleus to edge of electron cloud
 Decreases from left to right (along same energy level) = Due to increase in nuclear charge
 Increases top to bottom = Due to electron being placed in increasingly large energy levels
o Ionic Radius: Radius of an ion compared to its neutral counterpart
 Anions are always LARGER = Due to added repulsion among electrons
 Cations are always SMALLER: Due to fewer electrons attracted by the same # of protons in the
nucleus
o Ionization Energy: Energy required to remove an outer electron from an atom in its gaseous state
 Increases from Bottom to Top = Due to outermost electrons being closer to nucleus
 Increases Left to Right = Due to increase in nuclear attraction as protons are added
o Electro Negativity: Relative tendency for an atom in a bond to pull electrons toward it
 Increases Bottom to Top: Due to bonding valence electrons being closer to the nucleus
 Increases Left to Right: Due to increase in nuclear attractions as protons are added
Chapter 7
Covalent Bonding



Generally between non-metals
Electrons are shared to achieve a stable octet
Usually takes place between valance electrons (outermost s and p sublevels)
Lewis Dot Structure



Used to visualize electron structure
X Represents Element
AB represents electron pair
AB
AB

X
AB
AB


Writing Lewis Structures for Molecules:
o Determine the most likely arrangement of Atoms
 Lease EN atom is usually in the center
 Molecules are often symmetrical
 Hydrogen is never central and can only bond to one thing (duet)
o Determine the total number of valence electrons
o Fill in the octets of the outer atoms (place electrons in pairs around outer atoms)
o Check for an octet on the central atom
 If the central atom has an octet and there are no leftover electrons, then you are done
 If the central atom doesn’t contain an octet, consider a double or triple bond
 If there are electrons left over, place them as long pairs on the central atom
Extra Misc.
o Sometimes a line is used to represent a pair of electrons
Free Radicals



Molecules that have a lone electron (not paired)
Very reactive, harmful to health (cancer, aging, combated with antioxidants)
For example: NO2 has a free radical on the N
Resonance


Resonance occurs when there are more than one equivalent structure for the same molecule
For example, on NO3-, a double bond can be drawn from any one of the O’s
Bond Order

Bond order is the number of bonds shared between two atoms referring

For example, in NO3- since the double bond can be from any three of the O, the Bond Order is 1 and 1/3. This
is because there is a single bond between each O and N, and then there is one extra bond/3 O atoms to make it 1
and 1/3
Formal Charge





Helps you determine the best structure for a given compound
Formal charge must be determined for the each element in the molecule
Ideally, the formal charge would be zero. If it is not zero, it should still be a very small number (eg ± 1,2) with
the more negative charge on the more electronegative element
Formal Charge = Normal Valence – Nonbonding Electrons – ½ Bonding Electrons
For example, a double bonded oxygen would have the formal charge of 6-4-2=0
Electron Deficient Molecules



These are elements that are exceptions to the octet rule
Boron – B – Only needs three pairs of electrons to fulfill its valence shell
Beryllium – Be – Only needs two pairs of electrons to fulfill its valence shell
Expanded Valence

This is when a central element has more electron pairs around it than it needs to fill its valance shell
VSEPR


