A) \[\operatorname{CHC}{{l}_{3}}>C{{H}_{2}}C{{l}_{2}}>C{{H}_{3}}Cl>CC{{l}_{4}}\]
B) \[{{\operatorname{CH}}_{2}}C{{l}_{2}}>C{{H}_{3}}Cl>CHC{{l}_{3}},>CC{{l}_{4}}\]
C) \[{{\operatorname{CH}}_{3}}Cl>C{{H}_{2}}C{{l}_{2}}>CHC{{l}_{3}}>CC{{l}_{4}}\]
D) \[{{\operatorname{CH}}_{2}}C{{l}_{2}}>CHC{{l}_{3}}>C{{H}_{3}}Cl>CC{{l}_{4}}\]
Correct Answer: D
Solution :
\[CC{{l}_{4}}\] has zero dipole moment because of symmetric tetrahedral structure. \[C{{H}_{3}}Cl\] has slightly higher dipole moment which is equal to 1.86 D. Now \[C{{H}_{3}}Cl\] has less electronegativity than\[C{{H}_{2}}C{{l}_{2}}\]. But \[C{{H}_{2}}C{{l}_{2}}\] has greater dipole moment than\[CHC{{l}_{3}}\].
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