Dear Student, Given, q=0 167e-=0 167×1 602×10-19C (1e-=1 602×10-19C) = 2 675×10-20Cr=154 6pm=154 6×10-12mdipole moment = q×r= 2 675×10-12C × 154 6×10-12m =4 136×10-30CmNow, I Debye = 3 34×10-30CmSo, dipole moment = 4 136×10-30Cm3 34×10-30Cm=1 24D Hope this information cleared your doubts about the topic If you have any other doubt, just post it on the forum and our experts will help you out as soon as possible Regards,

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Dear Student,

G i v e n, q = 0.167 e^{-} = 0.167 \times 1.602 \times 10^{- 19} C (1 e^{-} = 1.602 \times 10^{- 19} C) = 2.675 \times 10^{- 20} C r = 154.6 p m = 154.6 \times 10^{- 12} m d i p o l e m o m e n t = q \times r = 2.675 \times 10^{- 12} C \times 154.6 \times 10^{- 12} m = 4.136 \times 10^{- 30} C m N o w, I D e b y e = 3.34 \times 10^{- 30} C m S o, d i p o l e m o m e n t = \frac{4.136 \times 10^{- 30} C m}{3.34 \times 10^{- 30} C m} = 1.24 D

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