• # question_answer The dipole moment of chlorobenzene is $1.5D$. The dipole moment of is A)  $2.86\,D$                    B)  $2.25\,\,D$C)  $1.5\,\,D$                   D)  $0\,\,D$

Dipole moments of $2Cl$ and $5Cl$are vectorically cancelled. It is duel $l$ $Cl$ and $3\,\,Cl$ ${{\mu }^{2}}$ $=\mu _{1}^{2}+\mu _{2}^{2}+2{{\mu }_{1}}{{\mu }_{2}}\cos \theta$ $={{(1.5)}^{2}}+{{(1.5)}^{2}}+2\times 1.5\times 1.5\times 1.5\cos \,120$$\therefore$ $\mu =1.5D$