Question

affirm that I did not have knowledge of this exam prior to taking it! Tunderstand that only the work I submit directly to the professor at the end of the exam will be graded it engage in prohibited activity such as cheating, talking, leaving the examination room or operating a phone during the exam, I accept that my grade for this exam will be entered as zero without question of molest hereby affirm that I have read all of the directions for this exam and understand what is expected of Student Signature Date Answer all the questions. Show all calculations in the space provided for full credit No credit will be given for answers only. Select the single correct response for questions 1-30 (2 points each 1. Which figure below illustrates ans atomic orbital? (1) 1 (2) 2 (3) 3 (4) 4 - Which, if any, is the ground-state electron configuration for Fe?? (2) [Ar]3d or 1s 2s 2p 3s 3p 3d (3) (Ar]30° or 1s 2s 2p 3s 3p 3dº (1) Ar]3d or 12 2p 3s 3p3 (4) [Ar]3d or 18 2203334 3. Which one of the following sets of quantum numbers is not possible? 3 1. 4 2. 3 3. 3 4. 4 2 0 1 -2 -3 0 1 +% % + 2 (1)4 (2)3 (3) 2 (4) 1 4. Which one of the following sets of quantum numbers represents an electron with the highest enerov!! (1) n=2=1, m = 0, m, = % (2) n = 3,7 = 2, m/=0,me=% (3) n 4.10 m m (4) n = 2. ( = 1, m, = 1, m, % 5. What is the maximum number of electrons in an atom that can have the following quantum numbers and t = 0? (1) 10 (2) 6 3 (4)2

Answer 1

0 Answer : (2) - "s" orbital is spherical in shape 2 Atomic Number of Fe = 26 Ground - State election configuration of te 1. = 15²2s²2p6 38² 388 43² 3d Fet means we should remove 2 elections from outermost shell that is 45 - shell. feth 15 25² 2p% 35² 3po 3d. [Ar] 3016 TNO OPTIONI. h Answer: OPTION (3) M=3 62 me.-3 mg = -42. y for any "n" value oto (nol) I values will 1. be present. for na 3 I values are l= 0, 1, 2) For given 'ľ value there will be l to tel "me values present s å for d=2 ml= -2,-1, 0, 1, 2 But it given imea -3/ which 1007 POSS(BLE/ Reason:... ve oto (n-1) I' I

Energy of orbital can be deteqmined by (ntl) value- → More (n+1) value meone is the energy option n=2 lie (ned) = 2+1=3 option ② na 3 l: 2 cntl) = 3+2 = 5 option ③ n. 4 loo (nal) = 4to = 4 Option @ no a lot (ned) = ate = 3. Option ③ having high (intl) values .8t cwill have highest energy. را م 5 Anscver: Toption 4 n=3 l=0 means the orbital is 35 . The capacity of s' orbital cie 2 elections.