South Pasadena AP Chemistry Name 8 Atomic Theory Period Date U N I T T E S T P R A C T I C E Part 1 Multiple Choice You should allocate 25 minutes to finish this portion of the test. No calculator should be used. A periodic table and data table will be provided. Select the answer that best responds to each question. 1. Sodium exists in only one naturally occurring isotope: Na-23. However, chlorine exists in two naturally occurring isotopes: Cl-35 and Cl-37. Which of the following can be the number of neutrons in one Na-Cl compound? (A) 30 (B) 58 (C) 58.5 (D) 60 2. Fluorine is one of nineteen elements that are said to be mononuclidic, meaning it has only one stable naturally occurring isotope. What are the numbers of protons, neutrons, and electrons in the most common ion of naturally-occurring F? (A) 9 protons, 10 neutrons, 10 electrons (B) 9 protons, 10 neutrons, 9 electrons (C) 9 protons, 9 neutrons, 10 electrons (D) 9 protons, 9 neutrons, 9 electrons 3. The mass spectrometry plot for the element boron, B, is shown above. The plot provides evidence for which of the following? (A) The B-11 isotope has a greater nuclear charge than the B-10 isotope. (B) The B-11 isotope has more electrons than the B-10 isotope. (C) The B-11 isotope is denser than the B-10 isotope. (D) The B-11 isotope is more abundant than the B-10 isotope. 4. Which of the following requires a photon of light with the shortest wavelength to break the bond? (A) The C O single bond in methanol, CH 3 OH. (B) The C...O bond in the carbonate ion, CO 3 2, which has a bond order of 1.3. (C) The C=O double bond in formaldehyde, CH 2 O. (D) The C O triple bond in carbon monoxide, CO. 5. When ultraviolet light of a particular wavelength shines upon a piece of zinc metal, several electrons are ejected, and their speeds measured. Electron A 5.0 10 5 m/s Electron B 8.0 10 5 m/s Which electron has a higher binding energy? (A) Electron A (B) Electron B (C) They have the same binding energy (D) Their binding energies cannot be compared with the data given 6. Ultraviolet light of the following wavelengths were detected from the hydrogen emission spectrum: 122 nm, 103 nm, 97 nm, 95 nm There were no ultraviolet light observed with longer wavelengths. The 122 nm line corresponds to electrons that drop between which energy levels? (A) n = 2 to n = 1 (B) n = 3 to n = 1 (C) n = 3 to n = 2 (D) n = 4 to n = 2
7. Which of the following statements about the 3p orbital in the aluminum atom is incorrect? (A) A 3p orbital has two nodes. (B) Electrons in the 3p orbital are always further away from the nucleus than electrons in the 3s orbital. (C) The 3p orbital is higher energy than the 3s orbital. (D) There are three degenerate 3p orbitals in this subshell. ψ radial distance 8. The wave function, ψ, for which atomic orbital is shown above? (A) 2s (B) 2p (C) 3s (D) 3p 9. How many electrons in the ground state of Te have l = 1? (A) 4 (B) 6 (C) 22 (D) 52 10. Which of the following could be the set of quantum numbers that describes an electron with the highest energy in the ground state electron configuration of Rh 2+? (A) n = 3, l = 2, m l = 0, m s = +½ (B) n = 4, l = 2, m l = 1, m s = +½ (C) n = 5, l = 0, m l = 0, m s = +½ (D) n = 5, l = 2, m l = 1, m s = +½ 11. The PES for lithium, Li, is shown above. Which of the following correctly identifies and explains the peak on the right? (A) The peak to the right represents electrons in the 1s orbital because electrons in lower energy shells have shorter peaks. (B) The peak to the right represents electrons in the 1s orbital because electrons in lower energy levels are closer to and more attracted to the nucleus. (C) The right peak represents electrons in the 2s orbital because electrons in higher energy shells have shorter peaks. (D) The right peak represents electrons in the 2s orbital because electrons in higher energy shells are further away from and less attracted to the nucleus. 12. The spectra for nitrogen (solid) and oxygen (dotted) shown above provide evidence for which of the following? (A) The first ionization energy of nitrogen is greater than that of oxygen. (B) The 2p electron of nitrogen is further from the nucleus than that of oxygen. (C) The nuclear charge of nitrogen is greater than that of oxygen. (D) The ionic radius of nitrogen is greater than that of oxygen.
