CSIR-NET Term Symbols and Electronic Spectroscopy — Deriving the Ground State

By Aniket Bhardwaj · 20 September 2026 · CSIR-NET Chemistry

A term symbol is a compact statement of how the electrons in an incomplete shell are arranged with respect to each other. Getting one right takes about forty seconds once you know the procedure, and it is the starting point for magnetic moments, for atomic spectra and for the whole of ligand-field spectroscopy. This article derives ground terms for p and d configurations, shows the microstate count that proves your answer is complete, and then connects the terms to the selection rules that decide how intense an absorption band will be.

What the symbol means

^2S+1L_J

S — total spin, the sum of the individual m_s values. 2S + 1 is the spin multiplicity: 1 = singlet, 2 = doublet, 3 = triplet and so on.
L — total orbital angular momentum, the maximum sum of m_l. It is written as a letter: L = 0, 1, 2, 3, 4, 5 → S, P, D, F, G, H.
J — total angular momentum, running from |L − S| to L + S in integer steps. Each J level has degeneracy 2J + 1.

This is the Russell–Saunders (or LS) coupling scheme, valid when electron–electron repulsion dominates spin–orbit coupling — true for lighter elements and for the first transition series. For heavy elements the jj coupling scheme takes over, because spin–orbit coupling grows roughly as Z⁴ and eventually beats the repulsion term.

Hund's rules, in the order you apply them

The term with the highest multiplicity (largest S) lies lowest.
Among those, the term with the largest L lies lowest.
For the J value: if the sub-shell is less than half full, J = |L − S| is lowest (a normal multiplet). If it is more than half full, J = L + S is lowest (an inverted multiplet). Exactly half full gives L = 0, so J = S and the question does not arise.

Rules 1 and 2 identify the ground term; rule 3 picks the ground level within it. A question asking only for the ground term does not need rule 3, but one asking for a magnetic moment always does.

The microstate count — your proof that nothing is missing

Number of microstates = ^(2(2l+1))C_n — the number of ways to place n electrons in the 2(2l+1) available spin-orbitals

For p, 2(2l+1) = 6; for d it is 10; for f it is 14. The sum of the degeneracies (2S + 1)(2L + 1) over all the terms you have found must equal that number exactly. If it does not, you have missed a term or invented one. Very few candidates use this check, and it is free.

Worked example 1 — p², the ground state of carbon.

Microstates = ⁶C₂ = (6 × 5) ÷ 2 = 15.

The allowed terms for p² are ¹S, ³P and ¹D. Check the count:
¹S: (1)(1) = 1
³P: (3)(3) = 9
¹D: (1)(5) = 5
Total = 1 + 9 + 5 = 15 ✓ — the list is complete.

Ground term. Highest multiplicity → ³P (S = 1, L = 1). p² is less than half of p⁶, so J = |L − S| = 0. Ground level = ³P₀, which is indeed the ground state of the carbon atom.

The ³P term splits into ³P₀, ³P₁ and ³P₂ with degeneracies 1, 3 and 5 — total 9, matching the term's own degeneracy.

Worked example 2 — d², as in Ti²⁺ or V³⁺.

Microstates = ¹⁰C₂ = (10 × 9) ÷ 2 = 45.

Terms: ³F, ³P, ¹G, ¹D, ¹S. Check:
³F: (3)(7) = 21 · ³P: (3)(3) = 9 · ¹G: (1)(9) = 9 · ¹D: (1)(5) = 5 · ¹S: (1)(1) = 1
Total = 21 + 9 + 9 + 5 + 1 = 45 ✓.

Ground term. Highest S first: ³F and ³P are both triplets, so rule 2 decides — F has L = 3, larger than P's L = 1, so ³F wins. Less than half filled, so J = |3 − 1| = 2. Ground level = ³F₂.

Worked example 3 — d⁵ high spin, as in Mn²⁺ or Fe³⁺.

Microstates = ¹⁰C₅ = 252 — a large number, which is why nobody enumerates d⁵ by hand. Fortunately the ground term needs no enumeration.

All five electrons parallel gives S = 5/2, multiplicity 6.
With one electron in each d orbital, L = 2 + 1 + 0 + (−1) + (−2) = 0, so the term is S.
L = 0 means J = S = 5/2. Ground level = ⁶S_5/2.

