# Determine the symmetry elements and point groups of "NF"_3, "NClF"_2, and "NBrClF". What symmetry elements are lost as we descend from "NF"_3 to "NCl"_2"F" and then to "NBrClF"?

Dec 2, 2017

Here's what I get.

#### Explanation:

Point group of ${\text{NF}}_{3}$

${\text{NF}}_{3}$ has a trigonal pyramidal geometry. It belongs to point group ${\text{C}}_{\textrm{3 v}}$, like ammonia.

It has a ${\text{C}}_{3}$ principal axis and three σ_text(v) vertical planes of symmetry.

The image below shows the three σ_text(v) planes. If you look at the molecule from the top, the ${\text{C}}_{\textrm{3}}$ axis becomes more apparent. You can rotate ${120}^{\circ}$ about the $z$ axis to return an indistinguishable configuration of the molecule. 1. ${\text{NF}}_{3}$ to ${\text{NClF}}_{2}$

${\text{NClF}}_{2}$ is also a trigonal pyramidal molecule. It belongs to point group ${\text{C}}_{\textrm{s}}$, however, just like if you slightly tilted one $\text{F}$ inwards on ${\text{NF}}_{3}$.

Its structure is Its only symmetry element is a single σ_text(h) plane that bisects the molecule into two mirror image halves.

Thus, it has lost two ${\sigma}_{\textrm{v}}$ mirror planes and its ${C}_{3}$ axis of rotation.

2. ${\text{NClF}}_{2}$ to $\text{NBrClF}$

$\text{NBrClF}$ is a trigonal pyramidal molecule, but it has no symmetry element except the trivial one of rotating 360° about a ${\text{C}}_{1}$ axis.

That now belongs to point group ${\text{C}}_{1}$ since it has only the identity element.

Its structure is Thus, it has lost even the ${\sigma}_{\textrm{h}}$ plane of ${\text{NClF}}_{2}$ .