Name. system is +/? +/? undetermined A Lost energy, Did work, B Gained energy, + Did work Undetermined. C Gained energy, + Did work Undetermined

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1 Name 1. [4 points] An electron in the hydrogen atom can undergo only set transitions. Calculate the wavelength for an electron transitioning from n =15 to n =1. Is this visible, infrared, or ultraviolet light? 1 λ = x107 m 1 1 λ = x107 m n 1 2 f n i λ = x107 m 1 l = nm this light is in the ultraviolet region. Even if I did not have to memorize the EM-spectrum, I remember that when n f = 1, we are in the Lyman region of the spectrum, which is UV. 2. (6 points total) Part 1 Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system, based on the following descriptions: (3 points) A. Surroundings get hotter and the system increases in volume. B. Surroundings get colder and the system expands in volume. C. Surroundings get colder and the system expands in volume. D. Surroundings get colder and the system does not change in volume. system The q of the The value of w is E is +/ or system is +/? +/? undetermined A Lost energy, Did work, B Gained energy, + Did work Undetermined C Gained energy, + Did work Undetermined D Gained energy, + Did no work, 0 +

2 Part 2: A mole of X reacts at a constant pressure of 84.0 atm via the reaction N 2(g) + 3 H 2(g) 2 NH 3(g) H = 91.8 kj The initial volume of the gaseous mixture was 3.00 L Calculate the final volume of the system if the value of the total energy change, ΔE, in kilojoules is 151.4kJ. (This will come in handy: 101.3J = 1L atm) [HINT: w= P V] (3 points) Express your answer with the appropriate units. E = q + w 84.0atm x 101.3J 151.4kJ = 91.8kJ+ x 1kJ x V 3.00L 1L atm 1000J 3. (5 points) Anxiety is attributed to heightened neuronal activity in the limbic system, the region of the brain associated with emotion. Valium (Diazepam C 16 H 13 ON 2 Cl (s) ) bind to proteins that helps GABA control this activity and reduces the feelings one might have of anxiety and generalized fear. Write the equation for the formation of one mole of diazepam from its elements. 16C (s) + 13/2H 2(g) + ½ O 2(g) + N 2(g) + 1/2 Cl 2(g) C 16 H 13 ON 2 Cl (s) 4. (6 points) Give the name or electron configuration for the following elements or ions. [You can give noble gas core]: a. Thallium (element 81) [Xe] 6s 2 4f 14 5d 10 6p 1 or [Xe]4f 14 5d 10 6s 2 6p 1 b. [Ar]4s 1 3d 10 This is the copper atom. There are 29 e, so 29p +. c. Cr 2+ the atom is Ar]4s 1 3d 5 the ion is [Ar]3d 4 d. Copper (II) ion [Ar] 3d 9 e. Te, tellurium (52) [Kr]4d 10 5s 2 5p 4 or [Kr] 5s 2 4d 10 5p 4 f. sulfide 1s 2 2s 2 2p 6 3s 2 3p 6 Exam 3 Fall 2017 Page 2 of 13

3 5. 5 points) Answer the [?] with the correct response: a. The number of un-paired electrons in a Cr 2+ ion is [?] Answer: 4 (see problem 4c the atom is Ar]4s 1 3d 5 the ion is [Ar]3d 4 box diagram of ion 3d 4 b. When n = 2, the angular momentum quantum number, l, can be what value(s) [?] Answer: 0, 1 c. The total number of electrons with n=4, l = 1 is [?] Answer: 6 d. The sub shell with the quantum numbers [7,2] is [?] Answer: 7d l = 2 which represents the d subshell e. Two electrons in the same [?] must have opposite spin. Answer: orbital f. The neutral fourth period atom having a total of six d electrons is [?] Answer: Fe, count over but remember the 4s 2 don t count g. Orbital with the same energy are said to be [?] Answer: degenerate h. The 2p orbitals of an atom have identical shapes but differ in their [?]. Answer: orientations in space, m l is okay, if you give numbers i. A 6f orbital has [?] angular nodes. Answer: 3 j. A 12d orbital has [?] radial nodes? Answer: = 9 nodes Exam 3 Fall 2017 Page 3 of 13

