Question

Fluorine is more electronegative than either boron or phosphorus. What conclusion can be drawn from the fact that $ B{{F}_{3}}$ has no dipole moment but $ P{{F}_{3}}$ does?
(A) $ B{{F}_{3}}$ is not spherically symmetrical but $ P{{F}_{3}}$ is spherically symmetrical
(B) $ B{{F}_{3}}$ molecule must be linear
(C) The atomic radius of P is larger than that of B
(D) The $ B{{F}_{3}}$ molecule must be planar triangular

Answer 1

Hint: Fluorine is the most electronegative element in the periodic table. It has the highest tendency to attract the shared pair of electrons in any compound. It has an electronegativity of 3.98. In the periodic table, electronegativity increases from left to right and decreases from top to bottom. Fluorine is an element of group 17. Boron belongs to group 13, and phosphorus belongs to group 15.

Complete Step by Step Answer:
$ B{{F}_{3}}$ has three dipoles from boron to fluorine at an angle of $ {{120}^{o}}$ to each other and are directed away from each other. It has trigonal planar geometry. Each dipole can be cancelled out by the dipole formed by combination of other two dipoles. So, it has no dipole moment. Whereas $ P{{F}_{3}}$ has three dipoles from phosphorus to fluorine and a lone pair of electrons.

Due to this lone pair, the molecule has a pyramidal shape. Thus, the three dipoles add up to give a net non-zero dipole moment.
Correct Option: (D) The $ B{{F}_{3}}$ molecule must be planar triangular.

Note: The symbols used to represent the charges on the two elements are $ \delta +$ and$ \delta -$ , where more electronegative elements get $ \delta -$ charge and less electronegative elements get $ \delta +$ charge.