I have to find the decreasing order of dipole moment of $\ce{CHCl3}$, $\ce{CH3Cl}$, $\ce{CH2Cl2}$, and $\ce{CCl4}$. Now I know that in $\ce{CCl4}$ there will be no dipole moment. Now to compare the other three compounds I don't have logical reasoning to compare the dipole moments but I have the data to compare but in the examination we won't be having these data so I was looking for any logical reasoning to compare the dipole moments of the remaining three compounds. Thank you
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asked Mar 17, 2016 at 18:38
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$\begingroup$ chemistry.stackexchange.com/questions/8977/… $\endgroup$– MithoronMar 17, 2016 at 19:05
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1 Answer
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Look at dipole moment as a vector with the head of vector pointing towards the more electronegative atom. Ccl4 has 4 vectors pointing outwards but these vectors cancel, hence 0 dipole. Chcl3, 1 vector from H towards c, 3 vectors from c to each cl atom. Note they have tetrahedral geometry. The net vector would be along one of cl atoms but as other cl atoms have components of their vectors opposite to the victorious cl atom, dipole would be reduced. Ch2cl2, net vector would be somewhere between the 2 c==>cl vectors but there are not much opposite components of vectors as compared to chcl3, it has greater dipole than the former. Ch3cl, net dipole is along the cl atom and further each H==>C dipole is increasing the magnitude of the vector. These problems are based on visualization and vectors help you visualize better.
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1$\begingroup$ The electronegativity difference between carbon and hydrogen is roughly the same as the difference between carbon and chlorine. So why is the magnitude of the dipole moment between methyl chloride and chloroform so different? $\endgroup$– ronMar 17, 2016 at 19:49
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$\begingroup$ Correct me if I'm wrong but as all the atoms around c are not the same, the structure won't be perfect tetrahedral. Cl atoms are much larger than the H atoms and would face greater repulsion from each other than 2 H atoms. The shape would be altered in space and cl atoms would be spaced away from each other. $\endgroup$– jatinMar 17, 2016 at 20:23
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$\begingroup$ Now the explanation. In trichloromethane the cl atoms would be spaced far and their bond angle would be increased, thus decreasing the net dipole. While in chloromethane H atoms won't face much repulsion from each other rather they would feel more repulsion from cl, hence they will be closer and increase the net dipole. $\endgroup$– jatinMar 17, 2016 at 20:24
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$\begingroup$ The bond angles in chloroform and methyl chloride are all very close to the ideal tetrahedral angle. Focus on the different bond lengths. No need to respond to me, just incorporate your thinking into your answer $\endgroup$– ronMar 17, 2016 at 21:21
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$\begingroup$ So c-cl bonds would be longer and have greater dipole. Then in trichloromethane you have 3 relatively bigger vectors contributing to the net vector and one small h-c dipole contributing. In chloromethane we would have only 1 bigger c-cl dipole and 3 h-c dipoles contributing towards it. The common vector arrangement is there are 3 vectors pointing upwards but away from each other and one pointing upwards. In one case upward pointing vector is the larger c-cl vector and smaller h-cl in the other. But we just switched their dipoles. Where am I wrong? $\endgroup$– jatinMar 17, 2016 at 21:37