The D3h Point Group in BF3

Introduction

Ideal boron trifluoride is trigonal planar: three equivalent fluorines lie around boron at approximately 120° in one plane. It shares a threefold principal axis with ammonia, yet its planarity supplies a horizontal mirror and perpendicular twofold axes that ammonia lacks. Its D₃h symmetry therefore gives different orbital and vibrational labels. Comparing the two molecules shows why a correct point-group assignment depends on three-dimensional geometry, not just the count of attached atoms.

Core explanation

Take the BF₃ molecular plane as xy and place the principal C₃ axis along z through boron. The group contains E, rotations C₃ and C₃², and three C₂′ rotations about axes in the plane that pass through B and one B–F direction. A 180° turn about one such axis leaves that F in place and exchanges the other two after reversing their out-of-plane coordinate. Reflection in the molecular plane is σh; it leaves all equilibrium nuclei in place because each has z = 0. Additional operations arise by combining these, including two S₃ operations and three vertical mirror planes. The full group has twelve operations in six conjugacy classes.

The three perpendicular C₂′ axes place BF₃ in the D₃ rather than C₃ family. The horizontal plane gives the h suffix. This is different from C₃v ammonia, which has vertical mirrors but no horizontal mirror or in-plane C₂ axes at its equilibrium pyramidal geometry. BF₃ also lacks an inversion centre: inversion through B would require an F at a position opposite each of the three F atoms, which is not occupied in a trigonal triangle.

The horizontal mirror immediately separates orbital types. In-plane coordinates x and y and the boron 2s orbital are unchanged under σh; their symmetry species carry a prime mark. The boron 2p z orbital changes sign because z → −z, so it has double-prime symmetry. In the D₃h character table, B 2s is A₁′, the (2p x,2p y) pair is E′, and 2p z is A₂″. The three F orbitals aimed along B–F bonds can form one A₁′ sum and one E′ pair. They can match B 2s and in-plane 2p orbitals respectively, while their sigma combinations cannot match the out-of-plane 2p z by symmetry.

This does not make the B 2p z orbital chemically irrelevant. Fluorine has other valence p orbitals, including components perpendicular to the plane, which can participate in pi-type interactions with matching symmetry. A sigma-only ligand basis is a deliberately limited model. Symmetry tells us what can couple within the chosen basis; it does not tell us whether a pi interaction is energetically dominant or whether a simple Lewis acidity description captures every bonding detail.

Prime and double-prime labels refer specifically to behaviour under σh, not to positive or negative electric charge. A function may change sign under a symmetry operation without changing its probability density ψ ². The sign is a phase property of a wavefunction or displacement pattern and is central to interference and matrix-element tests. In a D₃h problem, one should therefore mark the molecule's plane before reading prime symbols.

BF₃ has four atoms and is nonlinear, giving 3(4) − 6 = 6 normal-coordinate degrees of freedom. In the ideal group these can be organised into symmetry types including one-dimensional and doubly degenerate modes. The out-of-plane bending motion has different σh behaviour from in-plane stretches, so it belongs to a double-prime type. Detailed IR and Raman activity follow from matching the mode to dipole or polarizability components, not from an assertion that all vibrations of a polar bond must be IR active.

Step-by-step reasoning

Draw the trigonal plane and z axis. Confirm C₃, then rotate around each of three in-plane B–F axes to establish D₃. Reflect in the molecular plane to add h. Count E, two C₃, three C₂′, σh, two S₃ and three σv operations. Then classify any orbital by its behaviour under σh before comparing full symmetry species.

Visual explanation

Sketch B at the centre of an equilateral F triangle. Draw a vertical z axis and shade the xy molecular plane. Add three in-plane C₂′ axes through each B–F bond. Next draw an orbital lobe above and below B for 2p z; the horizontal reflection exchanges lobes and reverses orbital sign, illustrating the double-prime label.

