Homework 2 - lam-he (jpb2376) — Homework 2 - Sutcllfl'e...

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Unformatted text preview: lam-he (jpb2376) — Homework 2 - Sutcllfl'e - {53120) 1 This printout should have i8 quefilons. Multiplechoice questions may continue on the next column or page m find all choices before answering. This assignmmt is due at 11pm rm Lime date indicated in the assignment. menu on Quest. 001 10.0 :30th who, 31.2)16 related to 1. Bohr magnetoms. 2. the nocturnal confines of mercury lamps. 3. the probability ofap orbital becoming an s orbital 4. the Laplaciau transl'onn of the Bein- moa— lion. 5. the probability of finding an electron at the point. (2,352} in space with the origin (0) at; the nucleus. correct; Expkmtion: ‘1'! aim the probability of finding an 9139 tron in a partiouiar infinitesimale small vol— ume ofspaoe in an atom. one 10.0 points In a ouo-dimms‘rons] particle in a box, for ‘11.], how many nodes are predicted? 1. U 2. 2 B. 4 4. 1 5. 3 correct: Explanation: lav/3v! 003 10.0 pom Wlmt is the subshell notation and the nun» ber of eloctrom: that can have the quantum number-Sn =25, ewe? 1. 5p; 4 2.5.1,- '2 3. Noneoitliese 4. 5:", 1 5.537; 5 6. 55; floor-red: Explanation: The notation is we where 1" W 1'2331‘1’15t' -- £m0,1,2,--—,(n— 1) represented by alelzber: a flu m.= nz.w(c~1).--<tv-2>,---.o.-~ +(e w 2),+(e 1),+2 and m. = To find thenmibor of orbitelsthatcanhave the stated willie; era and t (and any allowed mince ofm; and rm), use 3 to find thenumber of different values oimg. To find the maximum number of electrons that can have the stated Values of n and. l (and any allowed value of m; 31111112,), double the number (AF different. values of mg. 004 10.0 points Thenandequantumnmnbasofflwlast electrenofanolmmnt Harrow-land fora The element is 1. an fitramniou metal. 2. a diminution metal. correct 3. a noble 33$. 4. a representative element. burke (jph‘zaTG) - iiomeworlc ‘2 w Subcllfi‘e -— (53120} 2 5. a nonmetal. Exploitation: Where :8 a 2, d electrons are filling. These are the transition metals. 905 10.0 points flow many d elem-om; doool (Memo member 53) mass? 1. 20 (nor-sect 2. 8 3- 5 4.10 5.0 6. 16 7. 53 Explanation: I has 10 eiectrom in the 3d orbital and 10 electh in the 4d orbital to give a total of 20 (i electrons. can 10.0 puma For wliid: llnatom wavefimcfiors aroyoumost. likely to find the cinch-on farthest from the nucleus? 1. 2;) 2. 39 3. dp eon-oat 4. 25 5. is Explanation: The 412 electron Will have the greatest elec- l.an demily farthest. from the nucleus. 0t)? 10.0 points Write the ground-state electron configluation oi a load atom. l. {Xe} 4;“ 54*" up“ 2. EXerltsdissl lspl'rs2 3.;Xe14f145d95336p3 4. file] 4f1‘5d‘06526p2 eon-em '5. “014;”de 6.1 6p? Explanation; The Anibal). order of electron filling is 1:, 25. 2P. 33, 3P. 4S. 34, 4F. 55. 4d. 5P, 65. 4f. 5d. 6:2, one. s orbitals can hold 2 ale-circus, p orbitals 5 oleclrnm, and d orbitals 10 electroma. N090 some Wtim do occur in the electron con- figuration of atoms because of the stability of either afull or halhfoll outmriost duorbital, so youmay need to account forthis by ‘sbufiiiug’ an elmlIOn hour the (n. - 1) s orbital. Finally use noble gas Kliortliand to got the armwar: [Xe-l 4}“ 5d“ 63* 69”. 0H8 10.0 points A comparison of the electron configurations: of nickel (Ni) and copper (Cu) Sndlcabefi that. i. Cu has one more of electron and one less 5 electron than Ni. 2. Cu luurmo more dduclxons and the same number of 5 electrom as Ni. 3. Cu has one more at electron and one more 5 electron than Ni. 4. Cu has two more d electrons and one has 5 electron than Ni. correct 5. Cu has one more a! elchan anti the same (rumba of 5‘ electromag- Ni. Explanation: The electron configuration for Ni is Mr] M3452, whereas the electron configuration for G!) to [fix] 3&1045‘, which indicates the dilfcrrmce between their clam-on diutrihufion. Cu has: it} delecth in the 3d orbital while burke (jpb2375) m Homework 2 -— Sutelifie - (53130) 3 NE has 8. fimilarly, Cu has 1, while NE has 2 dew-electrons. 059 10.0 poian We}: of tho follmving shows the correct quantum mmbers (w, dung) and electroncon— figmation for thehighest energy Plectrum-.1) in the given atoms? N= {WT-’93; '1 w 2; E 2; .m, a: 42,1.i.o.1,2 Rem l-‘u'lisam e“- 4; t 1m. m1.o+1 3.00: [Aroma :2 a 3; e x 2; m, m "2,-1,0, 1,2 correct: [Kr 59mm 5:3m0;mrno,1 \E‘o: [Arlfisgfidfi n t 4; a? 2 Cum z o Explanation: For Ca. the subsidiary quantum number 1 must be Zero became the higlawt endow electron is in an s-orbital. For N, the highea muggy electron is in n porbita}. Thu-eforct : 1 and m: a": A1,U,1. For Fe, the highest mam! electron is meadorbitalsonm3,£m2,im: m2, m1, 0, 1,2. For Rh, :1. and :8 are correct, but in! can only be ~12... -l- !. Therefore mg a: 0. Co is the only correct answer. 