Chem 211 Exam 3 Review Session PDF

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General Chemistry I Exam 3 Review Questions: Addtional Notes &  Answers  1. Estimate the standard enthalpy change ΔH° for the reaction below: C2H6(g) + Cl2(g) → C2H5Cl(g) + HCl(g) Data: Bond energies in kJ/mol: C-C, 348; C-H, 414; Cl-Cl, 242; C-Cl, 327; H-Cl, 431 Delta H rxn = Summation BE reactants - Summation BE products Reactants:

Products:

[1(C-C) + 6(C-H) + 1(Cl-Cl)] - [1(C-C) + 5(C-H) + 1(C-Cl) + 1(Cl-H)] Plug in the # associated with each bond a. b. c. d.

-113 kJ/mol -68 kJ/mol -102 kJ/mol -344 kJ/mol

2. Based on electronegativity trends in the periodic table, predict which of the following compounds will have the greatest % ionic character in its bonds Greatest % ionic character = great EN change Look for EN trends on the periodic table a. HCl b. CaI2 c. SrF2 d. CaCl2 3. The Lewis dot symbol for the oxide ion is: First find the valence electrons for oxygen (#VE: 6, found on the periodic table) Oxide ion needs to reach an octet (8 valence electrons = noble gas configurations = shell complete) Needs 2 additional electrons to fulfill octet

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4. The electronic configuration of the element whose atomic number is 26 is: Atomic # 26 is Iron Fe Must fill up each shell, S=2, p=6, d=10 Look for shell trends on the periodic table a. b. c. d.

1s2 2s2 2p6 3s2 3p6 4s0 3d8 1s2 2s2 2p6 3s2 3p6 4s2 3d6 1s2 2s2 2p6 3s2 3p6 4s2 3d4 4p2 none of the above

5. A sodium flame has a characteristic yellow color due to emissions of wavelength 589nm. What is the mass equivalence of one photon of this wavelength (1J=1kg*m2 /s2) Since it has yelow color and its visiable we know its in the visable light spectrum thus v is equal to c (3.0 x 10^8 m/s) Convert 589 nm to m Use E = h/mv We also know that 1 j = 1 kg*m^2/s^2, so then convert to kg ○

3.75 x 10^-36 kg

6. For the series of spectral lines of atomic hydrogen with n₁ = 3, which of the following wavelengths does not correspond to a wavelength predicted by the Rydberg equation? Rydberg equation: 1/𝝀= (1.097 x 107 m-1  ) x (1/n₁² - 1/n₂²) We know that n2 > n1 because wavelength cant be negative Test and trial question, test numbers higher than 3 for n2 and find what doesnt work a. 1875 nm ( n2 = 4) b. 1282 nm ( n2 = 5) c.  1184 nm doesnt work d. 1094 nm ( n2 = 6) 7. Which of the following elements has the largest ionization energy? Noble gasses have the highest ionization energy Look for ionization trends on the periodic table a. Na b.  Ne c. F d. K Which of the following statements about periodic properties is incorrect? ● IE, EN (increases up and right of the periodic table) ○ EN excluses noble gases ● EA (increases up and right of the periodic table) ○ Energy required to add an electron to an atom

● Atomic Radius (increases down and left of the periodic table) ● Effective Nuclear Charge: the ability of an atom to hold a proton closer (more +, then higher ENC)

a. b. c. d.

Both electron affinity and ionization energy decreases down a group  Atomic size increases to the right across a period Ionization energy increases to the right across a period Atomic size increases down a group

8. Lattice energies can also be calculated for covalent network solids using a B  orn-Haber cycle, and the network solid silicon dioxide has one of the highest ΔH°lattice values. Silicon dioxide is found in pure crystalline form as transparent rock quartz. Use theoretical values and the following data to calculate ΔH°lattice  of SiO2 . 

Delta h lattice = -13285 kJ Use delta h f = summation individual steps + delta h lattice 9. Which of the following lists has the bonds ordered from shortest  to longest? a. b. c. d.

C-Cl, C-I, C-F, C-Br C-Br, C-Cl, C-F, C-I C-F, C-Cl, C-Br, C-I C-I, C-Br, C-Cl, C-F

Use atomic radius and EN trends 10. ** Use the following data to calculate ΔH°f of MgCl. Use Hess’s Law to calculate for the   conversion of MgCl to MgCl2 and Mg (ΔH°f  of MgCl2 = -641.6 kJ/mol). Is MgCl favored relative to Mg and Cl2?

Use delta h f = summation individual steps + delta h lattice LA key was wrong **, plug into the calculator 11. The electrostatic force of attraction between two charges, A and B, is proportional to a. b. c. d.

