Ammonia Molecular Orbitals by Symmetry

Introduction

Ammonia's three equivalent N–H bonds suggest three hydrogen contributions, while nitrogen supplies four valence atomic orbitals. C₃v symmetry organises this seven-function basis before any matrix diagonalisation. One hydrogen combination is totally symmetric and two form an E pair. Matching them with nitrogen orbitals clarifies the bonding blocks and the origin of a largely nitrogen-centred occupied orbital often called a lone pair.

Core explanation

Place the C₃ axis along z through N and the centre of the hydrogen triangle. The three H 1s functions h₁, h₂ and h₃ reduce to A₁ + E. Their A₁ SALC is proportional to h₁+h₂+h₃. Two linearly independent difference combinations, such as 2h₁−h₂−h₃ and h₂−h₃, span E. Their detailed normalisation depends on overlap, and a rotation can mix the two E partners. They should be treated as one two-dimensional symmetry species, not two unrelated one-dimensional labels.

Nitrogen 2s and 2p z transform as A₁. Nitrogen 2p x and 2p y form an E pair. The A₁ block therefore contains three basis functions: N 2s, N 2p z and the H A₁ SALC. Its diagonalisation yields three A₁ molecular orbitals. The E block contains two N functions and two H SALCs, four functions total, organised as two E pairs of molecular orbitals. Adding dimensions gives 3 + 4 = 7 spatial molecular orbitals from the seven input valence functions.

Ammonia has eight valence electrons: five from nitrogen plus three from hydrogen. In a closed-shell description, four spatial orbitals are occupied. A common qualitative order has an A₁ bonding orbital and a doubly degenerate E bonding pair occupied, accounting for six electrons, followed by another occupied A₁ orbital with substantial nitrogen lone-pair character. Higher A₁ and E antibonding orbitals remain unoccupied. This is a qualitative picture; the exact order and mixing coefficients require energetic calculations or spectroscopic evidence. Symmetry supplies the permissible blocks, not the numerical orbital energies.

The lone-pair-bearing orbital is not simply identical to a pure atomic 2p z function. N 2s, N 2p z and the symmetric H SALC share A₁ symmetry and can mix. The occupied A₁ orbital's nitrogen localisation and directional character result from the effective Hamiltonian and molecular geometry. In a localised bonding model, one may describe a lone pair and three N–H bonds, but canonical MO functions distribute electron density in symmetry-adapted combinations. Both descriptions can be useful if the distinction is maintained.

The E block is particularly instructive. A C₃ rotation mixes x and y functions, so the two partners of an E molecular-orbital level remain degenerate under an exactly C₃v-symmetric Hamiltonian. Substituting one hydrogen or placing ammonia in an asymmetric environment can lower the symmetry and split the pair. The splitting is a direct sign that the original group operations no longer constrain the electronic problem.

Ammonia's pyramidal geometry matters for donor chemistry. An occupied nitrogen-centred A₁ orbital can donate electron density to an appropriate acceptor, forming an adduct. The resulting structure may have a different point group, so symmetry labels from isolated NH₃ should not be transferred unchanged to the product. Tracking geometry and electron count together prevents a static diagram from being overextended.

Step-by-step reasoning

Classify the equilibrium pyramid as C₃v and choose z along C₃. Reduce the three-H basis to A₁ + E. Classify N 2s, 2p z as A₁ and N 2p x, 2p y as E. Count three A₁ and four E-basis functions, then count eight valence electrons. Only after these checks construct a plausible energy diagram and discuss lone-pair character.

Visual explanation

Draw NH₃ as a pyramid and place h₁+h₂+h₃ below an A₁ heading. Place two difference phase patterns below E. Beside A₁ draw N 2s and 2p z; beside E draw N 2p x and 2p y. On an output side, show three A₁ lines and two doubly degenerate E levels, with eight electrons filling four spatial orbitals.

Real-world analogy

Three identical singers can combine into one common unison pattern and two independent contrast patterns. The nitrogen orbital functions can join only matching patterns under a symmetry-preserving arrangement. This resembles A₁ and E blocks, though electron wavefunctions involve phase, overlap and quantum energy rather than audible loudness.

Real-world example

Ammonia binds many Lewis acids through donation from its occupied nitrogen-centred electron density. The isolated-molecule C₃v picture helps identify the donor orbital's symmetry relative to the N–H framework. As a new bond forms, the geometry and point group may change, so the final bonding orbitals should be analysed in the product's actual structure.

Why?

Why are there two E molecular-orbital levels, each with two partner orbitals? The E block starts with one E pair from N 2p x/2p y and one E pair from hydrogen SALCs: four independent functions. Mixing two copies of E yields two E energy levels, and each level retains a two-component degeneracy under ideal C₃v symmetry.

Common misconception

An E label does not mean one orbital occupied by two electrons; it denotes a two-dimensional partner space. Nor is the ammonia lone pair necessarily a pure nitrogen p orbital. Same-symmetry N s and p functions and the symmetric ligand combination can all mix, with energy and overlap determining the final character.

Worked example

Start with N 2s, 2p x, 2p y, 2p z and three H 1s functions: seven spatial basis functions. The H functions contribute A₁ + E; N contributes 2A₁ + E. Thus the total is 3A₁ + 2E, whose dimension is 3(1) + 2(2) = 7. The eight valence electrons occupy four spatial orbitals in a closed-shell ground state, and the block count shows that a proposed eight-orbital valence diagram would be inconsistent with this basis.

Quick check

1. Which nitrogen valence orbitals match the H A₁ SALC? Answer: N 2s and N 2p z share A₁ symmetry in the chosen axis convention. 2. How many independent functions form one E species? Answer: Two partner functions form the two-dimensional E representation.

Exam focus

Show the total basis as 3A₁ + 2E and confirm dimension seven. Count eight valence electrons separately from orbital dimension. Explain the lone pair as an occupied nitrogen-centred MO with A₁ character, rather than assuming symmetry alone fixes its exact atomic composition.

Advanced insight

Within one symmetry block, diagonalisation can mix all basis functions of that species, but orthogonality of the resulting MOs prevents them from becoming redundant. A low-energy N 2s function and higher-energy N 2p z function may contribute differently to the occupied and unoccupied A₁ levels. This is one reason symmetry diagrams are best paired with energy and overlap information.

Summary

Pyramidal NH₃ is C₃v. Three H 1s orbitals give A₁ + E, nitrogen valence orbitals give 2A₁ + E, and the complete seven-function basis is 3A₁ + 2E. The A₁ and E blocks organise bonding and a nitrogen-centred occupied lone-pair-like orbital. Symmetry determines possible mixing and degeneracy, while detailed energies require calculation or experiment.

Practice questions

1. Why is the total dimension of 3A₁ + 2E seven rather than five? Answer: Each A₁ is one-dimensional, but each E is two-dimensional. The count is 3(1) + 2(2) = 7. 2. What happens to the ideal E degeneracy if one H is replaced by a distinguishable ligand and the C₃ axis is lost? Answer: The symmetry protection is removed, so the former E partner orbitals may have different energies and must be labelled in the lower group. 3. Does C₃v symmetry alone prove that the highest occupied NH₃ orbital is entirely N 2p z? Answer: No. N 2s, N 2p z and the symmetric H SALC all share A₁ symmetry and can mix; their proportions depend on energies and overlap.