Chemistry 103 Assignment No. 4 Summer 2010 Electron Configurations and Periodic Trends MULTIPLE CHOICE QUESTIONS Select the one best answer for each question. A. Which of the following is the full electron configuration for a neutral tin atom? (1) 1s22s22p63s23p64s23d2 (4) 1s22s22p63s23p64s23d10 (2) 1s22s22p63s23p63d4 (5) 1s22s22p63s23p64s23d104p65s24d105p2(3) 1s22s22p63p64s23d104p65s24d105p2 (6) 1s22s22p63s23p64s23d104p65s24d105p4 B. How many “p” electrons are in a neutral strontium atom? (1) 0 (5) 12 (2) 6 (6) 15 (3) 8 (7) 18 (4) 10 (8) 38 C. The highest energy electron in a neutral silver atom is located in which type of orbital? (1) 3p (5) 4d (2) 3d (6) 5s (3) 4s (7) 5p (4) 4p (8) 5d D. How many electrons are in the valence shell of a neutral antimony atom? (1) 3 (5) 10 (2) 5 (6) 13 (3) 7 (7) 51 (4) 8 (8) 85 E. Which of the following elements has the same number of unpaired electrons as germanium? (1) gallium (5) iodine (2) oxygen (6) Both (2) and (4) are correct (3) arsenic (7) Both (1) and (3) are correct (4) tin (8) All of the above are correct F. Which of the following is the correct electron configuration for the rubidium 1+ ion? (1) 1s22s22p63s23p64s23d104p6 (4) 1s22s22p63s23p64s24p64d105s1 (2) 1s22s22p63s23p64s23d104p65s2 (5) 1s22s22p63s23p64s23d104p45s2 (3) 1s22s22p63s23p64s23f104p6 (6) 1s22s22p63s23p64s24p64d10 G. Which of the following is the correct electron configuration for the arsenic 3- ion? (1) 1s22s22p63s23p64s23d1 (4) 1s22s22p63s23p3 (2) 1s22s22p63s23p64s23d104p6 (5) 1s22s22p63s23p64s23d10 (3) 1s22s23s22p63p6 (6) 1s22s22p63s23p62 H. Which of the following elements should have the smallest ionization energy? (1) cesium (5) phosphorus (2) krypton (6) oxygen (3) carbon (7) bromine (4) barium (8) calcium I. Which of the following elements should have the largest (most negative) electron affinity? (1) cesium (5) phosphorus (2) krypton (6) oxygen (3) carbon (7) bromine (4) barium (8) calcium J. Which of the following arrangements has the neutral atoms correctly sized from largest to smallest? (1) F > Al > S > Ca > K > Cs (5) F > S > Al > Ca > K > Cs (2) Cs > K > Ca > Al > S > F (6) Al > Ca > Cs > F > K > S (3) F > S > Ca > Al > K > Cs (7) Cs > F > Al > P > Ca > S (4) Cs > K > Ca > S > Al > F (8) F > S > Al > Cs > K > Ca K. The relative sizes of neutral rubidium and bromine atoms are shown on the right. Which picture below is the best representation for the relative sizes of the respective rubidium and bromide ions? L. Consider the ions Ba2+, Cs+, Se2¯, Br ¯. Which of the following arrangements would have the ions correctly sized from largest to smallest? (1) Cs+ > Ba2+ > Se2¯ > Br ¯ (4) Ba2+> Cs+ > Br ¯ > Se2¯ (2) Ba2+ > Br ¯ > Cs+ > Se2¯ (5) Se2¯> Br ¯> Cs+ > Ba2+ (3) Cs+ > Se2¯ > Br ¯ > Ba2+ (6) Ba2+> Se2¯> Cs+ > Br ¯ M. A second period element has the following successive ionization energies (in kJ/mol): IE1 = 1310; IE2 = 3390; IE3 = 5300; IE4 = 7470; IE5 = 10990; IE6 = 13330; IE7 = 71330; IE8 = 84080. Which element is this? (1) nitrogen (5) carbon (2) phosphorus (6) chlorine (3) silicon (7) fluorine (4) oxygen (8) sulfur Br¯Rb+Br¯Rb+Rb+Br¯Rb+Br¯(1) (2) (3) (4)RbBr3 PROBLEM QUESTION: You must show your work to receive any credit on this problem. This problem is worth 2 points. In Einstein’s interpretation of the photoelectric effect, the energy of the photon (E = h) absorbed by the metal is the sum of the minimum energy () needed to dislodge an electron from the metal surface and the kinetic energy (Ek) of the electron as it leaves the metal surface; therefore E = h=  + Ek. When light with  = 358.1 nm strikes the surface of potassium metal, the speed (u) of the dislodged electron is 6.40 x 105 m/s. If Ek = ½ mu2, what is  for potassium? Express your answer in J with the correct number of sig. figs. ANSWER LIST – List your answers here for grading. Check to make sure they are correct. Also, be sure to write your name in the space provided. Multiple choice: No. A B C D E F G H I J K L M Answer Problem Question: ________________ Name: