Question

Are bf3 and nf3 symmetrical or unsymmetrical molecules? explain.

Answer 1

Answer: $BF_3$ is symmetrical and $NH_3$ is unsymmetrical.

Explanation: Formula used  for calculating hybridization:

1. $BF_3$

$:{\text{Number of electrons}} =\frac{1}{2}[V+N-C+A]$

where, V = number of valence electrons present in central atom i.e. boron= 3

N = number of monovalent atoms bonded to central atom=3

C = charge of cation = 0

A = charge of anion = 0

${\text{Number of electrons}} =\frac{1}{2}[3+3-0+0]=3$

The number of electrons is 3 that means the hybridization will be $sp^2$ and the electronic geometry of the molecule will be trigonal planar as there are three atoms around the central atom, the molecule will be symmetrical.

2. $NF_3$

$:{\text{Number of electrons}} =\frac{1}{2}[V+N-C+A]$

where, V = number of valence electrons present in central atom i.e. nitrogen= 5

N = number of monovalent atoms bonded to central atom=3

C = charge of cation = 0

A = charge of anion = 0

${\text{Number of electrons}} =\frac{1}{2}[5+3-0+0]=4$

The number of electrons is 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are three atoms around the central atom the fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be trigonal pyramidal and hence will be unsymmetrical.

Answer 2

The correct option is this: BF3 IS SYMMETRICAL WHILE NF3 IS UNSYMMETRICAL.

Symmetry is used in chemistry to explain the structure of chemical substances, that is, it is used to explain how the particles in chemical substances are arranged. In chemistry, molecular symmetry is used to explain the behavior of chemical substances. A chemical substance is said to be symmetrical if its polar bonds are evenly distributed; it is described as unsymmetrical if its bonds are unequally distributed. In the compound BF3, the bond dipoles cancel one another out because the polar bonds are evenly distributed whereas in NF3, the bond dipoles were not able to cancel each other out because of unequal distribution of the polar bonds, the compound has a net dipole moment.