a=633×1039y1·C2·m2 4119 设OH键键矩为oH,则 2 JOH COS(a/2)=6.18×1030 OH=5.04×103C·m 4120 u=2HloHcoS(a/2) =2×504×103cos(104.5/2)C·m =6.17×1030C·m 4121 邻-二氯苯y=2×524×1030×cos30°C·m 9.08×1030C·m 间-二氯苯=5.24×103C·m 对-二氯苯=0 其中对-二氯苯μ=0更为可信,另外二个苯环可能变形。 4122 M 8-1 NA dE+23 E+23E。M a=4rE。a'=1.37×1039J1·C2·m1 2.222×10-39J-1·C 3kT 玩4=2.363×1038J·C2·m -=0.848 E=17.7 4123 = 6.33×10-39J -1·C 2·m2 4119 设 OH 键键矩为OH, 则 2OH cos (/2)= 6.18×10-30 OH = 5.04×10-30C·m 4120  = 2OHcos(/2) = 2×5.04×10-30cos(104.5/2) C·m = 6.17×10-30 C·m 4121 邻-二氯苯 = 2×5.24×10-30×cos30°C·m = 9.08×10-30C·m 间-二氯苯  = 5.24×10-30C·m 对-二氯苯  = 0 其中对-二氯苯 = 0 更为可信, 另外二个苯环可能变形。 4122 d M · 2 1 + −   = 0 3 NA (+ 3kT 2  ) 2 1 + −   = M N dA 3 0  (+ 3kT 2  )  = 4。’= 1.37×10-39J -1·C 2·m-1 3kT 2  = 2.222×10-39J -1·C 2·m-1 M N dA 3 0  = 2.363×1038J·C -2·m 2 1 + −   = 0.848  = 17.7 4123