13. The electron affinity of carbon ( 122 kj mol 1 ) is more negative than boron ( 27 kj mol 1 ). This can primarily be explained by: (A) Carbon has a smaller atomic radius than boron, so it is easier to add electrons. (B) Carbon has more core electrons than boron does, so the outer electrons experience greater shielding. (C) Carbon has more protons than boron, so it has greater effective nuclear charge. (D) Carbon has more valence electrons, so it has greater electron repulsions than boron. 15. Which element would most likely have the following ionization energies? First Ionization Energy 738 kj mol 1 Second Ionization Energy 1451 kj mol 1 Third Ionization Energy 7733 kj mol 1 Fourth Ionization Energy 10,540 kj mol 1 (A) Na (B) Mg (C) Al (D) Si 14. Which of the following quantities is larger for Mg than Al? I. Atomic Radius II. Ionic Radius III. Ionization Energy (A) I only (B) III only (C) II and III only (D) I, II and III Part 2 Free Response You should allocate 30 minutes to finish this portion of the test. You may use a scientific calculator. A periodic table and data table will be provided. Respond to each part of the questions completely. Be sure to show your work clearly for questions that involve calculations. (From AP Chemistry 1999 #2) 16. Answer the following questions regarding light and its interactions with molecules, atoms, and ions. (a) The longest wavelength of light with enough energy to break the Cl Cl bond in Cl 2 (g) is 495 nm. i. Calculate the frequency, in s 1, of the light. ν = c λ = 3.0 108 m/s 495 nm 10 9 nm 1 m = 6.06 1014 Hz ii. Calculate the energy, in J, of a photon of the light. E = h ν = (6.626 10 34 J s)(6.06 10 14 Hz) = 4.02 10 19 J iii. Calculate the minimum energy, in kj mol 1, of the Cl Cl bond. 4.02 10 19 J 1 bond 1 kj 1000 J 6.02 10 23 bonds = 242 kj/mol 1 mol
(b) A certain line in the spectrum of atomic hydrogen is associated with the electronic transition in the H 2.178 10 18 atom from the sixth energy level (n = 6) to the second energy level (n = 2). E n = n 2 J i. Indicate whether the H atom emits energy or whether it absorbs energy during the transition. Justify your answer. 2.178 10 18 2.178 10 18 E 2 E 6 = 2 J 2 6 2 J = 4.84 10 19 J Energy is emitted because E < 0. When electrons drop from a higher energy to lower energy level, energy is emitted. ii. Calculate the wavelength, in nm, of the radiation associated with the spectral line. λ = h c E = (6.626 10 34 J s)(3.0 10 8 m/s) (4.84 10 19 J) 10 9 nm = 411 nm 1 m iii. Account for the observation that the amount of energy associated with the same electronic transition (n = 6 to n = 2) in the He + ion is greater than that associated with the corresponding transition in the H atom. He + has 2 protons, so the electron shells are more attracted to the nucleus than those in H resulting in greater electron transition energies. (From AP Chemistry 2000 #7) 17. Answer the following questions about the element selenium, Se (atomic number 34). (a) Samples of natural selenium contain six stable isotopes. In terms of atomic structure, explain what these isotopes have in common, and how they differ. Isotopes have the same number of protons, but different number of neutrons. (b) Write the complete electron configuration (e.g., 1s 2 2s 2 etc.) for a selenium atom in the ground state. Indicate the number of unpaired electrons in the ground-state atom, and explain your reasoning. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4 (c) In terms of atomic structure, explain why the first ionization energy of selenium is i. less than that of bromine (atomic number 35), and Se and Br are both in the 4p subshell, but while Se has 34 protons, Br has 35 protons. Outer electrons are less attracted to a nucleus with fewer protons (lower Z eff ), resulting in a lower IE. Because Se has fewer protons than Br, it has a lower IE than Br. ii. greater than that of tellurium (atomic number 52). Outer electron of Se is in the 4p subshell, but that of Te is in the 5p subshell. Electrons in lower subshells are more attracted to the nucleus, and require more energy to remove. Because outer electron of Se is in a lower subshell than that of Te, Se has a higher IE than Te.
South Pasadena AP Chemistry Name 8 Atomic Theory Period Date U N I T T E S T B L U E P R I N T Part 1: Multiple Choice Format: 15 questions, four answer choices: (A)-(D) Expected time: 25 minutes Allowed resources: Periodic Table, Equations and Constants. No calculators. Q Lesson Topic Objective 1 8.0 Atomic Structure Describe an atom with isotopic notation. 2 8.0 Isotopes Describe an atom with isotopic notation. 3 8.1 Mass Spectroscopy Calculate the atomic mass of an element using isotopic abundance information from mass spectroscopy data. 4 8.1 Light Compare and calculate the wavelength, frequency, and energy of particular electromagnetic radiation. 5 8.1 Photoelectric Effect Explain how the photoelectric effect demonstrated the particle nature of light, and the diffraction of electrons demonstrated the wave nature of matter. 6 8.1 Hydrogen Spectrum 7 8.1 Atomic Orbitals Explain how the quantum numbers for a wave function determines the electron distribution of atomic orbitals 8 8.2 Atomic Orbitals Explain how the quantum numbers for a wave function determines the electron distribution of atomic orbitals 9 8.2 Electron Configuration Predict the ground state electron configuration of an element or ion based on its position on the periodic table. 10 8.2 Electron Predict properties of an element based on its valence shell electron configuration. Configuration 11 8.2 PES Explain an element s electron configuration using its photoelectron spectrograph. 12 8.3 PES Explain an element s electron configuration using its photoelectron spectrograph. 13 8.3 Periodic Trends Elements Use Coulomb s Law to explain how attractions between the nucleus and valence electrons change between elements on the Periodic Table or ions. 14 8.3 Periodic Trends Ions 15 8.3 Periodic Trends Successive IE Compare properties (i.e. size, ionization energy, electronegativity, ion charge, reactivity) between two elements or ions by comparing attractions between the nucleus and valence electrons. Compare properties (i.e. size, ionization energy, electronegativity, ion charge, reactivity) between two elements or ions by comparing attractions between the nucleus and valence electrons. Part 2: Free Response Format: o 1 long question (5-8 parts) o 1 short questions (2-4 parts) Expected time: 30 minutes Allowed resources: Periodic Table, Equations and Constants, and scientific calculators. Topics: Any