Stands for VALENCE SHELL ELECTRON PAIR REPULSION THEORY
Valence shells electron pairs will arrange themselves around a central atom to be as geometrically far apart as
possible for the lease repulsion
Major Classes
For an atom
with a Lewis
Dot Structure
of:
X A X
Less than an
octet
Octet
Greater than
an octet
X
X A X
X
X A X
X
X
X A X
X
X
X
X
X
A
X
X
X
The electronic
geometry will
be:
Linear
The bond
angle(s) will
be:
180o
(AX2)
Trigonal
Planar
120o
(AX3)
Tetrahedral
109.5o
(AX4)
Trigonal
Bipyramidal
Octahedral
Shape
120o, 90o
(AX5)
90o
(AX6)
*Electronic geometries refer to the arrangement of electron pairs around the central atom and will always be a major
class. The molecular geometries refer to the actual physical appearance of a molecule and may be a major class (no
nonbonding pairs on the central atom) or a subclass (one ore more nonbonding pairs on the central atom)
Subclass Geometries
Subclass Geometries: Those in which there are one or more nonbonding pairs of electrons on the central atom.
Trigonal Planar
Lewis Structure
Molecular Geometry
Bond Angle(s)
Bent
<120o
(AX2E)
Molecular Geometry
Bond Angle(s)
X A X
X
Trigonal pyramidal
<109.5o
(AX3E)
X A X
Bent
<109.5o
(AX2E2)
X A X
Tetrahedral
Lewis Structure
Trigonal bipyramidal
NOTE: Nonbonding electron pairs MUST be on the equatorial region of the molecule.
Lewis Structure
Molecular Geometry
Bond Angle(s)
X
See Saw
<120o, <90o
X A
(Distorted tetrahedron)
(AX4E)
X
X
X
X
A
T-shaped
<90o
(AX3E2)
Linear
180o
(AX2E3)
X
X A X
Octahedral
NOTE: Remove electrons from the axial regions BEFORE the equatorial regions.
Lewis Structure
Molecular Geometry
Bond Angle(s)
Square Pyramidal
<90o,<180o
(AX5E)
X
X
A
X
X
Square Planar
<90o
(AX4E2)
X
A
X
X
T-Shaped
<90o
(AX3E3)
A
Linear
180o
(AX2E4)
X
X
A
X
X
X
X
X
Hybridization



The blending of existing orbitals to produce a new set of orbitals that are consistent with the VESPR theory
Rules:
o Only 1 bond of a multiple bond is hybridized. The other bonds are unhybridized P-Orbitals
o Non-bonding electron pairs on the central atom are hybridized
o Sigma bonds , are bonds from hybridized orbitals. Pi bonds, , are unhybridized P-orbitals
Polarization



All polar molecules are considered dipole
Things to Look For:
o Nonbonding pairs on the central atom (usually polar)
o Symmetry about the central atom (usually non-polar)
o Differences in electro negativity (Bigger difference = greater polarity)
o THINK ABOUT THE MOLECULAR STRUCTURE
After deciding if polar or not, if polar draw charge arrow towards the most electronegative atom
Chapter 13/14 Acids, Bases, and Buffers
Review
 Arrhenius Definition:
o Acids that produce H+ ions in solution
o Bases that produce OH- ions in solution
 Bronstead-Lowry Definition
o Acid: Proton Donor
o Base: Proton Acceptor
o Ex.
 NH3 + H20  NH4+ + OH Water is called Amphiprotic
o Amphiprotic – Has a “proton” to donate or accept
 Strength of an Acid or Base depends on its ability to dissociate
 Strong Acids
o HNO3 = Nitric Acid
o HCl4 = Perchloric Acid

