Chem 111 Test II Study Guide
Chem 111 Test II Study Guide CHEM 111 - 01
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This 22 page Study Guide was uploaded by Lacy Marie Jones on Wednesday September 28, 2016. The Study Guide belongs to CHEM 111 - 01 at College of Charleston taught by Dr. Lavrich in Fall 2016. Since its upload, it has received 67 views. For similar materials see General Chemistry I in School of Science/ Mathematics at College of Charleston.
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Date Created: 09/28/16
1 Note Taker= Lacy Jones Professor/ institution=Dr. LavrichCofC Book= Atoms First 2nd edition by Julia Burdge and Jason Overby Chem 111Test II Study Guide Key: * = This means that Dr. Lavrich Stressed this during class, or the book made note of it. → This arrow means furthering information of the statement () This means that there is an AKA form or side note, or there is a source. My notes contain images from sources other than the Atoms First book, so I have to cite. This color means book topic Headlines This color is for definitions This color is for example problems This color with a high light is the answer This color means that the notes come from lecture and what day This color means Lecture Headlines This color is for the very end of the notes pages and covers key constants and Equations, and also difficult concepts. September 5th chapter 3 Lecture Notes~ For all atomic orbitals (s, p, d) there is a nucleus center and electrons orbiting→ there are different types of energy to describe the movements of these electrons. 2 (Weekes, Naomi) Types of Energy: 1. Kinetic Energy (Ek) → Energy of motion, the average speed of an electron is very fast so Ek aids in explaining. Ek=1/2mu^2 where u=the velocity 2. otential Energy → Stored Energy I. Chemical (Bonds) II. Electrostatic=Attraction of charged particles Eel ∝ ((q1)(q2))/d where qi=charge of particle, d= distance between them, and the quotient is in proportion to the electrostatic energy Energy Unit: Joule (J) ↳Ek of 2 kg mass moves at 1m/ s Ek= 1/2mu^2 → ½(2kg)(1m/s)^2 1J= (1Kg m^2)/ s^2 3.6: 3 a. 25kg moving 61.3 m/s Ek=½ (25kg) (61.3 m/s)^2 =4.7X10^4 J B. Be atom moving 275 m/s (Be atom)(1 mol/ 6.022X10^23 atom) (9.012 gBe/1 mol)(1kg/ 1000g) =1.497 X 10 ^26 Ek= ½(1.497X10^26)(275m/s) =4.7X10^4 J 3.8: Calculate the velocity of an electron with an Ek=3.14X10^21J u^2= 2Ek/m → u=+/ √2Ek/m→ u=+/√(2.34X10^21J/m) ↪u=8.65X10^4 Light: ↳Comes in waves * c=λν (Where c is the speed of light, λ is the wavelength, and v is the frequency. 1) Wavelength~ R epresented by λ, and is the distance between identical points. The unit is in nm (nanometers 1.0X109m). 2) Speed~ Represented by c, and is the speed of light constant= 3.0X10^8m/s 3) Frequency~Represented by V. The number of waves that pass a point of reference per second 1/s, s^1, hz → All of these units means Hertz 4) Amplitude~ The distance between the mid point and crest of a wave Electromagnetic spectrum λ(nm): Left is low energy and increases towards the right. 4 λ(nm) 10^11 10^7 10^4 700400 10 .1 .001 Radio Microwa Infrared Visible UV XRays Gamma ves (Sun) Rays (“Hubble Site”) 3.16: a. Calculate V of light having λ= 456 nm (3.0X10^8m/s)/456 X 10^9m =6.58X10^14 s^1 B. λ(nm) of light w/V= 2.45X10^8/ 2.45 X 10^9 hz= 1.22X10^1m 1.22X10^8 (1nm/1.oX10^9m) =1.22X10^8 nm 3.18: Calculate the time (sec) required for video feed on Mars rover to reach Earth. 