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1307HW08Key - Homework 8(Chapter 8 Name féx CHEM 1307 —...

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Unformatted text preview: Homework 8 (Chapter 8) Name féx CHEM 1307 — 003 (print your name) How many electrons can each of the following quantum number or sublevel designations have in an atom? (Q1 - 03) 1. n=2,l=1.rn,=-1 «(3; EML‘ aflfif-nj can I’m-Ve— MD More 79%;“ 2 gfecms w/ (B 23 g appostnj S in. (e) 8 2. 5d (VI-=5; 1:2) (a) 2 4; orA‘flJS = a: of possible we ulna 320 = 93+: = sum-{:5 C (d) 14 (e) 32 3. n=4,l=3 (4f) (a) 2 (b) 5 (c 10 14 go) 18 Write the full electron configurations for ground-state of the following element: (04 - 06) 4' 2:): 1321p“25"’2p‘53523p5 (b) 1522522p°3523p°3d3 E (c) 1322522p‘535234364523d3 (d) 1s’25’2p‘3523p54523d‘ 1522522p°3523p°4523d1 QQ—H = 2x34! =7 5. As ‘ 1522322p‘3523p‘54523d1"4:33 1522522p‘3523p64s23dw4p5 ,4 (c)1522522p63sz3p64824p64d7 (d) 1322522p63323p63da4324p3 (e) 1521p°2322ps2d1°3323p5 6. La a) 1522522p63523p64523d1°4p55524d1°5p66524f1 1s22s22p°3523p°4523d‘°4p°5s24d1°5p66525d‘ (c) 1s’2s’2p‘3s23p‘4s24p54d‘°5525p°5d‘°6s26p‘ (d) 1322522p53523p°3d1°4sz4p°4d1°4f9552 (e) 1521p62522p63523p63d1°4sz4p54d1°5525p3 Write the condensed electron configurations for ground-state of the following element: (QT — 08) 7. Ga (:0 [an4p (b) [Ca]3d° 494p [email protected] [Ar14s23d°4 (d) [Kr14s23d‘°4p (e) [Ne]35 23pfi4sz4p1 8. Bi (a) [X81693 (b) [Xe]6526p3 ”D c) [Xe16524f‘46p3 [Xe]6524f”5d1°6p3 (e) [)(e]~$sz$f1“Sclmfip3 Write a full set of quantum numbers for the following. (Assuming ms of 1St electron in an orbital is +112) (09 — Q12) 9. The outermost electron in an Na atom N“= ta‘as‘ir‘ Ls: (a)n=2.l=1,mr=1.m=+1/2 _ _ | @n= 3!=0,m,=o,m:=+1r2 'Yl=3_ 1-0, wuzo Wis-+/2 B (c)n= 3.I=1.m =Om5=+112 (d) n=31=1.m:=—1 ms=+112 (e) n=31=2m,,=2 ms=+112 1 l 10. The electron lost when a B atom becomes a B+ ion. 6, Isaas 1F (a n=2.l=0 m,= 0 m5=+1l2 +1 . (b;n=2,i=0. m,=1 ms=+112 n-a. 3-“ Y“: .3. “‘6" +/?- C ®n=2,i=1.mf=0.ms=+112 (d)n=2,l=1,m,.=-1.ms=+1 (e) n=3,l=1,m,=1, ms=-112 lo 11. The highest energy electron in the ground state Cu atom 04:]: Ar] 3‘1 19’. (a) n=31=0 rm: 0 ms=-112 R “L“: on“ {my}, Lin-DJ mi.“ km low C @n=4,r=o.m,=0.ms=+112 ““31 lord Nu hi. (d) n=4,l=1,m,=1.ms=-112 (e)n=4,l=1,m,=0,ms=+1l2 “=4, i=0. Wu- 0. Ms=+}é. 12. The electron gained when a Br atom becomes a Br ion. . (a) n= 4 [=0 m,= o ms=-112 Br:- fiflg 11.-H elm-u.“ :3de (b)n=4,l=0,m;=1.m5=-1l2 "'- '— (c) n=4.l=2,m,=1.m5=-112 4‘5 *l’ E (d)n=4,l=3,m,=1,m5=-1IZ “:4 1H m +1 @n=4,l=1,m,=1,ms=-1IZ ' " ‘ "31 \m,=-%. Which of the following sets of the quantum numbers of electrons are not allowed in the atoms? (Assuming ms of 1"" electron in an orbital is +1I2) (Q13 — Q14) 13.Catom : a. 1 (a)n=1,l=o,mi=0,ms=+112 C= “5 ’5 ‘l’ 4 4 (b)n=2.l=o.m,=0,ms=—1/2 flfl_.__.. D c)n=2,l=1.m,=-1.ms=+1lz is 25 2P n=2.r‘=1.m,=—1,ms=-1IZ (e)n=2,i=1,m,= ,ms=+112 14. Mn atom Mn.- EH] ads-‘5 @n=4,l=1,m;=-1ms=+112 llv (b)n=3,l=2,m;=-1,ms=+112 LLLLL "; A (c)n=3,l=1,m,=0.ms=+112 34 (d)n=3,l=0,mi=0,ms=+1l2 (e)n=2,i=0,mi=0,ms=-1I2 (Ti How many valence electrons are present in an atom of the following elements? (Q15 — Q16) 15. Si (a) 5 For Mn STOUT e\ewxe.vn— B (°)3 *1: valence 6‘ = (“5+"0 9'— (e) 1 _. _ Grow‘afi‘ik (c) 3 (d) 4 (e) 5 How many inner electrons are present in an atom of the following elements? (Q17 — Q19) 17. F . Balsa} 5 @\ 2 F. _ f 4 Pk (c) 6 (d) 8 (e) 9 18. Ti T; ; [fir] 34:45,. 