Lecture 8 - Lewis Structures

Wednesday, January 31, 2024

10:31 PM

"The beauty of chemistry is that I can design my own molecular world." - ﷟HYPERLINK "https://www.nobelprize.org/prizes/chemistry/1999/press-release/"Ahmed Zewail Class notes can now be found in an alternative location: https://bricejurban.github.io/CHEM101/ I will update this on Tues/Thurs evenings This is HTML based and compatible with all devices (no more issues!) Only downside is that hyperlinks do not work. Assignments due this week: ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28698/modules/items/2975632"HW 5 - Inorganic Nomenclature Part 1 (Aktiv Chemistry) (Fri 2/9) ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28698/modules/items/2975633"HW 6 - Lewis Structures, Polarity (Aktiv Chemistry) (Sun 2/11) Assignments due next week: ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28698/modules/items/2975637"HW 7 - Inorganic Nomenclature Part 2 (Aktiv Chemistry) (Wed 2/14) ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28698/modules/items/2975638"HW 8 - Resonance, VSEPR, IMFs (Fri 2/16) Office Hours: Friday 11-1 CIC ﷟HYPERLINK "https://calendly.com/bricejurban/office-hours"By appointment Midterm 1: The first midterm is graded and available for review using the Gradescope software which can be found on Canvas. Click through each question's rubric to see how you were graded. I recommend printing a copy then reviewing the answer key in the big display case outside SCNC 336 next to the chemistry computer lab. See map for location of answer key. After reviewing, if you notice any discrepancies in the free response you can submit a regrade request through Gradescope. Statistics: Mean 69% Median 70% Standard deviation 16% Top score 100% The test is broken up into two sections, here is how you can calculate the score I put on canvas Section 1 (Matching/Multiple Choice): Questions 1-15 (Matching) Multiply # of correct responses by 2 Questions 16-25 (Multiple Choice) Multiply # of correct responses by 3 Add these together to get Section 1 Score Section 2 (Free Response): Add 3 points to your score to get Section 2 Score. Midterm 1 Grade: Add Section 1 and Section 2 scores together to get your grade (the one on Canvas). Today's Schedule: Thursday (2/6) Polarity and Electronegativity Lewis Structures Octet Rule Formal Charge Not enough electrons Too many electrons Looking Ahead Thursday (2/8) Schedule Resonance Structures The Valence Shell Electron Pair Repulsion Theory of Molecular Geometry Intermolecular Forces Application to DNA AnswerKey.png SCALE: REV: / 04 Ill 2 3 CAMPUS LN 1 0 20 0 工 ○ C 二 丄 冖 凵 工 一 工 」 , qen Jalndtuoo s 一 go 山 00 」 00 ; ; S ! 山 0 0 0 O•h•ue 」 。 」 。 。 56 冖 冖 凵 POOuedPV
Section 1 is out of 60 points Section 2 is out of 40 points Total grade out of 100 points Your lowest midterm grade for the semester will be dropped from the gradebook. Midterms amount for 40% of your course grade. If you did not do as well as you would have hoped, you may need to adjust how you study for this class. Here are some ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28698/modules/items/2913696"suggestions. AnswerKey.png SCALE: REV: / 04 Ill 2 3 CAMPUS LN 1 0 20 0 工 ○ C 二 丄 冖 凵 工 一 工 」 , qen Jalndtuoo s 一 go 山 00 」 00 ; ; S ! 山 0 0 0 O•h•ue 」 。 」 。 。 56 冖 冖 凵 POOuedPV Bond Polarity The difference in strength two atoms have on shared valence electrons In an ionic bond, valence electrons are completely transferred to the anion. In a covalent bond, valence electrons are shared or mostly shared Ionic Bond = Unhappy baby Untitled picture.png Covalent Bond = Tug of war Untitled picture.png ﷟HYPERLINK "https://ptable.com/#Properties/Electronegativity"Electronegativity 𝝌 How chemists quantify and predict bond polarity "The tendency of an atom to attract a shared pair of electrons towards itself". Linus Pauling came up with electronegativity (𝜒) scale based on a geometric averaging of the ionization energy and electron affinities Scale of 0 to 4 What element has the highest tendency to attract electrons? Fluorine Which elements have no EN? Noble gases Which of these elements is the least EN? Francium Where does Hydrogen fall in terms of the other elements? Untitled picture.png 0.98 11 0.93 55 0.79 87 0.7 1.57 12 Mg 56 0.89 88 0.9 B 2.04 13 AI 81 2.04 c 2.55 14 1.90 3.04 15 P 2.19 3.44 16 s 2.58 F 3.98 CI 3.16 He 10 Ne 18 Ar 57-71 72 Hf 1.1-1.2 1.3 89+ 1.1-1.3 73 Ta 1.5 74 w 2.36 75 Re 1.9 76 os 2.2 77 ir 2.20 78 Pt 2.28 79 2.54 80 Hg 2.00 82 83 85 86 Pb Bi Po At Rn 2.33 2.02 2.0 2.2 Figure 7.13 Electroncgativities Of tine elements aş calculatecl Ink Drawings Ink Drawings Ink Drawings
Where does Hydrogen fall in terms of the other elements? Hydrogen is less than many of the nonmetals, but more than most of the metals. Hydrogen is similar to Carbon but less Untitled picture.png 0.98 11 0.93 55 0.79 87 0.7 1.57 12 Mg 56 0.89 88 0.9 B 2.04 13 AI 81 2.04 c 2.55 14 1.90 3.04 15 P 2.19 3.44 16 s 2.58 F 3.98 CI 3.16 He 10 Ne 18 Ar 57-71 72 Hf 1.1-1.2 1.3 89+ 1.1-1.3 73 Ta 1.5 74 w 2.36 75 Re 1.9 76 os 2.2 77 ir 2.20 78 Pt 2.28 79 2.54 80 Hg 2.00 82 83 85 86 Pb Bi Po At Rn 2.33 2.02 2.0 2.2 Figure 7.13 Electroncgativities Of tine elements aş calculatecl Electronegativity (EN) values as calculated by Linus Pauling Untitled picture.png > 3.98 O 44 Cl 3.16 > N 3.04 > S 2.58 > C 2.55 p 2.19 2.1 Most common EN Values Electronegativity differences Δ𝜒 determine bond polarity Type of Bond Example Bonded Elements Nonpolar Covalent Bond Also called Pure Covalent Cl-Cl Δ𝜒<0.4 Strong NM† & Strong NM or Weak NM & Weak NM Polar Covalent Bond H-Cl ﷐𝛿﷮+﷯ ﷐𝛿﷮−﷯ 0.4≤Δ𝜒≤﷐2.0﷮∗﷯ Weak NM & Strong NM Ionic Bond Na-Cl Δ𝜒>﷐2.0﷮∗﷯ Metal & NM Metallic Bond Na-Na Metal and Metal † NM = Non-metal * These are general guidelines. At the extremes, we tend to get a mix of ionic and covalent character (i.e. HF, AlI3) Bond polarities can be indicated with partial charges ﷐𝛿﷮+﷯ means slightly positive. ﷐𝛿﷮−﷯ means slightly negative Examples of calculating the polarity of a bond: C - N Na - O Mg - C Br - O (Br 𝜒 = 2.96) N - N S - O P - F Cr - Cl (Cr 𝜒 = 1.66) Fe - H (Fe 𝜒 = 1.83) Ca - O (Ca 𝜒 = 1.0) | 2.55 - 3.04 | = 0.49 Polar Covalent | 0.94 - 3.44 | = 2.50 Ionic | 1.31 - 2.55 | = 1.24 Polar Covalent | 2.96 - 3.44 | = 0.48 Polar Covalent | 3.04 - 3.04 | = 0.00 Pure Covalent | 2.58 - 3.44 | = 0.86 Polar Covalent | 2.19 - 3.98 | = 1.79 Polar Covalent | 1.66 - 3.16 | = 1.50 ? | 1.83 - 2.10 | = 0.27 Pure Covalent | 1.00 - 3.44 | = 2.44 Ionic Note that some metals and nonmetals have significant covalent character and these are just estimates of the bond character. Drawing Covalent Molecular Structures (Lewis Structures) Instead of the transfer of electrons to obtain a noble-gas electronic configuration, in covalent molecules there is a simultaneous sharing of electrons. For example the molecule Cl2: Untitled picture.png Untitled picture.png Untitled picture.png Untitled picture.png Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
Untitled picture.png Untitled picture.png Untitled picture.png Cl ([Ne]3s23p5) + Cl ([Ne]3s23p5) ⟶ Cl2 ([Ar][Ar]) Both atoms now have a noble gas outer electron configuration of eight electrons. G. N. Lewis generalized this result into what we call the octet rule: each element forms covalent bonds such that eight electrons occupy its outer shell. Unpaired electrons are known as lone pairs. Algorithm for Drawing Lewis Structures Arrange the symbols of the atoms that are bonded together in the molecule next to one another. Compute the total number of valence electrons in the molecule by adding the number of valence electrons for all the atoms in the molecule. If the species is an ion rather than a molecule, then you must take the charge of the ion into account