Calculate the dipole moment of the following compound
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Explanation

Per cent ionic character `= (mu_(O))/(mu_(C) xx 100`
`(2. 6 xx 10^(-30) CM)/(1.6 xx 10^(-9) C xx 1.14 xx 10^(-10) m ) xx 100 = 11.5%`
(b) `q = (mu)/(d) = (1.2D)/(1.0 xx 10^(-8) cm) = (1.2 xx 10^(-8) "esu cm")/(1.0 xx 10^(-8) cm )`
` =1. 2 xx 10^(-10)` esu
The fraction of an electronic charge
`=(1.2 xx 10^(-10)esu)/(4.8 xx 10^(-10) esu e^(-1)) = 0.25 e^(-) = 25% of e^(-)`
(c ) `mu = 1.85 D = 1.85 xx 10^(-8)` esu cm `q xx d`
`cos 52.5^(@) = (d)/(0.94 A) implies d =0.609 xx 0.94 A`
`= 0.572 A`
`:. mu = q xx d`
`q_(1) = (mu)/(d) = (1.85D)/(0.572A)`
`= (1.85 xx 10^(-8)esu cm)/(0.572 xx 10^(-8) cm)`
`3.2 xx 10^(-10)` esu
`q_(1) = 2q_(2)`
`q_(2) = (q1)/(2) = (3.2 xx 10^(-10))/(2)`
`1.6 xx 10^(-10) esu`
`x =d_(IB)`
`= (1.77)/(sin 60^(@)) = 2.04 A`
`r_(B) = (1.04A) - (1.33 A)`
`= 0.71 A`
`R^(2) = P^(2) + Q^(2) + 2PQ cos Q`
`P^(2) + Q^(2) + 2PQ cos 180`
`P6(2) + Q^(2) - 2PQ`
`R^(2) = (P -Q)^(2)`
`R =(P -Q)`
`R = 3.95 - 1.55 = 2.4 D`


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