Question

Identify the symmetry elements of the ${\mathbf{PF}}_{\mathbf{5}}$ molecule (see Fig. 3.20a).

Short answer

The given molecule ${\mathrm{PF}}_5$ contains 3${\mathrm{C}}_2$ axes of symmetry, 2${\mathrm{C}}_3$ axes of rotation, 3${\sigma }_v$ planes of symmetry, and 1 ${\sigma }_h$plane of symmetry.

Step-by-step solution

The symmetry elements.

The fundamental way of classification of different organic structures can be performed by finding their symmetry elements.

When an object, resulting from the rotation, reflection, or inversion of its original structure can be superimposed on the original; then the structure is said to contain the corresponding symmetry element.

Different symmetry elements are the axes of rotation (${\mathrm{C}}_{\mathrm{n}}$), a plane of reflection ($\sigma$), or a center of symmetry (i).

Symmetry elements of the PF5 molecule.

The IUPAC name of the given compound is phosphorus pentafluoride.

The geometry of the given compound is trigonal bipyramidal as shown below.

The given compound has 2-fold rotational axis. Here, when the molecule is rotated by $180°$it brings one F in place of another F-atom. This can be performed from all the three F-atoms present on the horizontal plane. Therefore, the molecule contains 3-${\mathrm{C}}_2$axes of symmetry. The compound also has ${\mathrm{C}}_3$axes of rotation. This rotation goes two times as shown below.

The molecule also contains 2 planes of symmetry.

One plane ${\mathrm{\sigma }}_{\mathrm{v}}$cutting three F-atoms; reflects the two F-atoms on the horizontal plane (shown with wedge and dash). This can be done three times; reflecting any two-fluorine present on the horizontal plane as shown below.

Another plane ${\mathrm{\sigma }}_{\mathrm{h}}$cutting three F-atoms present in one plane; reflects the two fluorine atoms present up and down.