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Chapter 8 & 9 Chem 109

by: mkennedy24

Chapter 8 & 9 Chem 109 Chem 109


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These notes cover material in lecture as well as the book. They also include several examples and visuals for better understanding.
General Chemistry
Eric Malina
Chemistry, General Chemistry, Chem 109
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This 18 page Bundle was uploaded by mkennedy24 on Thursday April 14, 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 48 views. For similar materials see General Chemistry in Chemistry at University of Nebraska Lincoln.


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Date Created: 04/14/16
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


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