Real-world analogy

A flat patterned plate may look unchanged when flipped through its own plane if the pattern has no front-versus-back distinction. An arrow sticking out of the plate reverses direction under the same flip. In-plane sigma orbitals resemble features on the plate, while p z has opposite-phase lobes above and below it. The analogy concerns transformation behaviour, not physical orbital rotation during a reaction.

Real-world example

Comparing BF₃ and NH₃ helps interpret their orbital roles. In planar BF₃, an unoccupied boron 2p z orbital has out-of-plane symmetry and can accept suitable electron density from a donor as geometry changes. In pyramidal NH₃, the lone-pair-containing orbital belongs to an A₁-type block in C₃v. Their donor–acceptor interaction changes the geometry and symmetry of the resulting adduct, so one should not keep the isolated molecules' point groups unchanged after bonding.

Why?

Why does the horizontal plane matter beyond adding a suffix? It partitions functions into those unchanged and those sign-reversed by reflection. Matrix elements between incompatible parity types can vanish for a symmetry-preserving operator. This immediately prevents a sigma-only in-plane ligand combination from mixing with B 2p z at ideal planar geometry.

Common misconception

Threefold rotation alone does not make BF₃ C₃v. One must also test its three in-plane C₂ axes and σh, which give D₃h. Another misconception is that 2p z's double-prime label means the orbital is negative or unimportant. It describes phase under horizontal reflection only.

Worked example

Use three F sigma orbitals f₁, f₂ and f₃ directed toward B. Their total sum f₁ + f₂ + f₃ remains unchanged by every D₃h operation and has A₁′ symmetry. Two independent combinations with coefficients summing to zero form an E′ pair. B 2s is A₁′, and its (2p x,2p y) pair is E′, so these can interact with the corresponding ligand combinations. B 2p z is A₂″ and cannot mix with this sigma-only set by symmetry.

Quick check

1. What distinguishes D₃h BF₃ from C₃v NH₃ in the point-group decision tree? Answer: BF₃ has three C₂ axes perpendicular to C₃ and a horizontal molecular mirror plane; pyramidal NH₃ does not. 2. Does B 2p z remain unchanged under σh? Answer: No. Its wavefunction changes sign under reflection through the molecular plane, so it has double-prime behaviour.

Exam focus

List all six classes and their operation counts if asked for group order. Distinguish σh from σv and explain prime marks as reflection parity. For an MO diagram, state the ligand basis being used before declaring that an orbital has no symmetry-matched partner.

Advanced insight

Degenerate E′ components transform into linear combinations of one another under C₃ rotations. If a distortion makes one B–F bond different, the threefold axis fails and this degeneracy can split. Such splitting can change predicted spectra and reactivity; symmetry labels are properties of a specified geometry, not immutable names for orbitals along an entire reaction coordinate.

Summary

Ideal planar BF₃ belongs to D₃h because it has a C₃ axis, three perpendicular C₂ axes and a horizontal molecular mirror plane. Its twelve operations organise orbitals into prime and double-prime symmetry types. The three F sigma functions supply A₁′ and E′ combinations that match B 2s and in-plane 2p orbitals, while out-of-plane B 2p z requires a different matching basis.

Practice questions

1. How many operations does ideal BF₃ have in D₃h when grouped as E, 2C₃, 3C₂′, σh, 2S₃ and 3σv? Answer: Add the counts 1 + 2 + 3 + 1 + 2 + 3 = 12. The six headings are classes, not six total operations. 2. Could an out-of-plane fluorine p orbital have a nonzero symmetry interaction with B 2p z? Answer: Potentially yes, if their full symmetry species match and overlap and energies permit. The prohibition derived here applies only to the in-plane sigma-ligand combinations. 3. If BF₃ is distorted into a pyramidal structure, can the original D₃h prime labels still be exact symmetry labels? Answer: No. The horizontal mirror plane is lost, so prime and double-prime parity under it no longer classifies exact states of the distorted geometry.