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Chemistry 101/103 Due date: Friday, November 14.2019 Problem Set 2: Atomic Structure and Chemical Bonding 1. (15 marks) When the excited electron in a hydrogen atom falls from an unknown level to n-5, a photon of 3740 nm wavelength is emitted. What is the initial position for this electron? 2. (15 marks) Answer and explain: a) How many electrons can have: i.n=3,1 = 2? ii.n=3,1-2, m/=0? iii. n=3,1=2, ms - +1/2? iv. n-3,1-2, m/ = 0 and ms - +1/2? b) Indicate the orbital (i.e. 3s, 4p, ...) to which each of the following sets of quantum numbers applies: i. n-3,1-1, m/ 0 ii. n-2,7-1, m = -1 iii.n=4,7 = 2, m/ - 1 iv. n= 5,1 = 3. m/= -2 c) Give the complete ground state electron configuration for the atom of radium (Ra, Z-88). 3. (20 marks) Arrange the following elements in the order of: a) Increasing radius: CI, K, Ar, Cat, b) Increasing radius: As, Br, P. Al; c) Increasing radius: N, S, F, Br; d) Increasing electron affinity: Si, P, CI, C; e) Increasing ionization energy: Al, Rb, Mg, Ca, Mg. 4. (20 marks) The following three Lewis structures can be drawn for N20: :N=N-O: IN-N=0: N=N=Ö: a) Show the correct formal charges on each atom. Considering these values, which resonance structure is likely to be the most important? Explain. b) The N-N bond length in N:O is 1.12 A, slightly longer than a typical N=N bond; and the N-O bond length is 1.19 A. slightly shorter than a typical N O bond. Bases on these data, which resonance structure best represents N30? Explain. 5. (30 marks) Using the Valence Shell Electron Pair Repulsion theory, predict the geometric arrangement of each of the following molecules around the central atom (the underlined atom). In each case give a clear three-dimensional picture to show the geometric arrangement of substituents and lone pair electrons about the central atom and also write: a) the Lewis structure: b) the AXE classification; c) the electronic structure; d) the molecular shape: e) the angle between the bonds; f) the polarity (polar or non-polar). KrCl2 ICI OF2 SEO CIO PO"

Answer 1

Wavelength of emitted Photon = 3740 nm 1 or 9 = 3740410 m. Oky 8 = 3740x107cm falls Let us suppose the excited elethon a from N = n level to n-5 level. We know the Rydberg's equation For Hatou is - where = ware J =R (12 number= - - - Oy, 1 = 109671640-4) R=Rydberg_caround (52 75%) n = 5 andre 00, 5 - 109677 ( ) =n 07. 3440x137 = 4387.08 - 109677 109677 - 4387.08 - (2.673x10 ) 72 22 07. 109677 = 1713.283 22 07: 109677 - 64 1713.20 or n = √64 = 8 Oo, n = 8 Therefore the excites eledion of H-atom n = 8 level to n = 5 level. falls from

27447 0 n=3, l=27 d=2 indicates d subshell. d subshell consist of five orbitals namely day, dyz, daz, dup, dz2 In each op bital no onaximum electron Can Stay is two eledrond. no. of So, in d Subshell dia remun electron have Ten electrons." So, n-3, 1-2 can have Ten electrons. (279) ③ n-3, 2-2, mero? 2.2 means d subshell and magnetic quantunno for ad subshell ane - 2, 1, 0, 1, 2 So, me to indicates only one orbital. In each orbital maximum two dectron can occupied. So, n-3 l=2, me to can have two electrons.

n-3, 1-2 mg = + - / l-2 indicates 2d subshell, 3d Subshell) Ih each op bital there is ma numum toro elections. Among them one neethon's Spin quantum no is + 1 and another is -12. In 3d. Subshell, there is maximum ten declans and among them five elections have ugott and are there five eled hons have ng =- I can have So, n=3, 1-2, maximum five ung = + 1 electrons. Covy 2 . n=3, 1-2, meco and uns=t1 ) l-2 indientes 3d Subshell and me co indicates one rpbital in 3d Sub Shell, m2 -1 0 1 2 So, in one orbital of 3d Subshell there is maximum two electrons having spin quantum no + I and another is - - So, n-3, 1-2 hero, bisa + 1 can hand

one electron. 2 676 n-3, l=1, me-o cn - Principal quantum no n - lal indicates & P gubahan ) l - Azimuthal quantum P Subshell has three op bitals. I I mes magnetic guantum 0 no mse spin quantum no having magnetic quantum no. 1, oti a ) 3px 3py 3 Pz meru 0 H The opbital is 3 Py or 32 © 252, lal, me = -1 N-2 0-1 indicates 2P subshell. 2 2py 2 P2 Mearl 0 +1 + X The opbital is 21 800, 2 Py 1 ) n = 4, 1-2, me =+1 N-4, 1-2 indicates 4 d subshell. 4d Subshell consist of five orbitals having magnetic quantum non - 2, 1, 0, 11, 12 [d27 dus dyz day d2 3 4d The orbital 4 day, azad 2 - 0 +1 +2.

2 n = 5, 1-3, me = -2 indicates 5 f Subshell. consist of seven opbitals, 5 f Subshell me = -3-27 0 1 2 3 So, the orbital is 5f (2) (c) Ground state electronic of Radiussum (Ra). configuration Raz 18 25 206 35 3p6 45 3 do 4 pb 5 5 4 do 5 po 65 44 45 do 6p6 752 tut ки ба 40e relation 2 2 - ☺ @ e, kt, ar, eå + Por, cl, 3 e = 176 Sauere, zis ato mie no e is no of electrons p=radius of ion for, Ari 7 = 18 AS (3) value increases the size of ious Fori cat, 3 = 20 decreases order of increasing radius will be c2+ {kt L Ar Lee- 2+

26 AS, Br, P, Al we know that size of atom of element up to down in a group of Periodic sy left to right in a period atomie from up to down in increased table and radi a grou decreases. GPS TA TETET7 3rd Al si es el 4th Ga Ge As Se Br I Period So, order of increasing radius will be Br <PLAS LAI N,S,F, Br radius will be So, order of increasing FLN LSL Br. Si,p, cl, C Electron affinity increases along period from left to right! and decreases along group from top to bottom Son of Periodic table. So, increasing electron affinity order will ( PLCL si Lel © All Rbing @ Al, Rb, ng, ca The I. P gradually left to right and decreases down the group of peo increases along aroup of Periodic table. period from

Increasing Io nezation energy or potential (I.P) order will be ionization Rb LA e Lea <ng.