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"#?

1 Answer

Answer:

Here's what I get.

Explanation:

Point group of #"NF"_3#

#"NF"_3# has a trigonal pyramidal geometry. It belongs to point group #"C"_text(3v)#, like ammonia.

It has a #"C"_3# principal axis and three #σ_text(v)# vertical planes of symmetry.

The image below shows the three #σ_text(v)# planes.

σh planes
(Adapted from SlidePlayer)

If you look at the molecule from the top, the #"C"_text(3)# axis becomes more apparent. You can rotate #120^@# about the #z# axis to return an indistinguishable configuration of the molecule.

C3 axis

1. #"NF"_3# to #"NClF"_2#

#"NClF"_2# is also a trigonal pyramidal molecule. It belongs to point group #"C"_text(s)#, however, just like if you slightly tilted one #"F"# inwards on #"NF"_3#.

Its structure is

NClF2

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_text(v)# mirror planes and its #C_3# axis of rotation.

2. #"NClF"_2# to #"NBrClF"#

#"NBrClF"# is a trigonal pyramidal molecule, but it has no symmetry element except the trivial one of rotating 360° about a #"C"_1# axis.

That now belongs to point group #"C"_1# since it has only the identity element.

Its structure is

NBrClF

Thus, it has lost even the #sigma_text(h)# plane of #"NClF"_2# .