5 1.3X10^8 mi Earth ↔Mars Distance 1 mi= 1.61km → 1.3X10^2mi (1.1 Km/ 1mi)(1000m/1km) 1sec/3.0X10^3 m/s =7.0X10^2 sec= 11 min Quantum Theory: ↳Laws of Physics for atoms and subatomic particles (Protons, neutrons, etc.) ~Rules (19001920) different for small particles 1) Quantization → noncontinuous discrete values 2) Wave/particle duality Electrons can move in wave like| destructive instructive 3) Heisenberg Uncertainty Principle X, Px (momentum) Mu [E, t] Energy time~ Can’t simultaneously measure → juggling, eclectic (1900) Planck Energy is transferred in discrete amounts(Quanta) 1) Absorbed 2) Emitted E∝ V ESingle Quantum Energy E=hV Where h is Planck’s Constant= 6.626X10^34 JS Binding Energy (W) 6 ↳Work Function Energy of attraction between the metal and the electron Characteristic to metal Zn, etc Photon: A quantum of light Photoelectric Effect: Shine light electron bounces off of surface 1) E Light= hV <W ~No electron will be ejected have to have enough energy to pull electron from binding. 2) E Light= W ~Ejected Electron have enough 3) E light> W ~Any energy in excess of binding energy required ↑ Ek of ejected E (Kinetic energy)=½ mu^2 Classical Physics Mechanics: E Light ∝ Intensity → E light is independent of V(Frequency) ↪Expected Behavior? 1) Light of any frequency will eject e as long as intense enough 2) ↑ intensity ↑ Ek of ejected e September 12th Chapter 3 Continuation~ Observed: 1) V Light ≥ V threshold 2) Ek of ejected e is independent of intensity Einstein (1905) ↳ applied quantization 7 E Photon (“particle” of light) =hv All have 1 equation Conservation of Energy: *Ek (e)=Elight W ↳1/2mu^2=hvW 3.36: Binding Energy of Mg=5.56 X 10^19J Calculate min V of light needed to eject an electron. ½ mu^2=0 → hv ≥ W → v ≥W/h ↪Excess Ek 5.86X10^19J/6.626X10^34 JS =8.84X10^14 Hz 3.37: Calculate the Ek of the ejected e if light W/V= 2.00X10^15 s^1 strikes Mg metal * Particular metals have certain work constants Ek= (6.626X10^34JS)(2.00X10^15)5.86X10^19J =7.39X10^19 J H atom stability and atomic line segment→ not explained by quantum H atom Stability: Pre 1900’s e circulating charge→ orbiting Spiral into the nucleus→ orbiting charge loses energy= 10^8 This model isn’t stable Atomic Line spectrum: Bridburg did a math fit: 8 1/λ=R∞(1/n1^2 1/nz^2) where R=Rydberg Constant and n1<n2 and n is a positive integer. *DON’T USE THIS EQUATION: this is a mathematical fit, use Bohr’s model mentioned below~ Bohr Model 1913: Energy of e is quantized Restricted to quantize orbit Energy in and energy out Takes away spiraling instability→ closest model to truth En=2.18X10^18 J (1/n^2) ↳ Where E is the energy of the quantized “orbit”, and n is positive whole integers n=1, 2, 3… ΔE=EfEi → This means the energy change between the states of an atom when the e moves from Ei→ Ef Elight= hv 3.44: What is the wavelength (nm) of a photon absorbed to excite from E1→ E4 ΔE=EfEi ΔE=E4E1 E=hc/λ =2.18X10^18 J(¼^21/1^2) ΔE= 2.04X10^18 J (The +/ sign demonstrates the direction of energy in or out, once determined don’t utilize the negative if negative sign into the next step, because there isn’t a such thing as negative wavelength) 9 λ=hc/E → λ=9.74X10^8m → 97.4 nm b) Calculate the Energy(J) of the photon absorbed to excite and e from E2 → E3 ΔE=EfEi → 2.18X10^18 (⅓^2 ½^2) =3.03X 10^19 J Atomic Orbitals: “Containers for electrons” (Weekes, Naomi) Centered around the nucleus, images represent the electrons in a region of probability Characterize quantum numbers( integer/ discrete values) ↳Heisenberg Uncertainty: uncertainty of location and momentum of a small particle will have a minimum value. 1) N P rincipal~ Describes the size of the atomic orbital → larger n larger orbital. Integer values. n=1,2,3….. 2) L ngular momentum~ Describes shape size of atomic orbital. The possible values are dictated by n. Ex: n=3 l=0, 1, 2. 0… (n1). l=0 is S, l=1 is P, and l=2 is D. *Always start with n and work your way through. 10 3) Ml M agnetic~ Describes possible orientation values. ml= l, l1,...0,...l. → l determines the ml values. 