010 HM} points How many ungiaired floctrons are there inure electron configuration of a nitrogen atom? 1. one 2 . two 3. three correct 4. none 5. four Explanation: 0!! 10.0 gmint-s One of the following does NOT represmt the ground state electron configuration for an atom. Which one? 1.4513077 current 2. [Al-1451 36° 3. [Ari 4s? 3d” 41.3 4. [Ar] 4.? 3d“ 5. [Ml-1:1 Explanation: 012 10.0 points 152 2s2 2;;6 39" 3525 in the electron configurat- Lion for which one of the {all owing ions? 1. Cl" 2. 10+ 3. S" 4. Pa‘ (tori-art 5.1-?” Explanation: We needanionthathafi 2+ 2+G+2+ 5 electrons; to, P9“. 013 10.0 points According to the general Hands of the peri— odic table, which of the foliowing elemflilo has the largest atomic radius? 1. Kr 2. K 3. Ca 4. 09 correct Explanation: Following the: general pattern, which is that: atomic size-r are lamest in the bottom left- kandoemer and smallest in the top rigid-hem! www'of the pernith table. the largest atomic buriw (ipb2376) w Homework 2 ~ Sutoliife - (53120) 6 radius would belong to Cu in this group. 014 10.0 points Phoiaahorus has a higher ionization energy than silicon bommo I. the effective nuclear charge of phoophonul is leg: that: that of siliom. 2. the effective nuclear charge of :dlicon is 1035‘ than that of phosphorus. correct 3. the outer orbitals of phosphorus are lo- cated farther from the nucleus than the outer orbitals of silicon. 4. False; phosiflionm ha a lower ionization may than ailicon. 5. the outer orbitals of phoophoms are low noted cloeer to the nucleus than the outer orbitals ofsllioorr. Explanation: 155’ : {Ne} . 3s 3p “Si : {Na} 1; 3s 31) L Li: bot]: more, the valunco electrons. are lo- cated in 3:: and 3p orhitalo. llowwm, P has a greater melee: dial-go due to its greamr rum:- ber ofprotons, co more may is required to remove an electron iron-1P réativete Si. 015 10.0 poinfi Which elements are correctly timed in order of INCREASING ionization comfy? 1.0 (c l" < Noemi-(act; 2. 0 < S < Se 3. N < E’ < 114; 4. N ( 0 < F &C(N(0 Ex pianat ion: Ionization (:11ng corresponds to how much (energy it trim to ionizean element. Elements with high ionization energies do not want. to g,ch up their electrons, low ionimtionmergios mean they do nut want to receive an electron, but would rather gave one up. In general, ionization energies increase as you move right along a period and u}; a group. Some atoms (like B, 0, Al and S) can achieve more fitable electronic configuration in filled or half-filled sheik. (mi 50.8 points The alkali metals have a low first. and a high second ionization potential. The alkaline earth metals have low values for both first. and wound potentialo These observations rugged: that . aikali metals should form stable +! ions, while alkaline earth metals should form stable -—2 ion-la. 2. both alkali metals and alkaline earth mell- cls should form stable ions. with charge +1. 3. alkali metals should form stable wl iom, while alkAIino (with metals should form stable —2 ions. 4. both mu metalsand alkali earth metals should form very stable tom with charge m l. 5. ailmlimetals should form stable +3 ions, whiiealhaline earth metals should form stable +2Eons. con-m Explanation: Enorder toattain the electron configuration of the nearest noble gas, ails-Iii metals loot one electron to form moncvalenl. cations and slimline earth metals lose 2 olectrom to form diwlem cations. Any additional electron loss wili be from filled orbitals and will require much larger energiw. 01? 10.0 points Let X be a hypothetical element. barks! pr2376) w Homework 2 vSubcliflcw (53220) Wham of the fullowing would be largest? 1. X. ‘ 2. X" 3. X1” toned; 4. X 5. x“ Explanation: X1" would be largefl. as it has ‘2 more electron-s than protons, and the prellens here would be at th: greatest ciisaclvanlage when trying to draw the electrons Lowartis Elm nu- clans. 013 10.0 points Which of the following atoms has the highwl. cleaner: affinity? 1. A] 5‘. Si curl-act 3.1’u' ALP Explanation: Electron affinlt’y is a meme afhaw readily a neutral atom will accept an electron to form a —1 lon, The higher the EA, the mom read. fly this occurs. “maker” may mean “more positiva“ a: "mom negative)” depending on how this has been defined by your profw sax. ‘3‘}20 GENERAL Mani is that EA 1le max-ease from left to right and from bottom Le: Lop. However, noble gases do not. want, to gain electrons since they already have a full octet. Aim, since there is a alight increase in stability when a gel; of parkllals is half full. an element which can achleve fills by (palm an elecsmu will tend to have a large-r electron affinity (but not as large as that of a group Vii clement}, ...
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This note was uploaded on 03/24/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.

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Homework 2 - lam-he (jpb2376) — Homework 2 - Sutcllfl'e...

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