Exam 3 Summary Session Key

CHEM 105 11/14/2023 Exam 3 Summary Session Reminder: Summary session sheets should not be your only preparation method for the exam. We do not see the exams or write them, so this sheet should be a supplement to any review you complete. Multiple Choice 1. Which pair of electron configurations represent atoms with the most similar chemical properties? a. 1s 2 2s 2 2p 6 3s 2 3p 5 and 1s 2 2s 2 2p 6 3s 2 3p 4 b. 1s 2 2s 2 and 1s 2 2s 2 2p 2 c. 1s 2 2s 2 2p 6 3s 2 3p 5 and 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 d. 1s 2 2s 2 2p 6 and 1s 2 2s 2 2p 4 2. What is the wavelength of light with a frequency of 6.59 x 10 5 s -1 ? a. 2.0 x 10 -25 m b. 2.0 x 10 14 m c. 2.2 x 10 -3 m d. 455 m 3. Which of the following are diamagnetic? I. Sulfur II. Magnesium III. Zinc IV. Argon

a. I and II b. II, III, and IV c. I, II, and III d. IV only 4. Rank S, O, and Al in order of increasing electronegativity. a. O< S< Al b. Al<S< O c. O<Al<S d. S<O<Al 5. How many possible electrons can be placed in the n=4 energy level? a. 12 b. 16 c. 32 d. 24 6. Which of the following is true regarding electron affinity? a. Electron affinity and electronegativity measure the same value b. Electron affinity increases across a row due to electron shielding c. Electron affinity is always positive d. All of the above are false 7. How many radial nodes would a 6f orbital have? a. 2 b. 3 c. 4 d. 5 Total nodes= n-1 = 5 Angular nodes = ℓ = f = 3

Free Response 11. The following experimental data is collected for the reaction 2HgCl 2 (aq) + C 2 O 4 2- (aq) → 2Cl 1- (aq) + 2CO 2 (g) + Hg 2 Cl 2 (aq) Experiment Initial [HgCl 2 ] Initial [C 2 O 4 2- ] Initial Rate 1 0.0836 M 0.202 M 5.20 x 10 − 5 M/s 2 0.0836 M 0.404 M 2.08 x 10 -4 M/s 3 0.0418 M 0.404 M 1.04 x 10 -4 M/s 4 0.0316 M 0.514 M a) Calculate the order of reaction with respect to both HgCl 2 and C 2 O 4 2- . HgCl2: Experiment 3 / Experiment 2 (C2O4 is constant) exp3 exp2 : k [ HgCl 2 ] x ¿¿¿ k [ 0.0418 M ] x ¿¿ [ 0.0418 M ] x [ 0.0836 M ] x = ½ [½] x = ½ x= 1 → first order with respect to HgCl2 C 2 O 4 2- : Exp 2 / Exp 1 k ¿¿ 2 y = 4

y= 2 → second order with respect to C 2 O 4 2- b) What is the overall reaction rate order? Third order overall = 2 + 1 c) What is the value of k (include units)? Can choose any experiment and plug in the numbers Experiment 1: 5.20 e -5 = k [0.0836 M] 1 [0.202M] 2 k= 0.015 M -2 s -1 d) What is the initial rate in experiment 4? Rate = k [HgCl2][C 2 O 4 2- ] 2 Rate = (0.015 M -2 s -1 ) * [0.0316 M] * [0.514 M] 2 = 1.27 e -4 M/s e) Which reactant could the graph below represent and why? HgCl 2 because graphing ln[A] results in a straight line which is the first order integrated rate law 13. A photon of light has a wavelength of 580 nm. a. Calculate the total energy (in kJ/mol) released when 1 mole of photons are present 580 nm x 1 e-9 m ----------- = 5.8 e-7 m 1 nm E= (hc) / λ = (6.626 e-34 Js) x (3.00 e 8 m/s) / 5.8 e -7 m = 3.4 e -19 J per photon

a. Calculate the energy emitted by the electron movement on the hydrogen atom. Electron moving from 3n orbital to 1n orbital Δ E =− 2.178 E − 18 J ( 1 2 1 2 − 1 2 3 2 ) = − 2.178 E − 18 J ( 1 − 1 9 )= ¿ -1.936E-18 J b. Suppose there was an additional electron which fell from the 6n shell to the 1n shell. How would the frequency of the light emitted by this change in energy compare to the electron movement in the previous question. The electron would emit more energy because it is falling from a higher energy orbital. A higher energy emitted would translate to a higher frequency light emitted c. Why is the accuracy of the bohr model limited to H atoms? H only has one electron so there are not electron electron interactions. Once multiple electrons added and electron shells increase, electron-electron interactions will affect the behavior of electrons and they will not behave as the Bohr model predicts.

16. Consider calcium and silicon. a) Write the quantum number sets for the 13th and 19th electrons in both species in their ground states 13th electron (same for Ca and Si): (3, 1, -1, 1/2) 19th electron (ONLY for Ca, Si does not have a 19th electron): (4, 0, 0, ½) b) Calculate the total nodes, angular nodes, and radial nodes for the highest energy electron for each of these species in their ground state Ca: highest energy electron = 4s Si: highest energy electron = 3p Total nodes = n - 1 = 3 Total nodes = n - 1 = 2 Angular nodes = ℓ = 0 Angular nodes = ℓ = 1 Radial nodes = n-ℓ-1 = 3 Radial nodes = n - ℓ - 1 = 1 c) Which species has a higher ionization energy? Explain. Silicon will have a higher ionization energy because it has fewer core electrons and therefore has less shielding. It also has a higher effective nuclear charge to hold onto electrons more tightly. Therefore, it will require more energy to remove an electron from silicon compared to calcium. d) Calcium and silicon become stable isoelectronic species. What is the electron configuration for this isoelectronic series? 1s 2 2s 2 2p 6 3s 2 3p 6 e) Will calcium or silicon have a larger radius when both are in an isoelectronic state? Explain. Silicon will form Si 4- while calcium will form Ca 2+ . Since calcium has 2 more protons than electrons, it the nucleus will have a stronger pull on the electrons and hold them more closely. Therefore, the atomic radius of Ca 2+ will be smaller than the atomic radius of Si 4- 16. Given the Following Values, Construct a lattice energy diagram for LiF. Ionization energy of Li = 520.14 kj/mol Enthalpy of Formation of LiF = -616.93 kj/mol Electron Affinity for F = 328.16 kj/mol Bond dissociation energy for F 2 = 156.96 kj/mol Enthalpy of Sublimation of Li = 165.83 kj/mol

KBr would have a lower melting point than LiF. Because the lattice energy of LiF is higher it would require more energy to break up solid LiF resulting in a higher melting point for LiF.