Question

The decreasing order of dipole moment of the following is :
$I.{CH}_{3}Cl$

$II.{CH}_{3}Br$

$III.{CH}_{3}F$

• A. $\left(I\right)>\left(II\right)>\left(III\right)$
• B. $\left(I\right)>\left(III\right)>\left(II\right)$
• C. $\left(II\right)>\left(I\right)>\left(III\right)$
• D. $\left(II\right)>\left(III\right)>\left(I\right)$

Answer 1

The correct option is B $\left(I\right)>\left(III\right)>\left(II\right)$

The decreasing order of the dipole moments is $I>III>II$.

The increasing order for the electronegativities of halogens is $Br<Cl<F$. The decreasing order of bond lengths is $C-Br>C-Cl>C-F$.

These two factors oppose each other and the overall effect leads to the above order of dipole moments.