4) Ms ~ Spin of an e, can only be +/½ Orbital (n, l, ml) certain spins Shell= A collection of orbitals with the same n value → S ubshells= collection of orbitals with the same n and l values The number of subshells in a shell=n, the number of orbitals in a shell= n^2 Ex: n=1 shell → l=0 ml=0 1s → expect 1 subshell on 1 orbital Electron Configuration: ↳ Distribution of e in available orbitals *energy ordering by increasing n values: n=1 <n=2 <n=3 (The rest of the notes on electron configurations are on the last pages by Key Equations, underneath difficult concepts) 3.76: An e in an atom is in n=3 level what possible values of l & ml are possible? l 0 1 2 ml 0 1, 0, 1 2, 1, 0, 1, 2 3p 3d Ex: Which sets of quantum numbers are unacceptable and why? (n, l, ml, ms) 11 a. (1,1,½,½) → the 2nd and 3rd don’t work because l≠1, ml≠half integer b. (3, 0, 1, ½) → the 3rd doesn’t work because when l=0 ml also equals 0 c. (2,0,1,½) → the 3rd doesn’t work because when l=0 ml also equals 0 d. (4,3,2,½) → These values work Ex 2: a. n=1 shell → l=0 ml=0 1s ↳expect 1 subshell on 1 orbital B. n=2 shell 2s and 2p3 ↳expect subshells and 4 orbitals l 0 1 ml 0 1, 0, 1 2s 2p, 2p, 2p C. n=3 shell ↳expect 3 subshells and 9 orbitals l 0 1 2 ml 0 1,0,1 2,1,0,1,2 3s 3p^3 3d^5 3.94: Following G round State configurations explain which of the following are incorrect and then fix. 12 a. Al=1s^2 2s^2 2p^6 3s^2 3p^3 Z=13 valid if excited but:1s^2 2s^2 2p^6 3s^2 3p^1 b. B= 1s^2 2s^2 2p^5 Z=5 too many electrons: 1s^2 2s^2 2p^1 3.100: Given electron configuration for 1s^2 2s^2 2p^6 3s^2 write complete set of quantum numbers for each e and identify element. ↑↓ 3s (3, 0, 0, ½) or (3, 0, 0, ½) ↑↓ ↑↓ ↑↓ 2p (2, 1, 1, ½) or (2, 1, 1, ½) or (2, 1, 1, ½) or (2, 1, 1, ½) ↑↓ 2s (2, 0, 0, ½) or (2, 0, 0, ½) ↑↓ 1s (1, 0, 0, ½) or (1, 0, 0, ½) Pauli Exclusion Rule: No 2 e in an atom can have the same set of (n, l, ml, ms) 3.104: Determine the number of unpaired e in each of the following and identify the element as diamagnetic (no unpaired e) or paramagnetic (1 or more unpaired e) a. C Z=6 1s^1 2s^2 2p^2 ↑ ↑ _ 2p^2 → The orbitals aren’t paired (2) so paramagnetic ↑↓ 2s^2 ↑↓ 1s^2 B. S Z=16 1s^1 2s^2 2p^6 3s^2 3p^4 the 3p^4 has 2 paired and 2 unpaired orbitals so paramagnetic C. Cu Z=29 [Ar] 4s^23d^9 → this is incorrect form due to the exception to the rule *you can take away ONLY 1 electron directly below from s to half fill/ fill the d orbital to d^5 or d^10 13 [Ar] 4s^13d^10 Chapter 4 Periodic Trends: Similar e configuration value tend to have similar chemical/ physical properties. The valence e are furthest from nucleus outermost e others are the core Group A Valence Determination [12 1318] Shell at highest n Ex: C 1s^2 2p^2 valence n=2 Group B Valence Determination  All e outside of noble gas core Ex: Z=22 [Ar] 4s^2 3d^2→ The last two orbitals s and d Z effective: ↳The magnitude of nuclear charge felt by the valence e. Usually lower than the nuclear charge due to shielding by core electrons. Increases left to right on the periodic table the degree of shielding remains fixed, the core shields the valence Li z=3, Z eff 1.28 Periodic Trends Continued: 1) Atomic Radius Size~ I ncreases left ←right and up↓down. Z eff decreasing larger radius. 2) Ionization Energy~ The energy needed to remove an e from an atom: Na→ Na+ + 1e. IE ↑ left to right, when the Z eff increases the valence electrons are more tightly held. 14 (Brown, Doc) Deviations: Be→ B 1s^2 2s^2 vs. 1s^2 2s^2 2p^1 it’s easier to move a 2p e than 2s e so IE B<Be 3) Electron Affinity: nergy released when electron is added, also increases left to right. (Petrucci) Deviations for period 2: N→ C 1s^2 2s^2 2p^3 vs 1s^22s^22p^2 Ea= 7 kJ/ mol Ea= 122 KJ/mol Carbon is more favorable because an unoccupied orbital is better than pairing 15 Ions: Charged elements Cation positively charged → anion is negatively charged. *Cation has a t so means positive Na→ Na+ + 1e cation Cl+1e → cl anion Determination of ions formed~ Main group 1a7a *always seek to become +1 charge Rb+ Cs+ H is the exception seek to be isoelectronic (same number of e) with the nearest noble gas Ex: Na (1s^2 2s^2 2p^6 3s^1) → Na+(1s^2 2s^2 2p^6) becomes Neon form Group 2a Ex: Mg (1s^2 2s^2 2p^6 3s^2) → Mg+2 (1s^2 2s^2 2p^6) Group 7a Ex: F(1s^2 2s^2 2p^5) → F(1s^2 2s^2 2p^6) Group 6a Ex: O (1s^2 2s^2 2p^4) → O2 (1s^2 2s^2 2p^6) Group 15 N (1s^2 2s^2 2p^3) → N3 (1s^2 2s^2 2p^6) Ions of dblock elements~ No real way to predict, must be told first Always from (+) cations 16 Don’t seek to be isoelectronic with noble gases can’t form multiple cations (+1, +2, +3… Remove e from highest n shell first Mn: [Ar] 4s^2 3d^5 Mn+2: [Ar] 3d^5 Mn+3: [Ar] 3d^4 4.71: Write the ground (stable) state e configurations a. Fe+2 Fe: [Ar] 4s^2 3d^6 Fe+2: [Ar] 3d^6 B. Cu+2 Cu: [Ar] 4s^1 3d^10 Cu+2: [Ar] 3d^9 Atomic vs. ionic radius: Decrease in size relative to their neutral parent → Na→ Na+ +1e (Cations) Increase in size relative to their parent → F+e → Fe e Isoelectronic Series: Identical e configs Different Z 4.107: 17 Given the following valence orbital diagrams, rank the elements according to ↑ 1. Atomic Size 2. IE A. ↑↓ ↑↓ _ 3s 3p B. ↑↓ ↑↓ ↑↓ ↑ 2s 2p C. ↑↓ _ _ _ 5s 5p D. ↑↓ ↑ ↑ ↑ 3s 3p 1. B<D<A<C → B has the smallest n, C has the largest. To distinguish between A and D you have to look at their electron configurations and Z effectives D=16 and A=14 and the smaller the Z effective the larger the atomic size. 2. C<A<D<B → C has the largest n and easier to remove an electron, B has the largest n so harder to remove an electron. To distinguish between and A and D look at the Z effective again A has a smaller Z effective so lower IE. 3.7: (My worked out example) ½ mu^2 *DON’T ROUND AMU 20.180 g (1kg/1000g)=.02 Kg/mol (1mol/ 6.022X10^23atoms)= 3.32X10^26 Kg/atom 18 1.12X10^20 J=½ (3.32X10^26) u^2= 667489 u=817 19 ~Difficult Concepts~ Electron Affinity: Both ionization energy and electron affinity increase left to right. It is harder to remove an electron from an element that is favorable to have more electrons, hence the energy required to remove an electron will increase left to right. Same goes for electron affinity with favorability, more energy will be released when it is more favorable to add an electron. Also the radius decreases left to right, making the protons more tightly held, and so will the electrons. Most atoms release energy when an electron is added, especially when more favorable. Electron Configurations Summary: Form: 3s^2 → 3=the row or orbital number, S=orbital type, 2=Number of electrons in orbital type. Electrons Per Orbital: s=2 p=6 d=10 f=14 Remember (Exceptions to rule): Orbitals prefer to be half filled, example: manganese 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5 20 ↳ Normally would think it would be: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^4 → It would be correct form, but the element is more stable with Z half filled orbitals than 1 filled and 1 partially filled. N s<(n1)<d → Reverse rules N s<(n1) d 4s<3d 5s<4d Same value of n ↑ energy with l, 3s(l=0)<3p(l=1)<3d(l=2) Filling atomic orbitals: 1) Each orbital can hold 2e 2) Fill the lowest orbitals 1st (aufbau) 3) Degenerate orbitals each contain 1e before pairing 4) Paired e must have opposite spin ms=+/ ½ N, l, ml, ms h=whole (Burdge, Overby) 21 ~Key Equations~ ● Ek= ½ mu^2 → kinetic energy of an object= ½ mass X velocity^ 2 ● c=λV → speed of light= wavelength X Frequency ● E= hV → Energy of the photon= Planck’s constant X Frequency ● hV= Ek + W → Energy of the photon= Kinetic energy + binding energy ● En= 2.18 X10^18J (1/n^2) → Energy of an e is inversely proportional to n ● F ∝ Q1Q2/d → The Force between two charged objects is proportional to 2 charges/ the distance. ~Constants~ ● Planck’s Constant (h): 6.626 X10^34 JS ● Speed of light (c): 3.0 X10^8 m/s ● Mass of the electron (the m in 1/2mu^2): 9.1 X10^31 Kg ● Rydberg constant for h (used in ΔE equation): 2.18 X10^18 J
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