Energy Unit Learning Goal 2: Examine the Placement of Electrons in Orbitals.

Copyright © by McDougal Littell. All rights reserved. 2 Bohr’s Atomic Model 1. Atom has small positive nucleus. 2. Electrons orbit like planets orbit the sun. 3. Electrons orbit in certain allowed energy levels. 4. Electrons can jump to different orbits but only by absorbing or emitting a photon of light with the correct energy content. Good start but some basic problems: Electrons do not “orbit” in circular paths. Could not explain why negative electrons didn’t get attracted into the nucleus This model only worked for Hydrogen

Copyright © by McDougal Littell. All rights reserved. 3 DeBroglie & the Wave Mechanical Model of the Atom DeBroglie & Schrodinger suggested that electrons exhibited both wave and particle characteristics. This is referred to the wave/particle duality Schrodinger came up with a model (wave mechanical model) that worked with atoms in addition to hydrogen.

Wave Mechanical Model Electron states are described as orbitals. Electrons are more like fireflies than planets. – An orbital is described as the probability map of an electron’s motion.

Copyright © by McDougal Littell. All rights reserved. 5 Orbitals vs. Orbits Orbitals are nothing like orbits. To picture orbitals, imagine a single male firefly in a room. In the center of the room is a vial of nectar. The room is dark with a camera with an open shutter in the corner. The developed picture will look something the diagram to the left.

Copyright © by McDougal Littell. All rights reserved. 6 Suppose while watching the room, you see a flash here. X Where will the firefly flash next? There’s no way of really knowing, but the likely- hood is that it will flash where the “film” had the densest concentration of flashes. Shrodinger’s model cannot predict the path of the electron, but can predict the probability of find the electron in a certain region.

Copyright © by McDougal Littell. All rights reserved. 7 The first quantum number: n Describes the energy level n = 1, 2, 3, 4, etc. Notice how the rows are numbered on the periodic table!

Copyright © by McDougal Littell. All rights reserved. 8

9 Sublevels The principle energy levels are divided into sublevels. The second quantum number describes the shape of these sublevels. l = 0, 1, 2, 3, 4, 5, 6, 7 However, we rarely refer to these sublevels by number. We usually use a letter: spdf

Copyright © by McDougal Littell. All rights reserved. 10 Notice the energy level determines the # of sublevels! n l

Principle Energy Levels Discrete Energy levels that are labeled with integers. – 1, 2, 3, 4, 5, 6, 7. Sublevels – Each Principle Energy Level is subdivided into sublevels and labeled with a letter. s (holds 2 e - ) p (holds 6 e - ) d (holds 10 e - ) f (holds 14 e - ) The letters tell the shape of the orbital.

Copyright © by McDougal Littell. All rights reserved. 12 Figure 11.20: The hydrogen 1s orbital.

Copyright © by McDougal Littell. All rights reserved. 13 Shapes of Sublevels Every energy level has an s sublevel. The only difference being the diameter! s sublevels are “spherical” in shape.

Copyright © by McDougal Littell. All rights reserved. 14 The p sublevels are dumbell shaped and are made of 3 orbitals or “lobes”. Each orbital can hold 2 electrons. p sublevels are found on energy levels 2 or greater

Copyright © by McDougal Littell. All rights reserved. 15 The d sublevels are four- petaled and are made of 5 orbitals or “lobes”. Each orbital can hold 2 electrons. d sublevels are found on energy levels 3 or greater

Copyright © by McDougal Littell. All rights reserved. 16 The f sublevels are made of 7 orbitals or “lobes”. Each orbital can hold 2 electrons. f sublevels are found on energy levels 4 or greater

Copyright © by McDougal Littell. All rights reserved. 17 Summary....so far sublevel# orbitals# electrons s p d f

Copyright © by McDougal Littell. All rights reserved. 18 Notice the # of columns in each group!

Copyright © by McDougal Littell. All rights reserved. 19 Pauli Exclusion Principle: An atomic orbital can hold a maximum of two electrons, and those two electrons must have opposite spin. Now let’s put it all together.....

