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Bundle for Exam 4

by: mkennedy24

Bundle for Exam 4 Chem 109


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These notes cover material in all 3 chapters for exam 4 with information from both the textbook and lecture
General Chemistry
Eric Malina
General Chemistry, Chemistry, Chem, 109
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This 26 page Bundle was uploaded by mkennedy24 on Thursday April 21, 2016. The Bundle belongs to Chem 109 at University of Nebraska Lincoln taught by Eric Malina in Spring 2016. Since its upload, it has received 50 views. For similar materials see General Chemistry in Chemistry at University of Nebraska Lincoln.

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Date Created: 04/21/16
Chapter 8: Periodic Properties of the Elements  Section 8.3 o Mendeleev’s periodic table is based on periodic law which is when elements are arranged in increasing mass certain properties reoccur periodically o Periodic Property: One that is predictable based on an element’s position within the periodic table First N=1 L=0 Ml0 M s+1/2 o Elect Electron ron Second N=1 L=0 Ml0 M s(-1/2) Electron Number of electrons in orbital 1s2 Orbital Configuration: Shoes the particular orbitals that electrons occupy for that atom  Helium  2 electrons  **Minimum Energy Principle (MEP): Always obtain the lowest energy  The above configuration is the ground state configuration of He, the lowest energy configuration for He o Ground State: Lowest Energy Configuration o Excited State Highest Energy Configuration  Another way to show the electron configuration for elements is using orbital diagrams: Symbolizes the electrons as an arrow and the orbital as a box o Orbital diagram for He: **Direction of the arrow corresponds to the electron spin  (+1/2) o Coulomb’s Law 1s corresponds to up arrow  (-1/2) corresponds Q’s are charges to down 1 Q 1 2 E= 4πε R R is distance o Orbital Stability  Shielding: Inner electrons shield the nucleus from the outer electrons  ZEFF: Effective Nuclear Charge (ZEFF= Protons – E sE >Ep> d - Inner e ) E f o Aufbau Principle: “Build-up” Principle Experiences net charge of about 1+ Penetration Shielding Experiences full 3+ charge e-- e-- e-- e-- 3+ Nucleus e-- Nucleus e--  1 electron: lowest energy nd  2rdelectron: next lowest energy  3 electron: next lowest energy  and so on. . .  The point of the principle is to fill the lowest energy orbital first:  Example: Lithium Li  1s 2s 1  The lowest energy orbital (1s) is filled up first before the rest o Hund’s Rule: Degenerate orbitals (same energy) fill with electrons of same spin first (Don’t pair up electrons until you have to!)  Example: What are the 4 quantum numbers for the highest energy electron in a ground state Oxygen atom?  **Oxygen has 8 electrons : 1s 2s 2p2 4 Box notation for a ground state oxygen atom 1s 2s 2p o Objective: Be able to translate to box notation  Summarizing Orbital Filling  Electrons occupy orbitals so as to minimize the energy of the atom; therefore, lower energy orbitals fill before higher energy orbitals (Diagonal Diagram )  Orbitals can hold no more than two electrons each. When two electrons occupy the same orbital, their spins are opposite. This is another way of expressing the Pauli Exclusion principle (no two electrons in one atom can have the same four quantum numbers)  When orbitals of identical energy are available, electrons first occupy these orbitals singly with parallel spins rather than in pairs. Once the orbitals of equal energy are half full, the electrons start to pair ( Hund’s Rule)  Section 8.4 o Valence Electrons: Electrons involved in bonding. Held most loosely to the nucleus which is why they are easier to share  Outer electrons AND UNFILLED d electrons  Trick: The group number of main group elements is the number of valence electrons the element has.  