A) \[C{{H}_{4}}<N{{F}_{3}}<N{{H}_{3}}<{{H}_{2}}O\]
B) \[N{{F}_{3}}<C{{H}_{4}}<N{{H}_{3}}<{{H}_{2}}O\]
C) \[N{{H}_{3}}<N{{F}_{3}}<C{{H}_{4}}<{{H}_{2}}O\]
D) \[{{H}_{2}}O<N{{H}_{3}}<N{{F}_{3}}<C{{H}_{4}}\]
Correct Answer: A
Solution :
In\[{{H}_{2}}O,\]electronegativity difference is highest. So, dipole moment is highest in this. \[C{{H}_{4}}\] is a symmetrical tetrahedral structure and its dipole moment is zero.
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