Why this explains a colour. There is no other sextet term anywhere in the d⁵ manifold. Every possible d–d transition from ⁶S must therefore change the spin, so it is spin-forbidden as well as Laporte-forbidden. That double prohibition is why high-spin Mn(II) salts are famously pale pink rather than strongly coloured.

Worked example 4 — the hole formalism, a genuine time-saver. A shell that is n electrons short of full gives the same set of terms as one containing n electrons. So d⁸ has the same terms as d², and d⁷ the same as d³ — you can check that ¹⁰C₇ = 120 = ¹⁰C₃. The J ordering flips, though: d⁸ is more than half filled, so its ground level is ³F₄ (J = L + S = 3 + 1) rather than d²'s ³F₂. Similarly d¹ gives ²D_3/2 while d⁹ gives ²D_5/2.

Configuration	Microstates	Ground term	Ground level
d¹	10	²D	²D_3/2
d²	45	³F	³F₂
d³	120	⁴F	⁴F_3/2
d⁴ (high spin)	210	⁵D	⁵D₀
d⁵ (high spin)	252	⁶S	⁶S_5/2
d⁶ (high spin)	210	⁵D	⁵D₄
d⁷ (high spin)	120	⁴F	⁴F_9/2
d⁸	45	³F	³F₄
d⁹	10	²D	²D_5/2
f⁷	3432	⁸S	⁸S_7/2

Spin–orbit coupling and the Landé interval rule

Within a term, the J levels are separated by spin–orbit coupling, and the spacing follows a simple pattern:

E(J) − E(J − 1) = A · J, where A is the spin–orbit coupling constant

Worked example 5. For a ³P term the levels are J = 0, 1, 2.
E(³P₁) − E(³P₀) = A × 1 = A.
E(³P₂) − E(³P₁) = A × 2 = 2A.
So the two gaps are in the ratio 1 : 2. If a question gives you one measured interval you can predict the other, and if the observed ratio departs badly from 1 : 2 that is evidence Russell–Saunders coupling is breaking down.

A is positive for a less-than-half-filled shell (normal multiplet, lowest J lowest in energy) and negative for more than half filled (inverted multiplet) — the same rule that decided J in Hund's third rule, seen from the energy side.

Selection rules — what makes a band strong or weak

Rule	Statement	How it is relaxed in practice
Spin	ΔS = 0	Spin–orbit coupling, strongly for heavy elements
Orbital	ΔL = 0, ±1 (but L = 0 → L = 0 is forbidden)	Configuration mixing
Total	ΔJ = 0, ±1 (but J = 0 → J = 0 is forbidden)	—
Laporte (parity)	In a centrosymmetric species, g ↔ u only; d → d is forbidden	Vibronic coupling, or loss of the centre of symmetry (as in a tetrahedral complex)

These rules explain intensity, and intensity is the observable. The molar absorptivities below are the typical orders of magnitude quoted in inorganic texts, not precise values:

Transition type	Typical ε / dm³ mol⁻¹ cm⁻¹	Example
Spin- and Laporte-forbidden d–d	Below about 1	High-spin Mn(II) — very pale
Spin-allowed but Laporte-forbidden d–d (octahedral)	Roughly 1 to 100	[Ti(H₂O)₆]³⁺
d–d in a non-centrosymmetric field (tetrahedral)	Roughly 100 to 1000	Tetrahedral Co(II) — deep blue
Fully allowed charge transfer	Roughly 10³ to 10⁵	MnO₄⁻, intense purple

The permanganate example is worth remembering precisely because manganese in MnO₄⁻ is d⁰ — there are no d electrons to undergo a d–d transition at all. The colour is ligand-to-metal charge transfer, which is symmetry-allowed and therefore hundreds of times more intense than any d–d band. Metal-to-ligand charge transfer runs the other way and is seen with π-acceptor ligands such as bipyridine on a low-oxidation-state metal.

The splitting of these free-ion terms by a ligand field, the Orgel and Tanabe–Sugano correlations and the extraction of Δ_o and the Racah parameter B from a spectrum are the natural next step, and they are treated separately in the crystal field article linked below. What matters here is that every one of those diagrams starts from the free-ion ground term you have just derived.