by adding electrons if it is a negative ion or subtracting electrons if it is a positive ion Represent a two-electron covalent bond by placing a line between the atoms that are assumed to be bonded to each other Arrange the remaining valence electrons as lone pairs about each atom so that the octet rule is satisfied for each one Check the formal charge on each atom to help with competing structures If there is not enough electrons, you may need to use double bonds, triple bonds or rings or the central atom(s) has a deficient octet or you have a radical If there are too many electrons, you may need to put electron pairs on the central atom and/or form double bonds to reduce formal charges (this will not be assessed in CHEM 101) Draw any valid resonance structures if requested Untitled picture.png G. N. Lewis - Introduced Lewis structures and covalent bonds in the 1920s. Tips: The central atom is usually the least electronegative unless hydrogen which is always terminal. Some compounds will not have a central atom, but several. Be, B, and Al have deficient octets Elements in the 3rd row (S, P . . . ) can expand their octet to have 10, 12, or more e-. If the molecule is an ion, bracket the structure and put a charge in the top right corner Molecules with an odd number of electrons will form a free radical, a reactive species. Examples of Lewis Structures (Steps 1-4 only) OF2 NH2– PCl4+ Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
CHCl3 (chloroform) N2H4 (hydrazine) CH3OH (methanol) Assigning Formal Charges help with determining the best structure (Step 5) Sometimes more than one possible structure may be possible. In that case it is necessary to assign a formal charge to the atoms in the structure to help aid us in choosing the correct one. We assume each pair of shared electrons are shared equally and assign one of the electrons to each atom. Lone electron pairs are assigned to the atom they are located on. Use this equation: Untitled picture.png formal charge on an atom In a Lewis formula total number of = valence electrons in the free atom total number of lone-pair electrons total number of shared 2 electrons Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
Untitled picture.png formal charge on an atom In a Lewis formula total number of = valence electrons in the free atom total number of lone-pair electrons total number of shared 2 electrons NH4+ (ammonium) H3O+ (hydronium) OF2 (written as OFF) Which structure is preferable for hydroxylamine? NH3O NH2OH Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
Which structure is preferable for hydrogen peroxide? H2OO HOOH When there are not enough electrons (Step 6) Some elements are electron deficient are do not require an octet when forming a covalent bond. This is restricted to H, Li, Mg, Be, B, and Al. What is more likely is that you need to use double or triple bonds, especially if there is C or O present. If the molecule is larger, you can also form ring structures to use up 2 electrons. Lastly, if you have an odd number of electrons radical (unpaired electron) compounds are also possible. H2 Li2 BeCl2 MgH2 BH3 AlBr3
C2H6 (ethane) C2H4 (ethene) C2H2 (ethyne) C6H6 (benzene) C2H3O2– (acetate) CH3• (methyl radical) (C6H5)3C• Trityl radical
(methyl radical) O2•– (superoxide anion radical) Trityl radical First organic radical discovered. Moses Gomberg in 1900 at University of Michigan (my alma mater) When there are too many electrons (Step 7) Some elements allow for an expanded octet when forming covalent bonds. This is restricted to elements in the third shell or higher including the elements: P, S, Cl, As, Se, Te, Br, I, and even Xe. Never will an element such as C, N, O, or F have an expanded octet I will not test on this (unless it is an extra credit problem)  SF6               XeF2 IF5
   

 

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