The Electronic Partition Function

Introduction

After translation, rotation and vibration, the final ingredient of the molecular partition function is electronic. Most textbooks dismiss it in one line — "q E = 1" — and for most closed-shell molecules that is right. But O₂, NO, free radicals and halogen atoms do not obey that rule, and ignoring their electronic degeneracy produces entropies that are wrong by several joules per kelvin per mole. This page shows when q E is trivial, when it is a constant larger than one, and when it varies with temperature.

Core explanation

Definition. With energies measured from the ground electronic level,

q E = g₀ + g₁e^(−ε₁/kT) + g₂e^(−ε₂/kT) + …

where g j is the degeneracy of level j and ε j its energy above the ground level.

Why q E is usually a constant. Excitation energies of typical molecules are several electronvolts, tens of thousands of cm⁻¹. At 298 K, kT/hc ≈ 207 cm⁻¹, so ε₁/kT is well above 100 and every excited term is utterly negligible. The partition function reduces to the ground-state degeneracy:

q E = g₀

Finding g₀. The degeneracy comes from the term symbol.

- Closed-shell molecules (¹Σ, such as N₂, H₂O, CH₄): g₀ = 1, so q E = 1. - O₂ has a ³Σg⁻ ground state: spin S = 1 gives 2S + 1 = 3 components, so q E = 3 at ordinary temperatures. Its first excited level (¹Δg) lies about 7900 cm⁻¹ higher and is thermally negligible. - Radicals with one unpaired electron in a Σ state (for example the ²Σ ground state of CN) have g₀ = 2. - For atoms, a level with total angular momentum J has g = 2J + 1. Sodium (²S₁/₂) has g₀ = 2; a chlorine atom (²P₃/₂) has g₀ = 4.

Low-lying excited levels. Spin–orbit coupling can split a term into levels only a few hundred cm⁻¹ apart. Then q E depends on temperature.

- NO : the ²Π ground term splits into ²Π₁/₂ (lower, g = 2) and ²Π₃/₂ at 121.1 cm⁻¹ (g = 2). So q E = 2 + 2e^(−121.1 cm⁻¹ hc/kT), which rises from 2 at very low temperature toward 4 at high temperature. - Cl atom : ²P₃/₂ (g = 4) with ²P₁/₂ (g = 2) at 882 cm⁻¹. At 298 K q E ≈ 4.03; in a 2000 K flame the upper level becomes important.

Thermodynamic consequences. A constant q E = g₀ contributes R ln g₀ to the molar entropy but nothing to U or C V, because its temperature derivative vanishes. For O₂ this is R ln 3 = 9.1 J K⁻¹ mol⁻¹. A temperature-dependent q E, as in NO, behaves like a two-level system and adds a Schottky-type bump to the heat capacity.

Formulae

q E = Σ g j e^(−ε j/kT). Constant case: q E = g₀, S contribution = R ln g₀. Atoms: g = 2J + 1. Two-level case: q E = g₀ + g₁e^(−ε/kT); fraction in upper level = g₁e^(−ε/kT)/q E.

Step-by-step reasoning

To evaluate q E:

1. Find the ground term symbol and its degeneracy g₀. 2. List excited levels with their energies and degeneracies. 3. Compare each ε j with kT; discard levels with ε j/kT greater than about 10. 4. Sum the remaining terms g j e^(−ε j/kT). 5. If only g₀ survives, record q E = g₀ as a constant.

Visual explanation

Draw two horizontal lines for NO: a lower double line labelled ²Π₁/₂, g = 2, and a second double line 121 cm⁻¹ above labelled ²Π₃/₂, g = 2. Draw an arrow of length kT beside them. At 298 K the arrow is longer than the gap, so a substantial fraction of molecules sits on the upper line. Several electronvolts higher, far off the page, lies the next electronic state.

Real-world analogy

Think of a hotel with two identical rooms on the ground floor and two more one short step up. Guests fill both floors readily. Compare that with a hotel whose other rooms are at the top of a mountain: nobody climbs there, and the number of rooms that matters is just the ground-floor count.

Real-world example

Oxygen's triplet ground state makes O₂ paramagnetic and is visible in the thermodynamic tables: the standard molar entropy of O₂ (205 J K⁻¹ mol⁻¹) exceeds that of N₂ (192 J K⁻¹ mol⁻¹) partly because of the R ln 3 electronic term. Paramagnetic oxygen analysers used in medicine and industry exploit the same electron spin.

Why?

Why does a constant q E affect entropy but not energy? Entropy counts how many states are accessible; three degenerate spin states give three times as many configurations. Energy depends on how populations shift with temperature, and when all states share the same energy there is nothing to shift.

Common misconception

"The electronic partition function is always 1." It is 1 only for non-degenerate ground states with no low-lying excited levels. Open-shell molecules, radicals and most atoms have q E > 1, and forgetting it causes errors in entropies and equilibrium constants.

Worked example

Question: Calculate q E for NO at 298 K and the fraction of molecules in the ²Π₃/₂ level. (ε/hc = 121.1 cm⁻¹; kT/hc = 207.2 cm⁻¹.)

Reasoning: ε/kT = 121.1/207.2 = 0.584. e^(−0.584) = 0.557. q E = 2 + 2 × 0.557 = 3.11. Upper fraction = 1.115/3.11 = 0.36.

Answer: q E ≈ 3.1, with about 36 % of NO molecules in the upper spin–orbit level at 298 K.

Quick check

1. What is q E for a ground-state sodium atom at room temperature, and why? Answer: q E = 2, because the ²S₁/₂ ground level has J = ½ and so 2J + 1 = 2, with excited levels far too high to contribute.

Exam focus

State the rule "q E = g₀ unless there are excited levels within a few kT". Read degeneracies from term symbols (2S + 1 for Σ molecular states, 2J + 1 for atomic levels). NO is the classic example of a temperature-dependent q E; expect to calculate populations and its contribution to entropy.

Advanced insight

When computing equilibrium constants, electronic energies must share a common zero across all species. This is where the dissociation energy D₀ enters: each species' partition function is referred to its own ground state, and the differences in ground-state energies appear as an exponential factor e^(−ΔE₀/RT). Ignoring that alignment of energy zeros is a common source of large errors.

Summary

The electronic partition function sums g j e^(−ε j/kT) over electronic levels. For most molecules only the ground level contributes, so q E = g₀: 1 for closed shells, 3 for O₂, 2J + 1 for atoms. Species with low-lying levels, such as NO and Cl atoms, have temperature-dependent q E. A constant q E adds R ln g₀ to entropy but nothing to energy or heat capacity.

Practice questions

1. What electronic contribution does O₂ make to its molar entropy at 298 K? Answer: R ln 3 = 8.314 × 1.099 ≈ 9.1 J K⁻¹ mol⁻¹. 2. Evaluate q E for a chlorine atom at 298 K (²P₁/₂, g = 2, at 882 cm⁻¹; ground ²P₃/₂, g = 4). Answer: 882/207.2 = 4.26, e^(−4.26) ≈ 0.014, so q E = 4 + 2 × 0.014 ≈ 4.03. 3. What values does q E of NO approach at very low and very high temperature? Answer: 2 at very low temperature (only ²Π₁/₂) and 4 at high temperature (both levels equally populated). 4. Why does a constant q E not contribute to C V? Answer: C V comes from the temperature derivative of U, and a constant q E has zero temperature derivative, so it adds no energy as T changes.