Question

The decreasing order of dipole moments of the ${\text{HF}},{{\text{H}}_2}{\text{O}},{\text{Be}}{{\text{F}}_2},{\text{N}}{{\text{F}}_3}$ is
A.${\text{HF}} > {{\text{H}}_2}{\text{O}} > {\text{Be}}{{\text{F}}_2} > {\text{N}}{{\text{F}}_3}$
B.${{\text{H}}_2}{\text{O > HF}} > {\text{N}}{{\text{F}}_3} > {\text{Be}}{{\text{F}}_2}$
C.${\text{N}}{{\text{F}}_3} > {\text{Be}}{{\text{F}}_2}{\text{ > HF}} > {{\text{H}}_2}{\text{O}}$
D.${\text{N}}{{\text{F}}_3} > {\text{Be}}{{\text{F}}_2}{\text{ > }}{{\text{H}}_2}{\text{O > HF}}$

Answer 1

Dipole moment is vector sum of all dipole moment of covalent bonds present in a molecule or compound. A polar compound does not have a zero net dipole moment and on the other hand, the dipole moment of a nonpolar compound is zero.

Complete answer:
Polarity of any covalent bond or molecule is measured in terms of dipole moment. As polarity of bond depends upon electronegativity difference between atoms but polarity of molecule depends upon dipole moment. The product of positive or negative charge and the distance between two poles is called dipole moment. Therefore the dipole moment is a vector quantity. Unit of dipole moment is Debye. In the diatomic molecule, the dipole moment depends upon the difference of electronegativity. For example order of dipole moment in some diatomic molecules is as follow: ${\text{HF}} > {\text{HCl}} > {\text{HBr}} > {\text{HI}}$ and for compounds like ${{\text{H}}_2},{{\text{F}}_2},{\text{C}}{{\text{l}}_2},{\text{B}}{{\text{r}}_2},{{\text{O}}_2}$ dipole moment is zero. For polyatomic molecules, dipole moment depends upon the vector sum of dipole moment of all the covalent bonds. If the vector sum is zero, then the compound is nonpolar or symmetrical.
For example in compounds like ${\text{BC}}{{\text{l}}_3},{\text{CC}}{{\text{l}}_4},{\text{C}}{{\text{H}}_4},{\text{PC}}{{\text{l}}_5}$ the bond between ${\text{B}} - {\text{F}},{\text{C}} - {\text{Cl}},{\text{C}} - {\text{H}},{\text{P}} - {\text{Cl}}$ are polar even though compounds are nonpolar due to zero vector sum of all dipole moments.
As ${\text{Be}}{{\text{F}}_2}$ is non polar due to net vector sum of dipole moments in ${\text{Be}}{{\text{F}}_2}$ is zero. Compound like ${{\text{H}}_2}{\text{O}}$ has V shape and thus it will be having maximum dipole moment among given compounds as ${\text{HF}}$ is diatomic and its dipole moment is mainly depend on electronegativity difference but dipole moment of ${\text{HF}}$ will still be greater than that of ${\text{N}}{{\text{F}}_3}$ as in case of ${\text{N}}{{\text{F}}_3}$ vector sum of all dipole moment is less than that of dipole moment of ${\text{HF}}$ .

Thus, correct order of decreasing dipole moment is ${{\text{H}}_2}{\text{O > HF}} > {\text{N}}{{\text{F}}_3} > {\text{Be}}{{\text{F}}_2}$ and thereby the correct option is B.

Note:
Generally if the central atom has only one lone pair of electrons the molecule will be polar except ${\text{P}}{{\text{H}}_3}$ due to Drago rule. If the central atom has no lone pair and also all terminal atoms are same then the molecule will be nonpolar.