4 6. (8 points) A sample of gold metal must absorb radiation with a minimum frequency of x s 1 before it can emit an electron from its surface via the photoelectric effect. a. (2 points) What is the minimum energy required to produce this effect? (work or threshold energy) b. (2 points) What wavelength radiation will provide a photon of this energy? c. (4 points) If the surface of the gold sample is radiated with light of wavelength 106 nm, what is the maximum possible velocity of the emitted electrons? (KE = ½ mv 2 ) x kg E = hν x 10 NO J s x x 10 QR s Q = x 10 QS J hc l = x 10 QS = nm J hc E = 106 x 10 9 = x10 18 J E photon = E electron absorbed E electron = minimum energy + KE x J x J = KE 2xQ.\N] ^ Q\ _`a x kg =1.51 x106 m/s 7. (5 points) Using your knowledge of the everyday and super hero world, place the following sources of light in order of increasing energy: [This means: start with the smallest (lowest) energy and end at the highest energy. Do not put numbers next to the letters; you need to write out the correct order of letters. You will receive no credit if you are not clear about your answer.]: a. Gamma rays that turned Bruce Banner into the Hulk b. The red color in red dye no. 28; a component of an insecticide used to kill Mediterranean fruit flies c. Infrared heat from the wires in your toaster used to burn toast d. Superman's x-ray vision used to see through walls and watch nefarious evil deed doers. e. The green color of the Green Lantern's lamp C, B, E, D, A Exam 3 Fall 2017 Page 4 of 13

5 8. (6 points) When a sample g sample of benzoic acid is combusted in a bomb calorimeter, the temperature of the calorimeter rises by C. The heat of combustion of benzoic acid is kJ/g (benzoic acid). Calculate the calorimeter constant, (C v ) for the calorimeter. If a g sample of caffeine is combusted in the same calorimeter, the temperature rises by C, calculate the heat of combustion per gram for caffeine. q (rxn) =Cv T (calorimeter) ( 26.38kJ/g HBz)(0.300g HBz) = C v (1.642 C) C v = 4.819kJ/ C C v = 4.82kJ/ C ( H (comb) )(0.600g caffeine) = kj/ C(1.532 C) H (comb) = 12.3/g caffeine 9. (6 points) Ammonia will burn in the presence of a platinum catalyst to produce nitric oxide, NO. Determine the heat of reaction at constant pressure? Show your work for full credit PATHWAY OF INTEREST 4 NH 3(g +5 O 2(g) 4 NO (g) + 6 H 2 O (g) PATHWAY 1: N 2(g) + O 2(g) 2 NO (g) H = kj PATHWAY 2: N 2(g) + 3 H 2(g) 2 NH 3(g) H = kj PATHWAY 3: 2 H 2(g) + O 2(g) 2 H 2 O (g) H = kj 1. (6 points) Ammonia will burn in the presence of a platinum catalyst to produce nitric oxide, NO. Determine the heat of reaction at constant pressure? Show your work for full credit PATHWAY OF INTEREST 4 NH3(g+5 O2(g) 4 NO(g) + 6 H2O(g) PATHWAY 1: N2(g) + O2(g) 2 NO(g) H = kj PATHWAY 2: N2(g) + 3 H2(g) 2 NH3(g) H = 91.8 kj PATHWAY 3: 2 H2(g) + O2(g) 2 H2O(g) H = kj 2N2(g) + 2 O2(g) 4 NO(g) H = kj x 2 Exam 3 Fall 2017 Page 5 of 13