Example: Nitrogen (N)  Group 5A  5 Valence electrons  Example: Fluorine (F)  Group 7A  7 Valence electrons o Core Electrons: All non-valence electrons o Inner Electrons: Lower n-value o Outer Electrons: Highest n-value o Example: Titanium (name valence, core, inner, and outer) 1s 2s 2p 3s 3p 4s6 2 3d 2  Inner  20 electrons (lowest n-value(s))  Outer  2 electrons ( highest n-value)  Valence: Highest n-value + any unfilled d orbital electrons 2 2  4s + 3d  2+2=4 valence electrons  Core: All the rest  22-4=18 core electrons o Summarizing Periodic Table Organization  The periodic table is divisible into four blocks corresponding to the filling of the four quantum sublevels (s, p, d, f)  The group number of main-group element is equal to the number of valence electrons for that element  The row number of main-group element is equal to the highest principle quantum number of that element o Objective: Determine electron configuration, valence electrons, and core electrons based on position in periodic table o Nobel Gas Notation i.e. Core Notation  [Ar] 4s  the core notation for Calcium (Ca)  Exceptions:  Cr  [Ar] 4s 3d BUT actual configuration is [Ar] 4s 3d because a half full and full orbital is more stable than something less or more  Mo  Cu  Ag  Section 8.5 2 2 6 1 o Na  1s 2s 2p 3s  unstable because subshell is not full, causing sodium to want to loose electrons + 2 2 6  Na  1s 2s 2p  Now by loosing an electron, all subshells are full, making it stable. There are not more protons ( since # of protons never change) than electrons resulting in a (+) charge  Section 8.6 o Key ideas . . .  N-value  Z EFF s i Increasing Atomic Radius d R i m o A g i a r n I o Van der Waals radius/ non-bonding atomic radius: radius of an atom when it is not bonded to another atom o Covalent Radius:  Nonmetals: ½ the distance between 2 of the atoms bonded together  Metals: ½ the distance between 2 of the atoms next to each other in a crystal of the metal  Example: Distance between Br atoms is Br is 2 228 pm Br Br 228pm o Core electrons efficiently shield electrons in the outermost principle energy level from nuclear charge, but outermost electrons do not efficiently shield one another from nuclear charge o Summarizing atomic radii for main-group elements  As we move down a column in the periodic table, the principal quantum number (n) of the electrons in the outermost principal energy level increases, resulting in larger orbitals and therefore larger atomic radii  As we move to the right across a row in the periodic table, the effective nuclear charge experienced by the electrons in the outermost principal energy level increases, resulting in a stronger attraction between the outermost electrons and the nucleus and smaller atomic radii  Section 8.7 o Objective: Write electron configurations for any ion o Negative ions follow the rules  Cl  [Ne] 3s 3p 5 - 2 6  Cl  [Ne] 3s 3p  (-) adds to exponent (electrons)/ subshell; (+) subtracts exponents (electrons)/ subshells o Positive ions  outermost first  Geranium  usually is a Ge 2+ and Ge 4+ 2 10 2  Just the element: [Ar] 4s 3d 4p o Transition Metals  Remove the electron in the highest n-value orbitals first, even if it does not correspond to the reverse order of filling  Example: Vanadium 2 3 o V: [Ar] 4s 3d o V : [Ar] 4s 3d 3 o Magnetic Properties  Objective: Determine if atom/ion has magnetic properties  Spin state of the electron produces a magnetic field  Any unpaired electron will produce a magnetic field  Diamagnetic: Does not have a magnetic field (all electrons are paired)  Paramagnetic: Does have a magnetic field (at least 1 unpaired electron) o Example: Silver (Ag)  47 electrons  Paramagnetic 1 10  [Kr] 5s 4d o **Just because there are an even number of electrons does not mean it is diamagnetic  Ionic Radius  Objective: Rank ions by size and explain the 2 2 6 trend 2 2 6 1 1s 2s 2p  Cation versus Atom (that cation