Net direction of dipole moment in oxygen difluoride

Question

What is the net direction of dipole moment in $\ce{F2O}$ molecule?

In the case of $\ce{NH3}$ and $\ce{NF3}$ molecules both have net dipole moment with direction reversed due to strong electronegativity of fluorine atoms, whereas in the case of $\ce{H2O}$ the direction of net dipole moment is from oxygen atom towards lone pair with a value $\mu(\ce{H_2O}) = \pu{1.84 D}.$

But in my textbook the value of net dipole moment of $\ce{F_2O}$ molecule is given $(\mu(\ce{F2O}) = \pu{0.3 D}),$ but not its direction. I have done a lot of research on google but I could not find any related information.

Answer 1

After going through a standard book on inorganic chemistry , I found the answer. The direction of dipole moment in $F_2O$ is just reversed as in the case of $NF_3$ i.e from oxygen atom towards fluorine atoms. Reference: Concise Inorganic Chemistry 5th edition by J.D.LEE; Chapter "CHEMICAL BONDING" topic "APPLICATION OF DIPOLE MOMENT".

Answer 2

In H2O, as O is more electronegative than hydrogen, the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and the dipole moment of H2O is high. (1.84 D)

On the other hand,

In the case of F2O, the bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are acting in opposition resulting in a low dipole (0.3 D)

So, basically, the dipole moment is towards fluorine atoms (which are more electronegative than oxygen atom). The reason why H2O has a higher dipole moment is that the difference between the electronegativity of O and H is higher than that of O and F, so water has a higher dipole moment even tho both water and oxygen difluoride have a similar bent molecular geometry.