4059 4060 4061 4062 4069 u357×10-0 r130×10C=275×102 q275×10-20 =0.17 4070 +2 对于CCl,=0,所以 8. M 38 4×154×103×3×8854×101 4.24×1.59×103×60:3JC2m2 =1.25×10-39JC2m 4071 1H,o6.17×10 C·m=5.04×1030C 2cos(a/2) 2cos5224059 D3d 4060 D2d 4061 D2h 4062 Oh` 4069 =rq q= r μ = -10 30 1.30 10 3.57 10   − C=2.75×10-20C e q = -19 20 1.602 10 2.75 10   − =0.17 4070 2 1 + − r r ε ε ·  M = 0 3 NA (+ 3kT 2  ) 对于 CCl4,=0,所以 = 2 1 + − r r ε ε ·  M · NA 0 3 = 3 23 -3 -12 4.24 1.59 10 6.02 10 1.24 154 10 3 8.854 10          J -1C 2m2 =1.25×10-39J -1C 2m2 4071 2O-H cos(/2)= H2O O-H = 2cos( /2) H2O α  =   2cos52.25 6.17 10-30 C·m =5.04×10-30C·m