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

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

Chemical Bonding and Molecular Structure

Solution:

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

$\therefore , \,\,\,\, \mu=0$

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

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

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

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

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

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

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