E01-2 - 4. with the smallest first ionization energy? 5....

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Additional Exam 1 Practice Problems (from textbook sections 8.0-8.4) ( Answers are on the next page, don’t peek! ) 1. Select the correct ground-state electron configuration of sulfur (S). (A) 1 s 2 2 s 2 2 p 8 3 s 2 3 p 2 (B) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 4 (C) 1 s 2 2 s 2 2 p 8 3 s 2 3 p 4 (D) 1 s 2 1 p 6 2 s 2 2 p 6 (E) 1 s 2 2 s 2 2 p 6 3 s 2 3 d 4 2. A potassium atom absorbs 767.5 nm wavelength light, causing an electron in the 4 s state to be excited into the 4 p state. The energy difference (in J) between the 4 p and 4 s states of potassium is (A) 2.18 x 10 –18 J (B) 5.09 x 10 –31 J (C) 4.92 x 10 –17 J (D) 1.36 x 10 –19 J (E) 2.59 x 10 –19 J Consider the electron configurations in answering questions 3 – 5: A) [Ar]4 s 1 B) [Ne]3 s 2 3 p 5 C) [Ar]4 s 2 3 d 3 D) [Ne] 3 s 2 3 p 1 E) [Ar]4 s 2 Which electron configuration belongs to the element … 3. … with the most negative electron affinity?
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Unformatted text preview: 4. with the smallest first ionization energy? 5. with the largest difference between its third and fourth ionization energies? 6. Which correctly lists the five atoms in order of increasing atomic radius (smallest to largest)? (A) Cl < I < Sn < Rb < Cs (B) Cs < Rb < Sn < I < Cl (C) Cl < Rb < Sn < I < Cs (D) Cs < I < Sn < Rb < Cl (E) Cs < Sn < I < Rb < Cl 7. For which process does E equal the second ionization energy of aluminum? (A) Al 3+ (g) + e Al 2+ (g) (B) Al 2+ (g) + e Al + (g) (C) Al(g) Al + (g) + e (D) Al + (g) Al 2+ (g) + e (E) Al(g) Al 2+ (g) + 2e Answers (1) B (2) E (3) B (4) A (5) D (6) A (7) D...
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E01-2 - 4. with the smallest first ionization energy? 5....

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