6 4NH3(g) 2N2(g) + 6H2(g) H = kj x 2 6 H2(g) + 3O2(g) 6 H2O(g) H = kj x3 As you can see the hydrogens cancel, as do the nitrogens kj 10. (6 points) Heats of formation are often used to determine the desirability of a fuel for rockets. Calculate the H rxn for the reaction of chlorine trifluoride with dimethyl hydrazine in the following reaction: H ƒ(hcl(g) = 92.30kJ/mol, H ƒ(hf(g)) = kJ/mol, H ƒcf4(g) = 679.9kJ/mol, H ƒclf3 = kJ/mol, H ƒ(ch3)2n2h2(l) = 48.3kJ/mol, 4ClF 3(l) + (CH 3 ) 2 N 2 H 2(l) 2CF 4(g) +N 2(g) +4HCl (g) +4HF (g) H rxn = 4molHCl (g) + 4 mol HF (g) + 2mol CF 4(g) (4 mol ClF 3(l) + 1mol (CH 3 ) 2 N 2 H 2(l) ) H rxn = 4mol H ƒ(hcl(g) +4mol H ƒ(hf(g)) +2mol H ƒcf4(g) (4mol H ƒclf3 +mo1l H ƒ(ch3)2n2h2(l) ) H rxn =4mol( 92.30kJ/mol)+4mol( kJ/mol)+2mol( 679.9kJ/mol) (4mol( kJ/mol)+1mol(48.3kJ/mol)= = 369.2kJ kJ kJ kJ+ 48.3kJ= kJ 2,216.3kJ 11. (6 points) A 20.0 g piece of metal at C is placed in a calorimeter containing 50.7 g of water at 22.0 C the final temperature of the mixture is 25.7 C. What is the specific heat capacity of the metal? q (metal) =C P m T (calorimeter) = 4.184J/g C x 50.7 g x (25.7 C 22.0 C) q (metal) = J C P = J/(20.0g x(25.7 C C) =0.528J/g C 0.53J/g C 12. (8 points) Given below are several electron configurations that might be correct for the nitrogen atom. Indicate (by circling) whether each of these representations are the ground state (GS), the excited state (ES), or un-allowed (forbidden) state (FS). Using Hund s rule, the Pauli principle, and Aufbau Exam 3 Fall 2017 Page 6 of 13

7 (building up), BRIEFLY explain your choices. [Some might violate more than one rule.] a) GS ES FS 1s 2s 2p 3s Reason for your choice: obeys auf bau, but the spins are not parallel. This means the spin is not maximized. Hund s rule maximizes spin and minimizes repulsions. 1 s and 2 s follow Hund s Rule and Pauli b) GS ES FS 1s 2s 2p 3s Reason for your choice: This looks good! Obeys Pauli, Hund s Rule, Aufbau. c) GS ES FS 1s 2s 2p 3s Reason for your choice: 2 p follows Hund s rule, 2s is forbidden b/c it does not follow Pauli. Pauli Exclusion principle: 2 electrons in the same orbital can t have same 4 quantum numbers. d) GS ES FS 1s 2s 2p 3s Reason for your choice: This violates Aufbau. It takes energy to move electrons into a higher energy state. Both electrons in the 3s orbitals are in an excited state. All electrons satisfy Pauli and Hund s Rule. 13. (6 points) Circle the best choice in the list and explain your choice based on shielding effects, quantum shielding, and/or Z eff. Use one or two sentences to explain (not <,>, =,³, or any with slashes, symbols, arrows.) your choices. I WROTE THE EXPLAINATIONS, ON THE TEST, YOU DID NOT HAVE TO. HOWEVER, YOU MIGHT NEED THIS FOR THE FINAL! a. Smallest radius: Bi 3, N 3, P 3, so the largest would be Bi 3 for similar reasons Exam 3 Fall 2017 Page 7 of 13

