Question

According to the VSEPR theory, what is the shape of a molecule that has a central atom bound to three other atoms with no lone pairs of electrons?

Answer 1

Explanation:

A molecule which contains three bond pairs but no lone pair of electrons will have a trigonal planar geometry.

Hybridization of such a geometry is $sp^{2}$.

For example, $BF_{3}$ is a trigonal planar molecule.

As, hybridization of $BF_{3}$ will be calculated as follows.

                Hybridization = $\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

              Hybridization = $\frac{1}{2}[V+N-C+A]$

                                    = $\frac{1}{2}[3 + 3]$

                                    = 3

So, the hybridization of $BF_{3}$ will be $sp^{2}$. As there are 3 bond pairs and no lone pair of electrons. Hence, the geometry of $BF_{3}$ is trigonal planar.

Thus, we can conclude that according to the VSEPR theory, the shape of a molecule that has a central atom bound to three other atoms with no lone pairs of electrons is trigonal planar.

Answer 2

Triangular planar structure