In my book, the stimulate of dipole moment for CH3Cl, CH2Cl2, CHCl3 and CCl4 is given as:

CH3Cl>CH2Cl2>CHCl3>CCl4

The method I know this is—

In instance of CCl4, tetrahedral the opposite exists. Therefore, the dipole moment of 3 C-Cl bonds on one side, gives a resultant minute that is equal and opposite to the dipole moment of the solitary C-Cl link on the opposite side. So, $cemu=0$.

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This explains CHCl3>CCl4.

Again, in case of CHCl3, the three C-Cl bonds suggest to one side, for this reason a moment equal to C-Cl shortcut is generated as a result. A moment of C-H link is added to it. In CH2Cl2, the moment of two C-Cl shortcut is included at $ce109^circ$(approx.) so the resultant is higher than on C-Cl bond.

See more: Convert 5/8 To Decimal Equivalent Of 5/8 Into A Decimal ? ⅝ As A Decimal

This describes CH2Cl2>CCl4

In instance of CH3Cl, the dipole minute of one C-Cl bond is added with moment of one C-H bond (resultant of 3 C-H bonds).

But this should mean moment the CHCl3$ceapprox$ moment of CH3Cl, but clearly this is no so.Where did ns go wrong?


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