Question

The correct order of decreasing dipole moment of $\left(I\right)$ toulene, $\left(II\right)$ m-dichlorobenzene $\left(III\right)$ o-dichlorobenzene and $\left(IV\right)$ p-dichlorobenzene is:

• A. $IV<II<I<III$
• B. $IV<I<II<III$
• C. $I<IV<II<III$
• D. $IV<I<III<II$

Answer 1

The correct option is B $IV<I<II<III$

Dipole moment of p-dichlorobenzene is zero, while that of o-dichlorobenzene is zero, while that of o-dichlorobenzene is $2.54D$ and for m-dichlorobenzene is $1.48D$. This shows that benzene ring is planar and in $p-C_6{H}_4{Cl}_2$, two $C-Cl$ bonds are directed in the plane of the ring, thus on adding, they becomes zero. In this, two $C-Cl$ bonds contain an angle of ${60}^o$ and in $m-C_6{H}_4{Cl}_2$, the two $C-Cl$ bonds contain an angle of ${60}^o$ and in $m-C_6{H}_4{Cl}_2$. the two $C-Cl$ bonds contain an angle of ${120}^o$ and their dipole moments on vector addition give a higher value of dipole moment.
Therefore, sequence of dipole moment is as follows
p-dichlorobenzene< toulene <m-dichlorobenzene< o-dichlorobenzene