(charge A) + (charge B) + (distance between the charges)2 (charge A) x (charge B) x (distance between the charges)2 (charge A) x (charge B) + (distance between the charges)2 (charge A) x (charge B) / (distance between the charges)2

Know the equation for force when charges are involved F = kQ1Q2/R^2 Where F = force, k = constant, Q1 = charge 1, Q2 = charge 2, R = distance 12. Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent? ○ EN > .5 is polar ○ EN < .5 is nonpolar ■ Use EN trends on periodic table ○ Nonmetal to Nonmetal = covalent bond ○ Metal to Nonmetal = ionic bond (ionic automatically means its polar) ○ Metal to metal = metallic bond (usually not asked if polar/nonpolar) ○ S8 ■ Nonmetal + Nonmetal, same element (nonpolar  covalent) ○ RbCl ■ Metal + non metal (ionic) ○ PF3

■ Nonmetal + Nonmetal, EN > .5 because bonded to a halogen (polar  covalent) ○ SCl2 ■ Nonmetal + Nonmetal, EN > .5 because bonded to a halogen (polar  covalent) ○ F2 ■ Nonmetal + Nonmetal, same element (nonpolar  covalent) ○ SF2 ■ Nonmetal + Nonmetal, EN >.5 (Polar  Covalent) 13. The shape of an atomic orbital is associated with Name ● ● ● ●

Symbol Principle Angular Momentum Magnetic Spin

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N L M sub L M sub S

Definition ● Orbital size and total energy ● Orbital shape ● Orbital Orientation ● Direction of Spin

The principal quantum number (n ) The angular momentum quantum number (l ) the magnetic quantum number (m l ) the spin quantum number (m s ) the magnetic and spin quantum numbers, together.

14. Select the compound with the highest (i.e. most negative) lattice energy Lattice energy is the energy between elements 1. Largest charge cation 2. Largest charge anions 3. Smallest ion size ○ CaS(s) ■ CaS = Ca2+ + S2○ BaO(s) ■ BaO = Ba2+ + O2○ NaI(s) ■ NaI = Na+ + I○ LiBr(s) ■ LiBr = Li+ + Br-

○ MgO(s) ■ MgO = Mg2+ + O2-

15. Given the three statements below, which answer is best? (a) Although the wavelength of light and the frequency of light can change, the energy remains constant. False, energy of light is not constant because E = hc/wavelength (b) An important consequence of quantum theory is that the energy of various electronic states for atoms are at fixed discrete levels. T  rue, electrons can only occur at certain energy levels, quantum mechanical model explains this (c) The energy of an electronic state for a hydrogen atom is dependent only on the principal quantum number n. T  rue because N is related to energy level according to the quantum mechanical model a. b. c. d. e.

A and B are true, C is false B and C are true, A is false A and C are true, B is false All are true All are false

16. Which element listed below has the most exothermic (negative, giving off energy) electron affinity? 1st electron affinity: (X + e- = X-, this is releasing energy which is negative, exothermic) 2nd electron affinity: (X- + e- = X2-, this is gaining energy which is positive, endothermic) Anything after 2nd will be endothermic Look at the periodic table to see the charges of the elements You can also use EA trends ○ Se 2 nd electron affinity, this is endothermic ○ Sn t his is ionization because it is a positive ion ○ As t his is pasted the 2nd electron affinity, this is endothermic ○ Cl this is because it falls on the 1st electron affinity ○ None of these

17. How many electrons (both lone and bond pairs) are shown in the correct Lewis structure of HClO? How to draw a lewis structure ● Chose a central atom ○ Pick one with the lowest electronegativity ● Arrange other atoms (ligands) around the central atom ● Find the sum of all the valence electrons for all the atoms in the compound ● Connect all the other atoms to the central atom first starting with single bonds, you can change to double/triple etc later if needed ● Complete the octet for all ligands ○ Anything extra goes in the central atom ● If central atom does not have octet use ligands’ lone pairs to form double/triple bonds

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14 12 16 20 None of these

18. What is the oxidation number of N in the compound NO? What is the oxidation number of C in CCl4? Use the formula: ONatom = Number of Valence Electrons - (Number of Bonding Electrons + Number of Lone Pairs).

ONn = # valence - (# of BE + # LP) ONn = 5 - (0+3) ● The more electronegative atom will claim double bond (this is case oxygen will) ONo = 6 - (4 + 4)

ONc = 4-(0-0) = 4 ONcl = 28-(8-24) = 44

19. Which of the following is true? ○ An s- orbital becomes more dense as the distance from the nucleus increases. False because if the distance to the nucleus increases there is more space for it to move around ○ A p- orbital has a spherical boundary surface. F  alse, because the spherical shape is S, P is a dumbbell shape, d looks like a four leaf clover/double dumbbell ○ An electron in an s - orbital has a nonzero probability of being found at the nucleus. True, because when you have an s orbital it(electrons) can move around because its sphere and doesnt have lobes however in a P orbital its cut off so it can move near the nucleus (bc of the lobes) ○ An s- orbital has two lobes on opposite sides of the nucleus. False because s is a sphere

○ An electron in a p - orbital has a nonzero probability of being found at the nucleus. False because P orbital is a dumbbell shape and have lobes ○ A 3d orbital is closer to the nucleus than a 4s orbital. T  rue because even though you fill a 4s orbital because a 3d orbital,a 3d orbital is STILL closer to the nucleus than a 4s, FILLING ORDER = will 4s before 3d, reference the periodic table

20. The wavelength of a beam of light is 24 mm. What is the energy of one of its photons? E = hc/wavelength First convert 24nm to m = 0.024m ○ 5.3 x 10-47  J ○ 8.3 x 10-24  J ○ 5.3 x 10-45  J ○ -8.3 x 10-21  J

Important Things to Study Periodic Trends

Electron Configuration

Sublevels

Quantum Numbers

Electronegativity Trends *Helps determine the nature of the bond...