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In pyramidal ammonia, the hydrogen symmetry orbitals span $A_1\oplus E$ of the C3v molecular point group. Nitrogen $2s$ and $2p_z$ both have $A_1$ symmetry and $(2p_x,2p_y)$ span $E$. In the planar D3h molecular point group, these become $A_1^{\prime }$, $A_2^{\prime \prime }$ and $E^{\prime }$, respectively, while the hydrogen combinations span $A_1^{\prime }\oplus E^{\prime }$. Thus the planar out-of-plane lone pair is a symmetry-isolated nonbonding orbital; pyramidalization permits mixing with the other $a_1$ functions.^[ref]

An isolated boron trifluoride molecule has trigonal planar molecular geometry and the D3h molecular point group. Its three equal fluorine contributions give $I_{\parallel }=3m_{\mathrm{F}}{R}^2$ and $I_{\perp }=3m_{\mathrm{F}}{R}^2/2$ in the rigid rotor approximation.

Ideal gaseous phosphorus pentachloride is a trigonal bipyramid with D3h molecular point group symmetry. There are five chlorine sites and three 3p orbitals per site, hence a fifteen-dimensional basis. A moved atom contributes zero to the character; each fixed chlorine contributes the trace of the three-dimensional polar-vector matrix acting on its local p orbitals.For example, a $C_2^{\prime }$ axis through one equatorial chlorine fixes that site but rotates its p-vector basis with trace $1-1-1=-1$. Counting that atom as $+1$ would be incorrect.

Apply molecular character reduction with class sizes $(1,2,3,1,2,3)$ and group order twelve:For $A_1^{\prime }$, this gives $(15-3+3+9)/12=2$; for $E^{\prime }$, $(30+6)/12=3$. The full result isThe dimension check is $2+1+6+2+4=15$. As an additional transparent separation, the two axial chlorines supply $A_1^{\prime }+E^{\prime }+A_2^{\prime \prime }+E^{\prime \prime }$, while the three equatorial ones supply $A_1^{\prime }+A_2^{\prime }+2E^{\prime }+A_2^{\prime \prime }+E^{\prime \prime }$. These are all chlorine 3p orbitals, including both radial and tangential functions; a five-orbital ligand-sigma basis would answer a different question.

Use the D3h molecular point group for phosphorus pentafluoride, with $z$ along the two axial P–F bonds. For each symmetry operation, only atoms left in place contribute to the character of the Cartesian displacement representation; each contributes the trace of the corresponding $3\times 3$ spatial matrix. In the class order $E,2C_3,3C_2^{\prime },{\sigma }_h,2S_3,3{\sigma }_v$ this givesFor example a $C_2^{\prime }$ fixes P and one equatorial F, and each fixed atom has Cartesian trace $-1$, giving $-2$. reflection in the equatorial plane fixes four atoms, each with trace $1$, giving $4$. Reduction with the character table givesSubtract translations $E^{\prime }+A_2^{\prime \prime }$ and rotations $A_2^{\prime }+E^{\prime \prime }$ to obtainIts dimension is $2+6+2+2=12=3(6)-6$, as required for a nonlinear six-atom molecule. Thus there are eight symmetry-distinct fundamental frequencies in the harmonic picture, with each E mode doubly degenerate. This analysis concerns vibrations about one trigonal-bipyramidal minimum; it does not treat large-amplitude rearrangements between minima.

The planar three-radialene structure has a threefold axis normal to the plane, three in-plane twofold axes and the molecular plane as a horizontal mirror. Therefore its molecular point group is $D_{3h}$, the D3h molecular point group. Its threefold connectivity does not supply a sixfold axis.

The isolated molecular form is an ideal trigonal bipyramid with D3h molecular point group symmetry. Its solid-state structures need not consist of the same isolated molecules.

The hydrocarbon with a three-membered carbon ring and an exocyclic methylene double bond at each vertex. The ideal planar structure belongs to the D3h molecular point group.

A molecular geometry in which three attached chemical atoms and their central atom lie in one plane. The ideal symmetric arrangement has three $120^{\circ }$ angles and D3h molecular point group symmetry, as in an isolated boron trifluoride molecule.