o H2SO4 = Sulfiric Acid
o HCl = Hydrochloric Acid
o HBr = Hydrobromic Acid
o HI – Hydroiodic Acid
Strong Bases
o LiOH CsOH
o Ca(OH)2
o Sr(OH)2
o Ba(OH)2
Neutralization Reactions
 Acid + Base  H2O + Salt
o Ex.
 HCl + NH4OH  H2O + NH4Cl
Conjugate Acid/Base Pairs
 “Differ by a hydrogen”
o Ex.
 NH3 + H2O  NH4- + OH Base Acid Conj A Conj B
Auto-Ionization of Water
 H2O + H2O  H3O+ + OH EQUATION: IONIZATION CONSTANT FOR WATER [H3O+][OH-] = 1.0 x 10-14
pH and pOH
 EQUATION: pH = -Log[H3O+]
[H3O+]=10-pH
-7
 For pure water, pH = -Log(1x10 ) = 7
 EQUATION: pOH = -Log[OH-]
[OH-]=10-pOH
 EQUATION: pH + pOH = 14
Dissociation
 EQUATION: %Dissociation = X/NM
o X is the dissociation of H3O+ or OH- and NM is the nominal molarity
 EQUATION: Ka = x2/NM-x
o Ka is the dissociation constant
Visual pH Tests
 Litmus Paper
o Blue Litmus
 If Blue Litmus goes from Blue  Red, the solution is acidic
 If it doesn’t change, the solution is basic
o Red Litmus
 If Red Litmus goes from Red  Blue, the solution is basic
 If it doesn’t change, the solution is acidic
 Wide-Range pH Paper
o Red ----------Blue
o Acidic
Basic
o Colors displayed correspond directly to pH
Acidic or Basic Anhydrides
 Metal Oxides tend to produce Basic solutions
o Na2O + H2O  2NaOH
 Non-Metal Oxides tend to produce Acidic Solutions
o SO3 + H2O  H2SO4
Polyprotic Acid
 These acids have more than one hydrogen capable of dissociating. Generally, each H+ on the acid has a
different Ka value
o Ka1>Ka2>Ka3
 Ex.
o H2SO4
o But not CH3COOH
Ka and Kb Values
 Both acids and bases have a Ka/Kb value
 EQUATION: KaKb=1x10-14
 *When using a Kb value to calculate pH, be aware that the molarity you find will be of the base ions and not
acid ions, therefore, find the pOH and then subtract it from 14 to get the pH
Acid – Base Neutralization Reactions
 Acid + Base  Salt + Water
 Different Cases:
o Ex.
 Write Balanced Equation: HCl + NaOH  NaCl + H2O
 Check for hydrolysis (Further dissociation of salt):
 Na+ + H2O ? NaOH + H+ It re-forms the original base, so it doesn’t count
 Cl- + H2O ? HCl + OHIt re-forms the original acid, so it doesn’t count
 If hydrolysis does occur, additional creation of H+ ions turns the solution acidic, whereas
OH- turns the solution basic
Buffer Systems
 A buffer system uses a combination of elements to create a system that resists dramatic changes in pH when an
additional acid or base is added
 The principal behind a buffer is if an acid is added, the base of the solution reacts with it, consuming an amount
of the base and adding that amount to the acid to balance it. Visa-versa with a added base
 EQUATION: Ka=(X*(Base M + x))/(Acid M –x)
 EQUATION: pH=pKa + Log ([B-]/[HB])
Chapter 18
Activity Series of Metals (Oxidation Potential)




Based on ability to lose electrons (oxidize)
The more potential an element has to oxidize, the higher its activity
Activity Series: (In decreasing order)
o K>Ca>Na>Mg>Al>Cr>Zn>Fe>Cd>Ni>Sn>Pb>H2>Cu>Ag>Hg>Au
Electrons Flow from more active to less active electrode
Standard Reduction Potentials (STP)


Standard Reduction Potential is the exact opposite of activity/oxidation potential. Therefore, a highly active
metal would have a low reduction potential, and a low activity metal would have a high reduction potential
For example, a metal with an activity of 2.5 would have an STP of –2.5
Standard Hydrogen Electrode (SHE) = 0.00 V
 Standard Conditions:
o Gases @ 1 ATM (=1 bar)
o Solutions @ 1 M
o Values measured at 298K (25C)
Voltaic/Galvanic Cell






Two electrodes are connected by a wire and a salt bride
The negative terminal is the anode, usually on the left. The anode is the more active metal, and is the place of
oxidation. (Oxidization is the taking away electrons)
The positive terminal is the cathode, usually on the right. The cathode is the less active metal (with a higher
reduction potential), and is the place of reduction. (Reduction is adding electrons)
*Electrons always flow from the more active metal to the less active, or else you have a negative voltage
Cell Diagram:
o Anode| Anode Solution || Cathode Solution | Cathode
o Double bar represents a salt bridge
Measuring Voltages
o Voltages are measured using standard reduction potentials.
o Etotal = Ecathode - Eanode OR
Etotal = Ereduction – Eoxidation
o For example, if the anode of a cell has an STP of -2 and the cathode has an STP of 3, then the total
voltage produced would be Etotal = 3-(-2) = 5
Concentration Cell



