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 Δ 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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