Pi Bond Orders and Electron Densities

Introduction

Hückel eigenvalues describe pi orbital energies, while eigenvectors describe how each orbital spreads over atoms. To connect the model to local chemical questions, the occupied-orbital coefficients can be combined into site pi-electron populations and intersite pi bond-order indices. These quantities help compare related molecules, but they are model-dependent. They do not by themselves give an atom's full charge, an experimental bond length or a complete chemical bond order.

Core explanation

Write a normalised pi MO k as ψ k = Σ i c ikφ i. In the simple orthonormal p basis, let n k be its electron occupancy, usually 0, 1 or 2. The pi-electron population at site i is q i = Σ k n k c ik ² over occupied orbitals. This quantity counts model pi electrons assigned to that site. Summing q i over all sites gives Σ k n k because each orbital is normalised. That sum rule is a valuable check on coefficient or occupancy errors.

A simple intersite pi bond-order index is p ij = Σ k n k c ik c jk for real Hückel coefficients. Same-sign amplitudes in an occupied orbital contribute positively between sites; opposite signs contribute negatively. An antibonding orbital's occupation can reduce the index. This is an off-diagonal density-matrix element in the chosen basis and convention. Some texts define bond-order measures with different factors, so formulas should be stated before numerical comparisons.

For ethene's lower MO ψ b=(φ₁+φ₂)/√2 occupied by two electrons, q₁=q₂=2(1/2)=1. The pi bond-order index is p 12=2(1/√2)(1/√2)=1. If the upper antisymmetric orbital were also doubly occupied without orbital relaxation, its contribution would be −1 and the net simple pi index would be zero. This algebra expresses cancellation of bonding and antibonding occupancy, though such a four-pi-electron ethene scenario is not a normal neutral ground state.

For allyl's low orbital with coefficients (1,√2,1)/2, two electrons give q=(0.5,1,0.5). The adjacent index p 12 is 2(1/2)(√2/2)=1/√2≈0.707, and p 23 is the same. The middle orbital (1,0,−1)/√2 has zero central coefficient, so occupying it changes terminal populations but makes no direct contribution to either adjacent p 12 or p 23 in this simplified formula. Thus cation, radical and anion share the same adjacent pi index when only their middle-orbital occupancy changes in the fixed model.

These numbers require careful interpretation. q i is a pi-only population, not a formal charge. To infer a net atomic charge, one would also need sigma electrons, nuclear charge and a chosen partitioning convention. Likewise p ij is a pi contribution, not the whole bond order. A sigma bond may exist between the atoms, and real bond strength depends on geometry, electron correlation and more than one density-matrix number.

Symmetry can simplify populations. In ideal benzene, every carbon is equivalent, so a closed-shell six-pi-electron calculation must assign one pi electron per site: six electrons divided equally among six sites. A calculation producing unequal site populations for an unsubstituted ideal ring signals a broken symmetry assumption or numerical setup error. However, equivalent q i does not mean every MO has equal coefficient magnitude at each site in every real basis; the sum over the occupied subspace is the invariant physical model quantity.

The sign of a coefficient can be reversed for an entire MO without changing q i or p ij, because a product c ikc jk gains two minus signs. Relative phase between sites does matter. Rotating among equally occupied degenerate orbitals can change individual coefficient tables, but the summed density matrix over the complete occupied subspace remains the same. This is why local descriptors should be computed from all occupied orbitals, not interpreted from a single arbitrary degenerate partner.

Step-by-step reasoning

List the occupied MOs and their occupancies. Check each eigenvector is normalised. For a site, square its coefficient in each occupied MO and multiply by occupancy, then sum. For a bond, multiply the two site coefficients in each occupied MO, multiply by occupancy and sum with signs. Verify total q equals total pi electrons and state that the indices are pi-only model quantities.

Visual explanation

Draw an allyl three-site chain with the low MO coefficients 1/2, √2/2 and 1/2 written above sites. Put small squares below them to show electron population weights and products along each adjacent edge to show bond-order contributions. Overlay the middle MO with a zero at the centre, making clear why it changes terminal population but not adjacent bond indices.

Real-world analogy

A shared project can be measured by how much work each person contributes and how strongly two people coordinate. Site population resembles individual contribution; an off-diagonal bond-order index resembles coordinated participation between two sites. The analogy helps separate diagonal from intersite information but cannot turn a model index into an experimental bond strength by itself.

Real-world example

In a qualitative analysis of allyl radical, the singly occupied nonbonding MO gives terminal spin-density weight in the simplest model. This helps motivate why reactions can occur at either terminal carbon. Substituents or an asymmetric environment shift coefficients and reactivity, so actual product distributions need more than a symmetric Hückel calculation.

Why?

Why must all occupied MOs be included? Electron density is built from the entire occupied electronic state, not just the frontier orbital. A single orbital can have a node at a site while other occupied orbitals provide substantial density there. Summing with occupancy prevents the mistaken conclusion that a node in one MO means an atom contains no pi electrons.

Common misconception

An atom with q i=1 is not necessarily electrically neutral; q i counts only the chosen pi basis. Also, p ij=0 in one model does not prove there is no chemical bond, because sigma bonding or a different basis can still contribute. Numerical Hückel indices are useful comparative descriptors, not exact observables.

Worked example

Use the allyl radical's occupied low MO with two electrons and middle MO with one electron. At terminal site 1, q₁=2(1/2)²+1(1/√2)²=0.5+0.5=1. At central site 2, q₂=2(√2/2)²+1(0)²=1. Site 3 is likewise 1, so q₁+q₂+q₃=3 pi electrons. For bond 1–2, p 12=2(1/2)(√2/2)+1(1/√2)(0)=1/√2.

Quick check

1. What must Σ i q i equal in a normalised three-pi-electron Hückel calculation? Answer: Three, the total pi-electron occupancy. 2. Does reversing the overall sign of one entire MO change its bond-order contribution? Answer: No. Both coefficients in each product reverse, leaving their product unchanged.

Exam focus

Write the q i and p ij formulas, show occupancies and retain coefficient signs for bond indices. Check total population against electron count. Label results as pi-only and specify the model convention before comparing them with formal charge or full bond order.

Advanced insight

The occupied-orbital density matrix P ij=Σ k n k c ik c jk collects all q i on its diagonal and intersite indices off diagonal for an orthonormal real basis. Unitary rotations among fully occupied orbitals leave this matrix unchanged, even though individual orbital pictures change. This makes the density matrix a more stable descriptor than one arbitrary canonical orbital in a degenerate subspace.

Summary

Hückel coefficients give model pi populations through occupied coefficient squares and pi bond-order indices through occupied coefficient products. Sum rules check the electron ledger. These descriptors reveal delocalisation and phase cancellation, but they omit sigma electrons and depend on the chosen approximation, so they cannot be read as exact charges or complete bonds.

Practice questions

1. Ethene has a doubly occupied bonding MO with coefficients (1/√2,1/√2). Calculate q₁ and p 12. Answer: q₁=2(1/2)=1 pi electron, and p 12=2(1/√2)(1/√2)=1 in the stated index convention. 2. Why does occupying allyl's middle orbital not change its adjacent p 12 and p 23 in the fixed simple model? Answer: The middle orbital has zero coefficient at site 2, so each adjacent coefficient product involving that site is zero. 3. A student reports a total pi population of 3.4 for an allyl radical with three pi electrons. What should be checked? Answer: Check eigenvector normalisation, occupancies and arithmetic. In an orthonormal basis the site populations must sum to the total occupancy of three.