Chap 8B - Chem 210/ WANG / Chapter 8B Lewis Structures Atom...

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Unformatted text preview: Chem 210/ WANG / Chapter 8B Lewis Structures Atom placement Sum of valence eLewis 5e- F 7e- X 3 = 21e- Total Hint: for main group elements group # = valence e- # 26e- • Ions consist of two or more atoms bonded together. • Within these ions, groups of atoms stay together by covalent bonding. covalent ammonium NO3- nitrate OH- hydroxide Fall 2009 / UM-SJTU JI : F: :Cl C For CH4O Atom placement Both O and C are “center” atoms . H Sum of valence e- : Cl : : F: H Follow the same rules as molecules. Must add or subtract electrons based on the ionic charge. charge Cation Subtract electrons from total. number subtracted = charge Anion Add electrons to the total. number added = charge H Lewis structure with Single bonds only 6e4 e- 4x 1e- = 4eTotal 32e- Lewis structures of polyatomic ions O O C 4H F & Cl 7e- X 4 = 28e- C H Remaining valence e14 - 10 = 4 e- C = 4e- Total Polyatomic ions NH4+ Lewis structure with single bonds only. Molecular formula Lewis structure with single bonds only. 14e- Molecular formula SO42- Atom placement O OS Remaining valence e32 - 8 = 24 e- O O Sum of valence e- -2 : N Remaining valence e32 - 8 = 24 e- Not enough to give 8e- for all make double / triple bond. Add e- pairs to each atom (except H) 8e-. 8e Lewis structure of a polyatomic ion polyatomic with single bonds only. :O : : Remaining valence e- F Sum of valence e- Step 4 :O S O: : Lewis structure with single bonds only. Add e- pairs to each atom to give 8e-. F : : : F: Cl Cl C : Sum of valence e- Atom placement : Connect atoms with single bonds. bonds (use 2e- per bond) 26 e – 6e in bonding = 20e left. 20e CCl2F2 : : : F: N Molecular formula Connect atoms with single bonds. bonds (use 2e- per bond). Remaining valence e- Lewis Structure with = or ≡ bonds. : : F: Place atom with lowest EN in center : Atom placement : For NF3 Step 3 : •Explain bonding in polyatomic ions. bonding http://www.chempractice.com/drills/java_Lewis.php http://www.chempractice.com/drills/java_Lewis.php Molecular formula Too many eOne One gets more than 8e- structure structure with expanded octet •Help determine molecular shape molecular shape. http://www.up.ac.za/academic/chem/mol_geom/mol_geometry.htm Hint: for A group elements group # = valence e- # Step 2 : 8.7 Exceptions to the Octet Rule Lewis structures show how electrons how are distributed among all the atoms in a molecule. Place atom with lowest EN in center Step 1 : O: : 8.6 Resonance Structures Molecular formula : 8.5 Drawing Lewis Structures The steps in converting a molecular formula into a Lewis Lewis structure. Lewis structures of Covalent Compounds : Chapter 8.5- -8.7 8.5Basic Concepts of Chemical Bonding S & O 6e- x 5 = 30eAnion extra 2eTotal 32e- 1 Chem 210/ WANG / Chapter 8B Lewis Structures Molecular formula For CO2 Atom placement Make C the “center”. Remaining valence e16 - 4 = 12 e- 2 X 6e- C 4 e16e- count e- 6 e- from S 12 e- from 2O 18 e- total OSO SO2 count e- 6 e- from S 12 e- from 2O 18 e- total Positon atoms OSO : Total SO2 Position atoms & connect them. C O: : :O : Sum of valence e- : 2O OC Distribute e- Distribute remaining e- O Only have 18 e! OSO NOT enough e- for C missing 4emake 4÷2 = 2 double bonds!! OC O FAILED! NEXT Make a double bond! OSO Missing 2e! All have octet now. Lewis structure with two two double bonds. SO2 Which Lewis structure is right? We can write two or more Lewis structures for this molecule. They both - satisfy the octet rule - have the same number of bonds - have the same types of bonds count e- 6 e- from S 12 e- from 2O 18 e- total Positon atoms OSO Distribute eFAILED! NEXT Consider again….. The double bond can just as well be on the other side. Resonance Resonance structures O- S = O O - S =O OSO They They both are! • The actual SO2 molecule is described by a resonance resonance hybrid of two equivalent Lewis structures, each equivalent Lewis contributing 50% to the hybrid. • Experimental proof : Two S-to-O bonds are found to be equal (≈ 1½ bond), NOT one single and the other equal a double bond. O = S- O O =S - O OSO O Another Another example of Resonance Structures Benzene C6H6 also has two also has equivalent equivalent Lewis structures, each contributing 50% to the resonance hybrid. resonance SAMPLE PROBLEM 1 Write resonance structures for the nitrate ion, NO3-. Nitrate has 1(5) + 3(6) + 1 = 24 valence eO O N O N O O O N O O O O Experimentally, the C-C bonds in benzene are Call the same length (~ 1½ bond). Fall 2009 / UM-SJTU JI O N = O N O O N does not have an octet; a pair of e- will pair move in to make a double bond. O N O O O 3 equivalent resonance structures, each contributing 33.3% to the resonance hybrid. N – O is ~ 1 bond bond S O SAMPLE PROBLEM 2 NCO- also has 3 possible resonance forms. Are these equivalent forms? NC A O NC B O N C O C The 3 possible NCO- resonance forms are NOT NCO NOT equivalent to each other… • N, C, and