18+: =20 (a) 2 (b) 8 (c) 18 @20 (e) there are no inner electrons in an atom of Ti element 19. Pb (a) 36 Pic: U414?” SA" 55"er (b) 54 . (c) 56 (mu e“= 5'4 HA. +10 = “[8 (d) 70 (e) 9978 20. Select the element and electron configuration that do not match. (a) P [Ne] 3523p3 _. (b) Hg [Xe] 6525dw4f“ (c) Se [Ar] 45:3d;°4p4 (d) Co [Ar] 45 3d .Pr [Xe] 6525d‘4f‘ slant-I In «F One reason spectroscopists study excited states is to gain information about the energies of orbitals that are unoccupied in an atom‘s ground state. Each of the following electron configurations represents an atom in an excited state. Identify the element. and write its condensed ground-state electron configuration. (021 — Q24) 21 . 1s.22s22p"3s1 g (a) Ne. [leszzpi simu be 611511? (b) Ne. [Be]2p6 C @Ne. [i-Ie12s22p6 (d) Na. [Ne]3s1 (e) Ne. [He]2522p5391 221522522 63523p‘3d‘ (a) K. p [Arl4s‘ stLA he. 6125339352??? (b) Sc. [Ar] 3d1 (0) Ar, [Ne] 3323p6 D {3) CI, [Ne] 3523p5 (e) CI, [He] 2522p°3523p5 23. [Ar]4523d‘55' z y e 3:323:32 5in e We 34 n, s (c) v, [Ar]4523d3 B (d) Se, [Ar14s24p‘ (e) Br, [ll-tr14524p5 24.1512322p' 1 a (a) c. [He]2322p2 sink} in is 25 (b) 3. [He]2322,r.)1 (c) Be, [H1232 “D (3 Be, [He]252 (e) Li,[He]Zs1 Using periodic trend of IE. arrange each set of atoms in order of increasing IE1. (025 —- 026) 25 8 Ga, Ga, F, He -.Ca<Ga<S<F<He soon-e furs; (“46K (b)Ga<Ca<F<He<s A (C)He<F<S<Ga<Ca G'Wf3" Glutm ‘d’s‘He‘Ga<F<Ca 3” west; Ail-<5 (e) F<Ca<Ga<He<S 26. N, Al. 0, RD, P (Eur-up IA Rb< NIL. (a)Al<Rb<P<N<0 W343: Na<Al<P _’_ (b)A|<N<O<P<Rb GrumpyA. P416 t (c)O<N<P<AI<Rb ”a“ 2' N40 d P<AI<Rb<O<N Rb<Al<P<N<O Using periodic trend of atomic size. arrange each set of atoms in order of decreasing atomic size. (027—028) 27. Cs. Cl. Ge. Ca. Ne (a)Ca>Cs>Ge>Ne>CI Gun-PIA C.'>K, PusA4. K>Q>Ge>Br B ®Cs>Ca>Ge>CI>Ne (c)C|>Ne>Ge>CS>Ca 5!an.- Br7C\?~F. Fwd?! F7Ne. (d) Ne>CI>Ge>Ca>Cs (e) Ge<Cs>Ca>Ne>CI 28. As, Ba, Se, S, Sn (a) Ba>Sn>As>Se>S Erwin: Bo.>Sr. PuroJE, Sv>Sn>SL (b)S>Se>As>Sn>Ba . r 5' A (c)As>Ba>S>Se>Sn 5"?!” 51”“ P”°'”‘ A" ‘ (d)Ba>As>Sn>S>Se Ewan-96A; 59.78 (c) Se>As>Sn>S>Ba 29. Write the full electron configuration of the Period 2 element with the following successive lEs (in MJImoI) 2?‘ as" as‘ is‘ 15' IE = 0.801 IE2 = 2.427 IE3 = 3.659 IE4 = 25.022 IE5 = 32.822 (a) 1322522p'5 (b) 1522522535 E (c) 1s22522p4 d) $22522»2 1522522p1 30. Determine the element in Period 3 with the following successive lEs (in MJImoI) IE1 = 0.738 IE2 = 1.450 IE3 = 7.732 IE4 = 10.539 IE5 = 13.628 (a) Na 38" as‘ 2.95 a?! “1’4 (b) Al @3 Mg (d) P (e) Si 31. Which element would you expect to be teast metallic using periodic trend of metallic behavior? CI ( ) Cs A- (c) P (d) Se (e) Ca What is the charge of the monatomic ion most likely to be formed by each of the following. (032 — 035) 32. O (a) +2 (b) +1 j) (c) -1 ® -2 (e) -3 33. Mg em (b) +1 A (c) -1 (d) -2 (e) -3 34. Pb (a) +1 +2 B (c) +3 (d) +4 (6) +5 35. Ga (a) +1 @252 +3 C ( ) +4 (e) +5 36. Which element contains most unpaired electrons in the ground state of its atom? (a) S (b) {:19 ‘" (C) t d Mn Cr 37’. Indicate which of the following ions are paramagnetic. 1. Hg” 2 Cu2+ 3 Mn2+ 4. Ti“ 5. Pb” v' v' ‘ (a) 1. and 4 fwkmaakbhf-zfi uw‘sadreé elect-tong (b) ‘l, 4, and 5 C (E) 2 and 3 (d) 2, 3 and 5 (e) all ions above are paramagnetic 38. Which of the following species are diamagnetic? 