# Question 68532

The ground state electronoic configuration of ""_5B is $1 {s}^{2} 2 {s}^{2} 2 {p}_{x}^{1} 2 {p}_{y}^{\circ} 2 {p}_{z}^{\circ}$.During chemical interaction with $F$ atoms one $2 s$ electron gets promoted to vacant $2 {p}_{y}$ level as shown in the figure below forming 3 unpaired electons in the valence shells and thus it becomes capable of forming 3 covalent bonds using its 3 $s {p}^{2}$ hybrid orbitals.
As a result the shape of the $B {F}_{3}$ molecule becomes triangular planer (as only two ${p}_{x} \mathmr{and} {p}_{y}$ are involved in hybridisation) where each $\angle F B F$ is ${120}^{\circ}$. Due to this symmetric structure of $B {F}_{3}$ the resultant bond moment of any two $B - F$ bond just becomes equal but opposite of the bond moment of the third one. This makes the net dipole moment of $B {F}_{3}$ molecule zero.
Again the ground state electronoic configuration of ""_7N# is $1 {s}^{2} 2 {s}^{2} 2 {p}_{x}^{1} 2 {p}_{y}^{1} 2 {p}_{z}^{1}$. Here central N atom uses 3 $s {p}^{3}$ hybrid orbitals to form 3 $\sigma$-bonds with three $F$ atoms rendering a pyramidal shape to $N {F}_{3}$ molecule where lone pair of electrons of $2 s$ level of N atom occupies the vacant residual 4th $s {p}^{3}$ hybrid orbital of N-atom as shown in the figure below.
Here the resultant bond moments of 3 $N - F$ bonds is somewhat counter balanced by the opposing moment of lone pair on N-atom and thus net dipole moment of the molecule gets diminished.
Thus we observe that $B {F}_{3}$ has no dipolemoment but $N {F}_{3}$ has got.