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In Exercises 2730, use the graph of to describe the

College Algebra | 9th Edition | ISBN: 9781133963028 | Authors: Ron Larson ISBN: 9781133963028 204

Solution for problem 5.1.27 Chapter 5

College Algebra | 9th Edition

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College Algebra | 9th Edition | ISBN: 9781133963028 | Authors: Ron Larson

College Algebra | 9th Edition

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Problem 5.1.27

In Exercises 2730, use the graph of to describe the transformation that yields the graph of g gx3 fx3 x 1 x,

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This study guide is a reference for the exam, not all material for the exam Electron Configuration: Shoes the particular orbitals that electrons occupy for that atom  Helium  2 electrons First N=1 L=0 M=l M s+1/2r of electrons in orbital Electron 1s2 Second N=1 L=0 M=l Ms=(-1/2) Electron Orbital  **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) corresponds 1s to up arrow o Coulomb’s Law  (-1/2) o Orbital Stability corresponds to down o o o Q’s are charges o Aufbau Principle: “Build-up” Principle R is distance E sE >Ep> d E f st  1 electron: lowest energy  2 ndelectron: 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 2 1 Li  1s 2s  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 2p 2 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 ) 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  [Kr] 5s 4d 10 o **Just because there are an even number of electrons does not mean it is diamagnetic  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)  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 4s 2 3d2  Inner  20 electrons (lowest n-value(s))  Outer  2 electrons ( highest n-value)  Valence: Highest n-value + any unfilled d orbital electron2 2  4s + 3d  2+2=4 valence electrons  Core: All the rest  22-4=18 core electrons Objective: Determine electron configuration, valence electrons, and core electrons based on position in periodic table  Exceptions:  Cr  [Ar] 4s 3d BUT actual configuration is [Ar] Positive ions have 1 5 electrons come from 4s 3d because a half full and full orbital is outermost orbital which more stable than something less or more is the largest coefficien Mo or largest n-value  Cu  Ag ** Key Idea Students usually forget ** 2s orbitals are more stable then 2p orbitals. WHY The closer to the nucleus, the more stable Size Sphere within a sphere (2s) iMetallic Character 2 lobes which is what 2p orbitals look like Increase: Ionization Energy Negative Electron Affinity 2p doesn’t have this inner space or the small blue bump made by the blue graph which indicates2s orbitals have higher probability of penetrating the nucleus s Increasing Atomic Radius d a R i m t A n s a r n I o Objective: Represent ionic compound with Lewis symbols NaCl Na + Cl Na Cl Wrong Structure! Sodium clears out 3rd orbitChlorine 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(s) the 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 find Δ HLatKBr (Example Continued onto next page) 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 o Bond Energy: Energy required to break 1 mole of a bond Cl2 2Cl Visually Breaking in ½ 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 H fsing bond energies for CH4+ Cl 2 CH C3 + HCl Bonds Formed + + Bonds Broken  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) = 1899 1 C-Cl bond (1)(339) = 2012 1 H-Cl bond (1)(431) Broken – Formed = Δ Hf 1899 – 2012 = (-113) kJ  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 o Size decreases F2 Cl2 F-F Cl-Cl Longe r When constructing shape, REMEMBER TO ALWAYS COUNT ELECTRONS Central Atom A X E # of lone pairs # of bonds o Summarizing VSEPR Theory BE ABLE TO  The geometry of a molecule is determine by the RECOGNIZE number of electron groups on the central atom (or on ELECTRON AND all interior atoms, if there is more than one) MOLECULAR  The number of electron groups is determined from GEOMETRY A the Lewis structure of the molecule. If Lewis structure WELL AS THE DIFFERENCE BETWEEN THE 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 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**

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Textbook: College Algebra
Edition: 9
Author: Ron Larson
ISBN: 9781133963028

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In Exercises 2730, use the graph of to describe the