1. N 2. N3 ‘3 N. 4 Zn 5.Y2+ (a) 1 and 3 V ‘ \f _ (b) 1 and 5 ‘Ahw‘aa‘c‘hfi -. w» uqunA e C 2, 3and4 2. Band 5 Rank the ions in each set in order of increasing size. (Q39 — Q41) 39. F', Br', Cl‘, l‘ (a F' < Br' < Cl' < l' éF' < cr < Br' < r c Br'<Cl'< |‘<F‘ B (d) r< F' < or < Br' (if; F‘ < or < I'< Br" 40. st“, 32'; P3"; cr @ 01' < 32- < P3' < sr- (b) er < P3' < 32' < or A (c) P3‘< 32. < or < sr“ (d) Si4‘< 133' < Cl' < 82' (e) 52. < or < Si"< P3' 41. AI3+ F‘, Mg“, 02', Na‘ (3) F' < 02' < AI3+ < Mg” < Na‘ (b) 02' < P < AI3+ < Mg2+ < Na+ (c) F' < 02‘< Na* < Mg” < AI“ @ Ali“ < Mg” < Na+ < F“ < 02* (6) Na" < Mg” < Ai‘“ < P < 02' 42. According to First Ionization Energy of the Main-Group Elements, (Figure 8.18) one can observe that elements in 2"d row have increasing ionization energy in general as crossing the period from left to right. However, boron and oxygen have slightly lower ionization energies than their neighboring atoms. Why? (a) That is because both boron and oxygen are more metallic than other nonmetal elements. (b) That is because both boron and oxygen have larger atomic sizes than other nonmetal elements. would Any? low energy {eve/adwouU C @ That is because both elements will have empty and half-filled p sublevel whiclye iminates the repulsion between electrons in the same orbitais after having lost one electron. (d) That is because both boron and oxygen have less Zen than their neighboring elements as 25 orbital has stronger tunneling effect. (c) That is because both boron and oxygen have less electron affinities than their neighboring elements. 43. According to Electron Affinities of the Main-Group Elements, (Figure 8.20) one can observe that nonmetal elements in 2“” row have less negative values in general than the elements within the same group in 3"1 period. Why? (a) It is because the nonmetal elements in 2"" period have smaller Zefl so that they do not attract electrons than those in period 3. (b) It is because the nonmetal elements in period 2 are generally larger in sizes than those in in period 3, which have less affinity to electrons. (c) It is because the nonmetal elements in period 2 have more IE1 than those in period 3, which makes them more difficult to accept electron. :b It is because the nonmetal elements in period 2 are generally smaller in their atomic sizes than those in period 3, which cause more repulsion when adding one electron. (e) It is because the nonmetal elements in period 2 have more IE1 than those in period 3, which makes them more difficult to accept electron. 44. In Electron Affinities of the Main-Group Elements, (Figure 8.20) one can observe that metals in Group 1A show stronger electron affinities than those in Group 2A. Why? (a It is because Group 1A elements are less metallic than elements in Group 2A. KB @ It is because Group 2A elements have filled s orbital. no more electron can be added. (0) It is because Group 2A elements have larger atomic sizes than those in Group 1A. adding more electron cause more repulsion. (d) It is because adding electron to atoms causes additional repulsion, Group 2A elements have less EA because there are one more electron than those in Group 1A. ...
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