A potential difference can be created with two half-cells having the same metals in solutions containing their
ions if there is a molarity difference between the two solutions
La Chateliers Principal: A chemical system will react in such a way as to reduce any stress placed on that
system. AKA A system will try to balance itself by moving electrons to relieve stress
In most concentration cells, the anode is the place of the more dilute solution. The cathode is the place of the
concentrated solution. The dilute solution is oxidized, and ions are added to balance. The concentrated solution
is reduced and the ions are transferred to the dilute solution.
Batteries






LeClanche
o A LeClanche battery is the average battery used in CD players, stereos, and flashlights.
o The anode is the zinc metal casing and the cathode is the NH4Cl and ZnCl2 paste.
o 2MnO2 + 2NH4+ + 2e-  Mn2O3 + 2NH3 + 2H2O
o Cons: Large currents cause the battery to produce a buildup of NH3 gas that lowers voltage
Alkaline Batteries
o KOH is used instead of NH4Cl
o Zn +2MnO2  ZnO + Mn2O3
Button Battery (Mercury HgO)
o The small, flat battery used in watches and calculators
o Zn + HgO + H2O  Zn(OH)2 + Hg
o If made with silver, there is no toxic mercury
Lead Acid (Car Battery)
o Produces a very large current, but contains lead and sulfuric acid, and it is very large and heavy
o Good because it recharges very well
o Pb + PbO2 + 2H+ + 2HSO4-  PbSO4 + 2H2O
Fuel Cell
o In a fuel cell, oxygen and hydrogen combine to drive a reaction that creates pure water as a byproduct
o 2H2 + O2  2H2O
NiCd
o NiCd batteries suffer from weight, are toxic, and have bad memory problems (ability to recharge
decreases over time)
Cathodic Protection

With cathodic protection, an active metal in a place is prevented from oxidizing away. A more active metal is
connected to the original piece and is “sacrificial” and is oxidized instead of the original piece. This saves the
original from oxidizing away.
Electrolysis






Using electrical energy to drive a reaction in a non-spontaneous direction. Instead of using chemistry to
produce electricity, electrolysis uses electricity to produce chemistry.
Anodes and cathode polarity is switched (anode is now positive and cathode is negative)
*In battery diagram, long line ( | ) is positive and short line ( | ) is negative
Follow Specific steps to determine aspects of electroplating problems
o Amps = Charge/Time. Usually Amps= Coulombs/second
o With the given information, solve the amps equation for the missing value.
o To continue to find moles or grams, divide Coulombs by the Faraday constant=96480 coulombs/mole e-.
o Use the molecular coefficients to find mole ratios
o Convert moles of a product to grams
Example: How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current
of 2.00 A is run through a solution of Cu(SO4) for a period of 20 minutes.
 2.00 A (20 min)(60 sec/min) = 2400 Coulombs
 2400 C /96480 C/mol e- = 5/201
 5/201(1/2) = .0124 (Copper is oxidized from Cu to Cu2+, which means it needs two electrons for
every molecule, so a mole of electrons can only do half a mold of copper, therefore we multiply
by ½)
 .0124(mm of Cu = 63.546)= .79g Cu
If given moles or grams of a substance, just work backwards to get back to amps, charge, or time
o Example:
 How many hours would it take to produce 75g of metallic Chromium by the electrolytic
reduction of Cr3+ with a current of 2.25 Amps
 Cr3+ +3e-  Cr
 75g/(mm of Cr = 51.9961) = 1.442
 1.442(3) = 4.32 (3 is the number of electrons needed for each molecule, notice it is the
reverse of the usually downward process)
 4.32 (96480) = 417492 C
 417492 C/2.25 A = 18552.37 Seconds/(60 sec/min * 60 min/hr) = 51.54 hrs
*This document is the work of a student, and is not reviewed by Mr. Feebeck. The information is not
necessarily accurate.