is related too) ZEFF= (11 protons – 2 + ZEFF= (11 protons – inner electrons) = o Na v.s. Na 10 inner electrons) +9 Therefore, cation is smaller than atom = +1 because of the difference in the effective nuclear charge  Cation. . . IF isoelectric o Isoelectric: exact same electron configuration BUT NOT the same number of electrons o Mg 2+  1s 2s 2p  Same electron configuration as Neon  The greater the charge the smaller the ion o Objective: Be able to compare a cation to another cation and a cation to its parent atom Anions. . . adding electrons o F  1s 2s 2p 5 o F  1s 2s 2p 6  The anion has more electron electron repulsion o Compare: anion to parent atom  anion is much larger! o If isoelectric. . .  Greater negative charge o Ionization Energy (endothermic + ∆ H)  Energy required to remove one electron from a gas phase atom or ion 1E = Mg(g)  Mg (g) + 1e ; IE 2 Mg (g) Mg 2+ — (g) + 1e Determine ionization energy (IE) (each removal of 1 electron is a different ionization energy)  Objective: Rank atoms/ions by ionization energy Ionization energy incr ZeasFs left to right (gets harder to pull electrons out because the effective nuclear charge increases) Ionization energy increase Effective nuclear charge 2 2 6 1 + is huge when trying to  IE1Na  1s 2s 2p 3s (+1 Z EFF  Na  take second IE because 1s 2s 2p (+9) sodium wants to be a  Summarize IE for Main-Group elements: +1 cation not a +2  Ionization energy generally decreases as we move down a column (or family) in the periodic table because electrons in the outermost principal level are increasingly farther away from the positively charged nucleus and are therefore held less tightly  Ionization energy generally increases as we move to the right across a row (or period) in the periodic table because electrons in the outermost principal energy level generally experience a greater effective nuclear charge  Section 8.8 o Electron Affinity: Energy change associated with adding an electron to a gas phase atom/ion (exothermic reaction: releases heat - ∆ H)  O(g) + 1e  O(g) EA — 1 -- -- 2-  O + 1e  O (g) EA 2 E1  closer to electron configuration; larger negative value o Summarizing Electron Affinity: Decrease: Size Metallic Character Increase: Ionization Energy Negative Electron Affinity  Most groups (columns) of the periodic table do not exhibit any definite trend in electron affinity. Among the group 1A metals, however, electron affinity become more positive as we move down the column (adding an electron becomes less exothermic)  Electron affinity generally becomes more negative (adding an electron becomes more exothermic) as we move to the right across a period (row) in the periodic table o Trends in Metallic Character  Metallic Character: Behave like a metal (decreases up and across periodic table) Chapter 9: Chemical Bonding I – Lewis Model  Section 9.1 o Lewis Model: Valence electrons represented as dots produces Lewis Electron Dot Structures (Simplest electron model) o Chemical Bonding: MEP (minimum energy principal); minimize the energy  Chemical bonds form because they lower the potential energy between the charged particles that compose atoms Types of Atoms Type of Bond Characteristic of Bond Metal + Nonmetal Ionic Electron transferred Nonmetal + Covalent Electron Shared Nonmetal Metal + minimize energytallic ProElectron Pooledpulsion Electron + Electron Repulsion maximize energy Electron + Proton Attraction  The closer to the nucleus the more stable  Section 9.2 o Objective: Describe each type of bonding o Ionic: Electron transfer (Metal + Nonmetal)  Example: Sodium Chloride Does not conduct electricit Na+[ Cl ]- y o Covalent: Electron sharing (Nonmetal + Nonmetal) Conducts  Example: H O2 electricity  Metallic: Electron sharing (Just metals are present i.e. (Delocaliz Metal + Metal or Metal by itself) ed o “Electron Sea Model” electrons)  All of the atoms in a metal lattice pool their valence electrons  Pooled electrons no longer localized on a single atom but delocalized over the entire metal  Section 9.3 o Objective: Draw Lewis Symbol for any main group element o Examples: Li Be B C N O F Ne s1 s2 s2p1 s2p2 s2p3 s2p4 s2p5 s2p6 o Octet Rule: Everything wants to have the noble gas configuration usually stable arrangements have 8 electrons o Duet: Hydrogen and Helium are stable with 2 electrons Section 9.4 o Objective: Represent ionic compound with Lewis symbols NaCl Na + Cl Na Cl Wrong Structure! Sodium clears out 3rd oChlorine takes sodium’s electron to get an octet Na+ + Cl-Na+[ Cl ]- Right Structure! o Lattice Energy: Energy change when gas phase ions form a solid; Energy associated with the formation of a crystalline lattice of alternating cations and anions from the gaseous ions  Example: NaCl + — Releases energy in Na (g) + Cl (g)  NaCl(sthe process But where does this energy come from?? The transfer of an electron from sodium to chlorine, by itself, actually absorbs energy  Objective: Use Born-Haber cycle to calculate lattice energy (an application of Hess’s Law) Steps: o 1. Start with elements and correct mole ratios o 2. Get everything into gas phase o 3. Get all to gas phase atoms  because with ionic substances, ionization energies are atom by atom o Get all to gas phase ions o Calculate Lattice Energy  All components including lattice energy add up to heat of the reaction  Ex: Construct Born-Haber Cycle and ΔiHdLatBr K: K(s) + ½ Br2(s) KBr(s) Hsub = 77.08 kJ/mol IE = 418.6 kJ/mol Hsublimation Hvaporization x ½ Br: BDE = 193 kJ/mol Solve for Lattice energy EA = -324.6 kJ/mol K(g) ½ Br2(g) Hvap = 29.96 kJ/mol BDE x ½ KBr: IE1 Hf= -393.8 kJ/mol Br(g) K+(g) Hf= Hsub + IE + ½ Hvap + ½ BDE + EA + HLat EA1 -393.8 = 77.08 + 418.6 + ½ (29.96) + ½ (193) + (-324.6) + HLat HLat= (-677) kJ/mol Br—(g)  Objective: Rank compounds by their lattice energies Example: Arrange these ionic compounds in order of increasing magnitude of lattice energy: CaO, KBr, KCl, SrO o KBr and KCl should have lattice energies smaller than SrO and CaO because of their charges. SrO and CaO have +2 and -2 and KBr and KCl have +1 and -1 charges o Br is bigger than Cl because of atomic size o CaO and SrO should have larger lattice energies because their charges are larger than the others o SrO has a larger ionic radius which equals a lower magnitude o KBr< KCl< SrO< CaO Highest Lattice Energy  Biggest negative number Lowest Lattice Energy  Smallest negative number Size is smaller. . . **Charges always trump o Lattice energy is larger negative value size. SO if it comes to o NaF  Na= +1 and F= -1 equal charge in two o CaO  Ca = +2 and O= -2 different compounds, size  The charges change the Q values in Coloumb’s equation  The larger the charge, the larger the lattice energy overall  Summarizing Lattice Energy Trends: Lattice energies become less exothermic (less negative) with increasing ionic radius Lattice energies become more exothermic (more negative) with increasing magnitude of ionic charge Section 9.5 o Lewis Structures: Representing valence electrons in molecules and polyatomic ions o Objective: Draw Lewis Structures  Example: 2 O  Example: 2 H O H H O H O O Lone pair/ nonbonding pair Single Bond  Example: N Double Bond 2 Bond Strength: Section 9.6  Single: lowest N N strength Double Increase  Triple: Better Triple Bond stability/ strongest Bond Length: o Electonegativity: Ability to ho Single: Longest attract shared electrons (“Gree Double factor”)  Triple: Shortest o Objective: Rank by electronegativity  Inversely related to atomic size: larger the atom, leNoble gases are not ability it has to attractincluded because they electrons to itself in a are stable covalent bond  Main-group element trends Electronegativity generally increases across a period in the periodic table Electronegativity generally decreases down a column in the periodic table