Updated Nov.16/04 ----------------------------------------------------------------------------------------- NUMERICAL ANSWERS TO ASSIGNED TUTORIAL PROBLEM SETS FOR CHEM205 FROM KOTZ & TREICHEL'S CHEMISTRY & CHEMICAL REACTIVITY ----------------------------------------------------------------------------------------- NOTE: none of the answers from Ch.8 have been verified NOTE: "^" means "raised to the power of..." to indicate scientific notation... NOTE: I cannot represent superscripts/subscripts here -- so...for polyatomic ions, I have left a space between the ion's formula and its charge Ch. Q# Answer ----------------------------------------------------------------------------------------- 8 2 no 2 e's in an atom can have same 4 Q.N.s 8 4 most stable config. has max # unpaired e-s C: 1s2 s2s 2p2 with 1e- each in 2 of the p orbitals 8 9 atom size decreases across period, increases down group I.E. & E.A.: increase across period, decrease down group 8 11 Mg: 1s2 2s2 2p6 3s2 Mg in group 2A, has 2 e-s in outer shell (2 valence e-s) Ar: 1s2 2s2 2p6 3s2 3p6 Ar in group 8A, has 8 e-s in outer shell (8 valence e-s) 8 15a Sr: [Kr]5s2 15b Zr: [Kr]4d2 5s2 15c Rh: [Kr]4d7 5s2 ...but actually observed as [Kr]4d8 5s1 15d Sn: [Kr]4d10 5s2 5p2 8 21a draw orbital boxes: Na+: 1s2 2s2 2p6 21b draw orbital boxes: Al3+: 1s2 2s2 2p6 21c draw orbital boxes: Ge2+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 21d draw orbital boxes: F-: 1s2 2s2 2p6 8 25 (boxes) Cu+: [Ar]3d10 4s0 i.e., lost e-s from 4s; not paramagnetic (boxes) Cu2+: [Ar] 3d9 4s0 lost e-s from both; paramagnetic 8 29a 18 e- 29b 10 e- 29c 1 e- 29d none; when l=0, m(l) can only have value of 0 8 31 P: [Ne]3s2 3p3 with 1e- in each of the p orbitals n=3, l=0, m(l)=0, m(s)= +1/2 n=3, l=0, m(l)=0, m(s)= -1/2 n=3, l=1, m(l)=-1, m(s)= +1/2 n=3, l=1, m(l)= 0, m(s)= +1/2 n=3, l=1, m(l)=+1, m(s)= +1/2 8 35 P < Ge < Ca < Sr < Rb 8 37a Cs > Rb 37b O2- > O 37c As > Br 8 41a Al 41b Al 41c C (based on periodic trends); Si (according to experimental data) 41d Al < B < C 8 43a F < O < S trend: increase across period, decrease down group 43b based on trends: predict highest I.E. for S. However, I.E. of O < N, and same thing observed for 3rd row --> real I.E. order = Si < Se < S < P 43c F- < O2- < N3- isoelectronic; size decreases as eff. nucl. charge increases 43d Cs < Ba < Sr trend: increase to right, decrease down table 8 45 read filling order off periodic table: U: [Rn]5f3 6d1 7s2 to form cation, lose from highest n values 1st: U4+: [Rn]5f2 8 47a Z = (2+8+8+2) = 20 47b total # s e-s = (2+2+2+2) = 8 47c total # p e-s = (6+6) = 16 47d total # d e-s = 0 47e Ca, calcium, a metal 8 49 (b) the maximum value of l is (n-1) 8 55a metal 55b B 55c B 55d A 8 57 In4+ 3 outer shell e-s, unlikely to form 4+ ion Fe6+ ions with charge > 3+ or 4+ unlikely to form Sn5+ tin has 4 outer shell e-s, unlikely to form 5+ ion 8 63 I.E. increases across P.T. and decreases down P.T. decrease occurs because e- farther from nucleus, & electrostatic attractions between + nucleus and - electrons is distance dependent 8 65a V, vanadium 65b group 5B, period 4 65c transition element 65d paramagnetic, 3 unpaired e-s 65e n=3, l=2, m(l)= -2, m(s)= +1/2 n=3, l=2, m(l)= -1, m(s)= +1/2 n=3, l=2, m(l)= 0, m(s)= +1/2 n=4, l=0, m(l)= 0, m(s)= +1/2 n=4, l=0, m(l)= 0, m(s)= -1/2 8 67 [Ne]3s2 3p6 4s1 --> [Ne]3s2 3p6 + 1e- K --> K+ [Ne]3s2 3p6 --> [Ne]3s2 3p5 + 1e- K+ --> K2+ 8 69 For Li+, one less e- to be attracted by same nuclear charge, but outer e-s are in n=1 shell, closer to nucleus. For F-, one additional e-, in 2p orbital. This adds e- e- repulsions while nuclear charge remains the same. 8 71 Many possible arguments why Mg/O composed of Mg2+ & O2- ions: 1) groups characteristically form ions with noble gas configns; Group 2A loses 2 e-, Group 6A gains 2 e-s... 2) look at other ionic compounds from same groups: MgS, CaO 3) energetically more favourable to form Mg2+/O2- than Mg+/O- Possible experiments: 1) determine 1st and 2nd I.E.s for Mg & 1st and 2nd E.A.s for O 2) determine melting point for MgO & compare to +1/-1 compound NaF (990C) and the +2/-2 compound CaO (2580C)...think Coulomb's law 8 79 Increasing effective nuclear charge across period...causes I.E. to increase. For S: 2 of its 4 "2p" e-s are paired; greater repulsion (pairing energy) makes it easier to remove one of them, so I.E. of S is a bit lower than expected. (same as explanation for oxygen vs. nitrogen in class)