Chem%20161-2009%20Exam%20III%20%2B%20answers

Chem%20161-2009%20Exam%20III%20%2B%20answers - Chem...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Chem 161-2009 Exam III + Answers Chem 161-2009 Exam III Hill, Petrucci et al. Chapter 8 – Electron Configurations & Periodic Table Trends Periodic table trends 1. Which of the following oxides would produce a basic solution when added to water? I Li 2 O II SO 2 III CaO IV P 4 O 10 A . I and III B. II and IV C. I, II, III, and IV D. I, II, and III E. II and III Basic solutions are produced by oxides from the left side of the periodic table, while acidic solutions are produced by oxides from the right side of the periodic table. Li 2 O + H 2 O 2LiOH, a basic solution SO 2 + H 2 O H 2 SO 3 , an acidic solution CaO + H 2 O Ca(OH) 2 , a basic solution P 4 O 10 + 6H 2 O 4H 3 PO 4 , an acidic solution Chem 161-2009 Exam III Hill, Petrucci et al. Chapter 10 – Bonding Theory and Molecular Structure Hybridization, σ - and π -bonds 2. Lithium forms a +1 ion and is in column 1. Which of the following is likely to form ions two units lower in charge than expected from the group number? (For example, an element in column 8 forming a +6 charge). A. Al B. Hg C . Sb D. Se E. Cd Group Number Ion Al 3 Al +1 Hg 2 Hg Sb 5 Sb +3 Se 6 Se +4 Cd 2 Cd Acceptable ions follow the formula: Group number to (group number – 8) Al: 3 to (3-8) = 3 to -5 = +3, +2, +1, 0, -1, -2, -3, -4, -5 Hg: 2 to (2-8) = 2 to -6 = +2, +1, 0, -1, -2, -3, -4, -5, -6 Sb: 5 to (5-8) = 5 to -3 = +5, +4, +3, +2, +1, 0, -1, -2, -3 Se: 6 to (6-8) = 6 to -2 = +6, +5, +4, +3, +2, +1, 0, -1, -2
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 Cd: 2 to (2-8) = 2 to -6 = +2, +1, 0, -1, -2, -3, -4, -5, -6 Based on the above approach, all of the atoms can form ions two units lower in charge, so this approach doesn’t rule out anything. One can argue, however, that Hg and Cd aren’t ions, so they can be ruled out. So that leaves Al+1, Sb+3 and Se+4 as options. Another approach would be to determine the answer by removing electrons from the Aufbau configuration. Only Al, Sb and Se should be considered, because Hg and Cd form zero charges. 13 Al = [Ne]3s 2 3p 1 . Does Al form a +1 ion? Possibly, by removal of a 3p 1 electron. 51 Sb = [Kr]5s 2 4d 10 5p 3 . Does Sb form a +3 ion? Possibly, by removal of three 5p electrons. 34 Se = [Ar]4s 2 3d 10 4p 4 . Does Se form a +2 ion? Possibly, by removal of two 4p electrons. Al is not likely to be the answer, because the students learned in an earlier chapter that Al only forms a +3 ion (i.e., type I nomenclature). Se is not a good answer because it is not likely that only two electrons will be removed from 4p 4 . That leaves only Sb. It’s reasonable for three electrons to be removed from antimony’s 5p subshell. The students might have difficulty solving this problem because the periodic table handed out has the group numbers from 1 to 18, rather than from 1A to 8A and 1B to 8B. However, it is my understanding that the students were told how to convert from the new group identifications to the old ones. Chem 161-2009 Exam III
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/08/2011 for the course CHEM 161 taught by Professor Vacillian during the Fall '08 term at Rutgers.

Page1 / 19

Chem%20161-2009%20Exam%20III%20%2B%20answers - Chem...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online