Fluorine is the most electronegative element Francium is the least electronegative element (sometimes called electropositive) o Bond Polarity: Partial charge across a bond due to unequal sharing  Classify Bond Polarity Pure covalent: equal sharing (Ex: C2 ) Cl Cl  Polar covalent: unequal sharing H Cl  Ionic (nonmetal + metal) o Electron almost completely transferred Na+ Cl--  Measure Dipole Moment: Anytime there is a separations of positive and negative charge Section 9.7 o Rules for Drawing Structures  1. Skeletal Structure H atoms are always terminal( i.e. at the ends) More electronegative atoms actually tend to go to the outside of the molecule Less electronegative atoms will be more central  2. Total valence electrons Adjust for charge o Polyatomic ion: (-) ion  add an electron; (+) ion  subtract an electron  3. Distribute electron into molecule Start with skeleton (1 step) Fill octets ( or duets) o Fill electrons in the order of outside to inside  4. Complete Octets (only if necessary)  5. Minimize formal charges Section 9.8 o Resonance: Half the time the bond is single and the other half its double or depending on the structure (Resonance structure for Ozone also the best lewis structure for ozone) o Hybrid resonance structure:  Uses dotted lines to explain which bonds switch o Formal Charge  Fictitous charge accounts for which element “claims” with eValence electrons as an element  Formal Charge = -Al( lone eec+tonnsing ele)trons Section 9.9 o Exceptions to the Octet rule:  Odd # of valence electrons  free radicals Only write the best lewis structure that you can! Everything in this structure has a formal charge of 0 therefore it is the best lewis structure.  Incomplete Octets  Expanded Octets: No expanded octets for row 1 and 2 on periodic table Section 9.10 o Bond Energy: Energy required to break 1 mole of a bond Cl2 2Cl Visually Breaking in ½ All fluorines have full octets and the formal charges are minimized therefore this is the best lewis structure even though sulfur has 12 o Objective: Estimate Δ H fased on average bond energy o **Key idea to remember: Not all bonds are the same, which is why we are given AVERAGE bond energies  Example: Estimate Δ Hfusing bond energies for CH 4 Cl 2 CH C3 + HCl Bonds Brokened + +  A reaction is exothermic when weak bonds break and strong bonds form  A reaction is endothermic when strong bonds break and weak bonds form Bonds (Left Side): Bonds (Right Side): 4 C-H bonds (4)(414) 3 C-H bonds (3)(414) 1 Cl-Cl bond (1)(243) 1 C-Cl bond (1)(339) = 1899 1 H-Cl bond (1)(431) = 2012 Broken – Formed = Δ H f1899 – 2012 =  Any bond formed is energy released into system  Any bond broken is energy absorbed o Bond Length  Objective: Rank bonds by length Bond Length: o  Single: Longest o  Double o  Triple: Shortest  How do you determine length differences in same bonds? Size decreases F2 Cl2 F-F Cl-Cl Longe r Chapter 10: Chemical Bonding Part II  VSEPR Theory: Based on the simple idea that electrons groups which we define as long pairs, single bonds, multiple bonds, and even single electron repel on another through columbic forces Valence Shell Electron Pairs Repulsions Lone pair: 1 group Single Bond: 1 group Double Bond: 1 group  Objective: Use VSEPR to predict: o Electron geometry: shape made by electron groups that are present o Molecular geometry: Shape made by nuclei that are present o Bond angles: Angle produced between bonds  Ideal: 180°, 120°,; i.e. symmetry  Non-Ideal: non-symmetrical  Two groups of electrons: o BeCl 2Linear Geometry with bond angles of 180°  Three groups of electrons: o Electron Geometry: Trigonal Planar with bond angles of 120° Double bonds need more space therefore have greater  Four groups of electrons: o Electron geometry is tetrahedral and has bond angles of 109.5°  Five groups of electrons: o Electron geometry is a trigonal bi-pyramid o Bond angles: Anything from equatorial to axial is 90°. Anything form axial is 120°  Six groups of electrons: o Electron geometry is octahedral with bond angles of 90°  But what does resonance do in a particular structure? - o NO 3 o Are bond angles equal? YES!  Effects of Lone Pairs o Key Idea: Lone pairs require more space Lone pair – Lone pair > Lone pair – Bonding pair > Bonding pair – Bonding pair  Lone Pairs  “Invisible” in molecular geometry o 3 groups of electrons  electron geometry that is Trigonal Central Atom A X E The formation of ozone is bent Ozone is a because of the lone pair that is on the central atom repulsing AX 2 model # of bondsthe other bonds further away because lone pairs want to maximize space Planar o 4 groups of electrons  electron geometry is Tetrahederal  1 Lone pair: Example/ NH 3 In the electron geometry, this lone pair is visible, but in the molecular geometry they are “invisible” meaning they are there causing difference in the shape. NH3 is a Trigonal pyramid because the lone pair repulses the other bonds. NH3 is an AX3E AX2E2  2 Lone pairs: Example/ H O2 o 5 Groups of electrons  Electron geometry is Bi-Pyramid  1 Lone pair: Example/ SF 4 AX4E Since sulfur is in period 3 and can obtain an expanded octect, this lone pair is aloud. REMEMBER TO ALWAYS MAKE SURE THERE ARE THE CORRECT AMOUNT OF ELECTRONS   2 Lone pairs: Example/ BF 3 AX3E2 AX2E  3 Lone pairs: Example/ XeF 2 o 6 Groups of electrons  Electron geometry is Octahedral  1 Lone pair: Example/ BrF 5 AX5E  2 Lone pairs: Example/ XeF 4 AX4E2 o Larger Molecules  Determining geometry o Summarizing VSEPR Theory Around this Carbon: AX4 Tetrahedral Bent Linear shape around Oxygen Around this Carbon: AX3 Trigonal Planar Around this Nitrogen: Trigonal Pyramid **Be able to notice the geometrical shapes in large molecules**  The geometry of a molecule is determine by the number of electron groups on the central atom (or on all interior atoms, if there is more than one)  The number of electron groups is determined from the Lewis structure of the molecule. If Lewis structure contains resonance structures, use any one of the resonance structures to determine number of electron groups  Each of the following counts as a single electron group: a lone pair, single bond, double bond, triple bond, or a single electron (as a free radical)  In general the electron group repulsions vary as follows: Lone pair – Lone pair > Lone pair – Bonding pair > Bonding pair – Bonding pair  Molecular Polarity o When the change is electronegativity is 0, it is perfectly nonpolar (no difference in how the atoms are sharing electrons) o Objective: Determine if molecule is polar based on bonds and shape Have a symmetry about that central atom and that symmetry has a cancel of bond polarities making it look nonpolar But because both sides cancel the overall molecule has a change in electronegativity of 0 o Example: CO 2 o Example H O 2 o Example: CCl 4 Everything in this molecule is pulling form the central atom equally because all the atoms around Carbon are the same making this a nonpolar molecule Point of symmetry= nonpolar molecule The electrons come at each other at an angle instead of head on making the electronegativities not cancel making it polar Because there is no symmetry due to the lone pairs in the electron geometry, this molecule is polar Because central atom(which is carbon even though that is not pictured) is not being pulled equally (due to different atoms) this molecule is polar because the electronegativities do not cancel like in CCl4 o Example: CCl F 2 2 o Determining Polarity:  1. Lewis Structure  2. Shape Ozone is an exception to these  Polar bonds? rules o No?  nonpolar o Yes?  is there a point of symmetry?  Yes? Nonpolar  No? Polar Trigonal Planar Trigonal Planar Point of symmetry (between nitrogen atoms)= nonpolar molecule o Example: N O 2 4


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