Copyright © by McDougal Littell. All rights reserved.22 H H 1s 1s 1 He 1s He has 2 electrons 1s

Copyright © by McDougal Littell. All rights reserved.23 H H 1s 1 He 1s has 3 electrons 1s 2 Li 1s2s 1s 2 2s

Copyright © by McDougal Littell. All rights reserved.24 H He has 4 electrons Li 1s2s 1s 2 2s 1 Be 2s1s 1s 2 2s 2 B B 2s1s 2p 1s 2 2s 2 2p

Copyright © by McDougal Littell. All rights reserved.25 H He LiBeB C 2s1s2p 1s 2 2s 2 2p 2 Li 1s2s 1s 2 2s 1 Be 1s2s 1s 2 2s 2 B 2p 1s 2 2s 2 2p 1 1s2s C Stop! Before adding the next electron, we have to know about Hund’s Rule: Hund’s Rule: we put 1 electron in each orbital before we pair them up!

Copyright © by McDougal Littell. All rights reserved.26 H He LiBeB C 1s 2 2s 2 2p 3 B 2p 1s 2 2s 2 2p 1 1s2s C N N 1s2p 1s 2 2s 2 2p 2 2p1s2s

Copyright © by McDougal Littell. All rights reserved.27 H He LiBeB C 1s 2 2s 2 2p 4 B 2p 1s 2 2s 2 2p 1 1s2s CN 1s 2 2s 2 2p 2 O 2s1s2p 1s2s N 2p1s2s 1s 2 2s 2 2p 3 O

Copyright © by McDougal Littell. All rights reserved.28 H He LiBeB C 1s 2 2s 2 2p 5 CN 1s 2 2s 2 2p 2 2p1s2s N 2p1s2s 1s 2 2s 2 2p 3 F 2s1s2p O F O 1s2s 1s 2 2s 2 2p

Copyright © by McDougal Littell. All rights reserved.29 H He LiBeB 1s 2 2s 2 2p 6 CN N 2p1s2s 1s 2 2s 2 2p 3 OF O 2p1s2s 1s 2 2s 2 2p 4 Ne 2s1s2p F 1s2s 1s 2 2s 2 2p 5 Ne This is a very important arrangement! With 8 electrons in the valence shell, we have a stable octet. Notice neon is a noble gas, very inert, and is at the end of its row!

Copyright © by McDougal Littell. All rights reserved. 30 H He O 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 4 2p1s2s F 2p1s2s 1s 2 2s 2 2p 5 Ne 2p1s2s 1s 2 2s 2 2p 6 Li Be B C N O F Ne Na 2s1s2p 3s [Ne] 3s

Copyright © by McDougal Littell. All rights reserved. 31 H He Na [Ne] 3s [Ne] 3s 1 Li Be B C N O F Ne Na Mg [Ne] 3s [Ne] 3s 2 Ne 2p1s2s 1s 2 2s 2 2p 6 Mg

Copyright © by McDougal Littell. All rights reserved. 32 H He Na [Ne] 3s [Ne] 3s 1 Li Be B C N O F Ne Na Mg [Ne] 3s 2 3p 1 Ne 2p1s2s 1s 2 2s 2 2p 6 Al Mg [Ne] 3s [Ne] 3s 2 Al [Ne] 3s3p

Copyright © by McDougal Littell. All rights reserved. 33 H He Li Be B C N O F Ne Na Mg Al [Ne] 3s 2 3p 2 3p Si Al [Ne] 3s [Ne] 3s 2 3p 1 Si [Ne] 3s3p

Copyright © by McDougal Littell. All rights reserved. 34 H He Li Be B C N O F Ne Na Mg Al Si [Ne] 3s 2 3p 3 3p P Al [Ne] 3s [Ne] 3s 2 3p 1 3p Si [Ne] 3s [Ne] 3s 2 3p 2 P [Ne] 3s3p

Copyright © by McDougal Littell. All rights reserved. 35 H He Li Be B C N O F Ne Na Mg Al Si P 3p S Al [Ne] 3s [Ne] 3s 2 3p 1 3p Si [Ne] 3s [Ne] 3s 2 3p 2 S [Ne] 3s3p P [Ne] 3s [Ne] 3s 2 3p 3 [Ne] 3s 2 3p

