151F09Disc 10 Lewis structures D2L

151F09Disc 10 Lewis structures D2L - Meeting 10 Welcome to...

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Unformatted text preview: Meeting 10 Welcome to your Chem 151 Discussion. Where chemistry comes to you! 1 Electron Distribution in Molecules PRACTICE,PRACTICE, PRACTICE Electron distribution is depicted with Electrons are distributed as shared or BOND PAIRS and unshared or LONE Lewis electron dot structures PAIRS. G. N. Lewis 1875 - 1946 Lewis Structure: HNO3 Step 1. Central atom Step 2. Count valence electrons H O O N N = 3 x O = O H = charge = ? 8 pairs of electrons are now left. TOTAL = Step 3. Form single bonds Lewis Structure: HNO3 Remaining pairs become lone pairs, first on outside atoms and then on central atom .. .. H O O: .. N .. :O : .. Lewis Structure: HNO3 Adjust by making double bonds to form octets. N cannot Exceed 8 so There are formal Charges of +1 on N And -1 on O .. .. H O O: .. N .. :O : REVIEW:Formal Charge Consider the Ozone molecule (O3): O-O=O Formal Charge Compare the number of valence electrons the atom would have by itself to the number of electrons it actually has (Assuming the bonding electrons are equally divided) 6-7=-1 Valence electrons from group number on periodic table 6-6=0 O-O=O 6-5=+1 Total formal charge -1 + 1 + 0 = 0 Lewis Structure: HNO3 Resonance forms .. .. H O O .. N .. :O : .. .. .. H O O: .. N .. :O : Lewis Structure: C3H8O Step 1. Central atom Step 2. Count valence electrons O = 3 x C = 8 x H = charge = ? TOTAL = Step 3. Form single bonds 26 e 13 pairs Lewis Structure: C3H8O C C C O Two Possible `chains Fill in with H's O C C C H H O H 11 pairs 2 pairs fill on most Negative. H C C C H H H H H H :O : H H C C C H H H H All have zero Formal charge H H H H C C C O H H H H 13-11 bonding Pairs 2 pairs to Make octets H H H H C C C O H H H H All have Zero formal charge : : H H H H C C C O H H H H 1-propanol STRUCTURAL ISOMERS Same formula C3H8O Different arrangements : : H H :O : H H C C C H H H H 2-propanol or iso-propanol or "rubbing" alcohol Is that all the possibilities? H H : : H H methyl ethyl ether H C C O C H H H All zero formal charge General bonding pattern to achieve both zero formal charge and an octet Element H O N C Bonds (pairs) 1 2 3 4 Lone pairs 0 2 1 0 Lewis Structure: C2HCl Carbon bonds to carbon Step 1. Central atom Step 2. Count valence electrons Cl = H C C Cl 2 x C = H = charge = ? 5 pairs of electrons are now left. TOTAL = Step 3. Form single bonds Lewis Structure: C2HCl Remaining pairs become lone pairs, first on outside atoms and then on central atom : : : : H C C Cl : Lewis Structure: C2HCl Adjust by making double bonds to form octets. : : : All have zero formal charge : H C C Cl : : H C C Cl : : Lewis Structure: SO4 2 Step 1. Central atom Step 2. Count valence electrons O O S O S = O 4 x O = charge = ? 12 pairs of electrons are now left. TOTAL = Step 3. Form single bonds Lewis Structure: SO4 : O: : 2 Remaining pairs become lone pairs, first on outside atoms and then on central atom :O S O : : : O: : : : : Lewis Structure: HNO3 Adjust by making double bonds and octets. 2- : :O : : : O: S can Exceed 8 so There are formal Charges of 0 on S And -1 on 2 O's O S O : : There are many resonance forms of SO4 2- S S S S S S In a resonance form the atoms stay in place but the electrons move to give an equivalent structure. None of these are correct the real Molecule is "hybrid" of all of them. In this form of SO4 2- the S formal charge is -2 and all the O's are zero formal charge. This is a possible resonance structure, but not the best since S Is -2 and S is more positive that the four oxygens. Negative formal charge Should be on the negative atoms. S Lewis Structure: XeF2 Step 1. Central atom Step 2. Count valence electrons Xe = 2 x F = 9 pairs of electrons charge = ? are now left. TOTAL = Step 3. Form single bonds F Xe F Lewis Structure: XeF2 Remaining pairs become lone pairs, first on outside atoms and then on central atom .. :F .. Xe .. F: .. Three pairs left. CNO or F can NEVER Have more than 8 electrons because They are 2s2p and no more Lewis Structure: XeF2 Adjust by making double bonds to form octets. The Xe gets All the remaining Electrons And has 10 BUT it has d orbitals to hold them. F.C. (Xe) = 8-6-2 = 0 All atoms have zero formal Charge. .. .. .. : F .Xe F: .. . .. .. Here is the single bond skeleton for phenylalanine. There are 18 electrons yet to be placed in the structure. Locate them and complete the structure. Check with formal charge. H H C H H C C C H C C C H H H C N C H O H H O Fill the octet Of the most Electronegative First. H H C H C C Make a double bond C=O 6 electron pairs H H :O: C C C C O H : : C H :N H C H H H : 3 electron pairs H H C H C C H H :O: C C C C O H : : C H :N H C H H H All atoms have zero formal charge PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PERFECT End See D2L for Lewis Structure Practice file which has over 100 structures to draw. ...
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