8 These are in Group 5A, size increases down a family due to quantum shielding, the outer electron is further from the nucleus. The larger the n value, the larger the ion or atom in the same family. b. Lowest second ionization energy: Al, Ne, F The lowest second ionization would belong to Al. when Al loses 2 electrons, the electron configuration is isoelectronic with Na. Even though, Al has a larger Z, it will have more shielding than the other two atoms, so it will be easier to remove the outer two electrons. c. Smallest atom: Sn, I, At Tin is larger than Iodine. They are in the same row. Elements in the same row have similar l-type shielding leading to a larger Z eff. When Z eff increases, the attraction to outer electrons increase, so radius decreases. Iodine is above At on the periodic table. Iodine is smaller than At due to quantum shielding, d. Impossible shell designation: 2f, 5d, 4p The angular quantum number restricts the values of the subshells to n-1 0. Since n= 2, the largest l value can be only 1. e. Largest negative electron affinity: C,O,S Electron affinity increases across the period (trend-wise) since these elements are in the same row (O & C), the Z eff is increasing across the periods. O would better accept the electron and bring the subshell closer to a noble gas core. Group 1A elements can accept an electron, but typically the electron affinity is lower than non metals. a. Smallest radius: Si, Ge, C, Br this is a repeat of [c] Si and Ge are in the same family but have different n values, r [Si] <r[ge]. C has the smallest n, so it is smallest of three in same family r [C] <r [Si] <r[ge]. This is a quantum effect; the outermost electron defining the cloud is further away from the nucleus as n increases. Br and Ge are in the same period. Atoms decrease from left to right due to Z increasing while shielding staying relatively constant. The Zeff increases across, increasing the attraction of the outer most electron to the nucleus, and decreasing the radius. The order is r [C] <r [Si] <r [Br]<r[Ge]. (as the n values increase, the change in the radii across a period become very slight, we are looking at the changes as a theoretical exercise.) Exam 3 Fall 2017 Page 8 of 13

9 14. (6 points) Ethanol (C 2 H 5 OH) melts at 114 C. The enthalpy of fusion of ethanol is 108J/g. The specific heat of ethanol is 0.97J/g C. Calculate the total energy needed to melt 42.0 g of ethanol at C to form liquid ethanol at C This is a two-step process; the solid ethanol (at C) must warm to the melting point ( 114 C), then melt to form liquid. The temperature does not exceed 114 C. The q total = q warm + q melt = Cp (solid ethanol) x mass ethanol T ethanol + H fus of ethanol 0.970J/g C x 42.0 g x( 114 C C)+ 109J/g x 42.0 g J+4578 J= 4822J= 4.8kJ 15. (8 points) The atoms and ions Na, Mg+, Al 2+, and Si 3+ all have the same number of electrons they are an iso-electronic series. (No, I did not make a mistake on the charges!) a. (2 points) Draw the box diagram of the valence electrons of Mg+ and Si 3+ 3s 3p 3s 3p they are the same because they are iso electronic. b. (1 point) For which of these atoms and ions (Na, Mg+, Al 2+, and Si 3+ ) will the effective nuclear charge acting on the outermost electron be the greatest? As the shielding decreases while the nuclear charge increases, Z eff increases. Since they are in an isoelectronic series, they have the same shielding but do not have the same nuclear charge; the ion with the largest Z will have the largest Z eff (Si 3+ ) c. (1 point) How will it affect the size? Explain in one or two sentences. As the Z eff increases, the attraction to the outer electron increases. Since the radius is related to the size of the electron cloud, the attraction of the nucleus to the outer most electron determines the size of the cloud. So the radius must decrease for the ions with the largest positive charge in the series. (Si 3+ ) Exam 3 Fall 2017 Page 9 of 13

