A) \[C{{H}_{3}}F>C{{H}_{3}}Cl>C{{H}_{3}}Br\]
B) \[C{{H}_{3}}F>C{{H}_{3}}Br>C{{H}_{3}}Cl\]
C) \[C{{H}_{3}}Cl>C{{H}_{3}}F>C{{H}_{3}}Br\]
D) \[C{{H}_{3}}Cl>C{{H}_{3}}Br>C{{H}_{3}}F\]
Correct Answer: C
Solution :
As we move down the group in the Periodic Table from F to I, the electronegativity of halogen decreases, therefore the polarity of the C?X bond and hence the dipole moment of the haloalkane should also decrease accordingly. But the dipole moment of \[C{{H}_{3}}F\] is slightly lower than that of \[C{{H}_{3}}Cl\]. The reason being that although the magnitude of -ve charge on the F atom is much higher than that on the \[Cl\] atom but due to small size of F as compared to \[Cl\], the C?F bond distance is so small that the product of charge and distance \[(\mu =q\times d)\], ie, dipole moment of \[C{{H}_{3}}F\] turns out to be slightly lower than that of \[C{{H}_{3}}Cl\]. Therefore, the order of dipole moment is \[\underset{1.860\,\,\,D}{\mathop{C{{H}_{3}}Cl}}\,\,\,\,>\,\,\,\underset{1.847\,\,D}{\mathop{C{{H}_{3}}F}}\,\,\,\,\,>\,\,\,\,\underset{1.830\,\,D}{\mathop{C{{H}_{3}}Br}}\,\]
You need to login to perform this action.
You will be redirected in
3 sec
