Question

Lewis structures Number of valence electrons Ressance structures Formula Remaining electrons Bond skeleton formal charges C2H4 402)+ = 12 L H11 C2H2 402) 2001:10 CE 1416 210 :EO: XeF4 18+462) Fuxe = 28 FX-F 10- C1047 416) o-ci-D U - 1 NO2 Page 140

Answer 1

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Date Page: o VE = valence electron ) BS = Bond RE = Remaining Election Skeletons LS = Lewis structure / Catly → VE = 2 X 4+ 4x1 = 12 S S RE= 0 ) Bond Skeletons H x Resonance of = C ICI★ J-- OR C - C C₂H2 VE = 2x4 + 2x1 = 10 RE = 0 4-6=C-4 or 6-c. Co VE = 4 +6=10 ) Resonance = ☺ C = 0; ORC=O OR EO BY :C FÖ :=o: TORCOLI I show Resonance RE = 6 and 4 Because it in I in II But you can write = RE -4 Xefy = 8 + 784 = 36 election = VE Bond F - Xef Skeleton RE= 28 Resonance = 0

Page: Clou 7 + 4 x 6 = = IS 7+24+1 32€ .. 1 B.S formal charge 0 CI- o =Cl =0: 02 RE = Resonance Not 18 35 I - I-I D RE = 18 e- VE = 7X3+1 22 e- Resonance = I-I- I Resonance No- IVE = 5 + 6x2 + 1 = 5+12+1= 18 No: NO B.S RE= 12 Resonance = No