Question

The dipole moment of $o,p$ and $m$ -dichlorobenzene will be in the order

• A. $o>p>m$
• B. $p>o>m$
• C. $m>o>p$
• D. $o>m>p$

Answer 1

Answer: (D)

In $p$ -dichlorobenzene, two $C$ - $Cl$ dipoles cancel each other,

$\therefore ,\mu =0$

In $o$ -dichlorobenzene, the two $C-Cl$ dipoles $(\mathrm{say}x)$ are inclined at an angle of $60^{\circ }.$ Therefore, according to parallelogram law of vectors, the resultant

$=\sqrt{x^2+x^2+2x^2\times \cos 60^{\circ }}$

$=\sqrt{x^2+x^2+2x^2\times 1/2}=\sqrt{3x^2}=\sqrt{3}x$

In $m$ -dichlorobenzene, the two dipoles are inclined to each other at an angle of $120^{\circ },$ therefore, resultant

$=\sqrt{x^2+x^2+2x^2\times \cos 120^{\circ }}$

$=\sqrt{x^2+x^2+2x^2\times (-1/2)}=\sqrt{x^2}=x$

Thus, the decreasing order of dipole moments is $o>m>p$.