Copyright © by McDougal Littell. All rights reserved. 36 H He Li Be B C N O F Ne Na Mg Al Si P S Cl 3p P [Ne] 3s [Ne] 3s 2 3p 3 Cl [Ne] 3s3p [Ne] 3s 2 3p 5 3p S [Ne] 3s [Ne] 3s 2 3p

Copyright © by McDougal Littell. All rights reserved. 37 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 3p P [Ne] 3s [Ne] 3s 2 3p 3 [Ne] 3s 2 3p 6 3p S [Ne] 3s [Ne] 3s 2 3p 4 Ar [Ne] 3s3p Cl [Ne] 3s [Ne] 3s 2 3p 5 Another Noble gas with a stable octet!

Copyright © by McDougal Littell. All rights reserved. 38 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K [Ar] 4s 1 K [Ar] 4s 3p Ar [Ne] 3s [Ne] 3s 2 3p 6 Notice that we have started filling the 4 th energy level before even starting the 3d!

Copyright © by McDougal Littell. All rights reserved. 39 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar Ca [Ar] 4s 2 Ca [Ar] 4s K [Ar] 4s [Ar] 4s 1 K

Copyright © by McDougal Littell. All rights reserved. 40 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar Sc [Ar] 4s 2 3d 1 K [Ar] 4s [Ar] 4s 1 K Ca Ca [Ar] 4s [Ar] 4s 2 3d Sc [Ar] 4s

Copyright © by McDougal Littell. All rights reserved. 41 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar Ti [Ar] 4s 2 3d 2 K [Ar] 4s [Ar] 4s 1 K Ca Sc Ca [Ar] 4s [Ar] 4s 2 Sc [Ar] 4s [Ar] 4s 2 3d 1 3d Ti [Ar] 4s

Copyright © by McDougal Littell. All rights reserved. 42 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar V [Ar] 4s 2 3d 3 K Ca Sc Ti Ti [Ar] 4s [Ar] 4s 2 3d 2 3d Sc [Ar][Ar] 4s 2 3d 1 3d 4s 3d V [Ar] 4s

Copyright © by McDougal Littell. All rights reserved. 43 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar Cr [Ar] 4s 2 3d 4 K Ca Sc Ti V Ti [Ar] 4s [Ar] 4s 2 3d 2 3d Sc [Ar][Ar] 4s 2 3d 1 3d 4s V [Ar][Ar] 4s 2 3d 3 3d 4s 3d Cr [Ar] 4s Stop! Chromium is a stealer! A more stable arrangement is formed when all orbitals are half- filled than one full & one empty [Ar] 4s 1 3d

Copyright © by McDougal Littell. All rights reserved. 44 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar Mn [Ar] 4s 2 3d 5 K Ca Sc Ti V Cr Ti [Ar] 4s [Ar] 4s 2 3d 2 3d V [Ar][Ar] 4s 2 3d 3 3d 4s 3d Mn [Ar] 4s Cr [Ar] 4s [Ar] 4s 1 3d 5 3d

Copyright © by McDougal Littell. All rights reserved. 45 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar Cu [Ar] 4s 2 3d 9 K Ca Sc Ti V Cr Mn Fe Co Ni 3d Cu [Ar] 4s Copper (and all in this column) steal an electron from the 4s orbital to fill its 3d! [Ar] 4s 1 3d

Copyright © by McDougal Littell. All rights reserved. 46 Zn H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Cu [Ar] 4s [Ar] 4s 1 3d 10 3d [Ar] 4s 2 3d 10 3d Zn [Ar] 4s

Copyright © by McDougal Littell. All rights reserved. 47 Ga H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Cu [Ar] 4s [Ar] 4s 1 3d 10 3d Zn[Ar] 4s [Ar] 4s 2 3d 10 3d [Ar] 4s 2 3d 10 4p 1 3d Ga [Ar] 4s 4p