10 d. (2 Points) Based on effective nuclear charge, which one will have the largest ionization energy? Explain in one or two sentences. Z eff increases, the attraction to the outer most electron increases. This makes it harder to remove the outermost electrons from the ion. Silicon (III) will have the largest ionization energy, because it has the largest Z eff, and the strongest attraction to the outermost electron. e. (2 Points)Which atom or ion will have the most stable 2s orbital? Explain Again, it is all based on Z eff. And Z. since the electron conficuations are the same 1s 2 2s 2 2p 6 3s 1, the 2s orbitals all experience the same shielding, but not the same nuclear charge, so not the same Z eff. The ion or atom with the largest Z will have the largest Z eff, and the strongest attraction to the electrons in the 2s orbital. This attraction lowers the energy of the subshell (orbital), making it more stable. 16. (11 points) Each drawing represents a type of atomic orbital a. (3 points) Give the angular momentum value (l) for each orbital. 2s 3dz 2 2pz b. (3 points) Give an appropriate value for ml for orbital each orbital Exam 3 Fall 2017 Page 10 of 13

11 2s 3dz 2 2pz c. (2 points) Provide two sets of quantum numbers (4 quantum numbers) for an electron in the 2s orbital Set 1 [2,0,0,1/2] Set 2 [2,0,0, 1/2] d. (3 points) Rank the orbital in order of stability, from most stable to least Most stable Least stable 2s 2pz 3dz (6 points) State which of the following sets of quantum numbers would be possible and which would not. Using one or two sentences (not <,>, =,³, or any with slashes, symbols, arrows) explain what is wrong with the quantum numbers that are not possible. Note: missing the spin quantum number is not an error. a. [0, 1, 1] n can t be zero b. n =4, l = 5, m l = 1 l can t be greater than n since l = n 1 c. n = 1, l l =0, m l = 0 this is okay, it s a 1s orbital d. n =9, l = 5, m l = 1 this is okay, it is a 9g orbital. You did not have to know that l =5 is the g-subshell, but you should know that l ranges from n-1 0; l can range from 9 0, therefore l =5 is a reasonable value. e. [ 3, 0, 1] n can t be negative f. [2, -1, 0] since l is based on n, l can t be a negative number since n can t be negative 18. Consider the following reaction: 2 Mg (s) + O 2(g) 2 MgO (s) H = 1204 kj. (a) Is the reaction exothermic or endothermic? The reaction is exothermic because the H of reaction is negative. This means heat is released by the process Exam 3 Fall 2017 Page 11 of 13

12 (b) Calculate the amount of heat released when 2.4 g of Mg (s) reacts with excess oxygen at constant pressure. 2.4g Mg 1 mol Mg 1204 kj = kJ = 59kJ g Mg 2 mol Mg (c) How many kilojoules of heat are absorbed when 7.50 g of MgO (s) are decomposed into Mg (s) and O 2(g) at constant pressure? 7.50g MgO 1 mol MgO kj (decomposition changes the sign!) = kJ = +112kJ g MgO 2 mol MgO Exam 3 Fall 2017 Page 12 of 13

13 EXTRA CREDIT 19. (5 points) Atomic sodium emits light at 389nm when an excited electron moves from a 4s orbital to a 3s orbital (this emission is, in fact, very weak), and at 300 nm when an electron moves from a 4p orbital to the same 3s orbital. What is the energy separation (in kilojoules/mole) when an electron moves between the 4s and the 4p orbital? Draw a picture of the process. 4s 389 nm 4p 300 nm???? 3s a. What is the energy of these two wavelengths? E (4s 3s) =6.626x10 34 Js x x 10 8 m/s = x J 389 x 10 9 m E (4p 3s) =6.626x10 34 Js x x 10 8 m/s = x J 300 x 10 9 m b. What is the energy separation (in kilojoules/mole) when an electron moves between the 4s and the 4p orbital? E (4s 3s) E (4p 3s)= 1.52 x J/electron E (kj/mol) 1.52 x J x10 23 electrons 1kJ 1 electron Mol electrons 1000J 91.2 kj/mol Exam 3 Fall 2017 Page 13 of 13