( ) ( s ) Answers to Practice Test Questions 4 Electrons, Orbitals and Quantum Numbers. Student Number:
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1 Anwer to Practice Tet Quetion 4 Electron, Orbital Quantu Nuber. Heienberg uncertaint principle tate tat te preciion of our knowledge about a particle poition it oentu are inverel related. If we ave ore inforation about it poition, we ut ave le inforation about it oentu ( vice vera). A a reult, te odel of te ato in wic an electron orbit a nucleu in circular faion cannot be correct a we could know te electron precie oentu poition at te ae tie (wic would violate Heienberg uncertaint principle).. A ubtance tat i diaagnetic a no unpaired electron. A ubtance tat i paraagnetic contain one or ore unpaired electron. A ubtance can be identified a eiter paraagnetic or diaagnetic b expoing it to a agnetic field. Paraagnetic ubtance will be attracted to te agnetic field wile diaagnetic ubtance will be ligtl repelled b te agnetic field. 3. Te Pauli excluion principle tate tat no two electron in a ceical pecie can ave te ae quantu tate. In oter word, no two electron in an ato or olecule can ave te exact ae et of quantu nuber. Becaue of te Pauli excluion principle, tere can onl be two electron per atoic orbital. Ti liit te nuber of electron allowed per ubell. e.g. are allowed, but 3 i not. 4. particle onl ligt onl (c) bot (d) particle onl 5. Step : Convert peed of bullet into SI unit k 000 in v = k 60in 60 = Step : Convert a of bullet into SI unit kg = = 000g 3.5g.5 0 kg Step 3: Calculate oentu of bullet 3 ( ) ( ) kg p = v = kg =
2 Step 4: Calculate wavelengt of bullet 34 J Hz kg 0.83 Hz λ = = = p Step 5: Ceck our work Doe our anwer ee reaonable? Are ig. fig. correct? Te wavelengt i uc, uc aller tan te accurac of an conceivable eaureent of te otion of te bullet, o quantu ecanical effect are not iportant for bullet Step : Calculate te energ of one poton E = ν c = λν terefore 34 J 8 9 c ( Hz )( ) 0 n Hz 9 E = = = J λ ( 55n) Step : Calculate te energ of one ole of poton 9 3 poton 5 E = E N = =.8 0 = 8 ( )( ) J J kj olar poton A poton ol ol ol Step 3: Ceck our work Doe our anwer ee reaonable? Are ig. fig. correct? p = p = v terefore λ J ( ) ( )( ) v = terefore λ 34 9 kg Hz n Hz u v = = = λ 55n u kg J 7. A 6f orbital correpond to te quantu nuber n = 6, l = 3. Te poible value of l for an electron in a 6f orbital range fro -3 to +3. Te poible value of for an electron in a 6f orbital range fro -½ to +½.
3 8. Quantu Nuber of Two Electron n = 3, l =, l = 0, = +½ n = 3, l =, l = 0, = -½ Could ti be a pair of electron in te ae orbital? Circle e or no. Briefl, jutif our anwer. n, l l are te ae, o te electron are in te ae orbital. Te value are oppoite, indicating oppoite pin. n = 3, l =, l =, = +½ n = 3, l =, l =, = -½ Te value for l are different, o te two electron decribed are in different orbital. n = 3, l =, l =, = +½ n = 3, l =, l =, = +½ Bot electron ave all four quantu nuber te ae. Te cannot terefore belong to te ae ato. n = 3, l =, l =, = +½ n = 3, l =, l =, = -½ Te value for l are different, o te two electron decribed are in different 3d orbital. 9. See our note/text for picture of te different tpe of orbital. 4dx, 4dxz, 4dz, 4dx-, or 4dz (c) 3px, 3p, or 3pz 6 (d) px, p, or pz 0. A coplete ( valid) et of four quantu nuber can onl decribe one electron in an ato. none l ut be aller tan n (c) 6 Ti et of quantu nuber decribe all te electron in a et of 5p orbital.
4 . (i) (ii) (iii) x x x nodal plane 0 nodal plane nodal plane + 0 radial node + radial node + 0 radial node node total node total node total Terefore n = + = 3 n = + = 3 n = + = label 3d x 3 p n 3 3 l 0. It would alo ave been acceptable to relabel te axe to avoid drawing in tree dienion (wic ti unartitic peron probabl ould ave done for te dxz orbital; ee coputergenerated iage on next page).
5 3. (c) 4. Or draw at an angle to ow dougnut p x p p z 5. (c) draw a p orbital draw a p orbital along a different axi tan our anwer to part draw a d orbital
6 6. 5 n = 5 l = l = -, -, 0, + or + = +½ or -½ (c) Te energ of te poton releaed wen H relaxe fro te n = 5 to te n = tate (ground tate) i equal to te energ difference between an electron wit n = 5 an electron wit n =. H a one proton o Z =. Step : Calculate te energ for H wit n = Z ( ) 8 En= = J n () Step : Calculate te energ for H wit n = 5 Z ( ) 0 En= 3 = RH = J n (5) 5 Step 3: Calculate te difference in energ between te n = n = 5 tate (ΔE = Efinal Einitial) E = En= En= 5 = ( J ) ( J ) = J Step 4: Calculate te energ of te poton releaed Since te ign of ΔE i negative, energ i releaed. Te energ of te poton releaed i J. Step 5: Calculate te frequenc of te poton releaed EE = νν νν = EE = JJ = HHHH JJ HHHH Step 6: Calculate te wavelengt of te poton releaed cc = νννν λλ = cc = HHHH νν HHHH = λλ = nnnn = nnnn Step 7: Ceck our work Doe our anwer ee reaonable? Are ig. fig. correct? Te energ of te poton a a iilar order of agnitude to RH te wavelengt i in te UV range (expect UV, viible or IR).
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