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Chemical bonding between two atoms is necessarily associated with an electrical moment arising out of the difference in electroegativity of two atoms. This means that every bond carries with it an electricalmoment called the ''bond moment''. The dipole moment of a molecule is really the vectorial sum of the individual bond moment present in it. To compute the dipole moment it is necessary to find out the values of various bond moment. In the following table, dipole moment of different bonds are as given. {:(,rarr,rarr,rarr),("Bond",H-C,C-Cl,C=0),("Bond moments",0.4 D,1.5 D,2.5 D):} The group moments of few groups as given {:("Group",NO_(2),OH,CN,CH_(3)),("Direction of dipole","Towards N","Towards O","Towards N","Away from "CH_(3)),("Dipole moment",4D,1.6D,3.8D,0.4D):} In CH_(3)"CCl"_(3)" (I)", CHCl_(3) (II) and CH_(3)Cl (III) the normal tetragedral bond angle is maintained. Also given cos70.5^(@)=(1)/(3). Therefore dipole moments of the given compounds are : (given due to -I effect of Cl. The bond moment of H-C bond directed towards the H in CHCl_(3)) |
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Answer» I = 1.9D, II = 1.9 D, III = 1.7 D `mu_(CHCl_(3)ltCH_(3)Cl` `Irarr1.9D"(LEAST)"` (given in options) Hence, no CALCULATIONS required. |
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