O are different elements • Single – double – triple bonds do not occur in the structures consistently. • They do not contribute equally to the overall do to resonance hybrid. (“Stable” ones contribute more than “unstable” ones.) 2 Chem 210/ WANG / Chapter 8B Lewis Structures More than one valid Lewis structure: Use “Formal Charges” to decide if one is more stable (preferred) than the other(s). “Formal charge” Is the charge assigned to each atom in a Lewis structure as if all atoms had the as all same electronegativity. = no. of valence electrons – sum of all the electrons assigned to the atom. • Assign all nonbonding electrons around the atom to the atom. • Assign half of the bonding electrons to each atom joined by the bond . Resonance Resonance Structures of O3 Explain why the O-to-O bond in O3 is longer than that in O2, but shorter than that in H2O2. Use “Formal Charges” to decide which Lewis structure is more stable A Lewis structure is more stable if it fits fits well with the electronegativity & number of electronegativity number valence electrons each element has, indicated by: • • • Formal charges on all atoms being zero or close to zero (low pos. or neg. number) Negative formal charge found on more formal more electronegative atom / postive formal atom postive charge on less electronegative atom. less atom No like-charges (+,+ or -,-) appearing on likeadjacent atoms. O H O The O-to-O bond in O3 O-tois between the double and the single. O Boron only has 3e … makes 3 bonds only. F BCl BCl2F = B lacks octet. B Cl O O O O formal charges: O (left) = 6 – (6+1) = -1 O (middle) = 6 – (2+3) = +1 O (right) = 6 – (4+2) = 0 -1 +1 O O O O H O OO PCl5 = P exceeds octet. Sulfur makes 6 bonds sometimes. SF6 = S exceeds octet. Fall 2009 / UM-SJTU JI A 0 -1 0 0 0 O N C O formal charges: N = 5 – (6+1) = -2 B formal charges: N = 5 – (4+2) = -1 C formal charges: N = 5 – (2+3) = 0 C = 4 – (4) = 0 O = 6 – (3+2) = +1 C = 4 – (4) = 0 O = 6 – (2+4) = 0 C = 4 – (4) = 0 O = 6 – (1+6) = -1 “A” has +1 formal charge on O, the most electronegative has +1 most atom, making it least preferred of the 3 forms. “C” contributes the most to the resonance hybrids because the to O has -1 formal charge and both C & N has 0 charge. has has “C” is the preferred Lewis Structure! O2 H H2O2 +1 -1 OO Exceptions to the octet rule Expanded octet occur with elements of the 3rd period and beyond: P, S and Cl, etc. P, Cl, – Extra electrons occupy the empty d orbitals (3d) in these atoms. Cl Phosphorus makes 5 bonds sometimes. NC O the double bond in O2 but shorter than the single bond in H2O2. H Exceptions to the octet rule -1 C Resonance Structures of O3 O2 has O=O double bond. H2O2 has O-O has Osingle bond. N 0 +1 has equivalent O-toANSWER O3 has 2 equivalent Lewis structures. O-to-O Conti. bond order being 1 , the bond is longer than ANSWER O SAMPLE PROBLEM 3 ANSWER -2 P These 5 orbitals can share electrons with five Cl atoms to make PCl5. Expanded octet also occurs more often with larger atoms: Br, I or Xe Br, – More room around these large atoms to accommodate extra electrons. (1.48+1.21)÷ 2 = 1.34 O-to-O bond in O3 = 1.278 Agree within 2 sig. figs = 1.3 Å 1.3 An example: ICl41. Total the electrons. 7e(I)+4×7e(Cl)+1e (charge) total =36e 2. Write a possible arrangement. 3. Spread the electrons so all get octet…expand Octet as necessary. -1 Cl I Cl Cl 3 Cl Chem 210/ WANG / Chapter 8B Lewis Structures 0 P P 5 – 4 = +1 0HO 0 6-(6) = 0 0 0 More important (preferred) Lewis Structure 0 0 O H0 O 0 0 more stable 0 0 O -1 O S O : : -1 O O 0 more stable Resonance forms possible : : O: -1 lower formal charges -1 O S O -1 How many resonance structures of PO4-3 are shown below? ______. Which is the preferred Lewis Structure? Why? -2 : More.. O: : O: : -2 : :O: :O S O: : : : O: : These are equivalent resonance forms less less formal charges more more stable Better Better Lewis Structure than than the one drew earlier! Fall 2009 / UM-SJTU JI O: : 0 S : : -1 O 0 0HO -2 :O : :O S : : :O S : : : : O: O: : :O S :O : : -2 :O : : : : O: O: O +2 O H0 O: SO4 -2 : : : :O S S lower formal charges Answer (continued): Re-consider the Lewis structure Regiven for the sulfate ion earlier. Is there another Lewis structure existing in resonance with it? If so, which is the better one? Why? 2:O: O 0 HO 0 : O H0 0 lower formal charges more stable 0 O 0 P O 0 H 0 -1 O 0 O H0 O H 0 : -1 O 0 :O : :O S : :O ANSWER 6-(6+1) = -1 0 HO -2 : ANSWER O Answer to: Re-consider the Lewis structure given for the Resulfate ion earlier. Is there another Lewis structure existing in resonance with it? If so, which is the more important one? Why? : Draw two Lewis structures for H2SO4. Identify the more important form based on formal charges. : SAMPLE PROBLEM 5 : Write two Lewis structures for H3PO4. Indicate the more important form based on formal charges. SAMPLE PROBLEM 4 http://wb.chem.lsu.edu/htdocs/people/sfwatkins/MERLOT/drawlewis/dls.html 4 +2 O -1 ...
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