Lacey_Che131_S2011_Lect-26-2(2)

Lacey_Che131_S2011_Lect-26-2(2) - Lec Lec-25-26: Lewis...

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Lec-25-26: Lewis structure, etc. Lec 25 26: Lewis structure, etc. Roy A. Lacey, Stony Brook University; Che 131, Springl 2011 1 Drawing Lewis Structures 1. Choose the central atom (the atom with the smallest ionization energy) and connect the other atoms to the central atom with a two electron bond (line). 1. Determine the number of valence electrons needed (N) by the atoms in the molecule. Sum the valence electrons (H) from all the 2. Sum the valence electrons (H) from all the atoms (consider the charge if it is an ion). 1. Calculate the number of available valence electrons that must be shared (S = N - H). Roy A. Lacey, Stony Brook University; Che 131, Springl 2011 2
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Drawing Lewis Structures 5. Determine the number of bonds in the molecule (# = S/2). 6. Add the needed bonds to the original structure. 7. Add lone pairs of electrons to obtain octets on the atoms (if the atom obeys octet rule). 8. Check total number of electrons in the structure. Roy A. Lacey, Stony Brook University; Che 131, Springl 2011 3 Drawing Lewis Structures Hydrogen and the halogens bond once Additional handy rules ± Hydrogen and the halogens bond once. ± The oxygen group bonds twice. ± The nitrogen group bonds three times. So does boron. The carbon group bonds four times ± The carbon group bonds four times. Roy A. Lacey, Stony Brook University; Che 131, Springl 2011 4
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Drawing Lewis Structures N=12 N = 12 H = 8 s = N – H = 4 Æ Two bonds N = 24 H = 16 s=N –H=8 s = N H = 8 Four bonds Roy A. Lacey, Stony Brook University; Che 131, Springl 2011 5 Drawing Lewis Structures N=16 N 2 N = 16 H = 10 s = N – H = 6 Æ Three bonds Triple bonds < double bonds < single bonds Roy A. Lacey, Stony Brook University; Che 131, Springl 2011 6
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A. 3 B 2 H H N = 24 H = 14 B. C. 1 D. 0 N-N H H s = N – H = 10 Æ five bonds 7 Roy A. Lacey, Stony Brook University; Che 131, Springl 2011 Drawing Lewis Structures N=24 NO 2 + ONO + N = 24 H = 16 s = N – H = 8 [ ] = = Æ four bonds Roy A. Lacey, Stony Brook University; Che 131, Springl 2011 8
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A. 1 nonbonding pair on carbon B 2 nonbonding pairs on nitrogen N = 18 H = 10 B. C. 1 nonbonding pair on nitrogen D. 2 nonbonding pairs on carbon s = N – H = 8 Æ four bonds 9 Roy A. Lacey, Stony Brook University; Che 131, Springl 2011 Resonance Structures Resonance is the condition shown by a O 3 Resonance is the condition shown by a molecule when the valence electrons can be arranged in two or more ways N = 24 H = 18 s = N – H = 6 Æ
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This note was uploaded on 02/21/2012 for the course CHE 131 taught by Professor Kerber during the Spring '08 term at SUNY Stony Brook.

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Lacey_Che131_S2011_Lect-26-2(2) - Lec Lec-25-26: Lewis...

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