4 aN 2 4汇×6.023×1023×1.5×1024×10 3.770×106m3·mol M+ 2Rd M-Rd 1.339 1.337 1.324 4124 u sco=ucs+ u cs=so+Hco=661×103C·m =[661×1030-2×501×1030cos(112/2)]C·m 01×10- Cl2C=S偶极矩方向由C指向S 4125 由水的偶极矩可求OH键键矩为 UH, 0=2u OH coS(a/2) uoH=1a2o2cos2)=5.04×10c·m 对于H2O2,设两OH键键角为α,则 LH202=AOH Sin9687 cos(e/2) cos(a/2)=42O2/( L OH sIn9687°)=0.6825 4126 由水之偶极矩可求OH键键矩oR = d M · 2 1 2 2 + − n n = 3 0  NA = 3 4π  NA  = 3 4π 6.023 10 1.5 10 10 23 -24 -6      m3·mol-1 = 3.770×10-6m3·mol-1 n = ( M Rd M Rd − + 2 ) 1/2 t/℃ 0 40 100 n 1.339 1.337 1.324 4124     ← → S == C == O   SCO =   CS +   CO   CS =   SCO +   CO = 6.61×10-30C·m     = [6.61×10-30 - 2×5.01×10-30cos (112/2) ]C·m = 1.01×10-30C·m Cl2C == S 偶极矩方向由 C 指向 S 。 4125 由水的偶极矩可求 OH 键键矩为 H2O = 2 OH cos (/2)  OH = H2O/[2cos(/2)] = 5.04×10-30C·m 对于 H2O2 , 设两 OH 键键角为 , 则 H2O2 =  OH sin96.87°cos(/2) cos (/2) = H2O2 /( OH sin96.87°) = 0.6825 /2= 46.96°  = 93.92° 4126 由水之偶极矩可求 OH 键键矩HO Cl Cl C S