Turning a band position into an energy

E = h c ν̃ per photon · E_molar = N_A h c ν̃ · ν̃ (cm⁻¹) = 10⁷ ÷ λ (nm)

Worked example 6 — a band at 500 nm.

ν̃ = 10⁷ ÷ 500 = 20 000 cm⁻¹.
E = (6.626 × 10⁻³⁴)(2.998 × 10¹⁰ cm s⁻¹)(20 000) = 3.973 × 10⁻¹⁹ J per photon.
Per mole: 3.973 × 10⁻¹⁹ × 6.022 × 10²³ = 2.393 × 10⁵ J mol⁻¹ = 239.3 kJ mol⁻¹.

Cross-check by a second route. Using 1239.84 eV·nm, the photon energy is 1239.84 ÷ 500 = 2.480 eV; multiplying by 96.485 kJ mol⁻¹ per eV gives 239.3 kJ mol⁻¹. The two routes agree exactly, so the answer is safe.

Organic electronic spectra in one paragraph

For organic chromophores the same selection-rule logic applies with different labels. n → π* transitions are symmetry-forbidden and weak (small ε), and they shift to shorter wavelength in polar protic solvents because hydrogen bonding stabilises the non-bonding lone pair and widens the gap. π → π* transitions are allowed and strong, and they shift to longer wavelength in polar solvents because the more polar excited state is stabilised more than the ground state. Learn the four words that describe these shifts: bathochromic (red shift, to longer λ), hypsochromic (blue shift), hyperchromic (greater intensity) and hypochromic (lesser).

Extending conjugation always lowers the transition energy, which is why polyenes absorb further into the visible as they lengthen. The Woodward–Fieser rules put numbers on this for dienes and enones: start from a base value and add tabulated increments for each substituent, each extra conjugated double bond and each exocyclic double bond. For example, a heteroannular diene with a base value of 214 nm carrying four alkyl or ring-residue substituents and one exocyclic double bond gives 214 + (4 × 5) + 5 = 239 nm. Use the increment table printed in your own textbook when you practise — the values differ slightly between editions, and the examiner's expected answer follows a standard table rather than any single source.

Mistakes that cost marks

Skipping the microstate check. It costs one line and catches a missing term immediately.
Applying Hund's rules out of order. Multiplicity first, then L, then J. A larger L never beats a larger S.
Getting the J rule backwards. J = |L − S| for less than half filled, J = L + S for more than half filled.
Assuming dⁿ and d¹⁰⁻ⁿ have identical ground levels. The terms match, but the multiplet inverts, so the J value differs.
Using L = 0 blindly for every half-filled shell. It is true for d⁵ and f⁷ because the m_l values cancel — say that, rather than quoting it as a rule with no reason.
Explaining permanganate's colour as a d–d transition. Mn(VII) is d⁰; it is charge transfer.
Ignoring the ε value the question gives you. A stated molar absorptivity is usually the clue that identifies the transition type.
Confusing hypsochromic with hyperchromic. One is about wavelength, the other about intensity.

Where this appears in the paper

Sub-topic	Typical question form
Term symbol derivation	Ground term and level for a stated pⁿ, dⁿ or fⁿ ion
Microstate counting	Number of microstates; verify a list of terms is complete
Hund's rules	Order two terms in energy and justify it
Spin–orbit coupling	Landé interval ratio; normal versus inverted multiplet
Selection rules	Is a stated transition allowed, and by which rule is it forbidden?
Band intensity	Identify the transition type from an ε value
Charge transfer	LMCT versus MLCT; why d⁰ ions can be intensely coloured
Organic UV–visible	n → π* versus π → π*; solvent shifts; conjugation effects

This is a map of the sub-topics, not a statement about mark distribution or question counts. For the structure of the paper you are actually sitting, read the current official notification.

Every term symbol starts from quantum numbers. If the n, l, m_l and m_s bookkeeping is shaky, the S and L values will be wrong before you begin. The Quantum Numbers tool checks which sets are allowed, how many electrons a sub-shell holds and the orbital each combination describes — the exact foundation these derivations rest on.

Open the Quantum Numbers Tool →

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